Physics Volume I Classical MechanicsChapter 0 Mathematical Preliminaries In this chapter, we review...
Transcript of Physics Volume I Classical MechanicsChapter 0 Mathematical Preliminaries In this chapter, we review...
Physics
Volume I
Classical Mechanics
Notes compiled by
Pau Roldan-Blanco
First version: March 2018
Latest version: September 2018
Preface
This volume covers Classical Mechanics. By this, we refer to the state
of physics at the end of the 19th century. It includes the physics of Newton,
Maxwell, Faraday, and their contemporaries.
Part I, called Newtonian mechanics, covers the set of rules that describe
the motion of bodies in the realm of the macroscopic. We discuss motion,
energy, and conservation laws, both in linear and rotational contexts, as well
as the behavior of waves, fluids, and gases. In the last part of Part I, we
present the Lagrangian and Hamiltonian approaches to these same topics.
[TBW]
Part II, called Electricity and Magnetism, discusses the physics of elec-
tric and magnetic fields. [TBW]
Disclaimer: Not one bit of the material included in this book is original.
Only some presentations and extended materials are mine. The core material
has been compiled from the following sources:
• Susskind, L. and G. Hrabovsky (2013). The Theoretical Minimum.
• The Feynman Lectures in Physics, by Richard Feynman.
• Walter Lewin’s lectures from MIT Open Courseware.
I have also drawn from numerous Wikipedia articles, and adapted TIKz
templates from various users on the LATEXStack Exchange websites. All
credit goes to these people. Any errors or omissions are strictly my own.
3
Contents
0 Mathematical Preliminaries 7
0.1 Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . 7
0.2 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
0.3 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . 11
0.4 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 13
0.5 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
0.6 Taylor Approximations . . . . . . . . . . . . . . . . . . . . . . 22
0.7 Differential Equations . . . . . . . . . . . . . . . . . . . . . . 25
I NEWTONIAN MECHANICS 51
1 Motion and Force 53
1.1 Newton’s Laws of Motion . . . . . . . . . . . . . . . . . . . . 53
1.2 Resistive Forces . . . . . . . . . . . . . . . . . . . . . . . . . . 77
1.3 Multi-Particle Systems . . . . . . . . . . . . . . . . . . . . . . 90
1.4 Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . 95
2 Energy 99
2.1 Work, Kinetic Energy, and Power . . . . . . . . . . . . . . . . 99
2.2 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . 105
2.3 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . 108
2.4 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
2.5 Impulse and Thrust . . . . . . . . . . . . . . . . . . . . . . . 125
2.6 Newton’s Universal Law of Gravitation . . . . . . . . . . . . . 132
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3 Rotation 139
3.1 Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . 139
3.2 Angular Momentum and Torques . . . . . . . . . . . . . . . . 143
3.3 Gyroscopic Motion . . . . . . . . . . . . . . . . . . . . . . . . 158
3.4 Elliptical Orbits and Kepler’s Laws . . . . . . . . . . . . . . . 162
4 Stability and Elasticity 169
4.1 Static Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . 169
4.2 Stress, Strain, and Elasticity . . . . . . . . . . . . . . . . . . 180
5 Waves, Fluids, and Oscillations 187
5.1 Waves and the Doppler shift . . . . . . . . . . . . . . . . . . . 187
5.2 Fluid Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
5.3 Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 209
5.4 Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
6 Heat, Temperature, and Thermodynamics 237
6.1 Thermal Expansion of Solids and Liquids . . . . . . . . . . . 238
6.2 The Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . 240
6.3 Phase Transitions . . . . . . . . . . . . . . . . . . . . . . . . . 244
6.4 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 246
7 Lagrangian Mechanics 249
7.1 The Euler-Lagrange Equation . . . . . . . . . . . . . . . . . . 250
7.2 Non-Inertial Reference Frames . . . . . . . . . . . . . . . . . 254
7.3 Generalized Coordinates . . . . . . . . . . . . . . . . . . . . . 257
7.4 Symmetry and Conservation Laws . . . . . . . . . . . . . . . 268
II ELECTRICITY
AND MAGNETISM 269
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Chapter 0
Mathematical Preliminaries
In this chapter, we review the basic mathematical toolkit that is indis-
pensable for working out problems in classical mechanics.
0.1 Dynamical Systems
Classical mechanics is built around the notion of determinism and re-
versibility.
Definition 0.1 (Determinism) The law stating that the evolution of a
system is fully predictable from its law of motion and a well-defined starting
point (or initial condition).
Definition 0.2 (Reversibility) The law stating that a system is deter-
ministic regardless of the direction of motion.
Example 0.1 Time is discrete, indexed by n ∈ Z. The state space is S =
{−1, 1}. Then, the law of motion:
σ(n+ 1) = −σ(n)
is deterministic (for any n, the state in (n + 1) can be predicted) and
reversible (for any n, both states (n + 1) and (n − 1) are known). The law
of motion:
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σ(n+ 1) = σ(n)2
is deterministic but not reversible. If σ(n) = 1, then σ(n + 1) = 1, but
if σ(n + 1) = 1, then both σ(n) = 1 and σ(n) = −1 are solutions. Since
the system is deterministic in the n → (n + 1) direction, but not in the
(n+ 1)→ n direction, it is not reversible.
Definition 0.3 (Information conservation) A system is said to conserve
information if it is both deterministic and reversible.
An implicit assumption in all of classical physics is that all dynamical
systems conserve information, that is, are both deterministic and reversible.
Definition 0.4 (Chaotic Behavior) A system is chaotic if a change in
the initial condition implies a change in the system’s outcome.
Nearly all systems are chaotic, in that they all rely on the value of
the initial state. Insofar as the law of motion of the system is know, our
inability to determine its initial value impedes that we can perfectly predict
its behavior.
0.2 Trigonometry
Trigonometry is present through all the disciplines of physics. Here we
cover the basic concepts. We start with some definitions.
Definition 0.5 (Radians) The radian is the standard measure of an angle,
defined by:
1 radian =180◦
π
For example:
• There are 2π radians in 360◦ (a full circle).
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• A right angle has π2 radians.1
Trigonometric functions are defined in terms of the properties of right
angles. Let a, b, and c be the altitude, base, and hypothenuse of a right
triangle. Let φ be the angle opposite the base, i.e. the angle formed by sides
a and c, or φ ≡ ∠ac. Let θ = ∠bc. The angle ∠ab is 90◦ because abc is a
right triangle, that is ∠ab = π/2 radians.
Definition 0.6 (Basic trigonometric functions) We define the sine (sin),
cosine (cos) and tangent (tan) functions as follows:
sin θ =a
c, cos θ =
b
c, tan θ =
a
b.
Note we can write:
tan θ =sin θ
cos θ
Here are some useful properties of sine and cosine functions:
π2
π 32π
2π
−1
1
sin θ
cos θ
θ
Figure 0.1: The sin and cos functions over [0, 2π].
• Both sin and cos functions are oscillating waves taking values in [−1, 1].
Since positions in circles range from 0◦ to 360◦ degrees, we will typi-
1 Incidentally, this clarifies why the diameter of the circle of radius R is 2πR. Ifwe roll the circle on a plane, the total distance it will have traveled after one fullrevolution is the product of the radius and its total number of radians.
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cally restrict these functions’ supports to [0, 2π] (see Figure 0.1). Be-
cause of their oscillatory shape, these functions are sometimes called
sinusoidal (of cosinusoidal) waves.
• A useful property relates right angles to circles. Take a circle of radius
c and centered at some (x0, y0). Suppose c is also the hypothenuse of
a right angle with origin at (x0, y0), and again denote θ = ∠bc. Then,
the position of any point (x, y) satisfies:
x = c cos θ y = c sin θ (0.1)
See Figure 0.2 for an illustration of equation (0.1) when c = 1.
a = c sin θ
b = c cos θc c
c
c
Figure 0.2: A right triangle in a circle of radius c.By equation (0.1), we have a = c sin θ and b = c cos θ.
• Pythagorean theorem:
sin2 θ + cos2 θ = 1 (0.2)
where sin2(θ) ≡ sin(θ) sin(θ), and similarly for cos2. Note this is sim-
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ply a special case of Pythagoras’ theorem for a right triangle of hy-
pothenuse c = 1, i.e. a2 + b2 = 1.
• Product rule:
sin θ cos θ = sin 2θ
• For any two angles (θ1, θ2), the following properties hold:
sin(θ1 + θ2) = sin θ1 cos θ2 + cos θ1 sin θ2
sin(θ1 − θ2) = sin θ1 cos θ2 − cos θ1 sin θ2
cos(θ1 + θ2) = cos θ1 cos θ2 − sin θ1 sin θ2
cos(θ1 − θ2) = cos θ1 cos θ2 + sin θ1 sin θ2 (0.3)
• Sine and cosine functions are related by their derivatives:
d
dθsin θ = cos θ
d
dθcos θ = − sin θ (0.4)
0.3 Complex Numbers
Definition 0.7 (Complex number) A complex number is a number which
can be expressed in the form:
p+ qi
where p and q are real numbers, and i is a solution to x2 = −1.
Since no real number can solve the equation x2 = −1, then i =√−1 is
called an imaginary number. Hence, in the number p + qi, p is called the
real part, and qi is called the imaginary part.
Using the definition generally, all numbers can be though of having the
form p + qi, with real numbers having q = 0 and pure imaginary numbers
having q 6= 0.
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Moreover, since i is definitional, the “+” sign in front of bi is just a
convention. Then, we adopt the following terminology:
Definition 0.8 (Complex conjugate) The complex conjugate of a com-
plex number p+ qi is p− qi.
Complex numbers have the following algebraic properties:
• Addition and subtraction: The real and imaginary parts must be
added/subtracted independently, that is:
(a+ bi) + (c+ di) = (a+ c) + (b+ d)i
(a+ bi)− (c+ di) = (a− c) + (b− d)i
• Multiplication and division: Since i2 = −1, we have:
(a+ bi)(c+ di) = (ac− bd) + (bc+ ad)i
a+ bi
c+ di=
(ac+ bd
c2 + d2
)+
(bc− adc2 + d2
)i (0.5)
Importantly, complex numbers are intimately related to trigonometric
functions through the so-called Euler’s formula:
Result 0.1 (Euler’s Formula) For any real number θ:
eiθ = cos θ + i sin θ (0.6)
where i =√−1 is the imaginary unit.2
2 We will provide a proof of this famous result in mathematics by using Taylorexpansions in Section 0.6.
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Recall that cos θ = xc and sin θ = y
c for a right triangle of base x, height
y, hypothenuse c, and angle θ = ∠xc (see Figure 0.3). Then, another way
of writing Euler’s formula is:
x+ iy = ceiθ (0.7)
In sum, equations (0.6)-(0.7) provide us with a one-to-one mapping from
complex numbers to exponential functions and sinusoidal functions: any
complex number, composed of a real part x and an imaginary part iy, can be
written (i) in the exponential form ceiθ, with c =√x2 + y2 and some angle
θ, or (ii) equivalently, in the sinusoidal form, c(cos θ + i sin θ).
c y
xθ
Figure 0.3: Geometric interpretation of Euler’s for-mula.
A special case of Euler’s formula is θ = π (i.e. 180◦), in which case:
eiπ + 1 = 0
This formula is known as Euler’s identity.
0.4 Linear Algebra
In physics, a vector is an object with both length (or magnitude) and
a direction in space. We denote it ~r, and its length is denoted |~r|. Since
we interpret vectors as direction, multiplying a vector by a negative scalar
changes its direction. For instance, −2~r is twice the length of ~r (i.e. it has
twice the magnitude), but it points in the opposite direction.
For now, let us work in a Cartesian coordinate space with three dimen-
sions: (x, y, z). Of course, the results below extend to any finite number of
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dimensions, and even more generally to any vector space.
Definition 0.9 (Linear independence) The vectors (~ex, ~ey, ~ez) are lin-
early independent if, for all ax, ay, az ∈ R, ax~ex + ay~ey + az~ez = ~0 implies
ax = ay = az = 0.
Definition 0.10 (Spanning property) The vectors (~ex, ~ey, ~ez) are said
to span ~V if there exist numbers Vx, Vy, Vz ∈ R such that ~V = Vx~ex +
Vy~ey + Vz~ez.
Definition 0.11 (Basis vectors) The vectors (~ex, ~ey, ~ez) form a basis of
~V if (i) they are linearly independent; (ii) they span ~V .
Thus, a vector ~V can be written in terms of a basis as ~V = Vx~ex +
Vy~ey + Vz~ez. The numbers Vx, Vy and Vz are the so-called components (or
coordinates) of ~V , and by the linear independence property they are uniquely
determined. By convention, we will denote a vector by the unique list of its
components, i.e. ~V = (Vx, Vy, Vz).
The simplest basis for any vector in is the standard basis, with ~ex =
(1, 0, 0), ~ey = (0, 1, 0), and ~ez = (0, 0, 1).
For any vector ~V = (Vx, Vy, Vz), Pythagoras’ Theorem relates the coeffi-
cients to the vector’s magnitude as follows:
|~V | =√V 2x + V 2
y + V 2z (0.8)
which may serve as a definition of magnitude. Then, the following prop-
erties hold:
• Scalar multiplication: For any α ∈ R,
α~V = (αVx, αVy, αVz)
• Addition: For any two vectors ~A and ~B, ~A+ ~B is obtained by adding
up their corresponding components:
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(~A+ ~B
)x
= Ax +Bx
and similarly for y and z, where(~A + ~B
)x
denotes the x component
of the vector ~A+ ~B.
The product of vectors can be performed using the dot product or the
cross product. The dot product (or scalar product) of two vectors results
in a scalar number. Where θ = ∠ ~A~B denotes the angle between the two
vectors, the dot product of ~A and ~B is:
~A · ~B = | ~A|| ~B| cos θ (0.9)
Equivalently, in terms of the components of ~A and ~B, we can write:
~A · ~B = AxBx +AyBy +AzBz
Note, for instance, that | ~A|2 = ~A · ~A, a direct implication of (0.8) and
(0.9). Moreover, note that the dot product is negative if cos θ < 0. From
Figure 0.1, this is the case, for instance, if θ ∈[π2 , π
](i.e. an angle of more
than 90◦ but less than 180◦).
Definition 0.12 (Orthogonality) Two vectors ~A and ~B are called orthog-
onal if they are perpendicular. We denote this by ~A ⊥ ~B.
An implication of orthogonality is that the dot product is zero. To see
this, let ~A and ~B be orthogonal. Then θ ≡ ∠ ~A~B = π2 radians (that is, 90◦),
so cos θ = 0. Using (0.9), then ~A · ~B = 0.
The cross product (or vector product) of two vectors results in another
vector. The cross product can only be performed in R3. Where θ = ∠ ~A~B
denotes the angle between the two vectors, the cross product of ~A and ~B is:
~A× ~B =(| ~A|| ~B| sin θ
)~e (0.10)
where ~e is a unit vector perpendicular to the plane containing ~A and ~B.
Thus, ~A× ~B is also perpendicular to the plane containing ~A and ~B.
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For (x, y, z) coordinates, if (x, y) is the plane containing ~A and ~B, then
~A× ~B is contained in z. In the standard representation where x is depth, y
is width, and z is height, the direction of ~A× ~B (i.e. the direction that ~e is
pointing to) is as follows:
• If θ < π (that is θ < 180◦), then sin θ > 0, and thus ~A × ~B points
toward z = +∞ (i.e. “upward direction”, that is toward more positive
z values).
• If θ > π (that is θ > 180◦), then sin θ < 0, and thus ~A × ~B points
toward z = −∞ (i.e. “downward direction”, that is toward more
negative z values).
• If θ = π or θ = 0 (that is, ~A and ~B are parallel), then sin θ = 0, and
thus ~e = ~A× ~B = ~0.
Because ~e is a unit vector, the magnitude of the cross product is:
| ~A× ~B| = | ~A|| ~B| sin θ
Therefore:
• If θ = π/2 (i.e. ~A ⊥ ~B), we have sin θ = 1, so | ~A × ~B| = | ~A|| ~B|.In words, the magnitude of the cross product of two perpendicular
vectors is the product of their lengths.
• If ~A and ~B are parallel (θ = 0 or θ = π), so that sin θ = 0, then
~A× ~B = ~0, and so | ~A× ~B| = 0. In words, if two vectors are parallel,
their cross product is the zero vector, and thus its magnitude is zero.
Some other useful properties of the cross product are as follows:
• Anticommutative: ~A× ~B = − ~B × ~A.
• Distributive over addition: ~A× ( ~B + ~C) = ~A× ~B + ~A× ~C.
• Scalar multiplication: (k ~A)× ~B = ~A× (k ~B) = k( ~A× ~B), ∀k ∈ R.
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• The product rule in differentiation applies:
d
dt( ~A× ~B) =
d
dt~A× ~B + ~A× d
dt~B
Definition 0.13 (Eigenvalues and Eigenvectors) A n× 1 column vec-
tor ~v 6= ~03 is an eigenvector of an n-dimensional square matrix A if:
A~v = λ~v
where λ is a scalar known as the eigenvalue associated to ~v.
Notice that the n × 1 vector A~v can be though of as a linear transfor-
mation of ~v. Therefore, ~v is an eigenvector if the linear transformation A~v
does not change the direction of the vector, but only scales it by some scalar
λ, called the eigenvalue.
By definition, to find an eigenvalue λ and the associated eigenvector ~v
of a squared matrix A, we may solve the linear equation:
(A− λIn
)~v = ~0
where ~0 is an n × 1 column vector of zeros, and In is an n-dimensional
identity matrix. Furthermore, since we are looking for a vector ~v 6= ~0, then
it must be that:
Det(A− λIn
)= 0
This is known as the characteristic polynomial of A. Since this is a
polynomial of degree n, we can factored into the product of n linear terms:
Det(A− λIn
)= (λ1 − λ)(λ2 − λ) · · · (λn − λ)
where λi, for i = 1, . . . , n, is the i-th root. The numbers λ1, . . . , λn may
be complex numbers, and may not all have distinct values.
3 By this notation we mean that ~v has at least one non-zero entry.
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Example 0.2 (Two dimensions example) Consider A =
(2 1
1 2
). Then,
A − λI2 =
(2− λ 1
1 2− λ
), so Det(A − λI2) = 3 − 4λ + λ2. If λ is an
eigenvalue, then 3−4λ+λ2 = 0, so λ = 1 and λ = 3 are the two eigenvalues
of A. The corresponding eigenvectors ~vλ satisfy A~v = λ~v, so:
2v1 + v2 =λv1
v1 + 2v2 =λv2
Thus, for λ = 1, ~vλ=1 = (1,−1)>, and for λ = 3, ~vλ=3 = (1, 1)>.
Example 0.3 (Three dimensions example) Consider A =
2 0 0
0 3 4
0 4 9
.
Then, A−λI3 =
2− λ 0 0
0 3− λ 4
0 4 9− λ
, so Det(A−λI3) = −λ3+14λ2−
35λ+ 22. If λ is an eigenvalue, then −λ3 + 14λ2 − 35λ+ 22 = 0, so λ = 1,
λ = 2, and λ = 11 are the three eigenvalues of A. The corresponding eigen-
vectors are ~vλ=1 = (0, 2,−1)>, ~vλ=2 = (1, 0, 0)>, and for ~vλ=11 = (0, 1, 2)>.
Definition 0.14 (Trace) The trace of an n-dimensional square matrix A =
(aij) is the sum of its diagonal elements, Tr(A) =∑n
i=1 aii.
Trace and determinants are related in different ways. One relation is
given by the eigenvalues of the matrix. Where λi is the i-th eigenvalue of a
square n-by-n matrix A, we have:
Tr(A) =
n∑i=1
λi Det(A) =n∏i=1
λi
The eigenvalues of a matrix can also be used for the following decompo-
sition. Suppose A has n linearly independent eigenvectors ~v1, . . . , ~vn, with
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corresponding eigenvalues λ1, . . . , λn (which need not be distinct). That is,
(~v1, . . . , ~vn) is a basis. Then, we have:
A = QΛQ−1
where Q = [~v1 · · ·~vn] is an n × n matrix whose columns are the n lin-
early independent eigenvectors of A, and Λ = diag(λ1, . . . , λn) is a diagonal
matrix with the corresponding eigenvalues on the diagonal.
0.5 Calculus
The derivative of a univariate function f(t) with respect to t will be
denoted:
df(t)
dt≡ lim
∆t→0
f(t+ ∆t)− f(t)
∆t
The integral is the area below a function f(t) in the interval [a, b], and
is denoted: ∫ b
af(t)dt ≡ lim
∆t→0
∑i
f(ti)∆t
which results from dividing and adding up the areas of small rectangles
of base ∆t.
Result 0.2 (Fundamental Theorem of Calculus) Differentiation and in-
tegration are reciprocal in the following sense:∫ b
af(t)dt = F (b)− F (a)
where F : T 7→∫ Ta f(t)dt and f(t) = dF (t)
dt .
Result 0.3 (Integration by parts) For any two functions f and g:∫ b
ag(x)
df(x)
dx= f(x)g(x)
∣∣∣ba−∫ b
af(x)
dg(x)
dx(0.11)
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Example 0.4 To compute the integral of∫ π/2
0 x cosxdx, note cosx = d sinxdx .
Using (0.11):
∫ π/2
0x cosxdx =
∫ π/2
0x
d sinx
dxdx
= x sinx∣∣∣π/20−∫ π/2
0
dx
dxsinxdx
=π
2sin
π
2−∫ π/2
0sinxdx
=π
2− (− cosx)
∣∣∣π/20
=π
2− 1
For multivariate functions, we may speak of partial differentiation. For
a function f(x1, . . . , xk), we define, for any i = 1, . . . , k:
∂f
∂xi≡ lim
∆xi→0
f(x1, . . . , xi + ∆xi, . . . , xk)− f(x1, . . . , xi, . . . , xk)
∆xi
We may collect partial derivatives into a gradient vector :4
Definition 0.15 (Gradient Vector) The gradient of a multi-variate func-
tion is the vector of partial derivatives with respect to each argument:
∂f
∂~x≡(∂f
∂x1, . . . ,
∂f
∂xk
)Equivalently, we may write ∂f
∂~x =∑k
j=1∂f∂xj
~ej , where ~ej is the standard
basis vector in the j-th coordinate.
Definition 0.16 (Stationary Point) A stationary point of a multivariate
function F satisfies ∂F∂~x = ~0 (i.e. ∂xiF = 0, ∀xi).
Definition 0.17 (Optima for univariate functions) For a univariate func-
tion F (x), a local maximum (minimum) is a stationary point for whichd2
dx2F (x) < (>) 0. If d2
dx2F (x) = 0, then x is called a point of inflection.4 For the gradient we will use the same notation as with the simple partial derivative,
except we denote the partial with respect to the whole vector ~x ≡ (x1, . . . , xk).
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Definition 0.18 (Hessian matrix) For a function F (x1, . . . , xk) with k ≥2, the Hessian matrix H is the matrix of second partial derivatives of F :
H ≡
∂2F∂x2
1
∂2F∂x1∂x2
. . . ∂2F∂x1∂xk
∂2F∂x2∂x1
∂2F∂x2
2. . . ∂2F
∂x2∂xk...
.... . .
...∂2F
∂xk∂x1
∂2F∂xk∂x2
. . . ∂2F∂x2k
Notice that the Hessian is symmetric along its diagonal, for ∂2F
∂xi∂xj=
∂2F∂xj∂xi
, ∀i, j = 1, . . . , k.
Result 0.4 (Stationary points for bivariate functions) Consider a bi-
variate function F (x, y) with Hessian matrix H. Suppose point (x, y) is a
stationary point, i.e. ∂xF = ∂yF = 0 at (x, y) = (x, y). Then, to determine
whether this point is a local maximum, a local minimum, or a saddle point,
we may use the following rules:
• If Tr(H) > 0 and Det(H) > 0, then (x, y) is a local minimum.
Equivalently, H is positive definite (i.e. strictly positive eigenvalues).
• If Tr(H) < 0 and Det(H) > 0, then (x, y) is a local maximum.
Equivalently, H is negative definite (i.e. strictly negative eigenvalues).
• If Det(H) < 0, then (x, y) is a saddle irrespective of the trace.
Equivalently, H is indefinite (i.e. eigenvalues have mixed signs).
These rules are only valid for bivariate functions (i.e. 2 × 2 Hessian
matrices). They extend non-trivially to functions of k ≥ 3 variables.
Example 0.5 (Finding Stationary Points) Consider F (x, y) = sinx +
sin y. Then ∂xF = cosx and ∂yF = cos y. The Hessian matrix is:
H =
(∂2F∂x2
∂2F∂x∂y
∂2F∂y∂x
∂2F∂y2
)=
(− sinx 0
0 − sin y
)Notice that Det(H) = sinx sin y and Tr(H) = − sinx− sin y.
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Classical Mechanics Pau Roldan-Blanco
• Clearly, (x, y) =(π2 ,
π2
)is a stationary point, as we recall from Figure
0.1 that cos(π/2) = 0. At (x, y) =(π2 ,
π2
), we have Det(H) = 1 > 0
(recall that sin(π/2) = 1) and Tr(H) = −2 < 0. Therefore, the point
is a maximum.
• Likewise, (x, y) =(π2 ,−
π2
)is also stationary, with Det(H) = −1 < 0
and Tr(H) = 0, so the point is a saddle. Similarly, (x, y) =(−π
2 ,π2
)is also a saddle point.
• Finally, (x, y) =(−π
2 ,−π2
)is a stationary point, with Det(H) = 1 > 0
and Tr(H) = 2 > 0, so this point is a minimum.
Now consider F (x, y) = cosx + cos y. Then ∂xF = − sinx and ∂yF =
− sin y. The Hessian matrix is:
H =
(− cosx 0
0 − cos y
)so Det(H) = cosx cos y and Tr(H) = − cosx− cos y.
• The point (x, y) = (π, π) is a stationary point, as we recall from Figure
0.1 that sinπ = 0. At (x, y) = (π, π), we have Det(H) = 1 > 0 (recall
that cosπ = −1) and Tr(H) = 2 > 0. Therefore, the point is a
minimum.
• It is easily checked that (π,−π), (−π, π), and (−π,−π) are all station-
ary and minima.
0.6 Taylor Approximations
In certain situations we will have to approximate functions around a
point to gain analytical tractability. For this, we will typically use Taylor
expansions:
Definition 0.19 (Taylor series) Let k ≥ 1 be an integer and let the func-
tion f : R → R be k-times differentiable at some point a ∈ R. Then, the
Taylor series (or expansion) of f to the k-th order is:
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Classical Mechanics Pau Roldan-Blanco
T k(x, a) ≡ f(a)+df(x)
dx
∣∣∣x=a
(x−a)+1
2!
d2f(x)
dx2
∣∣∣x=a
(x−a)2+· · ·+ 1
k!
dkf(x)
dxk
∣∣∣x=a
(x−a)k
Example 0.6 (Second-order expansions) The second-order Taylor ex-
pansion of f around x = a is f(a) + f ′(a)(x− a) + f ′′(a) (x−a)2
2 . Examples:
• For f(x) = lnx, then f ′(x) = 1/x, and f ′′(x) = −1/x2, so:
T 2(x, a) = ln a+x− aa− 1
2
(x− aa
)2
For instance, around a = 1, then T 2(x, 1) = (x− 1)− 12 (x− 1)2.
• For f(x) = ex, then f ′(x) = ex, and f ′′(x) = ex, so:
T 2(x, a) = ea(
1 + x− a+(x− a)2
2
)For instance, around a = 0, then T 2(x, 0) = 1 + x+ x2
2 .
• For f(x) = sinx, then f ′(x) = cosx, and f ′′(x) = − sinx, so:
T 2(x, a) = sin a+ (x− a) cos a− (x− a)2
2sin a
For instance, around a = 0, then T 2(x, 0) = x (using that sin 0 =
1− cos 0 = 0).
• For f(x) = cosx, then f ′(x) = − sinx, and f ′′(x) = − cosx, so:
T 2(x, a) = cos a− (x− a) sin a− (x− a)2
2cos a
For instance, around a = 0, then T 2(x, 0) = 1− x2
2 .
Result 0.5 (Taylor’s Theorem) Let k ≥ 1 be an integer and let the func-
tion f : R → R be k-times differentiable at some point a ∈ R. Then, there
exists a function hk : R→ R such that:
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Classical Mechanics Pau Roldan-Blanco
f(x) = T k(x, a) + hk(x)(x− a)k
where T k(x, a) is the k-th order Taylor expansion of f around x = a,
and limx→a hk(x) = 0.
Then, to approximate a function f(x) around some x = a, we may use
Taylor’s Theorem. In particular, f(x) ≈ T k(x, a) around x = a. Since the
error of approximation (the term hk(x) in Taylor’s Theorem) declines with
k, the approximation is better for higher-order expansions.
Example 0.7 (k-th order expansions) Generally, for expansions of or-
der k:
• For f(x) = lnx, then T k(x, a) = ln a +∑k
n=1(−1)n
(n+1)!
(x−aa
)n+1. For
instance, around x = 1, lnx ≈∑k
n=1(−1)n (x−1)n+1
(n+1)! .
• For f(x) = ex, then T k(x, a) = ea∑k
n=1(x−a)n
n! . For instance, around
x = 0, ex ≈∑k
n=1xn
n! .
• For f(x) = sinx, then T k(x, a) = sin a+ (x− a) cos a− (x−a)2
2 sin a−(x−a)3
3! cos a + (x−a)4
4! sin a + . . . For instance, around x = 0, sinx ≈x− x3
3! + x5
5! −x7
7! + . . .
• For f(x) = cosx, then T k(x, a) = cos a− (x− a) sin a− (x−a)2
2 cos a+(x−a)3
3! sin a + (x−a)4
4! cos a + . . . For instance, around x = 0, cosx ≈1− x2
2 + x4
4! −x6
6! + . . .
Example 0.8 (Euler’s Formula proof) One way of proving Euler’s For-
mula (Result 0.1), is to use an infinite Taylor expansion for eit around x = 0.
Indeed, note:
eit =+∞∑n=0
(it)n
n!=
+∞∑n=0
(−1)nt2n
(2n)!+ i
+∞∑n=0
(−1)n−1t2n−1
(2n− 1)!
We recognize the Taylor expansion for cos t in the first summation, and
that of sin t in the second one. Thus, eit = cos t+ i sin t.
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0.7 Differential Equations
Definition 0.20 (Differential equation) A differential equation (DE) is
an equation relating a function with its derivatives.
DEs may be distinguished along different dimensions. The most common
distinctions are the following:
1. Ordinary/Partial DEs:
• Ordinary differential equations (ODE) are equations containing
an unknown function y of a single independent variable t, its
derivatives, and some known functions of t.
• Partial differential equations (PDE) are equations containing un-
known multivariate functions and its (partial) derivatives.
2. Linear/Non-linear DEs:
Linear DEs involve exclusively a linear polynomial in the unknown
function and its derivative. A k-th order linear ODE has the form:
a0(t)y + a1(t)dy
dt+ a2(t)
d2y
dt2+ · · ·+ ak(t)
dky
dtk+ b(t) = 0 (0.12)
where a0(t), . . . , ak(t) and b(t) are arbitrary but known functions of t
(these functions need not be linear), and y is the unknown function
(of t) that we are solving for. Further, if a0, . . . , ak are constant in t,
then (0.12) is called a constant-coefficient linear ODE.
A linear PDE would be similar, with the difference that y (and possibly
a0, . . . , ak, b) would be multivariate, and the derivatives in (0.12) would
be partial derivatives.
3. Homogenous/nonhomogenous DEs:
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Classical Mechanics Pau Roldan-Blanco
Homogenous DEs involve functions that are homogenous of the same
degree.5 DEs are nonhomogenous if this requirement fails.
Some examples:
• dydt = cy + t2 is an nonhomogenous (first-order, linear, ordinary)
DE, while dydt = cy is homogenous.
• The second-order linear ODE sin(x)d2ydx2 +ady
dx +y = 0 is homoge-
nous, whereas 2x2 d2ydx2 + axdy
dx + y = 2 is nonhomogenous.
• The second-order linear ODE ad2ydx2 + bdy
dx + cy = f(x) is homoge-
nous only if f(x) = 0, ∀x. Otherwise, it is nonhomogenous.
Solution methods for DEs vary depending on the type of DE. Sometimes,
the same DE can be solved using different methods. Often, a DE can only be
solved numerically. Selecting the right method is very much case-specific.
Here, we will introduce some of the more popular methods for the most
common types of DEs that we encounter in physics.
• We will start with a general method that is valid for both ODEs and
PDEs, as long as they are linear and homogenous (Method I).
• We then present a method that works for linear, first-order ODEs,
even when not homogenous (Method II).
• Then, we will see methods for solving constant-coefficient, second-
order linear ODEs, both homogenous (Method III) and nonhomoge-
nous (Method IV).
• Finally, we will move to ODEs with non-constant coefficients for both
the homogenous (Method V) and the nonhomogenous (Method VI)
cases.
Remark 0.1 (Method I: Separation of Variables) .
Used for: Linear and homogenous ODEs and PDEs.
5 A multivariate function g(x1, . . . , xk) is said to be homogenous of degree n ifg(αx1, . . . , αxk) = αng(x1, . . . , xk), for all α 6= 0.
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Classical Mechanics Pau Roldan-Blanco
The method: The method of separation of variables (or Fourier method)
consists of conjecturing that the solution for a DE of the unknown function
y on t can be written as a product solution, that is:
dy
dt= g(t)h(y)
for some continuous functions g, h.
Example 0.9 (Heat Equation) Consider the bivariate function u ≡ u(x, t),
obeying the PDE:6
∂u
∂t= α
∂2u
∂x2
By the Method of Separation of Variables, we conjecture a solution of
the form:
u(x, t) = X(x)T (t)
Plugging back and using the product rule, we find T ′(t)αT (t) = X′′(x)
X(x) . Since
the left- (right-) hand side is constant in x (in t), both sides are equal to
some constant −λ, that is:
T ′(t) = −λαT (t) and X ′′(x) = −λX(x)
Here, λ is known as the separation constant.7 Importantly, we have
turned the problem into two ODEs:
• First, note T ′(t)T (t) = −λα. Taking integrals, lnT (t) = A−
∫λαdt, so
T (t) = Ae−λαt
where A ∈ R is the constant of integration.
6 This is in fact a famous equation in physics, called the heat equation, and describingthe variation in temperature u in a given region x over time t.7 The minus sign on λ will be convenient later on if we assume λ > 0.
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Classical Mechanics Pau Roldan-Blanco
• Second, to solve X ′′(x) = −λX(x), conjecture a solution of the form
X(x) = B sin(bx) + C cos(cx) for some coefficients B,C, and some
numbers b, c to be verified. Then, X ′(x) = Bb cos(bx)−Cc sin(cx) and
X ′′(x) = −Bb2 sin(bx)− Cc2 cos(cx), so
−λ =X ′′(x)
X(x)= −Bb
2 sin(bx) + Cc2 cos(cx)
B sin(bx) + C cos(cx)
Thus, λB sin(bx) + λC cos(cx) = Bb2 sin(bx) + Cc2 cos(cx), and thus
b = c =√λ. Hence:
X(x) = B sin(√λx) + C cos(
√λx)
In sum, we have found the solution for the heat equation:
u(x, t) = Ae−λαt(B sin(
√λx) + C cos(
√λx))
for some constants A,B,C, λ.8
Remark 0.2 (Method II: Integrating Factor) .
Used for: Linear, nonhomogenous, first-order ODEs:9
y′ + a(t)y = b(t)
when a(t), b(t) are continuous functions of t, and y′ ≡ dydt .
The method: First, we conjecture the existence of a function µ(t),
called the integrating factor, which solves:
µ′(t) = µ(t)a(t)
8 The precise values of these constants can be deduced from the four boundary con-ditions of the system. If x ∈ [0, x] and t ∈ [0, t], the four boundary conditions are thenumbers u(0, t), u(x, 0), u(x, t), and u(0, 0).9 We have normalized the coefficient on y′′ to one. This is with no loss of generality.
Indeed, consider c(t)y′+ a(t)y = b(t). Then, just divide both sides by c(t) and definea(t) = a(t)/c(t) and b(t) = b(t)/c(t), and we obtain the equivalent representationy′ + a(t)y = b(t).
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Multiplying both sides of our ODE by µ(t) yields µ(t)y′+µ′(t)y = µ(t)b(t).
Noting the product rule on the LHS, we can write the ODE as:(µ(t)y
)′=
µ(t)b(t). Integrating both sides:
y =c1 +
∫µ(t)b(t)dt
µ(t)(0.13)
where c1 is a constant of integration. Now, all we need is an expression
for µ(t) to plug into equation (0.13). Since µ′(t)µ(t) = a(t) by assumption, then
integrating both sides yields lnµ(t) = c2 +∫a(t)dt, where c2 is a constant of
integration. Therefore, we have:
µ(t) = c2e∫a(t)dt
where c2 ≡ ec2. Substituting this into (0.13) gives:
y(t) =c1 +
∫c2e
∫a(t)dtb(t)dt
c2e∫a(t)dt
=c+
∫e∫a(t)dtb(t)dt
e∫a(t)dt
where c ≡ c1c2
. Often, the constant of integration can be deduced from
initial conditions. For instance, if a(t) and b(t) are defined on R+, with
a(0) = a0 and b(0) = b0 (say, because t represents real time), then µ(0) = ea0
and y(0) = c+ea0b0ea0 , so c = ea0(y(0)− b0), and we can write the solution for
the ODE as:
y(t) = ea0−∫a(t)dt
(y(0)− b0
)+ e−
∫a(t)dt
∫e∫a(t)dtb(t)dt
Thus, we have found the following result:
Result 0.6 (Solution for first-order linear ODEs) The solution of first-
order linear ODEs of the form y′ + a(t)y = b(t) is:
y(t) =c+
∫µ(t)b(t)dt
µ(t)
where µ(t) ≡ e∫a(t)dt, and c is a constant of integration determined by
initial conditions.
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Classical Mechanics Pau Roldan-Blanco
The next two methods will show how to solve in practice for linear,
second-order, constant-coefficient ODEs. That is, we will deal with ODEs
of the type:
y′′ + by′ + cy = d(t) (0.14)
where b, c are constants, and d(t) is continuous in t.10 We will see both
the homogenous case (Method III, Remark 0.4), i.e. d(t) ≡ 0; and the
nonhomogenous case (Method IV, Remark 0.5), i.e. d(t) 6≡ 0. In Method V
(Remark 0.7), we will then relax the assumption of constant coefficients.
We begin by stating some general properties and definitions of second-
order linear ODEs:
Result 0.7 (Principle of superposition) If y1, y2 are solutions to a lin-
ear, second-order ODE with constant coefficients (equation (0.14), with d(t) ≡0), then the linear combination:
y = C1y1 + C2y2
is also a solution, for all coefficients C1, C2 ∈ R.
As C1, C2 are arbitrary constants, this implies that linear, homogenous,
second-order, constant-coefficient ODE exhibit infinitely many solutions.
The coefficients are then pinned down by the initial conditions.
Example 0.10 Consider y′′−y = 0. Then y = C1et+C2e
−t is a solution for
any value of C1, C2. Indeed, y′ = C1et −C2e
−t and y′′ = C1et +C2e
−t = y,
so y′′ = y, ∀C1, C2. For example, 3et is a solution, 5e−t is a solution, and
3et + 5e−t is a solution. If y(0) = 3 and y′(0) = 1, then using the solution
y(t) = C1et+C2e
−t at t = 0, we have 3 = c1 + c2 and 1 = c1− c2, so C1 = 2
and C2 = 1. Thus, y(t) = 2et + e−t.
More generally, the initial conditions are numbers:
10 Again, the coefficient on y′′ is normalized to one wlog.
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Classical Mechanics Pau Roldan-Blanco
y(t0) = y0 and y′(t0) = y′0
Therefore, where (y1, y2) are solutions, C1 and C2 solve the system of
equations:
C1y1(t0) + C2y2(t0) =y0 (0.15a)
C1y′1(t0) + C2y
′2(t0) =y′0 (0.15b)
Solving, we obtain:
C1 =y0y′2(t0)− y′0y2(t0)
y1(t0)y′2(t0)− y′1(t0)y2(t0)and C2 =
−y0y′1(t0) + y′0y1(t0)
y1(t0)y′2(t0)− y′1(t0)y2(t0)
Note we can write these in terms of determinants as C1 = Det(A1)W and
C2 = Det(A2)W , where:
A1 ≡
(y0 y2(t0)
y′0 y′2(t0)
); A2 ≡
(y0 y2(t0)
y′0 y′2(t0)
)and
W ≡ Det
(y1(t0) y2(t0)
y′1(t0) y′2(t0)
)= y1(t0)y′2(t0)− y′1(t0)y2(t0)
The number W ≡Wy1,y2(t0) is the so-called Wronskian determinant or,
simply, the Wronskian of functions y1 and y2.
Definition 0.21 (Wronskian) The Wronskian determinant of two func-
tions y1 and y2 is:
Wy1,y2(t) ≡ Det
(y1(t) y2(t)
y′1(t) y′2(t)
)= y1(t)y′2(t)− y′1(t)y2(t)
Crucially, for the constants C1, C2 to exist, we need W 6= 0 at t = t0.
Then, we have arrived at the following result:
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Classical Mechanics Pau Roldan-Blanco
Result 0.8 (Fundamental solutions) Suppose y1 and y2 are solutions to
a linear, second-order ODE (equation (0.14), with d(t) ≡ 0). If there is a
point t0 such that Wy1,y2(t0) 6= 0, then every solution to the ODE can be
expressed as:
yc = C1y1 + C2y2
for some constants C1, C2. The solutions y1 and y2 are called the fun-
damental solutions, and yc is called the general solution.
Example 0.11 (Example 0.10 cont’d) For y′′ − y = 0, we found that
y1 = et and y2 = e−t are two solutions. The Wronskian of y1 and y2 is
W = y1y′2 − y′1y2 = −ete−t − ete−t = −2e0 = −2 6= 0, ∀t
Thus, y1 and y2 are two fundamental solutions to y′′ − y = 0, and can
be used to construct the whole set of solutions. In particular, the set of
solutions is {yc : yc = C1et + C2e
−t;C1, C2 ∈ R}.
Remark 0.3 (Linear independence) The following statements are equiv-
alent:
• The functions y1, y2 are fundamental solutions.
• The functions y1, y2 are linearly independent.
• The Wronskian of y1 and y2 is non-zero, Wy1,y2 6= 0.
To see this, recall from Definition 0.9 that y1 and y2 are linearly in-
dependent if C1y1 + C2y2 = 0 implies C1 = C2 = 0. Above, we found
that (y1, y2) and (C1, C2) are related through the initial conditions (y0, y′0) ≡
(y(t0), y′(t0)) via the system of equations (0.15a)-(0.15b). Then, (y1, y2)
will be linearly independent if the only solution for the constants when y0 =
y′0 = 0 is C1 = C2 =. We derived that C1 =y0y′2(t0)−y′0y2(t0)
W and C2 =−y0y′1(t0)+y′0y1(t0)
W , where W ≡ y1(t0)y′2(t0)− y′1(t0)y2(t0) 6= 0 since y1, y2 are
fundamental solutions. Then, if y0 = y′0 = 0, the only solution to the system
is indeed C1 = C2 = 0. �
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Classical Mechanics Pau Roldan-Blanco
Then, the sense in which the solutions y1, y2 are fundamental is that any
linear combination between them is also a solution, but neither y1 nor y2
can be written as linear combinations of other solutions.
To explain the general toolkit to solve for these ODEs, we will also need
the following terminology:
Definition 0.22 (Auxiliary equation) The auxiliary (or characteristic)
equation of a linear, second-order ODE with constant coefficients (equation
(0.14), with d(t) ≡ 0) is the equation:
α2 + bα+ c = 0
It follows that, if (α1, α2) are the roots of the auxiliary equation, then
(α− α1)(α− α2) = 0.
We are now ready to explain the method:
Remark 0.4 (Method III: Exponential guess) .
Used for: Linear, homogenous, second-order ODEs with constant coef-
ficients (equation (0.14), with d(t) ≡ 0).
The method: By Result 0.8, we know there exist two fundamental
(linearly independent) solutions, (y1, y2). To find them, conjecture that yj =
eαjt, for j = 1, 2, where α1, α2 need to be found. Note that y′j = αjeαjt and
y′′j = α2jeαjt. The Wronskian is:
W = (α2 − α1)e(α1+α2)t
Therefore, if the roots are distinct, W 6= 0 and y1, y2 are fundamental
solutions (and linearly independent). Since yj is a solution, then they satisfy
the ODE, α2jeαjt + bαje
αjt + ceαjt = 0, that is:
eαjt(α2j + bαj + c
)= 0
This is satisfied for all t if, and only if:
α2j + bαj + c = 0
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Classical Mechanics Pau Roldan-Blanco
This is the auxiliary, or characteristic, equation (Definition 0.22). Thus,
to solve the ODE we must find the roots of the auxiliary equation. These
roots might be real or complex, depending on parameters. Thus, we must
consider different cases:
1. Case 1: If b2 > 4c, then the roots are real and distinct, given by
(α1, α2) = 12
(−b+
√b2 − 4c,−b−
√b2 − 4c
)∈ R2. Using Result 0.8,
then the general solution of (0.14) is:
y = C1e(−b+
√b2−4c) t2 + C2e
(−b−√b2−4c) t2
for some arbitrary constants C1, C2.
2. Case 2: If b2 = 4c, then the roots are real but identical, given by
α1 = α2 = − b2 . Thus, now W = 0, so the method is giving us two
(trivially) linearly dependent solutions which are, in fact, one and the
same: y1 = e−b2t.
Yet, we know (Result 0.8) that there must exist a second fundamental
solution, y2. To find it, Result 0.7 tells us that if y1 is a solution, then
y2 = vy1 is also a solution, for any function v ≡ v(t). Thus, we can
just figure out a functional form for v that satisfies the ODE. Note:
y′ = v′e−b2t − b
2ve−
b2t and y′′ = v′′e−
b2t − b
2v′e−
b2t +
b2
4ve−
b2t
Into the ODE y′′ + by′ + cy = 0, we get, after some algebra:
v′′ −(b2 − 4c
4
)v = 0
and, therefore, v′′ = 0. This implies v = k3t+k4, for some k3, k4 ∈ R,
so y2 = vy1 = (k4t + k4)e−b2t. Thus, the fundamental solutions are
y1 = e−b2t and y2 = te−
b2t, and general solution is y = k1e
− b2t +
k2(k3t+ k4)e−b2t, or (since the constants are arbitrary):
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Classical Mechanics Pau Roldan-Blanco
y = (C1 + C2t)e− b
2t
This is thus the general solution when the roots of the characteristic
equation are repeated.11
3. Case 3: If b2 < 4c, then the roots (α1, α2) are imaginary numbers,
and one root is the complex conjugate of the other. That is:
α1 = β + γi and α2 = β − γi
for some (β, γ), where i =√−1. Using Result 0.8, then the general
solution of (0.14) is:
y = C1e(β+γi)t + C2e
(β−γi)t
= eβt(C1e
γit + C2e−γit
)= eβt
[C1(cos γt+ i sin γt) + C2(cos γt− i sin γt)
]= eβt
[(C1 + C2) cos γt+ (C1i− C2i) sin γt
]where the third equality uses Euler’s formula (Result 0.1). Writing
A = C1 + C2 and B = C1i− C2i, we have found the solution:
y = eβt(A cos γt+B sin γt
)for some real number A and some imaginary number B.
The next method generalizes Method III when the second-order, linear
ODE with constant coefficients is nonhomogenous (i.e. d(t) 6≡ 0). First, we
introduce two more pieces of terminology:
11 One could further verify that y1 = e−b2t and y2 = te−
b2t are indeed linearly inde-
pendent. For this, we can simply check that Wy1,y2 6= 0 (Remark 0.3). Indeed, somealgebra shows that W = e−bt 6= 0, as expected.
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Classical Mechanics Pau Roldan-Blanco
Definition 0.23 (Complementary function) The complementary func-
tion of an nonhomogenous ODE is the solution yc to its homogenous coun-
terpart.
For linear, second-order, constant-coefficient ODEs, therefore, the com-
plementary function yc solves y′′ + by′ + cy = 0. To obtain yc, we can just
use Method III (Remark 0.4).
Definition 0.24 (Particular integral) A particular integral of an ODE
is any function yp which satisfies the equation. In other words, a particular
integral is any solution of a DE.
Finally, for our next solution method, we will make use of a general
property of ODEs:
Result 0.9 (General solution) The general solution y of a linear ODE
can be written as:
y = yc + yp
where yc is the complementary function and yp is a particular integral.
The language of this Result seems slightly circular, but it should not
lead to confusion. When applied to our context, the Result states that the
general solution of the nonhomogenous ODE is the sum of the complemen-
tary function (i.e. the general solution to its homogenous counterpart) and
any function that is a solution to the ODE.
Remark 0.5 (Method IV: Method of undetermined coefficients) .
Used for: Linear, nonhomogenous, second-order ODEs with constant
coefficients, that is, equation (0.14) when d(t) 6≡ 0.
The method: Result 0.9 has laid out the plan for this method. First,
find the complementary function. Then, find the particular integral. Finally,
add the two together.
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Classical Mechanics Pau Roldan-Blanco
• Step 1: The complementary function is the solution yc to the ho-
mogenous version of (0.14), y′′ + by′ + cy = 0. For this, we can just
use Method III. The solution is (derivations in Remark 0.4):
yc =
C1e
(−b+√b2−4c) t2 + C2e
(−b−√b2−4c) t2 if b2 > 4c
(C1 + C2t)e− b
2t if b2 = 4c
eβt[(C1 + C2) cos γt+ (C1i− C2i) sin γt
]if b2 < 4c
(0.16)
for some constants C1, C2, where β (resp. γ) is the real (resp. imagi-
nary) part of the roots of the auxiliary function, α2 + bα+ c = 0 (i.e.
such that α = β ± γi). Equation (0.16) is thus the complementary
function.
• Step 2: Now, we find a particular integral, yp. This is very much
case-specific, but a commonly fruitful approach is to guess that yp has
the same functional form as d(t), and then use the method of unde-
termined coefficients. Let’s see some examples:
1. d(t) = d (a constant).
Guess yp = A ∈ R. Then y′p = y′′p = 0. Substituting into the
ODE, we get 0 + 0 + cA = d. In sum:
yp =d
c
2. d(t) = d1 + d2t (a line).
Guess yp = At + B. Then y′p = A and y′′p = 0. Substituting
into the ODE, we get 0 + bA + cAt + cB = d1 + d2t. Matching
coefficients, we get cA = d2 (so A = d2/c), and bA+ cB = d1, so
B = 1c
(d1 − bd2
c
). In sum:
yp =1
c
[d1 + d2
(t− b
c
)]37
Classical Mechanics Pau Roldan-Blanco
3. d(t) = d1 cos(d2t).
Guess yp = A cos(d2t) + B sin(d2t). Then y′p = −Ad2 sin(d2t) +
Bd2 cos(d2t) and y′′p = −Ad22 cos(d2t)−Bd2
2 sin(d2t). Substituting
into the ODE, we get:
(Bbd2−Ad22+Ac) cos(d2t)−(Bd2
2+Abd2−Bc) sin(d2t) = d1 cos(d2t)
Matching coefficients, we get (after some algebra):
Bbd2 +A(c− d22) = d1
B(d22 − c) +Abd2 = 0
From the second equation, we get B = Abd2
c−d22. Into the first equa-
tion, we get A =d1(c−d2
2)
b2d22+(c−d2
2)2 . In sum:
yp =d1(c− d2
2) cos(d2t) + d1d2b sin(d2t)
b2d22 + (c− d2
2)2
A similar example is d(t) = d1 sin(d2t), for which the guess should
again be yp = A cos(d2t) +B sin(d2t).
4. d(t) = d1ed2t.
Guess yp = Aed2t. Then y′p = Ad2ed2t and y′′p = Ad2
2ed2t. Substi-
tuting into the ODE, we get:
Aed2t[d2
2 + bd2 + c]
= d1ed2t
Therefore, A = d1
d22+bd2+c
. In sum:
yp =d1
d22 + bd2 + c
ed2t
A similar example is d(t) = d1e−d2t, for which the guess should
be yp = Ae−d2t.
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Classical Mechanics Pau Roldan-Blanco
Whatever the case, the general solution to the ODE is, then:
y = yc + yp
by Result 0.9.
Example 0.12 Consider the ODE:
y′′ + 3y′ − 10y = 3t2
This is a second-order, linear, nonhomogenous ODE, of the form y′′ +
by′ + cy = d(t) with b = 3, c = −10, and d(t) = 3t2. We use Method IV to
solve.
• Complementary function (yc):
Letting y = ekt, so that y′ = kekt and y′′ = k2ekt, the auxiliary function
is k2 + 3k − 10 = 0, so (k − 2)(k + 5) = 0. Thus, the roots are k = 2
and k = −5, so the solutions are y1 = e2t and y2 = e−5t. The general
complementary solution is: yc = C1e2t + C2e
−5t, where C1, C2 are
arbitrary constants.
• Particular integral (yp):
To find a particular integral, note d(t) is quadratic, so we try a quadratic
candidate: yp = At2 +Bt+C. Thus, y′p = 2At+B and y′′p = 2A. Sub-
stituting into the ODE gives 2A+3(2At+B)−10(At2+Bt+C) = 3x2.
Matching coefficients, we have: 2A + 3B − 10C = 0, 6A − 10B = 0,
and −10A = 3, so A = − 310 , B = − 9
50 , and C = − 57500 . Thus,
yp = − 310 t
2 − 950 t−
57500
Thus, the general solution is
y = yc + yp = C1e2t + C2e
−5t − 3
10t2 − 9
50t− 57
500
for some arbitrary constants C1 and C2.
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Classical Mechanics Pau Roldan-Blanco
In the following example, we will see that the choice of the trial particular
integral is not always so straightforward. In particular, our choice here will
not replicate exactly the functional form of the independent function d(t).
The example will demonstrate the sense in which solving these type of DEs
is often very much case-specific.
Example 0.13 Consider the ODE:
y′′ − y′ − 6y = e3t
Now, b = −1, c = −6, and d(t) = e3t. We use Method IV to solve.
• Complementary function (yc):
Letting y = ekt, so that y′ = kekt and y′′ = k2ekt, the auxiliary function
is k2 − k − 6 = 0, so (k − 3)(k + 2) = 0. Thus, the roots are k = 3
and k = −2, so the solutions are y1 = e3t and y2 = e−2t. The general
complementary solution is: yc = C1e3t + C2e
−2t, where C1, C2 are
arbitrary constants.
• Particular integral (yp):
To find a particular integral, note d(t) is exponential, so let’s try an
exponential candidate and see why it fails: yp = Ae3t. Thus, y′p =
3Ae3t and y′′p = 9Ae3t. Substituting into the ODE gives Ae3t(9 − 3 −6) = e3t, a contradiction. Thus, this candidate does not work in this
case.
The reason why the candidate does not work is because the function
d(t) = e3t appeared explicitly in the complementary solution. When-
ever this is the case, it is useful to use the following alternative guess:
yp = Ate3t
that is, the same as before, times t. Now, y′p = Ae3t(3t + 1) and
y′′p = Ae3t(9t + 6). Substituting into the ODE and solving for A, we
will find A = 15 .
40
Classical Mechanics Pau Roldan-Blanco
Thus, the general solution is
y = yc + yp = C1e3t + C2e
−2t +1
5te3t
for some arbitrary constants C1 and C2.
Example 0.14 Consider the ODE:
y′′ − 3y′ − 4y = −8et cos 2t
• Complementary function (yc):
Letting y = ekt, so that y′ = kekt and y′′ = k2ekt, the auxiliary function
is k2 − 3k − 4 = 0, so (k − 4)(k + 1) = 0. Thus, the roots are k =
4 and k = −1, so the solutions are y1 = e4t and y2 = e−t. The
general complementary solution is: yc = C1e4t + C2e
−t, where C1, C2
are arbitrary constants.
• Particular integral (yp):
To find a particular integral, note d(t) is the product of an exponential
and a cosine function, so let’s try the candidate : yp = Aet cos 2t +
Bet sin 2t. Here:
y′p = Aet cos 2t− 2Aet sin 2t+Bet sin 2t+ 2Bet cos 2t
= (A+ 2B)et cos 2t+ (B − 2A)et sin 2t
y′′p = (A+ 2B)et cos 2t− 2(A+ 2B)et sin 2t+ (B − 2A)et sin 2t
+ 2(4B − 3A)et cos 2t
= (4B − 3A)et cos 2t− (4A+ 3B)et sin 2t
Substituting into the ODE and matching coefficients will then yield
A = 10/13, B = 2/13. Therefore, yp = 1013e
t cos 2t+ 213e
t sin 2t.
Thus, the general solution is
41
Classical Mechanics Pau Roldan-Blanco
y = yc + yp = C1e4t + C2e
−t +10
13et cos 2t+
2
13et sin 2t
for some arbitrary constants C1 and C2.
Next, we note that these methods extend naturally to linear, constant-
coefficient ODE of higher order.
Remark 0.6 (Higher-order ODEs) Consider a n-th order, linear, non-
homogenous, constant-coefficient ODE:
n∑j=0
bjdjy
dtj= d(t)
where {bj}nj=0 are constants, and d is continuous in t.12 Again, we use
that the generalized solution is:
y = yc + yp
and we find the complementary solution yc and the particular integral yp.
• Particular integral (yp):
The particular integral will depend on the specific problem, and typi-
cally we will use a method of undetermined coefficients (see above).
• Complementary function (yc):
The auxiliary equation is∑n
j=0 bjαj = 0, and thus the roots {αj}nj=1
solve the polynomial∏nj=1(α − αj) = 0. Then, the complementary
function is:
yc =n∑j=1
Cjeαjt (0.17)
Again, we may have three cases (which we saw already for second-order
ODEs, see Method III):
12 Here, we use the convention d0ydt0
= y.
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Classical Mechanics Pau Roldan-Blanco
1. If all roots αj are real and distinct, then the complementary func-
tion reads exactly as equation (0.17).
2. If each root αj is repeated kj < n times, then equation (0.17)
reads:13
yc =n∑j=1
kj∑`=1
Cj,`t`−1eαjt
3. If there exist complex roots, then we set αj = βj + γji for each of
those roots. Using Euler’s formula (Result 0.1), then:
Cjeαjt = Cje
βjt cos(γjt+ ϕj)
where ϕj is called a phase shift. That is, in this case we must re-
place the terms Cjeαjt in (0.17) on all complex roots by Cje
βjt cos(γjt+
ϕj).
Methods III and IV showed us ways to solve homogenous and nonho-
mogenous linear, second-order ODEs with constant coefficients. The next
two methods will deal with ODEs whose coefficients are non-constant, such
that:
y′′ + b(t)y′ + c(t)y = d(t) (0.18)
When the ODE is homogenous, we can use the following solution method:
Remark 0.7 (Method V: Reduction of order) .
Used for: Linear, homogenous, second-order ODEs with non-constant
coefficients.
The method: Consider a homogenous ODE with varying coefficients,
i.e. impose d(t) ≡ 0 in (0.18). Suppose that y1 is a solution to (0.18). By
13 We derived this equation within Method III for the case of n = 2 (and k = 1),arguing that if the two roots are identical, the root-finding method gives us onlyone of the two fundamental solutions, and the other one we can find via a simpleguess-and-verify approach. The derivation for higher order follows a similar logic.
43
Classical Mechanics Pau Roldan-Blanco
Remark 0.7, then, y2 = vy1 is also a solution, for any function v ≡ v(t).
Note:
y′2 = v′y1 + vy′1 and y′′2 = v′′y1 + 2v′y′1 + vy′′1
Replacing these into the ODE and rearranging terms, we get:
y1v′′ +
(2y′1 + b(t)y1
)v′ +
(y′′1 + b(t)y′ + c(t)y
)︸ ︷︷ ︸=0
= 0
where the last term is zero because y1 is a solution. Thus, we get:
y1u′ +(2y′1 + b(t)y1
)u = 0
a first-order ODE in u ≡ v′. Hence, we have reduced the order of the
problem, from a second-order ODE to a first-order ODE. Using a separation
of variables method (Remark 0.1), we have:
y1du
dt= −
(2y′1 + b(t)y1
)u ⇔
∫du
u= −
∫2y′1 + b(t)y1
y1dt
On the LHS,∫
duu =
∫u′
u dt = lnu. On the RHS:
−∫
2y′1 + b(t)y1
y1dt = −2
∫y′1y1
dt−∫b(t)dt = −2 ln y1 −
∫b(t)dt
Thus, lnu = −2 ln y1 −∫b(t)dt + c, or u = c
y21e−
∫b(t)dt, with c = ec.
Therefore:
v =
∫c
y21
e−∫b(t)dtdt
Thus, the general solution to the ODE is yc = (C1 + C2v)y1 for some
C1, C2.
Example 0.15 Consider:
t2y′′ + 3ty′ + y = 0
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Classical Mechanics Pau Roldan-Blanco
The function y1 = t−1 is a solution. (To see this, note y′1 = −t−2 and
y′′1 = 2t−3, so t2y′′1 + 3ty′1 + y1 = 2t−1 − 3t−1 + t−1 = (2 − 3 + 1)t−1 = 0.)
Now, we use the method of reduction of order to find a second solution. For
this, let y2 = vy1 = vt−1 be a solution. Then:
y′2 = v′t−1 − vt−2 and y′′2 = v′′t−1 − 2v′t−2 + 2vt−3
Substituting these into the ODE and collecting terms:
0 = t2(v′′t−1 − 2v′t−2 + 2vt−3
)+ 3t
(v′t−1 − vt−2
)+ vt−1
= v′′t− 2v′ + 2vt′ + 3v′ − 3vt−1 + vt−1
= tv′′ + v′
Letting u ≡ v′, we now have a first-order ODE: tu′ = u. Solving via
separation of variables:
tdu
dt= −u ⇔
∫du
u= −
∫1
tdt ⇔ lnu = ln t+ C ⇔ u = ct−1
and thus v′ = ct−1, or v = c ln t+ k, for some constants c, k. Thus, the
second solution is y2 = vt−1 = (c ln t+ k)t−1 = ct−1 ln t+ kt−1. Since y1 =
t−1 and y1, y2 are fundamental solutions (and hence linearly independent),
we can just ignore the second term in y2. Therefore, y2 = t−1 ln t, and the
general solution to the ODE is:
y = C1t−1 + C2t
−1 ln t
for arbitrary constants C1, C2.
If ODE (0.18) is nonhomogenous, we can again solve the homogenous
equation first, then find a particular integral, and add the two together. For
the particular integral, we can always try to use the method of undetermined
45
Classical Mechanics Pau Roldan-Blanco
coefficients: guess that the particular integral yp has a similar functional
form to d(t), plug it into the ODE, and then match the coefficients on y′′,
y′, and y.
Often, this method may be unfeasible. In those cases, the following
method may help:
Remark 0.8 (Method VI: Variation of parameters) .
Used for: Linear, nonhomogenous, second-order ODEs with non-constant
coefficients.
The method: Consider a homogenous ODE with varying coefficients,
and let y1 and y2 be fundamental solutions of the homogenous counterpart of
the ODE (found using the methods developed above). The general solution,
then, can be written as:
y = u1y1 + u2y2
for some functions u1 ≡ u1(t) and u2 ≡ u2(t). Then, y′ = u′1y1 +u1y′1 +
u′2y2 + u2y′2.
The key in this method is that, whatever u1, u2 may be, we make the
following assumption:
u′1y1 + u′2y2 = 0 (0.19)
Imposing this restriction, we get y′ = u1y′1+u2y
′2, and y′′ = u′1y
′1+u1y
′′1 +
u′2y′2 + u2y
′′2 . Plugging (y′′, y′, y) into the ODE, and rearranging terms:
u′1y′1 + u′2y
′2 + u1
(y′′1 + b(t)y′1 + c(t)y1
)+ u2
(y′′2 + b(t)y′2 + c(t)y2
)= d(t)
Since both y1 and y2 are complementary functions, both terms in paren-
theses equal zero. Imposing this and rearranging terms yields:
u′1y′1 + u′2y
′2 = d(t) (0.20)
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Classical Mechanics Pau Roldan-Blanco
Thus, to find (u1, u2), we need to solve the system of equations (0.19)-
(0.20). The two unknowns are (u′1, u′2). Solving the system, we find that we
can write (u′1, u′2) in terms of the Wronskian as:
u′1 = − y2d(t)
Wy1,y2
and u′2 =y1d(t)
Wy1,y2
where Wy1,y2 ≡ y1y′2 − y′1y2 6= 0 because (y1, y2) are fundamental solu-
tions. Thus:
u1 = −∫
y2d(t)
Wy1,y2
dt and u2 =
∫y1d(t)
Wy1,y2
dt
Provided that these integrals can actually be computed, we have found a
particular solution to the differential equation:
yp = y1u1 + y2u2 = −y1
∫y2d(t)
Wy1,y2
dt+ y2
∫y1d(t)
Wy1,y2
dt (0.21)
Example 0.16 Consider the ODE:
2y′′ + 18y = 6 tan 3t
where recall that tanx = sinxcosx .
• Complementary function (yc):
It is readily checked using our methods above that the fundamental
solutions (to 2y′′ + 18y = 0) are y1 = cos 3t and y2 = sin 3t, so the
complementary function is yc = C1 cos 3t+ C2 sin 3t.
• Particular integral (yp):
To find the particular integral, we could try the method of undeter-
mined coefficients. But because d(t) here does not involve a sum, a
polynomial, or a product, this method is likely to fail. We will use the
method of variation of parameters instead.
The general solution is y = u1y1 + u2y2, where y1 = cos 3t and y2 =
sin 3t. The Wronskian is:
47
Classical Mechanics Pau Roldan-Blanco
Wy1,y2 = y1y′2 − y′1y2 = 3 cos2(3t) + 3 sin2(3t) = 3 6= 0
where we have used the Pythagorean theorem, equation (0.2). Then,
using equation (0.21), a particular integral is:
yp = − cos(3t)
∫3 sin(3t) tan(3t)
3dt+ sin(3t)
∫3 cos(3t) tan(3t)
3dt
= − cos(3t)
∫sin2(3t)
cos(3t)dt+ sin(3t)
∫sin(3t)dt
= − cos(3t)
∫1− cos2(3t)
cos(3t)dt+ sin(3t)
∫sin(3t)dt
= −cos(3t)
3
[ln (sec(3t) + tan(3t))− sin(3t)
]+
sin(3t)
3(− cos(3t))
= −cos(3t)
3ln (sec(3t) + tan(3t))
where secx = 1cosx is the reciprocal of the cosine, often called the secant
function.
Thus, the general solution y = yc + yp is
y = C1 cos 3t+ C2 sin 3t− cos(3t)
3ln (sec(3t) + tan(3t))
Example 0.17 Consider the ODE:
y′′ − 2y′ + y =et
t2 + 1
• Complementary function (yc):
Again, we will not go through the derivation of the complementary
function, but using the methods above one should readily find that the
fundamental solutions to y′′ − 2y′ + y = 0 are y1 = et and y2 = tet.14
14 Here, the roots are identical, so we should follow Case 2 in Method III (Remark0.4).
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Classical Mechanics Pau Roldan-Blanco
• Particular integral (yp):
To find the particular integral using the variation of parameters method,
we again argue that the general solution is y = u1y1 +u2y2, where now
y1 = et and y2 = tet. The Wronskian is:
Wy1,y2 = y1y′2 − y′1y2 = et(et + tet)− te2t = e2t 6= 0
Then, using equation (0.21), a particular integral is:
yp = −et∫
tetet
e2t(t2 + 1)dt+ tet
∫etet
e2t(t2 + 1)dt
= −et∫
t
t2 + 1dt+ tet
∫1
t2 + 1dt
= −1
2et ln(1 + t2) + tet tan−1(t)
Thus, the general solution y = yc + yp is
y = C1et + C2te
t − 1
2et ln(1 + t2) + tet tan−1(t)
49
Part I
NEWTONIAN
MECHANICS
51
Chapter 1
Motion and Force
Classical mechanics focuses on describing the motion of particles. We
start with the description of systems with a single particle. Section 1.3 will
generalize the concepts to systems of two or more particles.
1.1 Newton’s Laws of Motion
To describe particle motion, we must first specify a position, i.e. the
value of each of the spatial coordinates at a point in time, and the motion,
or the change in the position when time advances.
1.1.1 Position, Velocity, and Acceleration
We start with some definitions:
Definition 1.1 (Position) The position of a particle is a vector ~r(t) =
(x(t), y(t), z(t)) specifying the location of a single particle on each coordinate
x, y, z at time t.
Definition 1.2 (Velocity) The velocity of a particle is the displacement of
its motion over infinitesimal time along each one of its coordinates:
~v(t) ≡ ~r(t) = (x(t), y(t), z(t))
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Classical Mechanics Pau Roldan-Blanco
where x(t) denotes the time derivative of x (and similarly for y and z).
Definition 1.3 (Speed) The speed of a particle is the magnitude of its
velocity, that is, the scalar |~v(t)|.
Definition 1.4 (Acceleration) The acceleration of a particle is the dis-
placement of its velocity over infinitesimal time along each one of its coor-
dinates:
~a(t) ≡ ~v(t) =(x(t), y(t), z(t)
)Example 1.1 (Simple Harmonic Oscillator) Consider an oscillating par-
ticle along a single dimension:
x(t) = sin(ωt), y(t) = z(t) = 0
where ω is a constant. The particle is constant in the second and third
dimensions. Along the first dimension, this is a simple harmonic motion,
with larger values of ω implying more rapid oscillations. Indeed, recall from
equation (0.1) and Figure 0.2 that sinωt is the position along one dimension
of a particle on a unit circle when the angle (in radians) is given by:
θ(t) = ωt
The assumption that the angle θ increases linearly with time implies that
motion is uniform.
• The velocity of the particle is:
v =d
dtsin(ωt) = cos(ωt)
d
dt(ωt) = ω cos(ωt)
using the chain rule, and equation (0.4).
Notice that when position x is at its maximum or minimum, the ve-
locity is zero. And vice versa, when the position is at x = 0, velocity
is either at its maximum or at its minimum. Technically, in this case
it is said that position and velocity are 90◦ out of phase.
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• The acceleration is:
a = −ω2 sin(ωt)
Note that acceleration is negative, meaning that whenever x is positive
(negative), the acceleration is positive (negative). That is, wherever the
particle is, it is accelerated back into the origin (indeed, it “oscillates”).
Technically, in this case it is said that position and acceleration are
180◦ out of phase.
For an example of motion in two dimensions, we need to introduce two
more concepts:
Definition 1.5 (Angular velocity) The number of radians that an angle
θ advances per unit of time. That is:
ω ≡ dθ
dt
Definition 1.6 (Period of motion) The time it takes for a particle to
complete one full cycle of motion. Typically denoted by T , measured in
seconds.
Definition 1.7 (Frequency of motion) The number of cycles a particle
takes per unit of time. Thus, f ≡ 1/T , measured in cycles per second, or
hertz (Hz).
Example 1.2 (Harmonic Oscillator: Circular Motion) Consider a par-
ticle moving in a perfect circle on a plane, i.e. along the x (horizontal) and
y (vertical) dimensions, but not along the z dimension. An example is the
motion of a planet orbiting the Sun in a perfect circular orbit.
Formally, the most general (counterclockwise) uniform circular motion
around an orbit of radius R > 0 is:
~r(t) ≡
(x(t)
y(t)
)=
(R cos(ωt)
R sin(ωt)
)
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Indeed, note from equation (0.1) and Figure 0.2 that R cos(ωt) and
R sinωt are the coordinates on a circle of radius R if we define:
θ(t) = ωt
as the angle (in radians) at time t. Similar to Example 1.1, the assump-
tion that the angle θ increases linearly with time implies that motion around
the orbit is uniform.1
Notice in such a motion the two coordinates are 90◦ out of phase: as
the particle revolves around the orbit, x oscillates between a maximum of R
and a minimum of −R, and at both of these points we have y = 0. And
vice versa: at its maximum and minimum points, y = R and y = −R,
respectively, but in both cases x = 0.
The notation ω for the coefficient is no coincidence, for it is exactly the
angular velocity of this system (indeed, since θ = ωt, then dθ = ωdt). For
orbital motions, the period of motion, or the time it takes to go one full
revolution (i.e. 360◦, or 2π radians), is:
T =2π
ω
(To see this, just plug in θ = 2π into θ = ωt at t = T ). The frequency of
the oscillation is thus f = ω2π Hz. Finally, using simple differentiation, we
can find the velocity and acceleration vectors for the simple uniform circular
motion:
~v(t) =
(−Rω sin(ωt)
Rω cos(ωt)
); ~a(t) =
(−Rω2 cos(ωt)
−Rω2 sin(ωt)
)
Note two interesting observations, first noticed by Newton when studying
the motion of the Moon (see Figure 1.1):
• For uniform circular orbits, the position and velocity vectors are or-
thogonal, ~r ⊥ ~v (i.e. ~r · ~v = 0).
1 In Section 3.1, we will generalize circular orbits to non-uniform motion.
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• The acceleration of uniform particle motion along a circular orbit is
parallel to the position vector (that is, ∠~r~a = π, or 180◦), but has op-
posite direction (i.e. toward the origin). This is called the centripetal
acceleration.
Thus, we define:
Definition 1.8 (Centripetal acceleration) In circular motion, the ac-
celeration that is perpendicular to the velocity of the particle.
Finally, let us compute magnitudes.
• The magnitude of the position is, of course, R.2
• The magnitude of the velocity (i.e. the speed) of the particle is:
|~v| =√R2ω2 sin2 ωt+R2ω2 cos2 ωt = Rω
where we have used equation (0.2).3 Because speed is proportional to
the angular velocity ω, when motion is uniform and circular we may
also refer to ω as angular frequency.4
• The magnitude of the acceleration is:
|~a| =√R2ω4 cos2 ωt+R2ω4 sin2 ωt = Rω2
That is, |~a| = |~v|2R .
In sum, in uniform circular orbits, the particle’s speed is proportional to
the angular frequency, and the acceleration is proportional to the product of
the speed and the angular frequency. For given R, higher angular frequency
increases the particle’s speed, and even more so its acceleration.2 Proof. Recall the notation sin2 x ≡ sin(x) sin(x), and similarly for cos2. Then,
|~r| =√R2 cos2 ωt+R2 sin2 ωt = R
√cos2 ωt+ sin2 ωt = R, using equation (0.2).
3 This makes sense. We have obtained |~v| = 2πRT
: the speed (in m/s) is the circum-ference (in m) divided by the time it takes for the particle to complete one full cycle(in s).4 Of course, angular frequency and angular velocity are not the same if the latter is
not constant. We will see one such case in Example 2.7.
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~v
~a
θ(t) = ωt
~v
~a
θ(t) = ωt
~v
~aθ(t) = ωt
Figure 1.1: In a uniform circular motion around anorbit, the velocity vector is orthogonal to the positionvector. The acceleration vector is parallel to the po-sition vector, but of pointing to the origin (so-calledcentripetal acceleration).
1.1.2 First and Second Laws
We are now ready to state our first fundamental principles: Newton’s
laws of motion. These laws are based on the idea that, in order to overcome
friction, force must be applied on a body to change its velocity. An isolated
object moving in free space, with no forces acting on it, does not need force
to keep it moving if it has inertia. But to change its trajectory and/or
velocity, one must apply force.
Principle 1.1 (Newton’s Second Law of Motion) Force equals the prod-
uct of mass and acceleration:5
~F = m~a (1.1)
Remark 1.1 Some remarks:
• Here, the mass of the object is a scalar describing the resistance of the
body to being moved. Heavier bodies experience lower acceleration for
5 Original text: “Mutationem motus proportionalem esse vi motrici impressae, etfieri secundum lineam rectam qua vis illa imprimitur.” In English: “The alterationof motion is ever proportional to the motive force impressed; and is made in thedirection of the right line in which that force is impressed.”
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given applied force. When no force is applied to an object, its velocity
does not change, whatever its mass.
• Notice force an acceleration are vectors, i.e. include all dimensions of
space, because these have not only magnitude but also direction.
• Equation (1.1) gives us the units of force. We use:
– Kilograms (kg) for mass.
– Meters per second (m/s) for velocity.
– Meters per second per second, or per second squared, (m/s2) for
acceleration, because acceleration is the change in velocity.
Thus, since F = ma, force is what it takes to accelerate one kilogram
by one meter per second per second, or “one kilogram meter per second
squared”.
Definition 1.9 (Newton units) A Newton (N) denotes one kilogram me-
ter per second squared (kg m/ s2), and is the unit of measure of force.
Example 1.3 (No Force) A particle with mass m, moving at velocity ~v,
and with no force acting on it, satisfies by Newton’s law:
m~v = ~0
Since m > 0, then ~v = ~0. Since velocity is constant in all components,
then ~v(t) = ~v0 ≡ ~v(0). Therefore, we have found:
~r = ~v0
That is, the change in the position of the particle is given by the initial
value of its velocity. This is a simple differential equation, with solution:
~r(t) = ~r0 + ~v0t
where ~r0, ~v0 ∈ R3. That is, the particle’s position at time t is the sum of
its initial position and the product of time and its initial velocity.
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Incidentally, in this example we have derived Newton’s First Law of
Motion as a special case of his Second Law when ~F = ~0. This configuration
has a name: inertial reference frame. To give a proper definition, let us first
introduce some more concepts:
Definition 1.10 (Reference frame) A reference frame is a coordinate
system and a set of points that uniquely fix the coordinate system and stan-
dardize measures.
A coordinate system is, of course, the system that is used to uniquely
determine the position of points on a manifold.
In classical mechanics, the manifold is often the Euclidean space (i.e. a
space with no curvature). The reference frame is often the Cartesian set of
coordinates, where vectors represent directions in flat space. Alternatively,
we will sometimes consider the polar coordinate system. Above we have
seen examples of both systems of coordinates. The Cartesian system is our
usual (x, y, z) system. The polar system is used when dealing with circular
motion:
Definition 1.11 (Polar coordinate system) A two-dimensional coordi-
nate system in which each point on a plane is determined by: (i) a distance
(called radius) from a reference point (called pole, analogous to the origin
in Cartesian coordinates); (ii) an angle from the reference direction.
As we shall see, however, in Einstein’s special relativity theory the ref-
erence frame depends on the observer, so we must use non-Euclidean spaces
and non-Cartesian coordinate systems.
Definition 1.12 (Inertial reference frame) A reference frame in which
bodies, whose net force acting upon them is zero, have no acceleration.
Newton’s First Law then states:
Principle 1.2 (Newton’s First Law of Motion) In an inertial reference
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frame, every object in a state of uniform motion remains in that state of mo-
tion unless an external force is applied to it.6
Example 1.4 (Constant Acceleration) In the previous example, we saw
the case of a particle of constant velocity (i.e. no acceleration). A slightly
more general particle has constant (zero or else) acceleration. The general
way to describe the position of such a particle is:
x(t) = c1 + c2t+ c3t2 (1.2)
for some constants c1, c2, c3 to be interpreted shortly. Indeed, note that
velocity and acceleration are:
v(t) ≡ x(t) = c2 + 2c3t a(t) ≡ x(t) = 2c3
respectively. Indeed, acceleration is constant. Moreover, note c1 = x(0),
c2 = v(0), and c3 = a(0)2 , so (c1, c2, c3) determine the initial conditions of
the system.
A specific example of constant acceleration is that of an object in free
fall.
Definition 1.13 (Free fall) The situation in which the only force exerted
on a body is exclusively that of gravity.
Example 1.5 (Free fall) Consider a particle of mass m, but now suppose
there is a constant force Fz > 0 being exerted along the z direction. Newton’s
law says that, in this case, acceleration along the z-axis is:
vz =Fzm
Again, vx(t) = vy(0), but now:
6 Original text: “Corpus omne perseverare in statu suo quiescendi vel movendi uni-formiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare.”In English: “Every body persists in its state of being at rest or of moving uniformlystraight forward, except insofar as it is compelled to change its state by force im-pressed.”
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z(t) = vz(t) = vz(0) +Fzmt
Solving this differential equation, we find the particles position along the
third dimension at time t:
z(t) = z(0) + vz(0)t+Fz2m
t2
In vector notation, and recalling the notation ~r(t) = (x(t), y(t), z(t)), we
have:
~r(t) = ~r0 + ~v0t+ ~ε(t)
where ~ε(t) =(0, 0, Fz2m t
2).
Incidentally, we have just derived the equation for the motion of a falling
object. If z(t) represents the object’s height above the surface of the earth
at time t, and we use the acceleration caused by gravity, az = vz = g, then:
z(t) = z(0) + vz(0)t− gt2
2
describes the object’s position at time t after falling from height z(0)
with initial velocity vz(0).
These examples also bring to bear the difference between mass and
weight. Though often used interchangeably, these are two very different
concepts:
Definition 1.14 (Mass) The mass of an object is the quantity of matter
in the body, regardless of its volume or of any forces acting on it. It is
usually expressed in kilograms (kg).
Definition 1.15 (Weight) The weight of a body is the force that is exerted
on the body, that is, the product of its mass and the acceleration experienced
by the body. Therefore, weight is expressed in Newtons (N).
In inertial reference frames, the weight is just the magnitude of the
gravitational force, so a body’s weight is simply (by Newton’s Second Law)
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W = mg, where g denotes the gravitational acceleration. On Earth, g =
9.80665m/s2, so a body of mass 10kg weighs approximately 98 Newtons. In
outer space, however, g = 0, so objects have a certain mass but they are
weightless.
More generally, a body’s weight depends not only on the gravitational
force, but all other forces presently acting upon it. Here are two classic
examples:
Example 1.6 (Weight on an Elevator) Suppose a man of mass m =
80kg enters an elevator. The elevator has a scale inside.
• If the man stands on the scale and the elevator is at rest, the man
exerts a force mg on the scale, and the scale exerts a force Fscale on
the man. By Newton’s laws, Fscale = mg gives the weight of this man.
In this case, the man weighs about 80× 9.80 = 784N .
• Now, imagine the elevator is accelerated upward, with acceleration a.
Again, let Fscale be the force that the scale exerts on the man. The
man still exerts a force mg on the scale. But now, additionally, there
is an upward acceleration, so clearly it must be that Fscale > mg,
or else the elevator would not move. In particular, by Newton’s law,
Fscale −mg = ma, or:
Fscale = m(g + a)
For instance, if the acceleration is 5 m/s2, the man weighs 80× (9.8 +
5) = 1184N . This is why we feel heavier when elevators accelerate
upward, even though our body mass has not changed.
• Finally, imagine the elevator is in free fall. Using the notation above,
the acceleration is now simply a = −g (recall Example 1.5).7 Thus,
Fscale = m(g + a) = 0. Therefore, objects in free fall are weightless,
whatever their mass.7 Here, we use the convention that the “plus” direction is upward, and the “minus”
direction is downward.
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Example 1.7 (Body hanging from a string) Consider a body of mass
m that is attached to a (massless) string. The string is in turn attached to
the ceiling, so that the body is hanging and at rest.8
As usual, there are two forces of opposite direction. On the one hand,
the string has a tension T which pulls upward on the body. On the other,
the body has a force mg, due to gravity, pushing downward on the body.
• If the system is at rest (no acceleration), then clearly T = mg.
• Now suppose the system is accelerated upward with acceleration a.
Then, T − mg = ma, so the tension in the string is T = m(g + a).
Just like in the elevator example.
• Conversely, if the system is accelerated downward, the body will weight
less, whatever its mass. If the string is cut and the object goes in free
fall, then T = 0 (the body becomes weightless).
Later on, in Example 1.12, we will revisit the notion of weight for a more
sophisticated example of motion.
Example 1.8 (Projectile motion) When an object is shot with an angle,
the force of gravity will bring it back down, and the body will describe a
parabolic trajectory (Figure 1.2).
To describe this motion, consider two dimensions, horizontal (x) and
vertical (y). At time t = 0, the (initial) velocity is denoted ~v0, with speed
v0 = |~v0|, in an angle of α radians. As per Equation (0.1), the x and y com-
ponents of the initial vector velocity are v0 cosα and v0 sinα, respectively.
At a later moment in time, vector ~r(t) = (x(t), y(t)) describes the posi-
tion of the projectile. Since acceleration is arguably constant on both (x, y)
dimensions, we may use equation (1.2) to write:
8 For the case of a moving body, see Example 1.10. For an example with two strings,see Example 1.11.
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~v 0
v0 cosα
v0 sinα tend2
~r(t)
α
x
y
Figure 1.2: Projectile motion.
x(t) = x(0) + vx(0)t+ ax(0)t2
2
y(t) = y(0) + vy(0)t+ ay(0)t2
2
First, we know (vx(0), vy(0)) = (v0 cosα, v0 sinα). Supposing the projec-
tile starts at the origin, (x(0), y(0)) = (0, 0). In the x direction there is no
acceleration (provided there is no air friction), so velocity is constant.9 In
the y direction, however, there is the (attractive) force of gravity, g. There-
fore, at the beginning of time, (ax(0), ay(0)) = (0,−g), so at time t:
x(t) = (v0 cosα)t y(t) = (v0 sinα)t− gt2
2(1.3)
The velocities in each direction are:
vx(t) = v0 cosα vy(t) = v0 sinα− gt
Note we have been able to decompose the complicated projectile motion
into two independent motions along each direction. In the case of a parabola,
9 This can be confirmed experimentally: if a ball is thrown vertically from a movingplatform, it will land exactly on that same platform after completing the parabolictrajectory.
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the x displacement happens to be completely independent of the y displace-
ment.
Let us derive more properties:
• First, the shape of the trajectory. For this, use equation (1.3) to solve
for t in x(t), and obtain t = xv0 cosα . Plugging this into y(t) gives:10
y = (tanα)x− 1
2
g
v20 cos2 α
x2
This now gives us the location of the particle along the y direction, for
given location along the x direction. Thus, we have an equation of the
form y = ax+ bx2. This is indeed the formula for a parabola.
• Next, we want to know the highest point reached by the projectile. For
this, just impose vy(t) = 0. Indeed, at the highest point, the projectile’s
vertical speed is zero before it picks up again when the body starts to
move along the downward portion of the arc. Using our formula:
0 = vy(tp) = v0 sinα− gtp
where tp denotes the time at which the maximum height is reached.
Thus, tp = v0 sinαg . Now we can just substitute time tp into the y law
of motion to obtain, after some algebra:
y(tp) =v2
0 sin2 α
2g
This is intuitive: the highest point is increasing in the initial angle
(steeper trajectories reach higher maxima) and the initial speed, and
decreasing in the gravitational force. For instance, for the same initial
speed and angle, the projectile will reach a higher maximum point if
the experiment is done on the surface of the Moon.11
10 Recall that sinαcosα
= tanα.11 The gravitational force on Earth is g = 9.80m/s2. On the Moon, it is g =1.625m/s2.
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• When will the projectile hit the ground? For this, we simply look for
the tend such that y(tend) = 0. Alternatively, we know time tp = 2tend
if there is no air friction. Thus,
tend =2v0 sinα
g
• What will the x position of the body be when it hits the ground? Know-
ing tend, we have:12
x(tend) =v2
0 sin 2α
g
and of course y(tend) = 0.
• From our last calculation we see that, for given initial velocity, the
farthest we can possibly throw the object is if we use an angle of exactly
45◦. Indeed, at this angle (α = π4 ), sin 2α reaches a maximum (recall
Figure 0.1).
Example 1.9 (Harmonic Oscillator: Hooke’s Law) Consider a parti-
cle that moves (for simplicity) only along a single dimension (x), subject to
a force that always pulls it back to the origin. The force can be represented
as:
F = −kx
where k > 0. The force being negative means that it is always a restoring
force (positive when distance is negative, and vice versa). It being propor-
tional to x means that, for each unit distance away from the origin, the force
increases by a fixed amount k. Thus, we can think of this as the motion of
a particle at the end of a spring (Figure 1.3).
When a force is restoring and proportional to the displacement, we say
that Hooke’s Law holds:
12 Recall: sinα cosα = sin 2α.
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x > 0x = 0
F = −kx
Figure 1.3: Spring motion as a harmonic oscillator.Above: Spring in a relaxed position (x = 0); Below :Stretched spring, in position x.
Principle 1.3 (Hooke’s Law) The force F needed to extend or compress
a spring by a distance x is proportional to the distance, or F = −kx, where
k > 0 is called the spring constant.
Newton’s Second Law says ma = −κx. Thus, acceleration is negative,
as expected. We can write this as:
x+k
mx = 0 (1.4)
This important formula generally describes the acceleration motion of
a simple harmonic oscillator, which are very prevalent in nature, from the
motion of the pendulum (see Example 1.10) to the oscillations of the electric
and magnetic fields in a light wave. We will see many examples of simple
harmonic oscillators throughout these notes.
For the example of a spring, think of the spring being pulled and then
released. As the spring returns to its resting (or equilibrium) position, it
oscillates between positive (respectively, negative) and negative (respectively,
positive) distance (respectively, force) with respect to its equilibrium point.
As it oscillates, its displacement describes a wave over time.
To see formally that this is an oscillator, let us solve the ODE (1.4) and
demonstrate that it is a sinusoidal (i.e. a wave-like) function in time. Here,
equation (1.4) is a second-order, constant-coefficient homogenous ODE, so
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we can use Method III (Remark 0.4). In the notation of Method III, the
coefficients are b = 0 and c = k/m. Since clearly b2 < 4c, then the solution
has the form:
x = A cos(ωt) +B sin(ωt)
for some numbers A, B, and ω. In physics, the following alternative
formulation is more commonly used:
x = xmax cos(ωt+ ϕ) (1.5)
To see that the two expressions are equivalent, note that cos(ωt + ϕ) =
cos(ωt) cosϕ − sin(ωt) sinϕ (as per equation (0.3)), so A ≡ xmax cosϕ and
B ≡ −xmax sinϕ.
In the formulation of equation (1.5), we have the following objects:
• xmax is the called the amplitude, corresponding to the maximum x
distance that the body reaches before oscillating back to the origin.
• ω is, after Example 1.2, the angular frequency of oscillation.
• ϕ is called the phase angle (in radians).
Figures 1.4-1.5 provide a graphical interpretation for these objects. In
Figure 1.4 we see that ϕ controls for the displacement of the wave in the x
direction. In the example, the thin line is out-of-phase by 180◦ (i.e. π radi-
ans) relative to the thick line. In Figure 1.5, we see that xmax corresponds
to the distance between the maximum and the minimum of the sinusoidal
waves at each given oscillation. If the amplitude is negative (dashed line),
the wave is out of phase by 180◦ (i.e. π radians).13 The exact realization
of these variables depends on the initial conditions of the system (more on
this in Example 2.9).
13 That is, for given ω, setting (xmax, ϕ) = (1, π) produces the same wave as setting(xmax, ϕ) = (−1, 0).
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π2
π 32π
2π
−1
1 Phase ϕ = 0
Phase ϕ = π
θ
Figure 1.4: Same amplitude, different phases.Graphs for x = xmax cos(ωt+ϕ), xmax = 1, with ϕ = 0[thick line] and ϕ = π [thin line].
π2
π 32π
2π
−1
1
Amplitude xmax = 12
Amplitude xmax = 1
Amplitude xmax = −1
θ
Figure 1.5: Same phase, different amplitudes.Graphs for x = xmax cos(ωt+ ϕ), ϕ = 0, with xmax =0.5 [thick line], xmax = 1 [thin line], and xmax = −1[dashed line]. When the amplitude is negative [dashedline], the wave is out of phase by 180◦.
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Recall that the time it takes for the oscillation to take one full cycle
(before it repeats itself) is called the period of motion, equal to T = 2π/ω.
The frequency of motion is f = 1/T hertz (Hz). For solution (1.5), we have:
x = − xmaxω sin(ωt+ ϕ)
x = − xmaxω2 cos(ωt+ ϕ) = −ω2x
Plugging into equation (1.4) gives −ω2x+ kmx = 0, and therefore:
ω =
√k
m
is the angular frequency of the spring. The period of oscillation is thus
T = 2π√
mk seconds. Remarkably, note the period is independent of the
amplitude xmax and the phase angle ϕ. This is characteristic of objects
whose motion obeys Hooke’s Law.
Example 1.10 (Harmonic Oscillator: Pendulum) The motion of a pen-
dulum can also be described as a harmonic oscillator. Consider two dimen-
sions, horizontal and vertical, denoted (x, y). At rest, the pendulum is per-
pendicular to the ground (x = 0). In motion, the pendulum arcs back and
forth in oscillation.
There is a ball of mass m attached at the end of the pendulum. The
length of the pendulum’s string (which is massless) is `. The angle of the
pendulum relative to its equilibrium position is θ. Two forces are acting on
this pendulum (in blue in Figure 1.6): first, the force of gravity, equal to
mg, pulling the pendulum downward; second, there is a tension Tθ, pulling
the ball upward along the string and which depends on the pendulum’s angle.
Using Figure 0.2 and equation (0.1), the tension vector can be decomposed
into x = Tθ sin θ and y = Tθ cos θ.
Let’s start with the equations of motion. In the x (horizontal) direc-
tion, the force is restoring just like with the spring (Example 1.9), so Fx =
−Tθ sin θ = −Tθx/`.14 In the y (vertical) direction, there is an upward force
14 In the second equality, we have Taylor-expanded sin θ about θ = 0 to obtain sin θ ≈
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x = 0
`
θ
Tθ
Tθ sin θ
Tθ cos θ
mg
Figure 1.6: The pendulum as a harmonic oscillator.
Tθ cos θ, and a downward force mg, so Fy = Tθ cos θ −mg. Thus, Newton’s
Second Law is:(mx
my
)=
(−Tθ 0
0 −Tθ
)(sin θ
cos θ
)−
(0
mg
)(1.6)
This is now a coupled system of differential equations that can be solved
with a numerical solver (such as Mathematica), but is otherwise hard to
solve by hand. To make progress, one can use a so-called small-angle ap-
proximation.
Definition 1.16 (Small-Angle Approximation) A useful simplification
of trigonometric functions that considers that the angle θ is small or, specif-
ically, that θ � 1 radian.
Remark 1.2 To test the validity of this approximation, we can use a second-
order Taylor expansion (using results from Example 0.7) about θ = 0 to see
θ. Since the total horizontal displacement is x = ` sin θ, then sin θ ≈ x/`.
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that sin θ ≈ θ = 0, cos θ ≈ 1 − θ2/2 = 1, and tan θ ≈ θ = 0, respectively.
These approximations are quite good: when θ = 0.0877 (i.e. about 5◦), then
cos θ = 0.996 (only .4% away from unity), and when θ = 0.1745 (i.e. about
10◦), then cos θ = 0.985 (only 1.5% away from unity).
The small-angle approximation is very useful for computing the pendu-
lum’s motion. In particular, is allows us to make two simplifications:
• First, we approximate cos θ ≈ 1.
• Second, notice that for small angles θ, the x displacement is much
larger than the y displacement: when swinging the pendulum, the hori-
zontal motion of the pendulum is overwhelmingly much larger than the
vertical motion.15 Hence, we say that y ≈ 0.
Imposing this approximation into our second law of motion in (1.6) gives:
Tθ ≈ mg
That is, when the angle is small, the tension force is invariant to the
angle’s size. In fact, it is approximately equal to the gravitational force.
This is intuitive from Figure 1.6: when there is no y acceleration, upward
(tension) and downward (gravitation) forces should cancel out. Using this
back into the first equation, we get:
x+g
`x = 0
Compared to (1.4), we see that this is a simple harmonic oscillation.
Using the results from Example 1.9, we then immediately know that:
• The position of the pendulum at time t is x = xmax cos(ωt+ ϕ).
• The angular frequency of the oscillation is ω =√
g` .
15 For example, at θ = 0.0877 (about 5◦), the y displacement is only 4% of the xdisplacement. At θ = 0.1745 (about 10◦), it is only 9%.
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• The period of the pendulum is 2π√
`g seconds.16
Once again, this is an approximately correct description of the pendu-
lum’s motion insofar as it does not swing “too much” (i.e. θ small).
Example 1.11 (Body hanging from two strings) Consider a body of
mass m that is attached to two strings. Each string is, in turn, attached
to the ceiling. The two strings are separated by a certain distance, so they
form an angle with the ceiling. The body is at rest.
Let us agree to call x the horizontal direction, and y the vertical direction.
Each string is exerting a pull on the body. We thus let Ti denote the tension
in string i = 1, 2 (similar to the pendulum case, Example 1.10). The two
tensional forces disagree in direction: while the left-hand side string pulls in
the north-west direction, the right-hand side string pulls in the north-east
direction. On the other hand, the gravitational force on the body is mg, a
downward pull.
As usual, we can decompose the forces into its directional components.
• For each string i = 1, 2, the x component of Ti is Ti cos θi, and the y
component is Ti sin θi, where θi denotes the angle between string i and
the vertical axis.
• Since each string pulls in opposite directions along the x direction, then
the force along this direction is:17
Fx = T1 cos θ1 − T2 cos θ2
• In the y direction, both strings are pulling up, and so they agree in
direction. However, gravity is pulling down. Thus:
Fy = T1 sin θ1 + T2 sin θ2 −mg16 For instance, when ` = 1 meter, a full cycle takes approximately 2 seconds. When` = 0.25 meters (4 times shorter), the period is 1 second.17 Notice here we agree that force increases to the right. This is just a convention.
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Since the body is hanging down and is not being accelerated, then ~a = ~0,
and Newton’s Second Law says ~F = m~a = ~0. Newton’s Third Law implies
that all forces must cancel out, i.e. the tension from the two strings must be
equal to the gravitational pull. Thus, Fx = Fy = 0, and we get a system of
two equations with unknowns (T1, T2):
T1 cos θ1 − T2 cos θ2 = 0
T1 sin θ1 + T2 sin θ2 = mg
Solving, we get:
(T1, T2) =
(mg
sin θ1 + tan θ2 cos θ1,
mg
sin θ2 + tan θ1 cos θ2
)For example, if m = 4 kg, θ1 = π/3 (i.e. 60◦), θ2 = π/4 (i.e. 45◦), and
using that g = 9.80m/s2 on Earth, we obtain (T1, T2) ≈ (28.696, 20.291)
Newtons.
Example 1.12 (Weight in circular motion) When a pendulum (Exam-
ple 1.10) is given a strong enough initial velocity, its trajectory will describe
a circular motion (Example 1.2).
Let R be the radius of the orbit, m be the mass of the object, and ω be
the angular velocity of motion, assumed to be constant (that is, the angle at
time t is just θ = ωt). From Example 1.2, we know that there must be a
centripetal acceleration, pulling the object toward the origin, equal to Rω2.
Let T denote the tension of the string at a given position.
We will now calculate the weight of the object in two different positions
along its orbital trajectory: point S (the peak position) and point P (the
trough position). See Figure 1.7.
• At point P , there is a gravitational force mg pushing down on the
object, and a centripetal acceleration ap = Rω2 creating tension on
the string and pulling in the object. Then, the tension of the string at
point P is Tp −mg = map by Newton’s Second Law, or:
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S
as
P
ap
mg
mg
R
Figure 1.7: Weight for a body describing a circularmotion.
Tp = m(ap + g)
• At point S, Newton’s law says that Ts +mg = mas, so:
Ts = m(as − g)
Notice these are very similar equations to the ones we found for the
weight in an elevator (Example 1.6). Thus, the object gains weight as it goes
down, and losing weight as it comes back up. For instance, if the centripetal
and the gravitational accelerations cancel one another (namely, as = g),
and the object is effectively weightless and the string has no tension.18 For
example, if the object being swung was a bucket full of water, the water would
not fall over, for it is weightless at that point.
18 If as < g, then of course the object could not have made it to point S in thefirst place –it was not given enough initial velocity (namely, ω was too small for thependulum to describe a full orbit).
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1.2 Resistive Forces
So far, we have assumed that objects move in a vacuum, so that the
force causing them to accelerate is not counteracted by any resistance. In
this section, we explore such resistive forces. We may refer to the frictional
force, or the drag force. Let us study each in turn.
1.2.1 Friction
We call friction the force resisting the motion of objects relative to others
that are in contact with it. When surfaces in contact accelerate relative to
one another, friction exerts a force that counteracts such acceleration.
To begin, consider an object (e.g. a brick) of mass m at rest on a surface
with no inclination. There is a gravitational pull mg on the object. The floor
exerts an upward force FN of equal magnitude (for there is no acceleration
in the vertical direction), or FN = mg. Since this force is perpendicular to
the surface, it is often called the normal force (i.e. normal in the sense of
orthogonal).
Definition 1.17 (Normal Force) The force that is perpendicular to the
surface that an object is in contact with.
Now, suppose we exert a force F on the brick in the horizontal direction.
The brick will move, though not without resistance. Indeed, a frictional
force Ff pulls back on the brick in a horizontal but opposite direction of
motion to that of the original force applied.
It is a fact of nature that the frictional force Ff has a limit, denoted by
the scalar µ > 0, whose exact value depends on the characteristics of the
materials that compose the objects considered.
Definition 1.18 (Coefficient of Friction (COF)) The ratio of the force
of friction between two objects and the force pressing them together. That
is, Ff = µFN .
There are two types of COFs:
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• The coefficient of static friction (µs), is the COF that corresponds to
objects that are at rest relative to each other. That is, µs is the COF
associated to objects when a force is applied to break them loose.
• The coefficient of kinetic friction (µk), is the COF that corresponds
to objects that are in motion relative to each other. That is, µk is the
COF associated to objects that are in motion relative to each other
and a force is applied that keeps them moving.
Intuitively, we have:
µk < µs
Indeed, less force must be applied when the brick is already sliding through
the surface than that which is needed to get it started.
To calculate the static and kinetic COFs between two objects, the eas-
iest way is to place an object on an inclined surface, and then change the
inclination angle. The following shows how:
Example 1.13 (Friction I: Brick on an incline) Suppose that a brick
of mass m is placed on an incline at angle θ (Figure 1.8). There is a gravita-
tional force mg pushing the object perpendicularly to the floor, and a normal
force FN pulling it perpendicularly to the inclined surface. For convenience,
consider that our (x, y) coordinate system is also “inclined”, with y indicat-
ing direction perpendicular to the surface of the incline (i.e. the hypothenuse
of the triangle, the same direction as FN ), and x indicating direction parallel
to the incline.
With this choice of coordinates, we can as usual decompose the forces
into its coordinate components. FN being a normal force, it does not need
to be decomposed (as it is zero in the x direction). The gravitational force
mg can be decomposed into mg cos θ in the y direction, and mg sin θ in the
x direction. Since there is no acceleration in the y direction, then obviously
FN = mg cos θ.
This object may or may not slide downhill, depending on the inclination
of the incline. For a certain inclination θ, a frictional force Ff pulls the
object back along the x direction. When the inclination is low, Ff is strong
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mg cos θ
~FN
mg sin θ
mg
θ
~Ff
Figure 1.8: Friction in a sliding object.
enough to prevent the object from moving at all. For a high enough θ,
however, Ff reaches its maximum value, friction gives in, and the object
starts to slide down the slope.
Let us calculate the critical value for θ, beyond which the object will slide
down. For this value, acceleration in the x direction is still zero (i.e. a = 0),
but Ff has reached its maximum. Thus, by Newton’s Second Law, at this
exact inclination we have:
mg sin θ − µsFN = 0
where we use µs for the COF because the object is currently at rest. Using
FN = mg cos θ, we then obtain:
µs = tan θ∗
where θ∗ denotes the critical angle. Thus, the angle at which an object
starts to slide down is related to the static COF (a value which depends on
the materials of the objects) by a very simple relationship. In particular,
this angle is independent of (i) the mass of the object, and (ii) the surface
area that the objects are in contact with.19 This is also a recipe to find the
19 For example, the static COF of rubber on concrete is about one, so for these objectsthe critical value of θ is about 45◦. This means that we can never park a car on aslope of more than 45◦ and hope that it will not slide down, no matter how heavy
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static COF of different objects. Experimentally, we can infer µs from µs =
tan θ∗ by trying different inclinations and recording the minimal inclination
at which sliding occurs.
We have found that, if θ < θ∗, then a = 0 (the frictional force will
adjust so that the object is not accelerated). Thus, suppose that θ > θ∗.
Then, the brick will slide downhill. The maximum friction that the brick
will experiment is then:
Ff = µkmg cos θ
Notice we have now used the kinetic COF because the object is already
in motion. By Newton’s Second Law:
mg sin θ − µkmg cos θ = ma
where a is the acceleration downhill. Thus, the acceleration that the brick
will experience is
a = g(sin θ − µk cos θ)
Interestingly, this is once again independent of the body’s mass. How-
ever, gravity now plays a role, as does the kinetic COF of the materials
used.
How much time does it take for the object to reach the bottom of the
incline once it starts accelerating, and at what speed will it arrive? Let ` be
the length from the bottom to the object. Since acceleration is constant and
the initial speed is zero, then ` = 12at
2, so the time it takes is t =√
2`/a.
The speed at that point is just v = at, so v =√
2a` =√
2g`(sin θ − µk cos θ).
Example 1.14 (Friction II: Pulleys) Consider a similar case to the pre-
vious example, but now suppose that the brick (of mass m1) is attached to
a hanging body (of mass m2) through a pulley (Figure 1.9). The pulley is
frictionless and massless, and so is the string connecting the two objects.
the car is or the width of its tires.
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θ
Figure 1.9: Friction in a sliding object attached to apulley and a hanging weight.
Let’s think of the forces at work. The gravitational forces acting on the
brick is m1g. The coordinate decomposition is m1g cos θ in the y direction,
and m1g sin θ in the x direction, just as in the previous example. Because
there is no acceleration in the y direction, the normal force cancels with the
gravitational force, so FN = m1g cos θ. Additionally, there is now a tension
T1 going up the string from the brick, and a tension T2 going up from the
hanging body, with T1 = T2 = T .20 If the system is at rest, then the tension
must compensate the force of gravity in the y direction on the hanging object,
so:
T = m2g
Because the pulley is frictionless, the frictional force is between the in-
cline and the brick. The direction of the frictional force is opposite to that
of the brick. Starting from a resting position, the maximum value of the
frictional force is:
Ff = µsFN = µsm1g cos θ
If the hanging body did not exist (or m2 = 0), then we would have the
previous example, with the friction pointing uphill. However, the presence
of a massive object at the other end of the string may now pull the object
uphill, in which case the frictional force would be pointing downhill. We
must therefore split the analysis into three limiting cases: uphill acceleration;
20 The tension must always same along the entire string. To see this, take an in-finitesimal (and thus massless) segment of the string. This segment is pulled by twoforces, one on each direction. If these forces differed in magnitude, the accelerationwould be infinite (by Newton’s Second Law), a contradiction. For this, of course, itis necessary that (i) the string is massless; (ii) the pulley is frictionless.
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downhill acceleration; and no acceleration at all.
Throughout, we take (θ,m1, g) as given, and treat m2 as the critical
attribute which determines which case occurs.21
• Upward acceleration: If the system is at the critical m2 where is just
about to accelerate upward (call it ms), then Ff = µsm1g cos θ, which is
pointing downhill. Since tension T is pointing uphill, then by Newton’s
Second Law:
T −m1g sin θ − µsm1g cos θ = 0
as the object has not accelerated yet. Replacing T = m2g and solving
for m2 = ms, we get:
ms = m1
(sin θ + µs cos θ
)(1.7)
Thus, if we use any mass m2 ≥ ms, the brick will slide up.
• Downhill acceleration: If the system is at the critical m2 where is just
about to accelerate downward (call it ms), then Ff = µsm1g cos θ,
which is now pointing uphill, agreeing with the direction of tension
T = m2g = msg. Therefore:
T −m1g sin θ + µsm1g cos θ = 0
Thus, now:
ms = m1
(sin θ − µs cos θ
)(1.8)
Thus, if we use any mass m2 ≤ ms, the brick will slide down.
• No acceleration: The brick will not move at all if neither (1.7) nor
(1.8) are met. That is, if m2 ∈ [ms,ms].
21 We may also do this with θ (as before), fixing m2. The analysis is similar.
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Let’s now look at a moving brick. Suppose m2 ≥ ms, so that the brick
moves uphill (the analysis for the other two cases is similar). The maximum
frictional force then becomes kinetic: Ff = µkm1g cos θ. Thus, Newton’s
Second Law in the x direction is now:
T −m1g sin θ − µkm1g cos θ = m1a (1.9)
if a is the acceleration uphill. Crucially, T is no longer m2g, for the
hanging body is being accelerated downward as a result of the brick’s motion.
In particular, the total force in the y direction on the hanging object is m2g−T , so Newton’s Second Law says:
m2g − T = m2a (1.10)
Solving the system of equations (1.9)-(1.10), we get:
a = gm2 −mk
m1 +m2
where mk ≡ m1(sin θ+µk cos θ). Note that since µs > µk (everything else
equal, static friction is always higher than kinetic friction), then ms > mk,
guaranteeing that m2 > mk and thus a > 0 (i.e. the brick is indeed moving
uphill). The associated tension on the string is then:
T = m2(g − a) = m2gm1 +mk
m1 +m2
and thus T < m2g, as expected. Unlike in our previous example, were
the brick’s mass played no role in its acceleration, now the acceleration does
depend on the mass of the objects, though only on the ratio m1/m2 (i.e. how
much more massive one object is relative to the other).
1.2.2 Drag
The drag force is the resistive force caused by the motion of an object
through a fluid, such as air or water. The drag force depends on the size
and shape of the body, as well as the medium through which the object is
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moved and the speed at which it is moved. For example, the drag force is
larger in water than in air.22
The drag force is very different from the frictional force. In frictional
forces, the coefficient of kinetic friction (Definition 1.18) remains constant
independently of the speed (only a function of the material composition of
the objects in contact). However, for drag forces, the object’s velocity is
key. In particular, it has been observed experimentally that:
Definition 1.19 (Drag force) The drag force on an object of velocity ~v
is given by:
~Fdrag = −(k1|~v|+ k2|~v|2
)~v
for some numbers k1 (in kg/msec ) and k2 (in kg
m3 ). Thus, the magnitude of
the resistive force is |~Fdrag| = k1|~v|+ k2|~v|2.
Note the resistive force has opposite direction to the velocity: faster
moving objects feels a stronger drag. Moreover, the coefficients k1, k2 depend
on the shape and size of the object, as well as the kind of medium.
Henceforth in this section, we will assume that the object is a perfect
sphere of radius R > 0. For these objects, it has been observed that:
k1 = c1R
k2 = c2R2
where c1, c2 are scalars. Each component of the drag force has a name:
• Viscous term: The term c1R|~v|, which relates to the stickiness of the
medium (e.g. the “thickness” of the fluid, in the example of liquid
mediums). For example, coefficient c1 is a function of temperature:
when water is heated to a gas, it loses viscosity, and thus the water’s
drag force due to viscosity diminishes.
22 Section 5.2 will be devoted to fluid mechanics. For now, we will take some propertiesof fluids as given.
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• Pressure term: The term c2R2|~v|2, which relates to the stress that
the body is exposed to.23 In this case, c2 is highly correlated to the
density of the body.
Thus, while viscosity depends on the temperature, the pressure terms
depends on the density, where:
Definition 1.20 (Density) The density (ρ) of an object is the ratio of its
mass (m) to its volume (V ), or:
ρ ≡ m
V
It is measured in kg/m3.
Consider a sphere of mass m and radius R in free fall. The sphere feels
a gravitational acceleration mg, and a drag force Fdrag in the opposite di-
rection as gravity, whose magnitude increases as the object picks up speed.
There comes a time when the drag force reaches its gravitational counter-
part, Fdrag = mg. At this point, the object experiences no further accel-
eration, and achieves a constant speed. This velocity is called the terminal
velocity of the object.
Definition 1.21 (Terminal velocity) The constant speed that is eventu-
ally achieved by an object in free fall through a medium that subjects it to
drag.
For spheres, the terminal velocity, denoted ~vterm, thus solves:
mg = c1R|~vterm|+ c2R2|~vterm|2
Therefore, the terminal velocity is a function of the mass, the radius, and
the attributes (viscosity and pressure) of the system that determine c1, c2.
Another key velocity is the following:
23 Intuitively, this stress is proportional to R2 because the force acts upon the area ofthe sphere, which itself is proportional to R2.
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Definition 1.22 (Critical velocity) The velocity at which the drag due
to viscosity equals that which is due to pressure.
Thus, the critical velocity of a sphere, denoted ~vcrit, solves c1R|~vcrit| =
c2R2|~vcrit|2 or, simplifying:
|~vcrit| =c1
c2R
Thus, |~vcrit| ∝ 1R . We can now split the drag forces that are experienced
by an object into two possible regimes:
• Regime I: Drag due to viscosity (c2 ≈ 0): When most drag is due to
the medium’s viscosity, then it must be that ~v � ~vcrit (reads “velocity
is much smaller than the critical velocity”). In this case, the sphere’s
terminal velocity solves mg = c1R|~vterm|, or:
|~vterm| =mg
c1R
If ρ is the density of the sphere, then m = 43πρR
3, so24 |~vterm| =4g3c1πρR2. Interestingly, |~vterm| ∝ R2. Thus, if two objects of the
same density are dropped into water, their terminal velocity will be
proportional to the square of their radius.
• Regime II: Drag due to pressure (c1 ≈ 0): When most drag is due to
the medium’s pressure, then it must be that ~v � ~vcrit (reads “velocity
is much greater than the critical velocity”). In this case, the sphere’s
terminal velocity solves mg = c2R2|~vterm|2, or:
|~vterm| =√
mg
c2R2
Again, for spheres, m = 43πρR
3, so |~vterm| =√
4g3c2πρR. Interestingly,
|~vterm| ∝√R.
24 Recall ρ ≡ mV
(Definition 1.20). For spheres, volume is V = 43πR3.
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Let’s looks at one example for each regime:
Example 1.15 (Regime I: Liquids) If we drop a ball into a thick liquid
(e.g. syrup), then all drag will be due to the liquid’s viscosity: the initial
velocity is zero, so the ball will approach its terminal velocity at a speed that
is proportional to the square of its radius. However, if we were to inject
the ball into the syrup medium at a great initial speed (in particular, at one
that is well above its terminal velocity), it would for a short while experience
drag due to pressure and its speed would slow down to the terminal velocity
proportionally to the square root of the object’s radius.
How long does it take for the terminal speed to be reached? Suppose we
drop the sphere into the fluid. A gravitational acceleration mg is counter-
acted by a drag force of magnitude c1R|~v| (since we are in Regime I). By
Newton’s Second Law:
md
dt|~v| = mg − c1R|~v|
a simple differential equation. To solve, write the equation as d|~v|dt +
c1Rm |~v| = g, and use the Method of the Integrating Factor (Remark 0.2) to
solve. Here, the integrating factor satisfies ddtµ = µ c1Rm , and thus µ = ke
c1Rmt
for some k > 0. Solving:
|~v| =c+ gk
∫ec1Rmtdt
ec1Rmt
=c+ gk m
c1Rec1Rmt
kec1Rmt
=mg
c1R+ ce−
c1Rmt
where c = ck . To find the constant c, set t = 0 and suppose that the initial
velocity is zero (because the object is dropped from a state of rest). Then,
0 = mgc1R
+ c. Finally, recall that mgc1R
= |~vterm| in Regime I, so c = −|~vterm|.Plugging back, we have found:
|~v| =(
1− e−c1Rmt)|~vterm|
Therefore, the speed of the sphere approaches the terminal velocity expo-
nentially from below. The speed of convergence to the terminal velocity is
higher when the sphere is smaller (low R) and heavier (higher m), which
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is intuitive. The convergence is also faster when the fluid is more viscous
(higher c1). Had there been no drag, the velocity would have approached the
terminal speed linearly.
Example 1.16 (Regime II: Air) In we drop the ball in air, then all drag
is due to air pressure. Indeed, for air, c1 ≈ 0. In particular, it has been
measured that, at one atmospheres and room temperature, c1 = 3.1×10−4 ≈0, while c2 = 0.85. The critical speed is |~vcrit| = 3.7×10−4
R , about 400 slower
than the critical speed in syrup. Thus, if we drop a ball in air, its velocity is
way above the critical speed, and so we are in Regime II: the pressure term
always dominates, and the terminal speed is proportional to the square root
of the radius of the sphere.25
How long does it take now for the terminal speed to be reached? As
in Regime I, we have gravity mg pulling down, but now a drag force of
magnitude c2R2|~v|2 (since c1 ≈ 0). By Newton’s Second Law:
md
dt|~v| = mg − c2R
2|~v|2
This is now a non-linear ODE, so it cannot be solved analytically. But
this equation is enough to understand how velocity will build up and con-
verge to the terminal velocity as a function of the sphere’s mass and radius
(alternatively, its density).
Example 1.17 (Projectiles revisited) Consider the experiment in Ex-
ample 1.8, where an object (e.g. a sphere) was shot at an angle. In the
absence of air drag, the parabola is perfectly symmetric (black dashed line in
Figure 1.10). This is because the velocity is not assumed to change. How-
ever, with air drag (red dashed line), there is a resistive force in both the y
direction (downward, as the object gets higher in the air) and the x direction
(leftward, as the object advances in the x direction).
The action of these forces will make the parabolic trajectory asymmetric,
with (i) a highest point that is strictly below and which is reached at an earlier
25 For example, if we drop a pebble of R = 1 cm from a high building, its speed willnever exceed 75 mph. For a sphere of mass m = 70 kg and radius R = 40 cm, theterminal velocity is 150 mph.
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time than in the case with no drag; and (ii) a downward trajectory that is
steeped than the upward one.26~v 0
v0 cosα
v0 sinαα
x
y
Figure 1.10: Projectile motion without drag (blackdashed) and with drag (red dashed).
Using Newton’s Second Law, we have (assuming c1 ≈ 0 in air drag):
mx = −c2R2x2 and my = −mg + c2R
2y2
where (x, y) = (vx, vy) are the x and y components of velocity ~v, with
initial values (x0, y0) = (v0,x cosα, v0,y sinα). If there was no drag force,
then mx = 0 (no acceleration in the x direction) and my = mg (the only
acceleration in the y direction is due to gravity), just what we got in Exam-
ple 1.8. With air drag, motion is characterized by non-linear, second-order
differential equations, which we cannot solve analytically.
At the highest point in the y direction, air drag and gravity cancel each
other out, so mg = c2R2y(tp)
2, or:
y(tp) =1
R
√mg
c2
where tp is the time at which the highest point is reached. Thus, every-
thing else equal, heavier and smaller spheres will reach a higher point.
26 Note (ii) is not reflected in Figure 1.10.
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1.3 Multi-Particle Systems
So far, we have focused on the laws of force and motion for isolated
systems with single objects. Ultimately, we wish to have a description of the
dynamical laws of motion of all particles of nature. The fundamental forces
of nature are those that act between particles. To provide a description of
these forces, we need to determine two objects. First, the intrinsic properties
of the particle, such as its electric charge and mass. Second, the location
and (by Newton’s law) velocity of the particle, as well as how the particle is
influenced by the location of all other particles.
1.3.1 Newton’s Laws in Multi-Particle Systems
Throughout, we consider N particles indexed i = 1, 2, . . . , N . Consider
a space with three dimensions labeled (x, y, z), and let:
~ri = (xi, yi, zi)
denote the coordinates of particle i. (The extension to D > 3 dimensions is
straightforward). We denote:
~r ≡ {~ri : i = 1, . . . , N}
as the collection of locations across all particles. Specifically, x ≡ {xi :
i = 1, . . . , N} denotes the collection of locations on the x dimension across
all particles, and y and z are defined similarly. Succinctly, ~r = (x,y, z).
Henceforth, we may refer to ~r as the configuration space.
Definition 1.23 (Configuration space) A (Cartesian) configuration space
is the 3N -dimensional space of particle locations, ~r ≡ {(xi, yi, zi) : i =
1, . . . , N}.
The force exerted along each dimension on a particular particle i is as-
sumed to be a function of the location of said particle as well as those of
all other particles, and is thus denoted ~Fi(~r). We consider that the force
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exerted on a particle i can be thought of as the sum of forces exerted by all
particles j 6= i on i. Because these forces occur between the particles within
the isolated system, they are sometimes called internal forces. A force that
acts upon the system and which does not involve the direct interaction of
particles is an external force.
Formally, let ~fij be the internal force on i due to j (with ~fii = ~0 without
loss of generality). Then:
~Fi(~r) ≡∑j 6=i
~fij(~r) (1.11)
With this notation at hand, we can first see that Newton’s law of motion
then extends naturally:
Remark 1.3 (Newton’s Second Law in Multivariate Systems) Force
on a particle equals its mass times its acceleration. In vector notation:
~Fi(~r) = mid2~ridt2
for each i = 1, . . . , N . Or, written in component form:
(Fx)i(x) = mid2xidt2
, (Fy)i(y) = mid2yidt2
, (Fz)i(z) = mid2zidt2
where (Fx)i(x), (Fy)i(y), and (Fz)i(z), denote the x, y, and z compo-
nents of the force on the i-th particle, as functions of the location of all
particles along the corresponding dimension.
Note there is one Newton equation for each coordinate of every particle.
With 3 dimensions and N particles, this means 3N equations.
Remark 1.4 (Is Newton’s dynamical law suitable?) Newton’s law of
motion is suitable because it conserves information. That is, the dynamical
law is both deterministic and reversible.
• By knowing the initial position of a particle and its initial velocity, we
can determine where it will be in the future (determinism).
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• Moreover, by knowing the current position and velocity of the particle,
we can know where it was an instant ago (reversibility).
Later on, we will see that quantum mechanics denies this very principle:
location and velocity cannot be both known with certainty. For now, since
position and velocity are all that is required for a full description of the
dynamical system, we may call this our state space:
Definition 1.24 (State space) The state space is the 6-dimensional space
composed of three spatial coordinates, ~r = (x, y, z), and velocities along each
coordinate, ~v = (vx, vy, vz).
Thus, a point in the state space tells us the position and velocity of a
particle along each spatial dimension.
1.3.2 Momentum and Newton’s Third Law
We will now introduce a convenient reformulation of this space.
Definition 1.25 (Momentum) The quantity of motion of a moving body,
measured as the product of the mass and the velocity:
~pi ≡ mi~vi
for each particle i = 1, . . . , N . Momentum is measured in kg m / sec.
Roughly speaking, the momentum measures how hard it is to stop a
moving object. For given velocity, heavier objects need more force to be
stopped. For instance, a moving ping-pong ball takes very little force to be
stopped compared to a locomotive that is moving with the same velocity.
Definition 1.26 (Phase space) The phase space is the 6-dimensional space
composed of three spatial coordinates, ~r = (x, y, z), and momentum along
each coordinate, ~p = (px, py, pz).
Thus, a point in the phase space tells us the position and momentum of
a particle along each spatial dimension. Changing the space from position-
velocity pairs to position-momentum pairs will become convenient later on.
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Remark 1.5 (Newton’s Second Law using momentum) Given that mass
mi is constant, we have that:
~pi = mi~vi
for each i = 1, . . . , N , by definition of momentum. Thus, we can write the
full description of the dynamical system as follows:
~pi = ~Fi(~r) =∑j 6=i
~fit(~r) (1.12a)
~ri =~pimi
(1.12b)
for each particle i = 1, . . . , N . The first equation is Newton’s law of
motion stated in terms of momentum, and the second follows by definition
of momentum. Written in component form, using ~p = (px, py, pz):
d(px)idt
= (Fx)i(x),dxidt
=(px)imi
for each particle i = 1, . . . , N , and similarly for coordinates y and z.
In words, if a force acts upon a particle, it changes its momentum. In
fact, the magnitude of the force is the magnitude of the change in momen-
tum.
We can now offer a full description of the phase space. For each particle
of a given mass, location, and momentum, we can determine the future
evolution of the object through differential equations (1.12a)-(1.12b).
Principle 1.4 (Newton’s Third Law of Motion) Every force exerted from
some particle j on i 6= j is equal and opposite to the force that i exerts on
j. That is, using the notation of equation (1.11):
~fij = −~fji
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for all i = 1, . . . , N and j 6= i.27
Another way to state Newton’s Third Law is the following: for every
action, there is a reaction of equal force and opposite direction. Internal
forces all cancel each other out.
Result 1.1 (Conservation of Momentum) Newton’s Third Law is of-
ten referred to as the Law of Conservation of Momentum. To see why, note
by equations (1.11) and (1.12a)-(1.12b) that:
~pi =∑j 6=i
~fij(~r)
In words, the rate of change in the momentum of a particle equals the sum
of the forces due to all other particles (i.e. the sum of all internal forces),
provided that there are no external forces. Newton’s Third Law implies that
all pairwise forces cancel out, that is:
N∑i=1
~Fi(~r) = ~0
Hence,∑N
i=1 ~pi =∑N
i=1
∑j 6=i
~fij(~r) = ~0. Interchanging derivative and sum-
mation, we get:
d
dt
N∑i=1
~pi = ~0
That is, the rate of change of total momentum (the sum of momenta
across particles) is nil.
In sum, the total momentum of an isolated system never changes, assum-
ing no external forces are present. Though individual particles may acquire
27 Original text: “Actioni contrariam semper et æqualem esse reactionem: sive corpo-rum duorum actiones in se mutuo semper esse æquales et in partes contrarias dirigi.”In English: “To every action there is always opposed an equal reaction: or the mu-tual actions of two bodies upon each other are always equal, and directed to contraryparts.”
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or lose momentum over their dynamic paths, all internal forces must cancel
out, and the total overall momentum of the system remains constant over
time.
Example 1.18 (Colliding particles) Consider a two-particle system, with
particles of mass m1 and m2 moving at velocities ~v1 and ~v2. Suppose one
particle is moving toward the other. Their momenta are ~p1 = m1~v1 and ~p2 =
−m2~v2, respectively. The system’s total momentum is ~p = m1~v1 −m2~v2.
Eventually, a collision occurs. At this point, assume the two particles
become virtually one, with mass (m1 +m2) and velocity ~v ′, where the prime
indicates “post-collision”. By the Conservation of momentum, the total mo-
mentum before the collision must be the same as the momentum right after
the collision, for there are no external forces on this system (all forces are
internal, as a result of the particle interaction implicit in the collision itself).
Therefore, m1~v1 −m2~v2 = (m1 +m2)~v ′, or:
~v ′ =m1~v1 −m2~v2
m1 +m2
For example, if the particles are going in opposite direction but at the
same speed (say ~v), the velocity after the collision will be ~v ′ = (m1−m2)~vm1+m2
.
In words, the newly formed particle will move in the same direction as that
which the more heavier particle had prior to the collision.
1.4 Center of Mass
Multi-particle systems (for example, material objects) can be compli-
cated collections of particles, i = 1, . . . , N . Each particle has a certain mass
mi and position ~ri. The center of mass is a convenient way to summarize
the properties of motion of multi-particle systems. We define the center of
mass as the unique position in space that is at the mean location of the
distribution of mass in space.
Definition 1.27 (Center of mass) The center of mass of a system of
i = 1, . . . , N particles, each with mass mi and in position ~ri, is the unique
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position vector ~rCM satisfying:
N∑i=1
mi(~ri − ~rCM ) = ~0 (1.13)
Solving the equation for ~rCM , we find:
~rCM =1
M
N∑i=1
mi~ri
where M ≡∑N
i=1mi is the total mass of the system. In words, the
center of mass is the unique position in space with the property that the
mass-weighted position vectors of all particles relative to this point sum
to zero (i.e. cancel out). Thus, the distribution of mass of the system is
balanced around the center of mass.
Taking the time derivative, we see:
~vCM =1
M
N∑i=1
mi~vi =1
M
N∑i=1
~pi ≡~P
M
where ~P ≡∑N
i=1 ~pi is the total momentum of the system. Therefore, we
have found the following important property:
Result 1.2 (Total momentum and center of mass) The total momen-
tum of a system of particles i = 1, . . . , N satisfies:
~P = M~vCM
Equivalently, taking the time derivative, and using ~pi = ~Fi(~r) by New-
ton’s Second Law (equation 1.12a), we have:
~P =
N∑i=1
~Fi(~r) = M~aCM (1.14)
Therefore:
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• If there are no external forces, then ~P = 0 by the conservation of
momentum, and ~aCM = ~0. In words, when only internal forces (i.e.
forces exclusively between particles within the system) are present, the
center of mass is not accelerated, has constant velocity.
• If there are external forces, then these forces equal ~F ≡∑N
i=1~Fi(~r) (as
internal forces all cancel out). Then, the laws of motion for the whole
system are condensed into those of the center of mass, as equation
(1.14) is simply Newton’s Second Law of motion for the position ~rCM .
This illustrates why it is convenient to talk about the center of mass in
multi-particle systems: it allows us to simplify the characterization of the
motion of each and every single particle within the system by just focusing
on one position, the one satisfying property (1.13). The behavior of the
body is predictable from the motion of this position.28 The center of mass
is thus the particle equivalent for application of Newton’s laws of motion to
the entire system.
Example 1.19 (A hammer in space) Suppose a hammer is floating in
outer space (i.e. no external forces). Conservation of momentum tells us
that the hammer’s center of mass (a unique point) must experience a con-
stant velocity. For example, if the hammer is rotating as it advances forward,
all of its particles will experience centripetal acceleration due to the uniform
circular motion, except for its center of mass, whose acceleration will always
be zero. Of course, here the center of mass is itself a particle of the system,
that about which the hammer is rotating.
Example 1.20 (Three particles) Consider three particles in space (x, y).
Suppose the masses are m1 = m2 = m and m3 = 2m. Consider the three
bodies are held together by massless strings, forming an equilateral triangle of
whose sides have length `. The strings hold the bodies in positions ~r1 = (0, 0)
(a normalization), ~r2 = (x2, y2) with x2, y2 > 0, and ~r3 = (`, 0). For the
28 Of course, the center of mass may or may not be itself a particle in the system.
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triangle to be equilateral, we need x2 = `/2. Moreover, y2 is the height of
the triangle, so by Pythagoras’ Theorem we have `2 = y22 + x2
2, or y2 =√
32 `.
The position ~rCM = (xCM , yCM ) of the center of mass satisfies:
4m~rCM =∑i
mi~ri
We can decompose this into the x and y components:
(x) : 4m · xCM = m · 0 + m · 1
2` + 2m · ` ⇔ xCM =
5
8` = 0.625`
(y) : 4m · yCM = m · 0 + m ·√
3
2` + 2m · 0 ⇔ yCM =
√3
8` ≈ 0.216`
Note that, in this case, the center of mass is not a particle in the system.
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Chapter 2
Energy
Though we often speak of many different types of energy (e.g. kinetic,
potential, nuclear, chemical, thermal), in the realm of particle motion there
exist only two fundamental forms of energy: kinetic energy and potential
energy.
2.1 Work, Kinetic Energy, and Power
The kinetic energy (T ) is the energy that a body possesses by virtue of
being in motion. It is defined as the work needed to accelerate a body from
some state (at, say, time t1) to another (at, say, time t2 > t1) with a certain
velocity.
Let us first define work and kinetic energy formally, and then state how
the two are related:
Definition 2.1 (Work) The work done by a force ~Fi(~r) on a particle i =
1, . . . , N of velocity ~vi over the time interval [t1, t2] is:
Wi(t1, t2) ≡∫ t2
t1
~Fi(~r) · ~vidt (2.1)
The unit of work is the joule (J).
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Definition 2.2 (Kinetic Energy) The kinetic energy of a particle of mass
mi at some time t is defined as follows:
Ti(t) ≡1
2mi|~vi(t)|2 (2.2)
where ~vi(t) denotes the velocity at time t.
First, from equation (2.1), notice that ~vidt = d~ri by definition (in words,
displacement equals velocity times time elapsed), and thus we can write
work as:
Wi =
∫X~Fi(~r) · d~ri (2.3)
where X denotes the trajectory from ~ri(t1) to ~ri(t2). Thus, the work
done by a force on a body is the dot product of the force and the body’s
displacement that has accumulated over time. Note that this means that
only forces that are non-perpendicular to the direction of motion contribute
to work.1
Work and kinetic energy are very closely related, and often (and loosely)
used interchangeably. The following result states the key equivalence be-
tween the two:
Result 2.1 (Work-Energy Theorem) The work done on an object by a
net force equals the change in the kinetic energy of the object. That is:
Wi(t1, t2) = Ti(t2)− Ti(t1)
for any two instants (t1, t2) in time.
Proof. To prove this equivalence, use Newton’s Second Law in terms of
momentum (equation (1.12a)) to write work (equation (2.1)) as:
Wi(t1, t2) =
∫ t2
t1
d~pidt· ~vidt =
∫ t2
t1
~vi · d(mi~vi) (2.4)
1 For example, as we shall see in the examples below, tension in pendular motion isperpendicular to the direction of motion, so the dot product is zero. Another exampleare normal forces (Definition 1.17)
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where the second equality follows by definition of momentum. By the
product rule, we know:
d(~vi · ~vi) = (d~vi) · ~vi + ~vi · (d~vi)
so ~vi ·(d~vi) = 12d(~vi ·~vi). Assuming that mass mi is constant (i.e. mi = 0),
we have that ~vi · d(mi~vi) = mi2 d(~vi · ~vi). Finally, ~vi · ~vi = |~vi|2. Plugging
everything back into (2.4), we find:
Wi(t1, t2) =
∫ t2
t1
d
(1
2mi|~vi|2
)=
1
2mi
(|~vi(t2)|2−|~vi(t1)|2
)= Ti(t2)−Ti(t1)
our desired result. �
Thus, the work done by a force causes the kinetic energy of the object
to change. Indeed, the fact that an object has kinetic energy implies that a
force must have accelerated it. If the work is positive, then kinetic energy
increases. If negative, the body loses kinetic energy. Finally, kinetic energy
is maintained unless a force changes the body’s velocity, that is, unless a
force works on it.
Henceforth, we will often only refer to kinetic energy, but it should always
be understood that the kinetic energy of a body results from work done on
it by one or more forces.
Finally, we shall define the total kinetic energy of the system (i.e. across
all particles) as follows:
T ≡N∑i=1
Ti =1
2
N∑i=1
mi|~vi|2
Note that, in equation (2.3), the integral is not over time, but over the
specific trajectory of displacement between times t1 and t2. In this definition
of work, therefore, the integral is potentially path-dependent.
When the work caused by a force does not depend on the path, but only
the starting and ending points, we say that the system has been exposed to
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a conservative force.
Definition 2.3 (Conservative force) A force whose work is independent
of the trajectory of displacement.
For example, gravity is a conservative force (as shown in Example 2.3).
So are spring forces. Resistive forces (friction and drag), however, are non-
conservative.
Therefore, the kinetic energy of a body that is put in motion by a con-
servative force is also path-independent. Henceforth, we will typically deal
with conservative forces. Indeed, it is no accident that we have denoted
kinetic energy simply by T as opposed to T (~r), for the latter is redundant.
Finally, note that, when forces are conservative, we can write the limits
of integration in work as just the start and end points of the path, so (2.3)
simplifies to:
Wi =
∫ ~ri(t2)
~ri(t1)
~Fi(~r) · d~ri
Example 2.1 (Shooting an object vertically) In dimensions (x, y), con-
sider a body of mass m going from point ~rA = (xA, yA) to ~rB = (xB, yB).
Suppose xA = xB and yB = yA +h, where h ∈ R. In words, the object expe-
riences no horizontal displacement, and moves only vertically until it reaches
some maximum height of h. Thus, the displacement is simply d~r = (0, h).
For instance, consider an object that is shot vertically into the air.
The object is given some velocity ~vA = (0, vA) at point A, and it comes
to a halt at point B, so ~vB = (0, 0). On the other hand, gravity acts in the
y direction, pushing downward with a force of −mg (with a minus sign to
indicate that the force acts in a magnitude opposite to ~vA).
The work done between points A and B, denoted WAB, is:
WAB =
∫ ~rB
~rA
~F · d~r = −mg(yB − yA) = −mgh
This work is only due to the force of gravity. By the Work-Energy The-
orem (Result 2.1), we know WAB = TB − TA, where TB = 0 (as ~vB = ~0)
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and TA = 12mv
2A. Thus, we have found mgh = 1
2mv2A, or:
h =v2A
2g
In words, if we throw an object vertically with velocity vA, the object will
gain kinetic energy as it reaches a maximum height of h =v2A
2g , at which
point it is attracted back to Earth and kinetic energy begins to decline. In-
terestingly, this height is independent of the object’s mass.
Example 2.2 (Lifting an object vertically) Consider the same case, ex-
cept now vA = 0. That is, the object is initially at rest, and it reaches point
~rB from ~rA by virtue of some force ~F = (0, Fy) that we apply along the way.
For instance, think of lifting a briefcase that it sitting on the floor.
The work now has two sources: that which is done by our own lifting
force ~F (denote it WF ), and that which is due to gravity (denote it Wg).
Clearly, to overcome gravity, we must exert work equal to WF = mgh. The
work due to gravity is, again, Wg = −mgh. The net work that is done is
then W = 0. By the Work-Energy Theorem, this means that there should
be no change in kinetic energy, or TA = TB. Indeed, TA = 12mv
2A = 0 and
TB = 12mv
2B = 0, so TB − TA = 0, as predicted.
Example 2.3 (Gravity as a conservative force) Consider a more gen-
eral example in coordinates (x, y, z). A body is displaced from position
~rA = (xA, yA, zA) to position ~rB = (xB, yB, zB). Call yA − yB = h the
height of the displacement. Gravity exerts a force ~Fg = (0,−mg, 0), that is,
only on the y direction.2 Since no other force is at work here, the work for
going from ~rA to ~rB is:
WAB =
∫ ~rB
~rA
~Fg · d~r = −mg(yB − yA) = −mgh
Thus, note that the work in this example is completely path independent:
2 Recall that we may also write ~Fg in terms of its basis vectors, such that ~Fg =(Fg)x~ex + (Fg)y~ey + (Fg)z~ez. In this case, (Fg)x = (Fg)z = 0 and (Fg)y = −mg, so~Fg = −mg~ey, where ~ey = (0, 1, 0).
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it does not matter how we get to B from A, all that matters is the height
between these two points.
In short, gravity is a conservative force (recall Definition 2.3).
Example 2.4 (Friction revisited) Consider again Example 1.13. At the
end of this example, we calculated that the speed at which the brick arrives
at the end of the incline is v =√
2g`(sin θ − µk cos θ).
Another way of deriving this is to use the Work-Energy Theorem. Let
A be the initial position of the brick (when at rest), and B be the bottom of
the incline. Since the brick starts at rest, TA = 0. The kinetic energy when
it arrives is TB = 12mv
2B, where vB is the speed at point B.
Letting h be the height of the object at point A, the amount of work gravity
is doing is mgh, or mg` sin θ.3 The work that friction does is `µkmg cos θ.
Since the frictional and the gravitational forces are in opposite directions,
we then get WAB = mg` sin θ − `µkmg cos θ.
Invoking the Work-Energy Theorem, we know WAB = TB − TA, or:
mg` sin θ − `µkmg cos θ =1
2mv2
B
Solving for vB yields vB =√
2g`(sin θ − µk cos θ), what we wanted to
obtain.
Finally, we will define a third and closely related concept: power.
Definition 2.4 (Power) The power (P ) is the rate of change of work, or:
P =dW
dt
It is measured in joules (J) per second, also known as watts.4
Recall that W =∫~F · d~r. Thus, the change in work is the dot product
of the force and the displacement which that force produces, dW = ~F · d~r.Thus, an alternative definition for power is:
3 The latter by definition of the sine function: sin θ = h`.
4 We may also use horsepower (hp), where 1 hp = 746 watts.
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P = ~F · ~v
where ~v ≡ d~rdt is the velocity. In particular, if the force is perpendicular
to the velocity vector, then power is zero.
2.2 Potential Energy
The second fundamental form of energy is the potential energy (V ). This
is the energy derived from a body’s position relative to others. The stronger
the force of one body on another, the lower the potential energy between
them. The key assertion is as follows:
Principle 2.1 (Potential Energy Principle) All forces are derived from
(i.e. are governed by) a potential energy function, denoted Vi(~r) for particle
i. In particular, for any system, there exists a differentiable function Vi(~r)
satisfying:
~Fi(~r) = −∂Vi(~r)
∂~r
for each i = 1, . . . , N . That is, in component-wise notation:
(Fx)i(x) = −∂Vi(~r)
∂xi(Fy)i(y) = −∂Vi(~r)
∂yi(Fz)i(z) = −∂Vi(~r)
∂zi
where recall that (Fx)i(x) denotes the x (similarly for y and z) coordinate
of the force on the i-th particle.
In short, a force exerted on a body always reduces its potential energy.5
Or, put differently, a force always works in the direction opposite to the
5 For illustration, take a single-particle, one-dimensional space. Let x denote the(scalar) position of the particle in space. Then, F (x) = −dV (x)
dx, or V (x) =
−∫F (x)dx. Thus, the particle’s potential energy equals the negative of the ac-
cumulated force. In fact, using definition (2.1), we recognize that the right-hand sideis nothing but −T (x), so V (x) + T (x) = 0. Thus, in single-particle systems, kineticand potential forces cancel each other out. This is nothing but a trivial applicationof the Energy Conservation principle, which we shall discuss shortly.
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increase in potential energy.
Example 2.5 (Gravitational potential energy) In dimensions (x, y, z),
the force due to gravity is ~F = (0,−mg, 0) on a body of mass m. The po-
tential energy due to this gravitational force, or in short the gravitational
potential energy, denoted Vg, is
Vg = −∫
~F · d~r = mgy
In should be noted that this value for the gravitational potential energy
holds for bodies that are close enough to each other (for example, the Earth
and an object standing on it). Example 2.19 will show that, when the dis-
tance between the bodies is small, this is indeed a good enough approximation
of Newton’s Universal Law of Gravitation.
The total potential energy of the system is the sum of potential energies
across all particles:
V (~r) ≡N∑i=1
Vi(~r)
Finally, we define the total energy of the system (E) as the sum of po-
tential and kinetic energy across all particles:
Definition 2.5 (Total Energy) The total energy of a system is the sum
of its kinetic and potential energies:
E(~r) ≡ T + V (~r)
Since in this chapter we are typically dealing with large visible objects
(e.g. planets or objects on Earth), in what follows we may also use the term
mechanical energy to refer to total energy. It should be noted, however, that
kinetic and potential energy are present not only in the motion of visible
objects, but also in that of heat, gases, and even atoms. Here is a (non-
exhaustive) list of different types of energy in nature:
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• Mechanical energy: Potential and kinetic energy present in large
visible objects (e.g. planets or large objects on earth). It usually
involves gravitational potential energy.
• Heat and chemical energy: Potential and kinetic energy contained
in gases or other collection of molecules. Heat energy (Q) is expressed
in calories (cal), and is given by:
Q = mC∆T
where C is called specific heat (in calories per gram per degree centi-
grade) and ∆T is the temperature increase, in degrees centigrade.6
Heat and mechanical energy are equivalent in the sense that mechan-
ical energy raises the temperatures of bodies. In particular:
1 cal ≈ 4.2 J
or, equivalently, doing work of 1 joule increases heat energy by 1/
4.2 ≈ .24 calories. For example, to warm up 100 kg of water by 50◦C,
Q = 5, 000 kcal ≈ 2× 107J are needed.
Other types of energy include:
• Atomic and nuclear energy: Potential and kinetic energy stored in
the bonds between the constituents of atoms and atom nuclei. Ruled
by the laws of quantum mechanics.
• Electrostatic energy: Potential energy associated with the forces of
attraction and repulsion between electrically charged particles.
• Magnetic energy: Potential energy between the poles of magnetized
objects.
6 For water, C = 1 cal per gram per degree centigrade. Thus, one calorie is definedas the energy that is required to increase the temperature of 1 gram of water by 1degree centigrade. For aluminum, C = 0.2. For ice, C = 0.5.
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• Electromagnetic radiation: Potential and kinetic energy stored
in radiation (e.g. the sun, radio waves, laser light, etc.). It is not
the energy of particles, but of fields. Ruled by the laws of quantum
mechanics.
All of these types of energy can be converted into one another. For in-
stance, mechanical energy can be turned into electric energy with a dynamo,
and chemical energy is turned into heat energy when gasoline is burned.
2.3 Conservation of Energy
Next, we show a fundamental principle: when forces are conservative,
total energy is always conserved. What this means is that, although indi-
vidual particles might experience different kinetic and potential energies as
they move, total energy is always constant over time. Intuitively, this means
that, overall, potential energy turns into kinetic energy as force is exerted
on particles and they experience motion.7
Result 2.2 (Conservation of Total Energy) When the force is conser-
vative, total energy is conserved. That is, if kinetic energy is path-independent,
then:
∂E(~r)
∂t= 0
Proof. By the Potential Energy principle and Newton’s Second Law, we
have that:
mi~vi = −∂Vi(~r)
∂~r
Left-multiplying by velocity and adding up across particles we find:
N∑i=1
mi~vi~vi = −N∑i=1
~vi∂Vi(~r)
∂~r
7 For example, an archer turns the bow’s potential energy from pulling the stringback into kinetic energy on the arrow as the string is released; overall, the potentialand kinetic energies of the bow, arrow, and the archer’s body all cancel out.
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Using the definition of kinetic energy, T ≡ 12
∑imi~vi · ~vi, note that the
left-hand side is just T (to see this, simply apply the chain rule, and notice
that we crucially need T to be path independent!). Moreover, ~vi∂Vi(~r)∂~r =
Vi(~r), again by the chain rule, so the right-hand side is just V (~r). There-
fore, T = −V (~r), or E(~r) = 0, our desired result. �
Once again, note that energy need not be conserved on each particle (i.e.
Ti + Vi(~r) = 0 need not be). Rather, it is total energy, i.e. the sum of the
energies across all particles within the system, which must be constant.
Moreover, we must once again stress that mechanical energy is conserved
only when the force is itself conservative, that is, when kinetic energy is
path-independent. This is often, though not always, the case. For example:
• The gravitational force is conservative (shown in Example 2.3), so the
mechanical energy of objects that experience gravitational acceleration
is always conserved.
• Spring motion is also conservative (shown in Example 2.10), so the
mechanical energy due to the motion of a spring is also conserved.
• However, frictional forces are (famously) not conservative. Intuitively,
this is clear: because of the COFs of the frictional force (Definition
1.18), the force that is needed to overcome friction is larger the longer
the path between two points. In short, kinetic energy depends on the
path that is taken between the starting and ending points. As a result,
mechanical energy is not conserved in that case.
Example 2.6 (Example 2.3, cont’d) Return to Example 2.3, where we
found WAB = −mgh. By the Work-Energy Theorem, we have −mg(yB −yA) = TB − TA or, rearranging terms:
mgyA + TA = mgyB + TB
Here, mgyA is the gravitational potential energy at point ~rA, and simi-
larly for point ~rB. Thus, the last equation says that total energy (the sum of
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potential and kinetic) does not change from point ~rA to point ~rB. In short,
total energy is conserved.
Example 2.7 (Harmonic Oscillator: Circular Motion) Consider uni-
form motion along circular orbits (Example 1.2), so that
~r ≡
(x
y
)=
(R cosωt
R sin(ωt)
)
for radius R > 0 and angular frequency ω when the angular displacement is
simply dθ = ωdt. Using the results for velocity and acceleration obtained in
Example 1.2, we readily obtain:8
T =m(Rω)2
2, V =
kR2
2
Clearly, then, T = V = 0, so energy is conserved. Indeed, note T is path-
independent, so the force is conservative and, therefore, energy is conserved
as well.
Example 2.8 (Rolling on a semicircle) Consider placing a marble in-
side a semicircle of radius R > 0 (Figure 2.1). The ball will roll down
and oscillate back and forth about the lowest-most point of the circle (call it
(x, y) = (0, 0)), until it comes to a rest at this point.
(0, 0)
R(1− cos θ)
mg
FNR
Figure 2.1: A marble (black dot) rolling down a semi-circular slope. Blue arrows are forces. In red, the ver-tical distance between y = 0 and the current positionof the marble on the y axis.
8 For both V and T , we use that sin2 ωt+ cos2 ωt = 1 by equation (0.2).
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Suppose the marble is in some position (x, y), at an angle θ about the
y axis. Relative to the circle’s center, the y component of the direction
is R cos θ. Relative to the origin y = 0, therefore, the distance is y =
R−R cos θ = R(1− cos θ). Thus, the gravitational potential energy is:9
Vg = mgR(1− cos θ)
For example:
• When θ = 0, then Vg = 0, which makes sense because θ = 0 means the
marble is positioned at the resting point (x, y) = (0, 0), and thus is has
no potential energy left.
• When θ = π/2 (the highest point in the semicircle, where the position
vector and the y axis are perpendicular), we have Vg = mgR, which is
exactly right since the marble in that case is at height R.
The speed is |~v| = Rθ, where θ ≡ dθdt is the angular frequency (Definition
1.5).10 Thus, the kinetic energy is:
T =1
2m|~v|2 =
1
2mR2θ2
The potential energy is:
V = mgR(1− cos θ)
Thus, the system’s total mechanical energy is:
E = T + V =1
2mR2θ2 +mgR(1− cos θ)
Now, as in the case of the pendulum, we use a Small-Angle approximation
(Definition 1.16) to make progress. Namely, we can (Taylor-)approximate
9 Note the potential energy includes forces only in the y direction, not the x direction.We explain this in the final paragraph of this example.10 The derivation of this velocity is similar to what we saw in Example 1.2, exceptthat now the angular velocity is changing with time (zero when the marble is released,and increasing as the marble rolls down, with a maximum at the lowest point).
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cos θ around θ = 0 by 1 − θ2
2 . With this approximation, V = mgR θ2
2 , so
V = mgRθθ. Therefore:
E = T + V = mR2θθ +mgRθθ = mRθ(Rθ + gθ)
Since gravity is a conservative force, we may invoke conservation of me-
chanical energy. Thus, setting E = 0, we obtain:
θ +g
Rθ = 0
Therefore, the angular motion obeys (up to the error in the Small-Angle
approximation method) a simple harmonic oscillation (recall equation (1.4)).
From Example 1.9, we then know that:
θ = θmax cos(ωt+ ϕ) (2.5)
where ω ≡√
gR is the angular frequency of motion,11 θmax is the am-
plitude (the maximum angle that is reached before the process oscillates
back), and ϕ is the phase angle (in radians). The period of motion is,
thus, T = 2πω = 2π
√Rg seconds, and the frequency of motion is f = 1/T Hz.
Importantly, we have found exactly the laws of motion that we found
for pendular motion (Example 1.10). The only difference between the two
set-ups is that, in the case of the pendulum, a string was adding tension
to the body, whereas for the present example a normal force plays this role
(assuming no friction). Yet, we have made no mention of these forces when
invoking the conservation of energy. Why do they make no difference?
The reason is that both of these forces are orthogonal to the direction of
motion (see Figure 2.1), and therefore ~FN · d~r = 0. As potential energy is
defined V = −∫~F · d~r, normal forces do not do any work, and thus they do
not contribute to potential energy.
11 Note that angular frequency and angular velocity are here different because thelatter is non-constant. When the circular motion is uniform, as in Example 1.9, thetwo coincide.
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Example 2.9 (Amplitude and phase angle) Consider the simple pen-
dulum once again (Example 1.10). Using the results there, as well as in
our previous example above these lines, the angle at any time t is given by
equation (2.5).
We now ask how to obtain the amplitude θmax, namely the maximum
angle that the pendulum can achieve, and the phase angle ϕ.
Let ~rA = (xA, yA) = (0, 0) be the position of the bob when at rest, ~rC =
(xC , yC) denote the position at the highest point (when θ = θmax), and ~rB =
(xB, yB) somewhere in between (at, say, angle θ0). If ` is the total length
of the string, then the heights at points C and B are, respectively, yC =
`(1− cos θmax) and yB = `(1− cos θ0).
• Amplitude: To find θmax, we invoke the conservation of mechanical
energy: TA + VA = TB + VB = TC + VC . Since the bob is at rest in
point A, we can normalize the potential energy to VA = 0. Since it has
come to a halt at C, it has no kinetic energy left, so TC = 0. Thus:
1
2m|~vA|2︸ ︷︷ ︸=TA
=1
2m|~vB|2︸ ︷︷ ︸=TB
+mgyB︸ ︷︷ ︸=VB
= mgyC︸ ︷︷ ︸=VC
(2.6)
Letting h = yC − yB, then the right side simplifies to 12m|~vB|
2 = mgh,
so h = |~vB |22g . We can now compute the amplitude. Since yC = yB+h =
`(1− cos θmax), then cos θmax = 1− yB+h` = cos θ0 + |~vB |2
2g` .
Another way to find θmax is to use the Work-Energy Theorem. First,
the change in kinetic energy between ~rA and ~rB is TB−TA = 12m|~vB|
2−12m|~vA|
2. On the other hand, the work done is just due to gravity,
WAB = −mgyB.12 By the Work-Energy Theorem, WAB = TB−TA, or
−mgyB = 12m|~vB|
2 − 12m|~vA|
2, exactly the left side of equation (2.6).
Similarly, WBC = −TB (as TC = 0), so mg(yC − yB) = 12m|~vB|
2,
exactly the right side of equation (2.6). This shows that the Work-
12 Again, even though there is also the tension force on the string, the force vector inthat case is perpendicular to the direction of motion, so the dot product of the twois zero. Thus, tension does not do any work.
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Energy theorem and the Conservation of Mechanical Energy are, in
fact, one and the same.
• Phase angle: To find ϕ, we need to consider the initial condition of
the system at t = 0. Suppose that the bob is in point B at that time,
with velocity |~vB,0| and angle θ0. We know, by equation (2.5), that
θ0 = θmax cosϕ. The angular velocity at t = 0 is:
dθ
dt
∣∣∣t=0
= −ωθmax sinϕ
(in radians / sec), but dθdt =
|~vB,0|` ,13 so |~vB,0| = −`ωθmax sinϕ, which
we can solve for sinϕ.
Example 2.10 (Harmonic Oscillator: Spring Motion) Consider a sin-
gle body with mass m that is attached to the end of a massless spring. Recall
the force in this case is proportional to the position (Hooke’s law, Princi-
ple 1.3), Fx = −kx. Similarly, if there’s a pull in the y direction, then
Fy = −ky.
• The work needed to bring the object from point A to point B is:
WAB =
∫ ~rB
~rA
~F · d~r = k
∫ ~rB
~rA
~r · d~r =1
2k|~r|2
where ~r = (x, y). We have then derived the potential energy:
V (x, y) =1
2k(x2 + y2)
• Conversely, we can recover the force from the potential energy. By the
Potential Energy principle, we have:
~F = −
(∂V (x,y)∂x
∂V (x,y)∂y
)= −
(kx
ky
)13 Take an infinitesimal angle dθ. The arc is some ds, and the hypothenuse is `. Byconstruction, then, dθ = ds
`. Dividing both sides by dt, we find dθ
dt= |~v|
`, where the
angular speed is |~v| = dsdt
.
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• Newton’s Second Law then says:
(x
y
)= −ω2
(x
y
)(2.7)
where ω =√
km is the spring’s angular frequency (for the derivation,
see Example 1.9). Again, this means that the spring is a simple har-
monic oscillator, so using our results for oscillators we have that:
x = xmax cos(ωt+ ϕx)
where xmax is the amplitude, and ϕ is the phase, and similarly for y.
• The system’s kinetic energy is:
T =1
2m|~v|2 =
1
2m(x2 + y2)
Note kinetic energy is path-independent (it depends on velocities, not
positions), showing that the force caused by springs is conservative.
• The system’s total energy is, therefore:
E = T (x, y) + V =1
2
[m(x2 + y2) + k(x2 + y2)
]=
1
2
(m|~r|2 + k|~r|2
)In words, the total energy is proportional to the squares of the velocity
and the position.
• Now we can compare these results with those of circular motion (Ex-
ample 1.2). For given velocity, the particle motion describes an orbit.
Along such orbit, potential energy remains constant. Thus, the particle
stays in orbits of constant energy. For particles of higher momentum,
the particle describes an orbit of larger radius, and the particle’s veloc-
ity picks up. However, though higher, potential energy remains con-
stant. This is illustrated in Figure 2.2, where outer orbits correspond
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to particles with higher momenta and, therefore, higher velocity. We
can thus think as each concentric orbit in the figure as nothing but con-
tours of constant potential energy, with energy increasing quadratically
as we move away from the origin.
#»v 1
#»a 1
#»v 1
#»a 1
#»v 2
#»a 2
#»v 2
#»a 2
Figure 2.2: Circular motion contours of constant po-tential energy. The particle describes orbits for a con-stant potential energy. For particles of higher velocities(e.g. ~v2 > ~v1), potential energy is higher and the orbitis larger.
For instance, planets in the Solar System that are farther from the Sun
orbit at higher velocities and exhibit higher potential energies.14 The
Sun, exerting a gravitational pull, provides the centripetal acceleration
that is necessary to keep the planets in orbit.
• Let us show formally that total energy is indeed conserved. First, the
change in total potential energy is:
V = k (xx+ yy)
The change in kinetic energy is:
14 The orbits of the planets in the Solar System are elliptical, not circular. We neglectthis detail here.
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T = m (xx+ yy) = −mω2 (xx+ yy) = −k (xx+ yy)
where the second equality uses (2.7), and the third equality uses ω2 =
k/m. Therefore, V + T = 0, as sought. Thus, all potential energy is
transformed into kinetic energy as the particle orbits around the circle.
• Finally, we can compute the amplitudes (xmax, ymax) and the phase
(ϕx, ϕy) of the spring, similarly to the way we did it in Example 2.9.
Pick three points: ~rA = (xA, yA) = (0, 0), where the spring is in a
relaxed position at some length `; ~rC = (xC , yC) = (xmax, ymax), where
the spring reaches its maximum amplitude before rebounding; and ~rB =
(xB, yB), somewhere in between. By the conservation of energy, TA +
V (xA, yA) = TB + V (xB, yB) = TC + V (xC , yC). Since V (x, y) =12k(x2 + y2), then V (xA, yA) = 0. Since T = 1
2m|~v|2 and the spring
comes to a halt at ~rC , then TC = 0. Thus:
1
2m|~vA|2 =
1
2m|~vB|2 +
1
2k(x2
B + y2B) =
1
2k(x2
max + y2max)
from which we can calculate directly the amplitudes from given speeds.
Similarly, the Work-Energy Theorem will give us the same result.
Example 2.11 (Roller Coaster) Mechanical energy is also conserved in
(frictionless) roller coasters. To show this, consider a roller coaster car of
mass m that slides down a incline with a loop at the end that has radius R
(Figure 2.3).
We will consider four positions for the car: ~rA = (0, h) (i.e. at the very
top of the incline, at height h > 0); ~rB = (xB, yB) with xB > 0, yB < h
(i.e. mid-way down the incline); ~rC = (xC , 0) with xC > xB (i.e. at the
lowest point of the loop); and ~rD = (xD, yD) with xD = xC and yD = 2R
(i.e. at the highest point in the loop, if the object ever ever makes it there).
The velocity vectors are ~vA = (0, 0), ~vB = (vB,x, vB,y), ~vC = (vC,x, 0), and
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Figure 2.3: A roller coaster.
~vD = (vD,x, 0).15 By the Conservation of Mechanical Energy, we have:
TA + VA = TB + VB = TC + VC = TD + VD (2.8)
where Ti and Vi denote kinetic and potential energy at point ~ri for i =
A,B,C,D. In particular:
• Point ~rA: At this point, since there is no velocity, TA = 0. Potential
energy is purely gravitational, so VA = mgh.
• Point ~rB: At this point, TB = 12m(v2B,x + v2
B,y
). Potential energy is
gravitational, so VB = mgyB.
• Point ~rC : At this point, TC = 12mv
2C,x. Potential energy is gravita-
tional, so VC = mgyC = 0.
• Point ~rD: At this point, TD = 12mv
2D,x. Potential energy is gravita-
tional, so VD = mgyD = 2mgR.
Note that kinetic energy is always path independent, so the principle of
Conservation of Mechanical Energy holds. This is because only gravitational
forces, which are conservative, are at play here. If friction was added to
15 In point A, the car is released with no acceleration; at points C and D there isclearly no vertical displacement.
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the picture, energy conservation could not be invoked (as friction is a non-
conservative force).
Here, equation (2.8) reads:
gh =1
2
(v2B,x + v2
B,y
)+ gyB =
1
2v2C,x =
1
2v2D,x + 2gR
Thus, vC,x =√
2gh and vD,x =√
2g(h− 2R). At point ~rB = (xB, yB),
half-way down the incline, the car is moving at speed
|~vB| =√
2g(h− yB)
We also know, from Example 1.2, that at point ~rD, if reached, there
must be a centripetal acceleration equal to aD ≡ |~aD| =v2D,x
R . Moreover,
from Example 1.12, we know that, if ~rD is reached, then aD ≥ g, or else
the car’s acceleration could not have overcome the force of gravity. Using
vD,x =√
2g(h− 2R), then we have 2g(h− 2R) ≥ gR, which simplifies to:
h ≥ 2.5R
This is a classic result: if we release the car down the hill, it will not
make it to the top of the loop unless we drop it from a height that is, at least,
two-and-a-half times the loop’s radius. In the real world, where there is both
air and car-to-rails friction, the Conservation of Energy will fail, and the
factor becomes even higher. However, 2.5 may be used as a lower bound, as
it holds in a frictionless environment.
2.4 Collisions
Collisions of bodies are an example of the conservation of total energy
(Result 2.2) and the conservation of momentum (Result 1.1) in practice.
In this section, we are interested in understanding the change in the ki-
netic energy of particles when total momentum and total energy must be
conserved.
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Example 2.12 (Colliding particles, cont’d) Recall our example with col-
liding particles (Example 1.18). The particles had masses m1 and m2, ve-
locities ~v1 and ~v2, and eventually collided and stuck together into a single
particle of mass (m1 +m2) and velocity ~v ′.
By assuming no external forces, we invoked conservation of momentum
to derive:
~v ′ =m1~v1 −m2~v2
m1 +m2
Now, we can study the energy involved before and after the collision:
• Before the collision, kinetic energy is T = 12m1|~v1|2 + 1
2m1|~v2|2.
• After the collision, kinetic energy is T ′ = 12(m1 +m2)|~v ′|2. Or, using
our formula for ~v ′:
T ′ =1
2
m21|~v1|2 +m2
2|~v2|2
m1 +m2− m1m2
m1 +m2|~v1||~v2| < T
In words, the kinetic energy goes down as a result of the collision (it
turns into heat energy, so that total energy remains constant). Thus, in the
absence of external forces, kinetic energy must be destroyed if the system’s
total momentum is to be conserved.
Example 2.13 (Splitting particles) While kinetic energy has been lost
in the above example, it would increase in the opposite scenario: the (sudden)
separation of particles. Consider a particle of mass m which suddenly splits
into two sub-particles. Initially, ~v = ~0 and thus momentum ~p = ~0. After the
separation, the sub-particles have mass m1 and m2, and acquire velocities
~v ′1 and ~v ′2 , in opposite directions as they fly away from each other.
Because no external forces act upon the system, momentum is conserved,
so ~0 = m1~v′
1 − m2~v′
2 . Interestingly, for momentum to be conserved, the
relative speed of each particle must be constant, with the relatively lighter
object acquiring a relatively faster speed:
|~v ′1 ||~v ′2 |
=m2
m1
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For instance, if one sub-particle is twice the mass of the other, after the
split it will move at half the speed as the other one. It must, or else the
system’s total momentum would not be conserved.
Finally, clearly, kinetic energy has increased: it was T = 0 initially, and
after the collision it becomes T ′ = T ′1 + T ′2 with T ′j = 12mj |~v ′j |2 for each
sub-particle j = 1, 2.
We have just seen two examples of collisions in which kinetic energy
decreases and increases, respectively. Let us analyze these cases more gen-
erally.
Suppose that the two particles collide and, as a result, the particles
bounce off. After the collision, the particles have velocities ~v ′1 and ~v ′2 .
Therefore, invoking conservation of momentum (because there are no exter-
nal forces) we have:
m1~v1 −m2~v2 = m1~v′
2 +m2~v′
2 (2.9)
On the other hand, total energy must be conserved. Here:
• Before the collision, kinetic energy is T = 12m1|~v1|2 + 1
2m1|~v2|2.
• After the collision, kinetic energy is T ′ = 12m1|~v ′1 |2 + 1
2m1|~v ′2 |2.
In any case, by the conservation of total energy, there must exist a Q
such that:
T +Q = T ′
Then, we distinguish the following cases:
Case 1: Q > 0⇒ Super-elastic collision.
In this case, kinetic energy increases as a result of the collision. This
is the case, for example, of explosions.
Case 2: Q = 0⇒ (Completely) elastic collision.
In this case, kinetic energy remains constant as a result of the collision.
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Case 3: Q < 0⇒ Inelastic collision.
In this case, kinetic energy decreases as a result of the collision. In
this case, kinetic energy is generally transferred into heat energy.
Therefore, Example 2.12 if an example of an inelastic collision, while
Example 2.13 showed a super-elastic collision. Let us now see an example
of Case 2.
Example 2.14 (Completely elastic collision) Since Q = 0, then T =
T ′, so:
m1|~v1|2 +m1|~v2|2 = m1|~v ′1 |2 +m1|~v ′2 |2 (2.10)
by the conservation of total energy. Now, we can use equations (2.9)-
(2.10) to solve for (~v ′1 , ~v′
2 ). The solution is:
~v ′1 =(m1 −m2)~v1 − 2m2~v2
m1 +m2and ~v ′2 =
(m1 −m2)~v2 + 2m1~v1
m1 +m2
Suppose for illustration that ~v2 = ~0, so that ~v ′1 = m1−m2m1+m2
~v1 and ~v ′2 =2m1
m1+m2~v1. For instance, think of a ping-pong ball (object 1) colliding with
a billiards ball (object 2) which is at rest. Then, ~v ′2 has the same sign
as ~v1 (the billiards ball will start moving in the same direction as the one
that the ping-pong had originally), but ~v ′1 has the opposite direction as ~v1 if
m1 < m2 (the ping-pong ball will “bounce off” the billiards ball since it is
less massive).
A few special cases:
• When m2 ≈ 0, we get:
~v ′1 ≈ ~v1 and ~v ′2 ≈ ~v2 + 2~v1
For example, if a bowling ball (here object 1) collides with a ping-pong
ball (object 2) which is at rest (~v2 = ~0), the bowling ball’s velocity will
hardly be altered. However, after the collision, the ping-pong ball will
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see its velocity increase to 2~v1. The latter is not intuitive, but it is
confirmed experimentally.
• When m1 ≈ 0, we get:
~v ′1 ≈ −(~v1 + 2~v2) and ~v ′2 ≈ −~v2
For example, if a ping-pong ball (now object 1) collides with a bowling
ball (object 2) which is at rest (~v2 = ~0), the ping-pong ball will bounce
off with the same speed but opposite direction, while the bowling ball
will stay at rest.
• When m1 = m2, we get:
~v ′1 = −~v2 and ~v ′2 = ~v1
Now, two billiards balls collide. For example, if ball 2 was originally
at rest (~v2 = ~0) and ball 1 collides into it, then ball 1 stops abruptly
after the collision and sets ball 2 in motion at an equal speed. This is
also perfectly illustrated with a so-called Newton cradle.
Let us now look at collisions from the frame of reference of the center
of mass (Definition 1.27). Recall that, in the absence of external forces, the
center of mass will always have the same velocity (Result 1.2). From the
point of view of the center of mass, the center of mass is at rest. Therefore,
from the point of view of the center of mass, particles have no momentum.
Consider two particles with masses m1 and m2 moving in opposite di-
rection toward the center of mass, i.e. with some velocities ~u1 and −~u2.16
Suppose kinetic energy is conserved, so that the collision is completely elas-
tic (Q = 0). After the collision, the velocities are ~u ′1 and ~u ′2 . Since total
momentum is zero before the collision, we have:
16 We use the notation ~u instead of ~v to emphasize that these velocities are strictlyfrom the point of view of the reference frame of the center of mass. The velocity ~v isfrom the general reference frame (e.g. the laboratory).
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m1~u′
1 +m2~u′
2 = ~0
Because the collision is perfectly elastic, conservation of energy says:
1
2m1|~u1|2 +
1
2m1|~u2|2 =
1
2m1|~u ′1 |2 +
1
2m1|~u ′2 |2
The solution of the system of two equations is ~u ′1 = −~u1 and ~u ′2 = −~u2.
This is thus a remarkable property: at the center of mass, colliding particles
reverse direction, but the speeds remain the same.
Outside the frame of reference of the center of mass (e.g. the laboratory’s
reference frame), the center of mass has some (constant) velocity ~vCM . As
we derived in Section 1.4, the velocity of the center of mass is:
~vCM =m1~v1 −m2~v2
m1 +m2
Of course, from the point of view of the general reference frame:
~u1 = ~v1 − ~vCM and ~u2 = ~v2 − ~vCM (2.11)
(These identities allows us to go back and forth between the two reference
frames). Suppose that the collision is perfectly inelastic, so that the particles
will stick together (kinetic energy is lost and transferred into heat). For
illustration, say particle 2 was at rest (~v2 = ~0). After the collision, a single
particle of mass m1+m2 emerges with velocity ~v ′. In the absence of external
forces, total momentum is conserved, i.e. m1~v1 = (m1 +m2)~v ′, so:
~v ′ =m1
m1 +m2~v1 = ~vCM
In words, the new particle acquires the velocity of the center of mass.
The change in kinetic energy from the lab’s reference frame, call it Qv, is
negative (as the collision is inelastic) and given by:
Qv = T ′v − Tv = −1
2
m1m2
m1 +m2|~v1|2 < 0
What about from the center of mass reference frame? Using the identities
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in (2.11), we have:
~u1 = ~v1 − ~vCM = ~v1 −m1
m1 +m2~v1 =
m2
m1 +m2~v1
~u2 = ~v2 − ~vCM = ~0− m1
m1 +m2~v1 = − m1
m1 +m2~v1
The kinetic energy in the center of mass frame before the collision is
Tu = 12m1|~u1|2 + 1
2m2|~u2|2. After some algebra, we obtain:
Tu =1
2
m1m2
m1 +m2|~v1|2 = −Qv
In words, all the kinetic energy at the center of mass is transferred to heat
when the collision occurs. The number Tu is therefore the maximum kinetic
energy that can ever be lost in an inelastic collision, and it is sometimes
called the internal kinetic energy of the system.
Definition 2.6 (Internal Kinetic Energy) The maximum kinetic energy
that can result from a collision, i.e. the change in total kinetic energy of a
system relative to the center of mass.
2.5 Impulse and Thrust
When acted upon by forces, an object is given an impulse. Impulse
is, thus, nothing but the accumulation of all these forces over a certain
interval of time. For example, when two particles collide, some or all of them
experience an impulse for the duration of the collision. In this section we
explore the connection between impulse, momentum, energy, and collisions.
Definition 2.7 (Impulse) The impulse is the accumulated force on an ob-
ject i = 1, . . . , N over a certain interval [t1, t2] of time. Formally:
~Ii ≡∫ t2
t1
~Fi(~r)dt
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Since ~Fi = mi~ai by Newton’s Law, and mi~ai = d~pidt by the definition of
momentum, we can write impulse as:
~Ii =
∫ t2
t1
d~pidt
dt =
∫ ~pi(t2)
~pi(t1)d~p = ~pi(t2)− ~pi(t1)
This may thus serve as an alternative definition: impulse is the change
in an object’s momentum. A force acting upon an object causes a change in
the object’s momentum, and this change is what we call the impulse.
Note, of course, that if the force is conservative, then the conservation
of total momentum (Newton’s Third Law) must hold, and thus
N∑i=1
~Ii = ~0
Let’s see some examples:
Example 2.15 (Ballistic pendulum) A ballistic pendulum (Figure 2.4)
is a device used to measure the speed of a bullet. Suppose that a bullet
of mass m is fired at speed v0 against an object of mass M , which hangs
from a massless string of length L and whose location before the collision is
normalized to (x, y) = (0, 0).
The collision is completely inelastic, with the resulting object, of mass
(m + M), speeding forward at velocity ~v′ along an arc, coming to a halt at
some angle θ, and swinging back down according to the motion of the simple
pendulum (Example 1.10).
This setting can be used to calculate the initial speed v0. By momentum
conservation, we know that
m~v0 = (m+M)~v ′
When the pendulum comes to a halt, ~v = ~0, the potential energy is
zero, so the object’s kinetic energy is completely converted into gravitational
potential energy. Thus, by conservation of energy, we have 12(m+M)|~v ′|2 =
(m+M)gh or, simplifying:
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Figure 2.4: The ballistic pendulum.
|~v ′| =√
2gh
where h denotes the vertical displacement of the object. In particular,
h = L(1− cos θ). As we know from Example 1.10, h is typically very small,
and hence hard to measure experimentally. But we can again use the Small-
Angle approximation, and say that h ≈ x2
2L .17 Therefore:
|~v ′|2 ≈ gx2
L
and so the velocity of the bullet is:
|~v0| =m+M
mx
√g
L
Finally, the impulse of the bullet on the massive object is ~I = (m+M)~v ′
(as the object’s momentum before the impact was zero), and the impulse of
the object on the bullet is ~I = (m + M)~v ′ −m~v0 = ~0, as momentum con-
17 First, perform a Taylor expansion of cos θ about θ = 0 to obtain cos θ ≈ 1− θ2/2,
so that h = L θ2
2. Next, we know that the horizontal displacement is x = L sin θ.
Since sin θ ≈ θ about θ = 0, then x ≈ Lθ, or θ ≈ x/L. In sum, h ≈ L θ2
2≈ x2
2L.
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servation holds. Indeed, since the collision is perfectly inelastic, all kinetic
energy is lost, and the newly formed object absorbs all momentum.
Example 2.16 (Falling object) Consider an object of mass m that drops
to the floor from height h and no initial speed. Using a similar reasoning to
above, the speed at the time that the object hits the floor is |~v| =√
2gh, as
~v = −(0,√
2gh). Its momentum is ~p = m~v.
Let’s look at the impulse for different types of collisions:
• If the collision is completely elastic (no kinetic energy is lost), the
ball would bounce back up with the same speed and in the opposite
direction, ~v ′ = −~v (for a derivation, see Example 2.14). Thus, the
new momentum would be ~p ′ = −~p, and the change in momentum (or
impulse) is:
~I = ~p ′ − ~p = 2m~v
• If the collision is completely inelastic (all kinetic energy is lost), the
ball would lose all speed, ~v ′ = ~0 (for a derivation, see Example 2.14).
Thus, the new momentum would be ~p ′ = 0, and the change in mo-
mentum (or impulse) is:
~I = ~p ′ − ~p = m~v
Interestingly, if one knows the impulse, then by Definition 2.7 an average
force can be calculated via:
~Faverage ≈~I
∆t
where ∆t ≡ t1 − t2 is the time during which the force is exerted, and
~I ≡ ∆~p is the change in momentum. For example, for elastic collisions (e.g.
a fast-moving tennis ball against a player’s racket), the collision time ∆t
is often extremely short, so the average force that the objects experience
can be extremely high. That is, during the brief instants during which the
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tennis ball changes its course, it may experience an extremely high increase
in weight.
Now suppose the impulse is exerted, more generally, by N ≥ 2 objects of
equal mass m traveling at equal velocity ~v (though at potentially different
times within the period ∆t). Then, the average force exerted on the body
satisfies ~Faverage =~I
∆t = Nm~v, where now Nm, in kg/sec, is the total mass
over the period ∆t. Thus, more generally, as ∆t→ 0, we have:
~F =dm
dt~v (2.12)
where dmdt is the rate of change in the mass of the colliding or expelling
objects. For instance, if many equal objects are thrown onto or propelled
out of another object consecutively over a short period of time, the latter
object will feel an impulse. This is the basic idea behind rockets, where the
force is called thrust.
Definition 2.8 (Thrust) The force, given by equation (2.12), that acts
upon a body and is typically due to the propulsion of particles (e.g. gas).
Let’s study rockets in more detail:
Example 2.17 (Rockets in outer space) Rockets experience an impulse
from their engines, which expel a huge number of gas particles over an ex-
tremely short period of time. As a result, the rocket experiences a thrust
given by equation (2.12), which points in the direction opposite to that of
the gas particles.
Consider a rocket in outer space (so that there are no external forces
and momentum must be conserved). The rocket is burning chemical energy
and expelling gas at some velocity ~u, which is fixed relative to the rocket.
Knowing the rate at which gas comes out (the object dmdt ) would then allow
us to calculate the thrust force:18
~Fthrust =dm
dt~u (2.13)
18 For instance, for the Saturn rocket in the Apollo missions, u = 2.5 km/sec anddmdt
= 15× 103 kg/sec, and so the rocket experienced a thrust of F = 35× 106 N.
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The thrust acts upon the rocket for a certain amount of time, called
the burn time, and as the fuel burns the rocket’s mass goes down, so the
acceleration during the burn goes up as a result.
Let us derive this change in velocity. The derivations are from the ref-
erence frame of the lab, not the rocket’s. Let m be the mass of the rocket,
and ~v be the velocity at time t from the lab’s reference frame (recall ~u is
the velocity from the rocket’s reference frame). At time t+ ∆t, velocity has
changed to ~v + ∆~v, and the mass is now m − ∆m. Always from the lab’s
reference frame, the cloud of gas particles with total mass ∆m that has been
expelled has velocity ~v − ~u. If ~v > ~u, we will see the exhaust of gas go up
from our frame of reference, otherwise we will see it go down.
Now, we can compare momenta at both moments in time:
~p(t) = m~v
~p(t+ ∆t) = (m−∆m)(~v + ∆~v)︸ ︷︷ ︸Rocket
+ ∆m(~v − ~u)︸ ︷︷ ︸Expelled gas
= m~v +m∆~v − ~u∆m+ ∆m∆~v︸ ︷︷ ︸≈0
Thus, the change in total momentum (zero, by the conservation of total
momentum) is:
~0 = ~p(t+ ∆t)− ~p(t) = m~v − ~u∆m ⇔ m~v = ~u∆m
Diving through by ∆t, and letting ∆t→ 0, we obtain:
m~a = ~udm
dt(2.14)
We recognize on the right-hand side the thrust on the rocket (equation
(2.13)). Thus, this is nothing but Newton’s Second Law for rockets.
Example 2.18 (Rocket on Earth) In the previous example, we have seen
rockets with no external forces (e.g. in outer space). What if the rocket suf-
fers the gravitational force (e.g. as it is launched from Earth)?
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Consider a standard vertical launch (i.e. ignore the x dimension so the
problem is one-dimensional). Now, a force mg goes opposite to the thrust
force Fthrust = udmdt . In this case, equation (2.14) must be modified to:
ma = udm
dt−mg
commonly known as the rocket equation. If the rocket has initial velocity
v(t1) and final velocity v(t2), then integration yields:
v(t2)− v(t1) = −u ln
(m(t2)
m(t1)
)− g(t2 − t1)
Note that because m(t2)m(t1) ≤ 1 due to the rocket burning fuel, the first term
will always be positive.
For instance, in a vertical launch, where the rocket starts at rest (v(t1) =
0 at t1 = 0), then the rocket’s speed after some time t (the burn time) is:
v(t) = −u ln
(m(t)
m0
)− gt
where m0 denotes the initial mass of the rocket before the exhaustion of
gas begins.19 For the rocket to lift off, then, the thrust must be strong enough
in the sense −u ln(m(t)m0
)> gt.
This has shown that the change in a rocket’s velocity for a given amount
of fuel and a given burn time is fixed. However, as we will now show, the
change in kinetic energy is not fixed. In particular, two rockets that burn the
same amount of fuel over the same amount of burn time (thereby achieving
the same increase in velocity) do not experience the same change in kinetic
energy if their initial velocities differ. For instance, the change in kinetic
energy at launch is different than that when the rocket is already in the air.
To see this, fix the burn time t and the exhaust m(t)m0
, so that ∆v ≡ v−v0
is fixed (where v0 is the initial velocity and v is the velocity at time t).
Consider two scenarios: a rocket at rest, and one that is already in the air.
• At launch (v0 = 0), the increase in kinetic energy is ∆T = 12m(∆v)2.
19 As a special case, note that if the rocket’s mass did not change, v(t) = v0 − gt,exactly what we derived for falling objects (e.g. Example 1.13).
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• Once in the air (v0 > 0), the increase in kinetic energy is ∆T =12m[v2
0 − (∆v)2].
Thus, the change in kinetic energy depends on the initial velocity of the
rocket. The increase in kinetic energy is higher if the rocket was in motion
to begin with.
2.6 Newton’s Universal Law of Gravitation
So far, we have assumed that a force of gravity mg acts on each body
of mass m, where g is the acceleration that the body feels in free fall. The
potential energy of this fall, we have argued, is mgh (recall Example 2.5),
where h is the height from which the object is dropped.
In considering this, we have made a simplification: the body and the
surface of the Earth are separated by a relatively small distance. But what
are the gravitational forces acting between two distant bodies, say the Earth
and the Sun?
Consider two bodies, of masses m1 and m2, which are at a distance r > 0
apart from each other (that is, r = |~r1−~r2|). These two bodies are attracted
by the force of gravity. Let Fmimj be the force exerted on body i by body
j 6= i. By Newton’s Third Law, Fm1m2 = Fm2m1 ≡ F (i.e. same magnitude
but opposite in direction).
Newton postulated a value for F . His statement has since become one
of the most famous formulas in all of physics:
Principle 2.2 (Newton’s Universal Law of Gravitation) The gravita-
tional force between two bodies of masses m1 and m2, separated by distance
r, is:
F = Gm1m2
r2
where G = 6.674× 10−11 is called the gravitational constant.
In words, the attractive force between two bodies is directly proportional
to the product of their masses, and inversely proportional to the square of
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the distance between them. This makes the Universal Law of Gravitation
an inverse-square law.
The gravitational constant is an extremely small number, showing that
gravitation is an extremely weak force. For example, two bodies of 1kg each
that are separated by 1m are attracted by a force of just G Newtons, and
extremely small value.
By Newton’s Second Law, Gm1m2r2 = m1a1 = m2a2, where ai is the
gravitational acceleration experienced by body i = 1, 2. Thus:
ai = Gm−ir2
for each i. Thus, the gravitational acceleration of a body is inversely
proportional to the squared distance to the other body. If the objects are
ten times apart, the gravitational acceleration falls by a factor of 100. This
is reminiscent of another inverse law: the centripetal acceleration experi-
enced by a body in uniform circular motion (Example 1.2), such as those
(approximately) described by the orbits of the planets.20
Remark 2.1 (Deriving g) Throughout, we have called g the gravitational
acceleration of bodies on free-fall toward the Earth (Definition 1.13). If the
body has mass m, then the gravitational force is mg (by Newton’s Second
Law). Experimentally, it has been found that g = 9.8m/s2.
Interestingly, this is nothing but a special case of Newton’s Universal
Law of Gravitation. In particular, let mEarth be the mass of the Earth.
Assuming the body is close enough to the surface of the Earth, the distance
r is approximately equal to the radius of the Earth, call it rEarth. Thus,
F = GmEarthmr2Earth
≈ mg, and therefore:
g ≈ mEarthG
r2Earth
(2.15)
Plugging in values for the Earth’s mass (mEarth = 6 × 1024 kg), the
Earth’s radius (rEarth = 6.4× 103 km), and the gravitational constant (G =
20 This is not strictly true, for planets describe elliptical (not circular) orbits aroundthe Sun. However, the connection between the two should be clear.
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6.674× 10−11), we will obtain the famous g = 9.8.
Example 2.19 (Example 2.5, cont’d) In Example 2.5, we argued that
the gravitational potential energy is Vg = mgh when the object is at height h
from the Earth’s surface. Let us now derive the gravitational potential energy
for the general case of distant bodies, and argue that it is well approximated
by mgh when the bodies are sufficiently close to one another.
Consider two bodies, with masses m (e.g. an object in space) and M
(e.g. the Earth). Body m moves from some point A to some point B, where
A and B are at distance RA and RB of M , respectively. Along the way from
A to B, body m feels the gravitational attraction of M . Let r denote the
distance between the bodies at some point along the path. The force acting
on m at that time is, thus, F = GmMr2 .
Since gravity is a conservative force (Example 2.3), then the gravitational
potential energy is independent of the path taken by m, so we may assume,
without loss of generality and for simplicity, that the path is a straight line.
Then, the work done by the gravitational force F = GmMr2 from A to B
is:21
WAB =
∫ RB
RA
GmM
r2dr = −GmM
r
∣∣∣RBRA
= −GmM 1−RB/RARB
Using the Work-Energy Theorem, WAB = TB − TA, so TB = GmMRA
and
TA = GmMRB
. As for potential energy, we can invoke the Conservation of
Mechanical Energy to say:22
VB − VA = −WAB = GmM1−RB/RA
RB
For instance, if point A (the starting point) is infinitely far (RA = +∞),
then TB = 0. That is, the body has no kinetic energy left by the time it
21 Path-independence allows us to use the actual distances RA and RB as the limits ofintegration in W (recall gravity is a conservative force). If work was path-dependent,we would need the more general specification in equation (2.3).22 By conservation of mechanical energy, TA + VA = TB + VB . Thus, VA − VB =TB − TA = WAB , the second equality by the Work-Energy Theorem.
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arrives in point B. In this case, work is given by
W∞B = −GmMRB
and the gravitational potential energy is, therefore:
VB − V∞ = VB = GmM
RB
(where we have set V∞ = 0 as A is infinitely far).23 Another way of
deriving this same result is by just using the Principle of Potential Energy
(Principle 2.1). Recalling that F = −∂V∂R (recall that R is the distance
between the objects), and using F = GmMR2 by Newton’s Law of Gravitation,
we have ∂V∂R = −GmM
R2 , i.e. V = −GmM∫
1R2 dR = GmM
R .
Generally, suppose A and B are separated by a distance h, i.e. RB =
RA + h (where h may be positive or negative). Then:
VB − VA = −WAB = GmMh
RB(h−RB)
This means that:
• If h < 0 (i.e. A is further from Earth than B, so the body is mov-
ing toward the Earth), then WAB < 0 and VB − VA > 0. In words,
as the body approaches the Earth, potential energy increases and (by
mechanical energy conservation), kinetic energy decreases.
• If the body moves away from the Earth (h > 0), the opposite is true:
the work of the gravitational force is positive, so potential energy is
turned into kinetic energy as the body drifts away.
Finally, let us argue that, for objects on Earth, the gravitational potential
energy is well approximated by mgh (as found in Example 2.5), where h is
the distance between the body and the surface of the Earth. To show this,
23 As a result of setting V∞ = 0, all potential energies are negative. The minus signhere is just a result of setting ∞ as our reference point. As usual, we are free tochoose this point, so the fact that potential energies are negative is meaningless perse. We only care about potential energies relative to the reference point.
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consider point A to be on the Earth’s surface, so that RA = REarth (the
radius of the Earth), and let RB = REarth + h, where h is tiny compared to
REarth (i.e. the body is just above the surface).
Then, the potential energy between points A and B is:
VA − VB = WAB = GmMh
(REarth + h)REarth≈ mgh
(1
1 + h/REarth
)where the last equality uses g ≈ MG
R2Earth
by equation (2.15). Then, using
a simple Taylor expansion (Result 0.5), we see that h(
11+h/REarth
)≈ h
around hREarth
= 0. Therefore:
WAB = VA − VB ≈ mgh
what we wanted to show. �
Having the Newtonian Laws of Gravitation also allows us to understand
escape velocities.
Definition 2.9 (Escape velocity) The lowest velocity that a body must
have in order to escape the gravitational attraction of another body.
To derive the escape velocity formula, consider a (relatively small) object
of mass m standing on Earth (of mass M), so that the distance between
them is R, the Earth’s radius. For simplicity, suppose the Earth has no
atmosphere. If at time t = 0 the object is given a velocity vesc ≡ v(0), then
the total energy at that moment is:
E = T + V =1
2mv2
esc −GmM
R
Because gravity is a conservative force, E = 0. Where r is the distance
at a certain time tr > 0 from the Earth’s center, E = 0 implies
1
2mv2
esc −GmM
R=
1
2mv(tr)
2 −GmMr
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Suppose the initial velocity vesc is just enough for the body to escape the
Earth’s gravitational pull. Since this body escapes, and because we assume
no other objects in space, it will eventually reach infinity (r → +∞). At
infinity, potential energy is zero. Further, because vesc is the minimum speed
at which this can happen, the body must abandon the gravitational pull of
the Earth at nearly zero “residual” velocity, so v(t∞) = 0. Imposing these
above, we get 12mv
2esc −GmM
R = 0. Solving for vesc:
vesc =
√G
2M
R
Thus, the escape velocity (e.g. from a planet) is higher for heavier and/or
larger planets. For Earth, M = 6 × 1024 kg and R = 6.4 × 103 km, so
vesc = 11.2 km/sec.
Example 2.20 (Gravity and circular orbits) Consider a satellite of mass
m orbiting in circular motion around a body of mass M � m. Let R > 0
be the radius of the circular orbit. As discussed before in these notes, object
m has some tangential velocity vorb, which we now call the orbital velocity,
and a centripetal acceleration that is perpendicular to it.
If the m object is in orbit, then the centripetal acceleration, with magni-
tudemv2
orbR , must all be due to the gravitational attraction between the two
objects. Thus,mv2
orbR = GmM
R2 , or:
vorb =
√GM
R
Note vesc =√
2vorb, so a body in orbit must increase its velocity by a
factor of√
2 in order to escape the orbit. The period of the orbit (the time
it takes to complete one full revolution) is T = 2π Rvorb
= 2πR3/2√GM
.24 Note
24 For instance, for a satellite about 400 km above the Earth’s surface, R ≈ 6, 800km, so one full revolution will be completed in about T ≈ 90 minutes, at a speed ofvorb ≈ 8 km/sec. For the Moon, the period is about T ≈ 27.5 days, and the orbitalspeed is vorb ≈ 1 km/sec (slower than satellites, as R is larger). For the Earth’smotion around the Sun (mass M = 21030 kg), at distance R ≈ 150× 106 km, we finda period of T = 365.5 days, i.e. one year. The orbital speed in that case is vorb ≈ 30km/sec.
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the period is independent of m: a feather and a space shuttle would both
complete a revolution around the Earth at the same time.
Interestingly, the total energy at the orbital velocity is:
E = T + V =1
2mv2
orb −GmM
r=
1
2GmM
r−GmM
r= −1
2GmM
r
That is, E = 12V = −T . This is a remarkable property of objects in
orbit (when the orbits are circular): the total energy of objects in orbit is
negative. For the Earth and the Sun, E = −2.7× 1033 Joules, an extremely
large negative number.
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Chapter 3
Rotation
So far, we have explored the laws of linear motion of bodies. In this
chapter, we will explore the dynamics of motion in rotational frameworks.
3.1 Moment of Inertia
In many examples above, we have explored motion along orbits when
velocity is constant (i.e. when the object is not being accelerated). In this
section, we generalize rotational motion to potentially non-uniform motion,
show how to work in polar coordinates (Definition 1.11), and introduce a
new concept: the moment of inertia.
Let R be the radius of the orbit, θ the angle at a given time, ω ≡ dθdt the
angular velocity (Definition 1.5), which is now possibly non-constant, and ~v
be the velocity of the particle, also possibly non-constant. In Example 1.2,
we derived that |~v| = ωR = θR. The centripetal acceleration is |~acent| =
Rω2.
Because the particle is now being accelerated (for example, it rotates
along a disk that is itself rotating due to a torque1), there is an additional
acceleration, called the tangential acceleration, in the direction of the cir-
cumference. The tangential acceleration is given by:
1 A torque is a rotational force, i.e. the rotational equivalent of a linear force. Amore general definition will be given in Definition 3.5.
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|~atang| =d
dt|~v| = ωR = θR = αR
where α ≡ θ is the so-called angular acceleration. We have then intro-
duced two more concepts:
Definition 3.1 (Tangential acceleration) The acceleration along the ro-
tational perimeter of a particle in rotational motion.
Definition 3.2 (Angular acceleration) The acceleration of the angle de-
scribed by a particle in rotational motion.
Of course, in the special case of uniform motion, ω is constant and the
tangential and angular acceleration are zero.
Now, we can apply the usual equations of motion in polar coordinates.
For instance, if we consider that the acceleration is constant, we can use the
equations from Example 1.4 and translate them into polar coordinates, so
that:
θ = θ0 + ω0t+1
2αt2 and ω = ω0 + αt
Consider now an object i of mass mi that is standing on the rotating
disk. Let ri be the distance of the particle relative to the disk’s center, point
C, which acts here as the axis of rotation. Importantly, we assume that this
axis is perpendicular to the disk. Then, the kinetic energy for object i is:
Ti =1
2mi|~vi|2 =
1
2miω
2r2i
The kinetic energy of the entire disk is, therefore, T = 12ω
2∑N
i=1mir2i ,
for all the N elements that compose the disk. More generally, when there
is a distribution of particles composing the body, we may instead integrate
over the entire mass, and say
T =1
2ω2
∫r2dm︸ ︷︷ ︸≡I
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where r ≡ r(m) is the distance to the (perpendicular) axis of rotation
of particle of mass m. Here, T is rotational kinetic energy, and the object
I ≡∫r2dm is called the moment of inertia of the system.
Note the similarity with linear motion. In linear motion, T = 12m|~v|
2.
In rotational motion, T = 12Iω
2. In both cases, to obtain the kinetic energy
(that is, a given force) we multiply the squared velocity (linear or angular)
by m, if motion is linear, or I, if motion is rotational. Therefore, just as mass
determines the force needed for a desired linear acceleration, the moment of
inertia determines the torque needed for a desired angular acceleration.
Hence, our definition:
Definition 3.3 (Moment of inertia) In rotational motion, the moment
of inertia is the torque needed to achieve a certain angular acceleration, given
by:
I ≡∫r2dm (3.1)
The moment of inertia depends on the properties of the system of consid-
eration. For some objects, especially symmetric and solid ones, the integral
in equation (3.1) can be solved analytically. For example:
• For a disk rotating through the center about an axis that is perpendic-
ular to it, the moment of inertia is I = 12mR
2, where R is the radius
about the center and m is the mass of the disk.
• For a solid sphere rotating about an axis through its center, the mo-
ment of inertia is I = 25mR
2.
• For a rod of length ` and mass m, the moment of inertia is I = 112m`
2.
The moment of inertia depends on the axis about which the rotation
is taken. Above we have considered that the axis goes through the center
of mass. However, this may not be. What is indispensable is for this axis
to be perpendicular to the direction of rotation. So long as this is true, a
convenient theorem allows us to calculate the moment of inertia about any
other axis:
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Result 3.1 (Parallel axis theorem) Let ICM be the moment of inertia
of a body of mass m that is rotating about an axis going through the center
of mass. Then, if the body is made to rotate about a different axis which is
parallel to the first one and separated by distance d from it, the moment of
inertia is given by:
I = ICM +md2
Proof. The proof is easiest in Cartesian coordinates, (x, y). Suppose,
without loss of generality, that the perpendicular distance d between the
two axes lies strictly along the x-axis, and that the center of mass is at the
origin. The moment of inertia relative to the axis going through the origin
is ICM =∫
(x2 + y2)dm. The moment of inertia relative to the alternative
axis is:
I =
∫ [(x+d)2+y2
]dm =
∫(x2+y2)dm+d2
∫dm+2d
∫xdm = ICM+md2
as∫xdm = 0 because the center of mass lies at the origin. �
Notice that, by this theorem, the moment of inertia is always lowest
when the axis goes through the center of mass.
A second result is useful for rigid objects that lie entirely within a plane
(i.e. think objects such as a piece of paper):
Result 3.2 (Perpendicular axis theorem) Define perpendicular axes x,
y, and z, all of which meet at the origin O. Suppose that a body lies entirely
on the xy plane, and the z axis is perpendicular to the plane of the body. Let
Ix, Iy, and Iz be the moments of inertia about axis x, y, and z, respectively.
Then:
Iz = Ix + Iz
Proof. In Cartesian coordinates, the moment of inertia of the body about
the z axis is:
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Iz =
∫(x2 + y2)dm =
∫x2dm+
∫y2dm
Since on the plane we have that z = 0, then the right-hand side equals
Ix + Iy. �
3.2 Angular Momentum and Torques
Angular momentum is the rotational equivalent of momentum in linear
motion. As we shall see, it too is conserved in isolated systems that are
subject to no external forces.
Let us first introduce it formally. Consider an object of mass m, velocity
~v, and momentum ~p = m~v. Fix an arbitrary point O in space, and let ~r be
the position of the object relative to O. Then:
Definition 3.4 (Angular momentum) The angular momentum relative
to point O is defined as the vector:
~L ≡ ~r × ~p = (~r × ~v)m
Therefore, the angular momentum is the vector resulting from the cross
product of the position vector and the momentum vector. Using the prop-
erties of the cross product (see equation (0.10)), the magnitude L ≡ |~L| of
the angular momentum is:
L = vmr sin θ︸ ︷︷ ︸≡r⊥
where θ = ∠~r~v is the angle between the position and the velocity vectors,
r ≡ |~r| is the distance between O and the body, and v ≡ |~v| is the speed.
Here, the number r⊥ ≡ r sin θ corresponds to the perpendicular distance
between the position vector and the point of reference, O (see Figure 3.1).
The direction of the angular momentum vector is perpendicular to the (x, y)
plane.
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Figure 3.1: The angular momentum.
Note that angular momentum is not an intrinsic property of a moving
object. For instance, if we chose O to be exactly so that ~r ⊥ ~v, then ~L = ~0.
But if O is chosen so that ~r and ~v are not perpendicular (as in Figure 3.1),
then ~L 6= ~0. Henceforth, we will call this point the reference point.
Thus, while the momentum of an object is the same regardless of the
reference point, the angular momentum is different for different reference
points.
Example 3.1 (Projectiles revisited) Consider again the parabolic mo-
tion of Example 1.8. At time t = 0, velocity is ~v0 and the position vector
is ~r0 = ~0 if we choose the origin (x, y) = (0, 0) as our point of reference.
Therefore, the object has no angular momentum: ~L = ~0. At some time
t > 0, however, when ~r 6= ~0, the angular momentum is clearly not the zero
vector. In this example, this is because the velocity vector is changing along
the trajectory of the body.
Example 3.2 (Circular motion) Consider the Earth (mass m) orbiting
around the Sun in a perfect circular orbit. Suppose our point of reference
is the Sun (point O), and let ~r be the Earth’s position vector relative to the
Sun at some point in time. The Earth has a certain tangential velocity ~v.
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Then, the angular momentum of the Earth is ~L = m~r × ~v. However, for
circular motion, ~r ⊥ ~v (recall results from Example 1.2), so sin θ = 1 for
θ ≡ ∠~r~v, and so the magnitude of the angular momentum is simply:
L = mrv
where r ≡ |~r| and v ≡ |v|. Therefore, although the velocity vector of the
Earth is changing direction all the time, its angular momentum relative to
the Sun remains constant in magnitude (because the speed is constant). In
short, angular momentum is conserved.
Again, this is true only relative to the Sun. Clearly, if we took a point O
that is right on the Earth’s path as our point of reference, then the magnitude
of the angular momentum would depend on the Earth’s position along this
path. It would be zero as the Earth goes through O, and non-zero otherwise.
To see this point more generally, consider taking the time derivative of
the angular momentum:
d~L
dt=
d~r
dt× ~p+ ~r × d~p
dt= ~v × ~p+ ~r × ~F = ~r × ~F (3.2)
where we have used ~v×~p = ~0, as ~v and ~p are always parallel, and d~pdt = ~F
by Newton’s Second Law. The right-hand side of this equation is what we
call the torque:
Definition 3.5 (Torque) The torque vector is defined by:
~τ ≡ ~r × ~F
that is, the cross product of the position vector of an object (relative to
some point O), and the force applied on the object.
Thus, in equation (3.2) we have found:
Result 3.3 (Torque and Angular Momentum) The torque equals the
rate of change in angular momentum, ~τ = d~Ldt .
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Thus, if there is a torque on an object, the angular momentum must
change over time. If there is no torque, angular momentum must be con-
served. This is reminiscent of linear momentum, which is conserved (by
Netwon’s Third Law) in the absence of external forces. Since torque is
nothing but an (angular) force, a direct corollary of this result is that an-
gular momentum is conserved unless a net torque is applied on the system.
Hence:
Result 3.4 (Conservation of angular momentum) Angular momentum
is conserved, d~Ldt = ~0, unless acted upon by a net external torque.
Now, it is clear why the Earth’s angular momentum with respect to the
Sun is conserved (Example 3.2): the force of gravity between the Earth and
the Sun is exactly parallel (at 180◦) to the position vector, so ~τ = ~Fg×~r = ~0.
Since there are no other forces acting upon the system, d~Ldt = ~0. However,
relative to any other point in the system, there will be a torque, and thus
angular momentum is not conserved.
Example 3.3 (Disk) Consider a disk of mass M and radius R, with center
of mass at point C. It rotates about point C with angular velocity ω. What
is the angular momentum of the disk as a whole?
Consider a particle i of the disk, with mass mi, position ri relative to C,
and speed vi. As in Example 3.2, since the velocity and the position vectors
are here perpendicular, we simply have that Li ≡ |~Li| = mirivi or, since
vi = ωri for all particles, then Li = mir2i ω. Thus, the magnitude of the
angular momentum of the entire disk is (that is, integrating across the mass
distribution of the disk) is:
L = ωI
where I ≡∫r2dm is the disk’s moment of inertia.
Therefore, we have derived the following result:
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Result 3.5 (Moment of inertia and angular momentum) For rotational
motion, the moment of inertia equals the ratio of the angular momentum to
the angular velocity: I = Lω .
Since a torque implies a change in the angular momentum (Result 3.3),
then this result makes clear how an external torque, by causing an angular
acceleration, implies a change in the moment of inertia.
Remarkably, if rotation is about the system’s center of mass (which we
happened to choose in Example 3.3 as our reference point), then the angular
momentum always has magnitude L = ωI regardless of the reference point.
That is, if an object is spinning about its center of mass, then the angular
momentum is uniquely determined, and its magnitude is given by L = ωI.
We call this the spin angular momentum of the system, and it is therefore
an intrinsic property of the system (i.e. independent of the reference point).
Definition 3.6 (Spin angular momentum) The intrinsic angular mo-
mentum of a spinning object that is rotating about a stationary axis going
through the object’s center of mass.
For example, the Earth spins about its center of mass, so it has an
intrinsic spin angular momentum. It also has an orbital angular momentum,
but the magnitude of the latter depends on which point of reference is chosen,
while the former does not.
We can now sum up our results for torque and angular momentum in
rotational motion:
Result 3.6 (Summary: Rotational motion) Consider the rotation of an
object with angular velocity ω ≡ θ and angular acceleration α ≡ θ, about
some axis that goes through point Q. Let C denote the object’s center of
mass. Then:
• The angular momentum vector is defined by ~LQ = ~rQ× ~p, where ~rQ is
the position relative to Q. The magnitude of the angular momentum
about Q, LQ ≡ |LQ|, is:
LQ = IQω (3.3)
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• The torque vector is defined by ~τ = ~rQ× ~F , where ~F is the force vector
relative to point Q. The magnitude of the torque about Q, τQ ≡ |~τQ|,is:
τQ = IQα (3.4)
• An external torque changes the angular momentum of the system, for
~τQ = ddt~LQ. If there is no net torque, therefore, the angular momentum
is conserved.
• If the rotation is instead about a stationary axis going through the
center of mass, the angular momentum LC = ICω is called the spin
angular momentum, and is fixed relative to the reference point.
Let’s apply these principles to a few examples:
Example 3.4 (Revolving chair experiment) Consider a man sitting on
a round revolving chair that rotates about its center of mass. The man holds
one weight on each hand. As the chair turns, the man will pull his arms out
and back in. When his arms are pulled out, the chair’s spin slower, and as
when the arms are pulled in, the chair spins faster. Why does this happen?
When the arms are pulling in, the moment of inertia is I = 12MR2,
where M is the mass of the man and R is its radius (here, for simplicity,
suppose we approximate the man’s shape by that of a cylinder). When the
arms are pulled out, the moment of inertia goes up both because the radius
is higher and, especially, because mass increases.
Clearly, since I increases but L = Iω must be conserved (as clearly no
external torque is at play here), then ω must decrease by the same proportion.
In words, because angular momentum cannot change, an increase in the
system’s moment of inertia when the arms are pulled out must translate
into a decrease in the angular velocity of the same proportion.2
2 Similarly, spinning figure skaters pull their arms in so as to reduce their momentof inertia and, in that way, increase their angular speed.
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The last example is true for all spinning objects. For example, when a
star shrinks, its radius goes down, its moment of inertia decreases, and so
its angular velocity must go up. In particular, if a star shrinks by 10-fold
in a certain time, the moment of inertia will decrease by 100-fold,3 and the
angular velocity will increase by 100-fold as well.
Example 3.5 (A spinning rod) Consider a rod of mass m and length `.
Suppose that it rotates about some point P , which is at distance d from the
rod’s center of mass, C. Let ω be the angular velocity of rotation.
The magnitude of the angular momentum about point P , denoted LP , is:
LP = ωIP = ω(IC +md2)
where the second equation follows from the Parallel Axis theorem (Result
3.1), and IC = 112m`
2 for rods. The torque relative to point P is ~τP =
~rP × ~FP , where ~rP and ~FP are the position and force vectors of P relative to
C. Clearly, ~r and ~F are parallel (a centripetal force is pushing the rod on P
just in the same direction as the position vector, as the rod is shaped like a
straight line), so ~τ = ~0. Since no external torque exists, angular momentum
relative to point P is conserved.
Clearly, however, angular momentum is not conserved anywhere other
than P (the point about which the rod rotates). Indeed, for any point Q, ~rQ
and ~FQ are not parallel, so a torque exists relative to Q.
If the rotation is relative to C, the center of mass, then clearly there is no
force at all, ~F = ~0. Thus, ~τ = ~0 relative to any point (inside or outside of the
rod). Because rotating about the center of mass has the special property that
no torque exists relative to any point of origin, the angular momentum about
the center of mass is the spin angular momentum, an intrinsic property of
the rod. In particular, it is LC = ICω = 112m`
2ω.
Example 3.6 (Translation and Rotation) Here is a classic problem that
illustrates well the concepts above.
3 Recall I = 25mR2 for spheres such as stars.
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Consider a rod (e.g. a ruler) of mass m and length ` lying at rest on a
frictionless surface (e.g. a frictionless table). We give the rod an impulse
(i.e. a force for a short period of time; see Definition 2.7) at some point P
on the rod. Point P is a distance d away from the rod’s center of mass, C.
The impulse vector ~I is perpendicular to the length of the rod.
Because the center of mass is a point that behaves like a single point where
all the mass is concentrated, it must experience a certain velocity ~vC which
will never change direction or magnitude. We call this the translational
velocity. The rod will also rotate about C at some angular speed, ωC .
Now, we want to derive ~vC (the translational velocity) and ωC (the an-
gular speed). The impulse is ~I =∫~Fdt =
∫ d~pdt = ∆~p, where ~p is the
momentum of the system. By the properties of the center of mass (Result
1.2), we have m~aC∆t = ∆~p = ~I. Since the initial velocity is zero, then
~aC∆t = ~vC . Thus:
~vC =~I
m
is the translational velocity of the center of mass. Remarkably, it is
independent of d: no matter where on the rod we give the impulse, the
velocity of the center of mass will always be the same (everything else equal).
For the angular velocity about the center of mass, C, we have to choose
the origin. We are of course free to choose its location, so let us do two
cases:
• Take C itself as our reference point. The torque relative to C is clearly
not zero, as ~rC and ~F are not parallel (in fact, ~rC ⊥ ~F ). Since there is
a torque, the angular momentum relative to point C must be changing.
In particular, the torque about C is ~τC = ~rC × ~F . Assuming the
hit occurs over a short enough time so that the position vector barely
changes, we have:
∫~τCdt = ~rC ×
∫~Fdt = ~rC × ~I
Since the object is initially at rest, then the angular momentum before
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the hit is zero, and after the hit it equals the accumulated torque, so
~LC = ~rC × ~I. Thus, the angular velocity about C, ωC , satisfies LC =
ICωC , where IC = 112m`
2 for rods, and LC = |~rC × ~I| = |~rC ||~I| = dI
is the magnitude of the angular momentum,4 where I ≡ |~I| is the
magnitude of the impulse.5 Thus, we get 112m`
2ωC = dI, so ωC = 12dIm`2
.
• Take P as our reference point. Now, the position vector is in the same
direction as the impulse, since the origin is trivially the same as the
point where the force is exerted. Thus, there is no torque, and angu-
lar momentum is conserved, about point P . The angular momentum
remains ~LP = ~0 before and after the hit. Note here we cannot invoke
the Parallel Axis theorem, for the point of rotation is still at the cen-
ter of mass.6 Thus, the moment of inertia is still IC = 112m`
2, and
ωC = LC/IC = 12LCm`2
. To compute LC using P as the reference point,
we note once again that ~rC ⊥ ~I, except now ~rC is going in the oppo-
site direction as before (from P to C instead of C to P ). The same
derivation as before then follows.
Example 3.7 (A Ruler) Take a ruler of mass m with center of mass C
and make it rotate around a perpendicular pin that is placed at some point
P , at distance b from C. We take P as our reference point.
The force vector is now due to gravity at point C, whose position vector
relative to P we denote by ~rP . Therefore, the torque at C relative to P
is ~τP = ~Fg × ~rP , with magnitude τP = mgb sin θ, where θ is the angle of
the ruler about the vertical axis. Using equation (3.4), it must also be that
τP = −IPα, where IP is the moment of inertia about a perpendicular axis
going through P , and α ≡ θ is the angular acceleration. The minus sign
is there because, just like in a spring (where ~F = −kx), the torque is a
restoring force.
4 Here, we use that ~rC ⊥ ~I to say that sin θ = 1, where θ ≡ ∠~rC~I.5 The notation of this problem should not lead to confusion: ~I denotes the impulse,
and ~IC denotes the moment of inertia about point C. Similarly, I and IC are theirrespective magnitudes.6 The theorem could be invoked if P , our new reference point, was also the new point
of rotation, which it is not.
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Thus, τP = mgb sin θ = −IP θ. Using a Small Angle approximation and
Taylor-expanding sin θ about θ = 0 we get sin θ ≈ θ. Putting things together,
we have found mgbθ + IP θ = 0, or:
θ +mgb
IPθ = 0
Clearly, this is a simple harmonic oscillation in θ. Thus, the solution is:
θ = θmax(ωt+ ϕ)
where the angular frequency ω is a constant,7 and (as we have derived
before) it is given by ω =√
mgbIP
. The period is T = 2π√
IPmgb . Moreover, we
know IP = IC +mb2 = 112m`
2 +mb2 by the Parallel Axis theorem. Thus:
ω =
√gb
112`
2 + b2and T = 2π
√112`
2 + b2
gb
The kinetic energy of rotation of the ruler is:
Trot =1
2IPω
2
where ω = θ and IP = 112m`
2 + mb2. Using θmax and ϕ as computed
in Example 2.9, the rotational kinetic energy will readily follow. Note the
kinetic energy changes with time: it is zero when the ruler comes to a halt
(at θ = θmax), and it is maximum as the ruler is exactly at θ = 0 (i.e. in a
vertical position).
Example 3.8 (A Hula-Hoop) Hang a hula-hoop vertically from a pin at
point P . Let C be the center of mass at a certain time. Note that the center
of mass will change as the hoop moves (in Figure 3.2, the center of mass
changes from C when the hoop is at rest, to C ′ at a later time). The mass
of the hoop is m, it has a radius R, and we let θ be the angle between ~rP
and the vertical axis.
7 Typically, for the solution of a simple harmonic oscillator we will denote the angularfrequency by ω. Here, we place a tilde to avoid confusion with the angular velocityof the problem, ω, which is not constant.
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P
CC ′
mg
~rP
Figure 3.2: The hula-hoop.
Fix P , the axis of rotation, as our reference point. The magnitude
of the torque relative to point P is τP = mgR sin θ = −IP θ, as before.
Now, the moment of inertia about point P is, by the Parallel Axis theorem,
IP = IC + mR2, where the moment of inertia about the center of mass is
IC = mR2, since all the mass is distributed along the hoop’s circumfer-
ence, which is at distance R. Thus, IP = 2mR2. In sum, using the Small
Angle approximation as before (i.e. sin θ ≈ θ for θ ≈ 0), we have found
θ + mgRIP
θ = 0, that is:
θ +g
2Rθ = 0
This is, again, a simple harmonic oscillator in θ, whose solution is θ =
θmax(ωt+ ϕ), where ω =√
g2R . The period of oscillation is T = 2π
√2Rg .
Remarkably, these are the same results (recall Example 1.10) as a pen-
dulum with length ` = 2R. Note, again, that the mass m is irrelevant for
the period of the hoop, as well as that of the pendulum.
Next, we examine rolling. Rolling is a type of rotational motion that
involves the translation of a moving object along a surface (e.g. the wheels
of cars). In particular, we will consider situations where there is pure rolling,
which we define as rolling motion that does not involve any sliding of the
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object. That is, for a round object with radius R, pure roll is the situation
where, after one full rotation, the object has travelled a distance equal to
the perimeter of its circumference, 2πR. As a consequence, in pure rolling,
the velocity of the center of the object equals that of the circumference.
Therefore:
Definition 3.7 (Pure Rolling) Rolling motion that does not involve slid-
ing, so that the velocity of the center of the circle (point Q) is vQ = ωR,
where ω is the angular velocity.8
If there is no friction, the object would rotate on its own axis, there
would be no translation, and vQ = 0. Thus, it is friction which permits pure
rolling. Let’s look at a famous example: rolling cylinders on an incline.
Example 3.9 (Rolling cylinders) Place a round object (e.g. a cylinder
or a sphere) on an incline (see Figure 3.3). The object will (pure-)roll
downhill. The cylinder has mass m, length `, radius R, and center of mass
Q.
Q
mg cosβ
~FN
mg sinβ
mg
β
~Ff
Figure 3.3: A rolling cylinder.
The forces involved in this problem are similar to those of Example 1.13.
First, we decompose the force of gravity mg into its x and y components,
mg sinβ and mg cosβ, respectively, where β is the angle of the incline. A
8 Recall that ωR is the velocity of the circumference. If there is pure roll, it coincideswith that of the center point.
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normal force from the incline’s surface onto the object has magnitude FN =
mg cosβ (as there is no acceleration in the y direction). Finally, there is a
frictional force Ff .
At any moment in time, there is a (time-varying) angular velocity of ω.
Point Q (the center of mass), travels at velocity vQ = ωR (because of pure
roll). Thus, the acceleration is a = ωR = αR, where α denotes the angular
acceleration.
When calculating the torque about point Q, the only contributing force
is ~Ff (for ~FN and mg both go through Q, so the position vector for them is
zero), so the magnitude of the torque is τQ = RFf . Using equation (3.4),
we know τQ = IQα, where IQ is the moment of inertia for axis through the
center of mass. Using a = αR, we have found τQ = RFf = IQaR , or:
a =R2FfIQ
On the other hand, if Q is the center of mass, we have ma = mg sinβ−Ffby Newton’s Second Law, or:
Ff = m(g sinβ − a)
Substituting this into our first equation, we find a = R2
IQm(g sinβ−a) or,
solving for a, we have a = mR2g sinβmR2+IQ
, and Ff =IQR2a. Then:
• If the cylinder is solid (the mass is evenly distributed on the area), the
moment of inertia about the center of mass is IQ = 12mR
2, and so we
obtain an acceleration of:9
asolid =2
3g sinβ
• If the cylinder is hollow (all the mass is at the circumference), the
moment of inertia is IQ = mR2, so:
9 If we considered a sphere instead of a cylinder, then IQ = 25mR2, in which case
a = 57g sinβ.
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ahollow =1
2g sinβ
These are surprising results. First, note that the acceleration is always
lower than in the case with no friction (where a = g sinβ). Second, the
acceleration of the cylinder when rolling downhill is completely unaffected
by the cylinder’s mass, length, or radius. Downhill races between rolling
cylinders of different masses, lengths, or radii, will always end in a tie if the
starting positions (and velocities) are the same. However, solid cylinders
have higher acceleration than hollow cylinders, so the former will always hit
the bottom of the incline before the latter does (again, independent of mass,
length, or radii).
Finally, to check that there is indeed pure roll (i.e. no slipping), we
need Ff < µsFN , where µs is the coefficient of static friction (recall e.g.
Example 1.13). Here, FN = Mg cosβ and Ff =IQR2a. For a solid cylinder,
for example, the condition for pure roll becomes µs >23 tanβ. For a hollow
cylinder, the condition for pure roll is µs >12 tanβ.
Example 3.10 (Atwood machine) Figure 3.4 shows a so-called Atwood
machine: a disk of mass M and radius R with center of mass P serves as a
pulley for a massless rope which goes around it and sustains, on each end,
two objects of masses m1 and m2, respectively.
Each hanging body is dragged down by gravity. There is tension in the
strings, both for supporting the bodies and downward from the pulley. Fi-
nally, the pulley itself suffers the gravitational force on its center of mass P ,
and a normal force supports the structure in balance, so FN = T1 +T2 +Mg.
Suppose the system is being accelerated so that the pulley rotates clock-
wise. Moreover, assume that the rope does not slip around the circumference
of the disk, i.e. that friction is strong enough for the rotational of the disk to
translate one for one into the rope’s motion. Since there is no slip, the veloc-
ity of the rope is v = ωR, where ω is the angular velocity. The acceleration
of the rope is thus a = αR, where α = ω is the angular acceleration.
For object 1 and 2, Newton’s Second Law reads:
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P
Mg
m1g
m2g
T1
T2
T1
T2
FN
R
a a
ωR
~r1~r2
Figure 3.4: The Atwood machine.
T1 −m1g = m1a m2g − T2 = m2a
respectively. On the pulley there are no linear forces, but there is a
torque. The torque is only due to forces T1 and T2, since FN and Mg go
through point P and thus do not contribute to the torque (their position
vectors being the zero vector). Relative to point P , the position vectors for
forces ~T1 and ~T2 are ~r1 and ~r2, respectively. Notice ~r1 ⊥ ~T1 and ~r2 ⊥ ~T2, so
in both cases sin θ = 1. Therefore, the magnitude of the total torque relative
to point P is simply:10 τP = RT2−RT1. Recalling τP = IPα, where IP is the
moment of inertia. Since this is a rotating disk with rotation axis through
the center of mass, IP = 12MR2, so we have, putting things together:
10 Note the two torques have opposite sign because the direction of the torque vectoris into the plane for the left-hand torque, but out of the plane for the right-handtorque.
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T2 − T1 =1
2Ma
Thus, we have three equations for three unknowns, (a, T1, T2). Solving,
we get a =(
m2−m112M+m1+m2
)g, T1 = m1(g + a), and T2 = m2(g − a).
The solution reveals that for rotation to be clockwise, it is necessary that
m2 > m1. Only then do we have that a > 0, so that m2g > T2 (body 2’s
weight is stronger than the tension holding it up), T1 > m1g (body 1’s weight
is insufficient to counteract the force pulling it up), and T2 > T1 (a stronger
tension on the right side of the rope makes it rotate clockwise).
3.3 Gyroscopic Motion
Up until now, we have studied rotational motion for objects which rotate
about a fixed axis. The relative position of this axis to the point of reference
has been crucial for determining the torques and the angular momentum
involved in the system. Now, we study the rotation of objects about an axis
which may itself not be at rest. This is often called gyroscopic motion.
A gyroscope is a device that consists of a wheel or disk that can rotate
about an axis which is free to change in orientation. When a gyroscope spins
and its axis of rotation is altered due to a one-time external torque, the
conservation of angular momentum (which must hold because no additional
external torque is applied on the system) forces a change in the orientation
of the system’s axis. We call this a precession.
Definition 3.8 (Precession) The change in the orientation of the rota-
tional axis of a rotating body.
Let’s see this by means of example:
Example 3.11 (Spinning Wheel I) Consider spinning a bicycle wheel in
outer space and, at the same time, applying a torque that sets the axis of
rotation in motion. Importantly, the wheel cannot continue spinning about
its axis and, at the same time, have said axis rotate in the same direction
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as that of the initial torque vector (thereby describing a circular motion
on the same plane as the torque vector). The reason is that, if that were
the case, the direction of the angular momentum vector would be changing
and, as a result, angular momentum would not be conserved.11 But angular
momentum must be conserved, because there are no external torques on the
system once the initial torque has been given! So how does nature resolve
this?
To understand what will happen, consider the upper panel of Figure 3.5.
Let x be the axis of rotation, and (x, y) be the plane on which the observer
is located. Suppose that two torques of equal magnitude F and opposite
direction are simultaneously applied for some duration ∆t at points A and
B, which lie strictly on the (x, y) plane. Let b denote the separation between
these two points, i.e. |~r| = b.
The torque relative to the center of mass is:12
~τ ≡ ~r × ~F = bF~e+
by definition of the cross product (equation (0.10)), where ~e+ is a unit
vector perpendicular to the (x, y) plane and pointing in the z+ direction (i.e.
pointing toward more positive values of z). Therefore, the torque vector is
pointing upward.13 Therefore, magnitude of the torque is τ = bF . The
angular momentum is ~L, perpendicular to the wheel’s spin.
When the new torque is applied for a period of length ∆t, the angular
momentum changes by ∆~L = ~τ∆t (by Result 3.3). Crucially, the direction
of ∆~L is the same as that of ~τ (that is, upward toward z = +∞). In words,
the spin angular momentum will change in the direction of the torque. After
the torque ceases to exist, the angular momentum of the system can no
longer change. Thus, in order for the angular momentum to be conserved,
11 Here, whenever we talk about angular momentum, we really mean spin angularmomentum (Definition 3.6), because we are computing the angular momentum takingthe axis of rotation that goes through the system’s center of mass.12 Here, we have used that ~r ⊥ ~F (so sin θ = 1).13 As usual, we are free to choose the direction of our axis so long as we are consistentthroughout. The convention we will use is that the z+ direction is out of the page,and the z− direction is into the page.
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Figure 3.5: A spinning bicycle wheel in outer space.
the wheel must tilt. After the torque has occurred, the spinning wheel must
keep its tilted position (Figure 3.5, lower panel). In short, the spinning wheel
precesses.
Here, the torque is pointing upward because of the direction of the forces
applied at points A and B. If we flip the directions of these forces, the
torque will point downward, the change in the angular momentum will also
point downward, and that will tilt the wheel in the other direction (to the
right). Similarly, if the forces at A and B lived on the (x, z) plane (i.e. they
were pointing up and down), then the torque would point perpendicular to
(x, z), the change in the angular momentum vector would point along the
(x, y) plane, and this would tilt the wheel right and left along the horizontal
dimension.14
The above example ignored the effects of gravity, because the wheel was
spinning in outer space. The following example examines how the wheel’s
14 A famous experiment has the experimenter holding a spinning wheel, sitting on arevolving chair, and experiencing a spin in the chair from left to right and vice versaas the axis of rotation of the wheel is moved vertically.
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precession will interact with gravity.
Example 3.12 (Spinning Wheel II) Consider now mounting the wheel
of mass m and radius R mounted on a structure similar to what is shown in
Figure 3.6. At point O (our point of reference), the wheel is attached to a rod
of length a, which serves as the axis of rotation. The position of the wheel
relative to point O is ~r. Clearly, the spin angular momentum vector is now
pointing in the y+ direction (in the figure, it is denoted by ~ω). On the other
hand, gravity acts upon the wheel in the z− direction, and has magnitude
mg. In the figure, the gravitational force vector is marked by ~W .
Relative to point O, the torque vector is:
~τ = ~r × ~W = amg~e
(note sin θ = 1 because ~r ⊥ ~W ), where ~e is pointing in the direction
of x+. In the figure, the torque vector is denoted ~Γgrav. Therefore, since
∆~L = ~τ∆t, the spin angular momentum vector (~ω, in the figure) will move
in the direction of the torque, and the system will precess. In particular, the
system will rotate about the axis of precession counterclockwise, as marked
in the figure.
Finally, the angular frequency of the precession (i.e. the velocity at
which the system rotates around its axis of precession) is given by:15
ωprecession =τ
LS
where τ and LS are, respectively, the magnitudes of the torque and the
spin angular momentum of the system. In our case, τ = amg (see above)
and LS = IOωspin (see equation 3.3), where IO is the moment of inertia
about point O, and ωspin here denotes the angular velocity of the spinning
wheel. If we assume that all the mass of the wheel is at its circumference (a
rough approximation), then IO = mR2, so ωprecession = agR2ωspin
. The period
of the precession is, as usual, Tprecession = 2πωprecession
.
15 This result is being stated without proof.
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Figure 3.6: Gyroscopic motion. Our usual notationis slightly different for this figure. ~ω here is the spinangular momentum vector, and ~W is the gravitationalforce.
3.4 Elliptical Orbits and Kepler’s Laws
So far, when discussing rotational motion (for example, the motion of
the Earth around the Sun), we have narrowed attention to circular motion.
In reality, however, planets describe elliptical orbits. In this section, we
explore aspects of this type of motion.
Let us first sum up the results we have found for circular motion:
Result 3.7 (Summary: Circular Motion) When an object of mass m
orbits around another object of mass M and describes a circumference of
radius R, then:
• The period of rotation is T = 2π√
R3
GM , where G is the gravitational
constant.
• The orbital speed of motion is v = 2πRT =
√MGR .
• Total energy is E = T + V = 12mv
2 − mMGR = −mMG
2R .16
16 Recall that, here, we normalized potential energy to be zero at infinity, V∞ = 0.This explains why all bound orbits have negative total energy.
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• The escape velocity is vesc =√
2MGR .17
In general, bound orbits are ellipsis, though in our Solar System most
are close to circular. The laws of motion for these types of planetary motion
were first famously formulated by Johannes Kepler (1571 – 1630). His laws
are the following:
Law 1 The orbits of planets are elliptical, and the Sun is at one focus.
Law 2 The line segment joining a planet and the Sun sweeps out the same
area at any point in the ellipse for any given fixed amount of time.
Law 3 The square of the orbital period of the ellipse is proportional to the
cube of the distance to the Sun.
Figure 3.7 illustrates what is meant by Law 2: if in a given amount of
time, a planet experiences a motion along the arc of any of the three shaded
regions, then these regions all have equal area. For this reason, this law is
commonly known as the Equal areas – Equal times law.
M
Q
m
P A
~r~v
C a
Figure 3.7: An elliptical orbit. An object of massm orbits about a mass M located at point Q. C isthe center of the ellipsis, a its semi-major axis, A theapogee, and P the perigee.
Let us thus study elliptical orbits formally. Consider (as in Figure 3.7)
an ellipsis about an object of mass M in location Q. The perigee/perihelium
and apogee/aphelium are clearly marked in Figure 3.7 by points P and A,
17 Recall that we calculated vesc by setting E = 0.
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respectively.18 Let the distance between the two be 2a, where a is the semi-
major axis of the ellipsis and point C marks the center of the ellipsis.19
Finally, let ~r be the distance vector of object m from M , and ~v be the
former’s object velocity vector.
The critical difference between elliptical and circular motion is that the
distance between the objects, r = |~rQ|, now changes with time, whereas for
circular orbits it is some constant R. Therefore, the speed v now changes
with time. In particular, the period of motion, total energy, and escape
velocity are as follows:
Result 3.8 (Summary: Elliptical Motion) Consider the motion of m
around M along an ellipsis with semi-major axis a. Then:20
• The period of rotation is T = 2π√
a3
GM .21
• Total energy is E = T + V = 12mv
2 − mMGr = −mMG
2a .
• The escape velocity is vesc =√
2MGr .
Compare these with the results for circular orbits (Result 3.7). Note
that the functional forms are all similar, but where we had an R in circular
motion (a constant), now we have an r (a time-varying distance).
• Both kinetic and potential energy are time-varying, though by the
conservation of mechanical energy E is constant.
• The period in elliptical orbits is also very similar to that of circular
orbits, except that the radius R is replaced by the semi-major axis,
18 The perigee (if M is the Earth and m is the moon), or perihelium (if M is the Sunand m is the Earth), is the point on the ellipsis that is closest to the object M . Theapogee (for Earth and moon), or aphelium (for Earth and Sun), is the point that isfurthest.19 The major axis of an ellipse is its longest diameter, i.e. the length of the longestsegment running through the center and both foci, with ends at the widest points ofthe perimeter. The semi-major axis is half of the major axis, and thus runs from thecenter of the ellipsis, through a focus and to the perimeter.20 These results are being stated without proof.21 Indeed, T 2 ∝ a3, Kepler’s third law.
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a. Remarkably, both period and total energy of a circular and an
elliptical orbit are the same if a = R.
• Finally, the escape velocity is again calculated by setting E = 0 and
solving for v (intuitively, the speed necessary to attain infinity with
no energy left). For elliptical motion, this speed is a function of the
orbital speed that the object is experiencing, which is now changing
in time.
Let us now calculate the semi-major axis a, and velocities at various
points. Let ϕ denote the angle between position and velocity, ϕ = ∠~r~v.
Take the initial conditions (r0, v0, ϕ0) as given. The total energy at time
t = 0 is E = T +V = 12mv
20− mMG
r0. Since energy is conserved, E will never
change. According to our formula above, E = −mMG2a , so 1
2v20−MG
r0= −MG
2a .
Solving for a, we obtain the semi-major axis that the orbit will attain:
a =MGr0
2MG− v20r0
(3.5)
Next, we calculate the velocities as the object passes through the perigee
and apogee, ~vP and ~vA. At both these points, ~v ⊥ ~r, so sinϕA = sinϕP = 1.
At t = 0, the magnitude of the angular momentum about point Q (where
the mass M is located), is then LQ = |~LQ|, is LQ(0) = mv0r0 sinϕ0. At
a later time t, when the object is at point P in Figure 3.7 (the perigee),
the angular momentum about point Q is LQ(t) = mvP rP , where rP is
the distance between Q and P (and we have used sinϕP = 1). By the
conservation of angular momentum about point Q, it must then be that
LQ(0) = LQ(t), so:
v0r0 sinϕ0 = vP rP (3.6)
On the other hand, the total energy must be conserved. At time t, when
at point P , we have E(t) = 12mv
2P −
mMGrP
. Using Result 3.8, then:
1
2v2P −
MG
rP= −MG
2a(3.7)
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where a is given by equation (3.5). Thus, we have a system of two
equations (3.6)-(3.7) and two unknowns, (vP , rP ). Note equation (3.7) is
quadratic, so it will give us two solutions for rP . One solution will indeed
be the distance to the perigee, rP . The other one will be the distance to
the apogee, rA. Indeed, note that we could have replaced rP by rA above,
and we would have obtained the same equations and, thus, the same two
solutions.
The two solutions should verify that rAvA = rP vP , a direct consequence
of the conservation of angular momentum, and rA + rP = 2a, by construc-
tion.
Example 3.13 (A satellite) Consider a satellite orbiting around the Earth.
The Earth is M = 6 × 1024 kg, and suppose the satellite is r0 = 9, 000
km above the Earth, with initial speed of v0 = 9 km/sec and angle ϕ0 =
∠~r0~v0 = 120◦. Then, using our results above, the major axis of its orbit
will be a = 50× 103 km, a very large number relative to the initial position.
To understand this huge displacement from its initial position, notice that
vesc =√
2MGr = 9.4 km/sec, so the initial velocity of the object is just shy
of the speed it would take for the satellite to fly out of orbit. One orbit is
completed in T = 31 hours.
On the other hand, it can be checked that rP = 6.6×103 km and vP = 10.7
km/sec, while rA = 9.3× 104 km and vA = 0.75 km/sec. That is, the object
would pass very close to the Earth (at the perigee) at high speed, then lose
speed until it reached the furthest point, and pick it up again on its way back.
Next, we study orbital changes, that is situations in which an object in
orbit is given a force along the orbit (e.g. a rocket in orbit that fires its
engines). If the object is initially in a circular orbit, then clearly the thrust
will produce kinetic energy, increase the velocity, and change the orbital
path from a circle to an ellipse. Let’s see an example:
Example 3.14 (Two astronauts) Consider two astronauts, A and B, or-
biting Earth along the same circular orbit of radius R. Let A and B also
index their positions. Astronaut A is a fraction f of the circumference of
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the circle behind astronaut B. Since the perimeter of the circumference is
2πR, this means that the distance along the arc between A and B is f2πR
in one direction, and (1− f)2πR in the other direction.
Suppose A holds a tennis ball and wants to throw it out in such a way
for astronaut B to catch it exactly as she passes through A at a later time,
using an impulse that it exactly tangential to the circumference. The ball will
experience an elliptical orbit. For a catch to occur, the throw has to satisfy
Tball = (1−f)Tastronauts. Since the astronauts are orbiting in circular motion
and the ball will describe an ellipsis, we have 2π√
a3
MG = (1−f)2π√
R3
MG or,
simplifying:
a = (1− f)23R
In words, A will have to ensure that his throw describes an ellipsis with
semi-major axis equal to a = (1− f)23R. Now we can work out the speed at
which the throw will have to be made. The total energy satisfies −mMG2a =
12mv
2ball −
mMGR , so we can solve for vball because a is known. Since A is
moving at some speed vA herself, the speed she will have to give the ball is
vball − vA.22
This is only one of many solutions, as the catch could occur at some other
point after the ball has already made nball revolutions and astronaut B has
already made nB revolutions. A similar derivation delivers a =(nB−fnball
) 23R.
For instance (nB, nball) = (1, 1) is trivially a solution (the one derived
above).
22 It is possible that this number be negative. This would mean that A would haveto throw the ball back for B to catch it.
167
Chapter 4
Stability and Elasticity
4.1 Static Equilibrium
In the previous chapters we have argued that the motion of bodies re-
quires the presence of forces and/or torques. By Netwon’s Laws, when the
(vector-)sum of forces acting on a particle is zero, the particle will be at
rest.1 For a collection of many particles, it is sufficient to consider the forces
that alter the object’s center of mass. If the sum of forces on the center of
mass is zero, the collection of particles as a whole remains at rest.
For rigid bodies, forces can also be treated as acting upon the object’s
center of mass. However, whether or not the body stays at rest ultimately
depends on where these forces are applied. In general, if the vectorial sum
of forces is zero, this may not guarantee that the body will be at rest. The
reason is that a torque (a rotational force) may be at play. For example, if
we pull two corners of a table in opposite directions but equal force, the sum
of forces is zero, but the table is not at rest (frictional forces permitting)
because a torque makes it rotate around the axis that goes through its center
of mass.
Therefore, for an object to be at rest, we require that the sum of all
forces as well as the sum of all torques be zero. This is the notion of a static
1 Alternatively, it may be in a uniform linear motion, but we can always change thereference frame so it remains at rest.
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equilibrium.
Definition 4.1 (Static Equilibrium) A body is in static equilibrium if:
• Translational stability: All the forces applied on the body cancel out.
That is: ∑~F = ~0
• Rotational stability: All the torques about any point Q in the body
cancel out. That is: ∑~τQ = ~0
for all reference points Q.
By definition, if an object is in a static equilibrium, it has no linear
acceleration (a = 0) nor angular acceleration (α = 0), so it remains at rest.
Example 4.1 (A Ladder) Consider a ladder that is reclined on a wall,
forming a right triangle (Figure 4.1, panel (a)). Suppose there is no friction
at point P , and let the coefficient of static friction at point Q be denoted by
µ (whose magnitude depends on the attributes of the ladder and the floor on
which it stands). Let C be the ladder’s center of mass. The ladder (the line
PQ) has length ` and mass m. If the angle θ is too small, the ladder will
slide due to gravity. If it is large enough, it will stay upright, and be in a
static equilibrium.
The force of gravity acts upon the ladder’s center of mass. On the other
hand, at points Q and P , normal forces FNQ and FNP push the floor and
the wall, respectively, against the ladder. Finally, a frictional force Ff keeps
the ladder upright.
For the ladder to be in static equilibrium, the sum of all forces must be
zero in both x and y dimensions. Thus:
∑Fx = FNP − Ff = 0∑Fy = FNQ −mg = 0
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θ
C
P
Qmg
FNQ
FNP
Ff
(a) A ladder reclined on a wall.
C
P
Q
D
Mg
(b) A mass M on the ladder.
Figure 4.1: A ladder (line PQ) on the floor, reclinedon a wall.
Moreover, the sum of all torques must be zero about any point of refer-
ence (e.g. P , C, Q, or any other point for that matter). The most conve-
nient point for the calculation of the torque is Q, because for Q the forces
~FNQ and ~Ff do not contribute to the torque (as the position vectors are
trivially zero for those forces act on Q itself).
The torques about point Q are the normal force on P and the gravitational
force on C. Thus, we require that the total sum of torques is zero, or:
∑~τC = ~FNP × ~rP + ~Fgrav × ~rC = ~0
where ~rP and ~rC are the position vectors from point Q to points P and
C, respectively. For point P , |~rP | = ` (as ~rP traces out the hypothenuse of
the triangle) and ∠~rP ~FNP = θ, so ~FNP × ~rP = (FNP ` sin θ)~e+, where ~e+ is
a unit vector from point Q that is orthogonal to the (x, y) plane and in the
plus direction (of the z axis, i.e. out of the page). For point C, |~rC | = 12`
(as the center of mass is exactly half-way between points P and Q) and
∠~rC ~Fgrav = π2 −θ, so ~Fgrav×~rP = 1
2(mg` cos θ)~e−, where ~e− is a unit vector
from Q that is orthogonal to (x, y) but into the minus direction (i.e. into the
page).2 Therefore, in magnitudes, we require FNP ` sin θ−mg` cos θ = 0, or:
2 We have used that sin(π2− θ)
= cos θ, a direct consequence of equation (0.3). For
another way to see this, note that the angle ∠~rC ~Fgrav is defined by the sides PQ and
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FNP =1
2mg cot θ = Ff
where cot θ ≡ cos θsin θ = 1
tan θ is the reciprocal of the tangent, often called
the cotangent, and the second equality comes from∑Fx = 0 above.
By stability, the ladder does not slide. Therefore, the frictional force
cannot exceed the product of µ (the coefficient of static friction at Q) times
the normal force (recall Definition 1.18), or Ff ≤ µmg. Using the result
above, 12mg cot θ ≤ µmg, that is cot θ ≤ 2µ. Thus, the critical angle satisfies:
tan θ∗ =1
2µ(4.1)
Thus, the condition that ensures that the ladder will be stable is θ ≥ θ∗.The smaller µ is (e.g. a slippery floor), the more unstable the system is, in
the sense that θ∗ is high: only if the ladder is at a high angle relative to the
slippery floor will it remain at rest.
Now suppose that a body of mass M is placed at some point D between Q
and P (see Figure 4.1, panel (b)).3 For instance, suppose a person of mass
M starts going up the ladder. Let d be the distance between Q and D. Now,
the sum of forces is:
∑Fx = FNP − Ff = 0∑Fy = FNQ − (M +m)g = 0
on each dimension. The magnitude in the torque about point Q is:
FNP ` sin θ −mg` cos θ −Mgd cos θ = 0
where the last term is new and due to the heavy body at point D. There-
OP (where O is the origin), and the ratio of the two equals cos θ by Definition 0.6.3 In the figure, D is placed between Q and C, but this need not be the case. As we
shall see shortly, the stability of the system is highly affected by the location of Drelative to C.
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fore, FNP = g cos θ` sin θ
(12m`+Md
), or:
FNP = g cot θ
(1
2m+
Md
`
)= Ff
where the second equality comes from ∼ Fx = 0. Again, for the ladder
not to slide we need Ff ≤ µFNQ = µ(m + M)g, so g cot θ(
12m+ Md
`
)≤
µ(m+M)g. Simplifying, we find the critical angle to satisfy:
tan θ∗∗ = tan θ∗
(m+ 2Md
`
m+M
)︸ ︷︷ ︸
≶1
(4.2)
where tan θ∗ = 12µ by equation (4.1). Therefore, whether the ladder
becomes more or less stable when an object of mass M is added depends on
whether the term in parenthesis in equation (4.2) is greater or lower than
one. Thus:
• If d ≥ `2 , then θ∗∗ ≥ θ∗.
In words, if the object is placed above the ladder’s center of mass (i.e.
between points P and C), then the ladder becomes more unstable, in
that it will remain at rest for a smaller range of angles. For instance,
if the ladder was reclined with θ = θ∗ and the object is placed above
the center of mass, the ladder will always topple.
• If d < `2 , then θ∗∗ < θ∗.
In words, if the object is placed below the ladder’s center of mass (i.e.
between points Q and C), the ladder becomes more stable, in that it
remains at rest for a wider range of angles. For instance, if the ladder
was reclined with θ = θ∗∗, it will topple. But if the object is placed, it
regain stability and stay upright.
The critical observation is that, while the friction Ff increases with the
distance d, the maximum friction, µFNP (i.e. µ(m + M)g) is independent
of d. If the person starts climbing at point Q (i.e. d = 0) and goes up the
ladder, the friction will increase but the maximum friction will stay constant,
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so the ladder will remain stable. But once the person gets passed the ladder’s
center of mass (say, the central rung), there will be a point where the ladder
will slide and fall. This point will be the center of mass itself if θ = θ∗, or
a little bit further up if θ > θ∗.
Example 4.2 (A Pulley) Take two objects, of masses M and m, hanging
from the same string which is itself wrapped around a cylinder of radius R.
This structure is called a single pulley (e.g. the result of removing the incline
in Figure 1.9).
On each string there is a tension T , given by T1 = mg and T2 = Mg.
Letting M > m, so that T2 > T1, in the absence of friction between the rope
and the cylinder we would see the system accelerate: the more massive body
would accelerate downward as the less massive body accelerates upward.
Suppose, however, that the contact between the string and the cylinder
is frictional. Moreover, suppose that the cylinder is fixed and cannot rotate.
Since T2 > T1, the system may start slipping and accelerate as in the fric-
tionless case. For every bit of rope that is in contact with the cylinder, a
frictional force exists pointing in the opposite direction. In particular, if the
static friction coefficient is µ, then one can show that a sufficient condition
for stability is:4
T2
T1= eµθ0
where θ0 is the angle of the arc on the cylinder’s circumference where
the rope and the cylinder’s surface are in contact. (Surprisingly, this factor
is completely independent of the pulley’s radius.)
If the force of gravity is present, the two bodies will hang vertically down,
so θ0 ≈ π (i.e. 180◦), meaning T2T1≈ eµπ. We can also wrap the string
around the cylinder any number of times we want (provided the string is long
enough). For example, θ0 = 6π means that the string is wrapped 3 times
around the cylinder; θ0 = 12π means 6 turns. If, for instance, µ = 0.2, then
4 This is stated without proof. To derive this result, we must integrate the frictionalforce for each infinitesimal distance over the total length of the arc on the cylinder’scircumference which is in contact with the rope.
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6 turns means T2T1
= 2, 000. This illustrates why pulleys are used to balance
heavy objects up in the air: by holding on to the lighter object (e.g. grabbing
one end of the string), the tension created on the heavier object is so strong
that one can balance very heavy objects up in the air with hardly any effort,
provided the rope is wrapped around the pulley enough times.
Example 4.3 (Hanging from a pin) Consider hanging an object of mass
m from some point P (for example, a piece of paper on a wall using a pin).
Figure 4.2 shows an example, though the shape of the object is not relevant.
Let C be the center of mass, separated by distance b from P .
P
C
mg
~rP
Figure 4.2: An object hanging from P with center ofmass C.
The torque relative to point P is ~τP = ~rP × ~Fgrav, of magnitude τP =
bmg sin θ, where θ is the angle between ~rP and ~Fgrav. Clearly, this object is
going to rotate about P , so the torque’s magnitude is τP = IPα, where IP is
the moment of inertia relative to P , and α is the angular acceleration.
The object will be in a stable equilibrium if the sum of all forces and that
of all torques (relative to any point) are both zero. Note that the only case
in which that can happen is if P and C are along the same vertical line.
Indeed, in this case, there is a downward force on C equal to mg, and (by
Newton’s Third Law) a force of the same magnitude and opposite direction
on P , so forces are zero. There is also no torque, because ~rP and ~Fgrav are
parallel, so ~τP = ~rP × ~Fgrav = ~0.
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For example, a pendulum lies at rest in a stable equilibrium when the
point about which it rotates is aligned along a vertical line with the center
or mass (the center of the bob, if the bob is solid).
This last example also gives us a recipe for finding the center of mass
of an object in practice. Suspend the object from some arbitrary point,
wait for it to become stable, and mark a straight vertical line going down
along the y direction. Repeat the exercise from another arbitrary point of
suspension. The intersection of the two resulting straight lines will then
mark the location of the center of mass.
Example 4.4 (The Tightrope Walker) Consider a walker on a tightrope.
As it is, the point of suspension (the contact point between the rope and
the walker’s feet) is below the system’s center of mass, which is about the
walker’s chest. However, the walker could balance two weights, hanging down
from his/her hands along long strings and toward the floor (suppose the
tightrope is highly elevated). By adding mass onto the system, the center of
mass would now shift below the tightrope, so the walker can keep his/her
balance.
This is one reason why tightrope walkers carry a pole while doing their
stunt: the pole is often curved down, which adds weight and shifts the center
of mass down closer to (and ideally below) the suspension point. Another
reason is that the pole helps add rotational inertia to the system (as in the
case of the spinning figure skater, Example 3.4), which helps keep balance.
Example 4.5 (The Lever Law) Consider a lever, that is, a rigid rod of
mass m that pivots about a fixed hinge (see Figure 4.3). Suppose that two
forces, ~F1 and ~F2, are applied on each end of the rod.5 The center of mass
is at C, it is subject to a gravitational force mg, and di denotes the distance
between C and the force i = 1, 2. Suppose that the lever is pivoted about C.
5 These forces would be, for example, normal forces with magnitude M1g and M2g,respectively, if two bodies of masses (M1,M2) were placed one on each side of thelever. We will analyze the more general case where the forces need not be normal(i.e. perpendicular to the surface of the rod).
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For a (one-dimensional) solid rod of length `, the center of mass is located
at the middle: d1 = d2 = `/2.
Figure 4.3: The generalized lever.
The system will be at equilibrium if ~F1 + ~F2 = ~0 and the sum of torques
about any point is zero. Decomposing the forces into its (x, y) components
in the usual way, we have:
∑Fx = 0 = F1 cos θ1 − F2 cos θ2 (4.3)∑Fy = 0 = F1 sin θ1 + F2 sin θ2 −mg + Fpivot (4.4)
where θi, i = 1, 2, is the angle between the position vector ~ri,C (i.e.
relative to the center of mass) and ~Fi, and Fpivot is the magnitude of the
normal force on the center of mass.
On the other hand, the torque about point C (the pivot point) has several
components. The torque due to ~Fpivot and that due to gravity are both zero,
because our point of reference is precisely the point C where these forces are
applied. Thus, the torque is only due to the ~F1 and ~F2 forces:
~τC = ~F1 × ~r1,C + ~F2 × ~r2,C
The first torque equals ~τ1,C ≡ ~F1 × ~r1,C = F1d1 sin θ1~e+, where ~e+ is
orthogonal to (x, y) and pointing toward positive values of z (i.e. pointing
out of the page). The second torque equals ~τ2,C ≡ ~F2×~r2,C = F2d2 sin θ2~e−,
where ~e− is orthogonal to (x, y) and pointing toward negative values of z (i.e.
pointing into the page). In magnitude, we have, F1d1 sin θ1−F2d2 sin θ2 = 0.
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Therefore:
d1F1 sin θ1 = d2F2 sin θ2 (4.5)
Equations (4.3)-(4.4)-(4.5) then guarantee that, if we pivot the lever
about its center of mass, the lever will be at rest (it will not move linearly
or rotate). These equations compose the often called Lever Laws.
For example, if two bodies of masses M1 and M2 are placed on each end
of a lever that pivots about its center of mass, then Fi = Mig and θi = π/2
(so that sin θ1 = sin θ2 = 1), and the condition for stability becomes:
d1M1 = d2M2
with Fpivot = mg by Newton’s Third Law.
Example 4.6 (Standing on an incline) A person of mass m stands on
an incline at an angle θ relative to the floor.6 The person’s legs are separated
by a distance d, with one foot uphill and the other one downhill. The center
of mass is at a distance h above the ground, perpendicular to the hillside, and
it is midway between the person’s feet (around his/her chest). We assume
that the coefficient of friction is high enough for the person not to slide down.
The center of mass is subject to a force mg, and each leg has a normal
force FN,L and FN,R (for left and right legs). These forces are perpendicular
to the incline (our x axis). Frictional forces Ff,L and Ff,R are pointing
uphill from each leg, preventing the person to slide.
Stability requires that both (x, y) components of these forces add up to
zero. If positive values of x run “uphill”, then:
∑Fx = 0 = Ff,L + Ff,R −mg sin θ (4.6)∑Fy = 0 = FN,L + FN,R −mg cos θ (4.7)
6 As in Example 1.13, let coordinate system be “tilted” so that the direction of thex axis is parallel to the incline, not the floor.
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For the torques, let us again pick the center of mass as the reference
point, for convenience. At the center of mass, the force of gravity does not
contribute to the torque (its position relative to C being the zero vector). The
torque is due only to the two normal forces and the two frictional forces.
For the left leg, and always relative to the center of mass C, we have:
~τL,C = ~rL,C × ~FN,L︸ ︷︷ ︸Normal force
+ ~rL,C × ~Ff,L︸ ︷︷ ︸Frictional force
=(rLFN,L sinαN
)~e++
(rLFf,L sinαf
)~e+
where rL ≡ |~rL,C | =√h2 + d2/4,7 (αL, αf ) are the angles between ~rL,C
(the position vector of the left left relative to the center of mass) and ~FN,L
and Ff,L, respectively, and ~e+ is the unit vector orthogonal to (x, y) which is
pointing toward z+ (i.e. out of the paper) because, if the person is facing us,
his/her left leg is uphill and therefore αR < 90◦. Next, by definition of the
sine function, rL sinαL = d2 and rL sinαf = h (note ~r, ~F and the height of
altitude h form a right rectangle!). Therefore, we can write the torque due
to the left leg as ~τL,C =(d2FN,R + hFf,L
)~e+. The magnitude of the torque
on the left leg is, therefore:
τL,C =d
2FN,L + hFf,L
The total torque on the right leg, similarly, is:
~τR,C =d
2FN,R~e− + hFf,R~e+
Notice that the two components of the right-leg torque now point in op-
posite z directions. Therefore, the magnitude of the torque on the right leg
is:
τR,C = −d2FN,R + hFf,R
7 If the center of mass is at height h and midway between the legs, then rL is thehypothenuse of the triangle with sides h and d/2 (and we use Pythagoras’ theorem).
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The total torque must be zero, so τL,C + τR,C = 0, or:
d
2(FN,L − FN,R) + h(Ff,L + Ff,R) = 0 (4.8)
From equation (4.8), we have FN,R − FN,L = 2hd (Ff,L + Ff,R). Substitu-
tion into equation (4.6) yields:
FN,R − FN,L =2h
dmg sin θ
Adding this to equation (4.7) then gives FN,R, and subtracting it from
(4.7) will give FN,L. The result is:
FN,R = mg
(1
2cos θ +
h
dsin θ
)FN,L = mg
(1
2cos θ − h
dsin θ
)The moment at which the person will start to roll over and fall downhill
occurs when the normal force on the upper foot (the left foot) is zero. From
our results, this is true whenever 12 cos θ = h
d sin θ, that is:
d∗ = 2h tan θ
The distance d∗ marks the minimum distance that the person’s feet have
to be separated for him/her not to rotate and fall, given a height h and an
inclination angle θ.
4.2 Stress, Strain, and Elasticity
In this section, we introduce the concept of elasticity. In physics, elastic-
ity is an object’s ability to return to its original shape and size after being
subject to a distorting force. A solid object will deform while being un-
der such force, but if the material is sufficiently elastic it will return to its
original state.
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The elasticity of an object depends on the materials of the object as well
as external conditions (e.g. temperature), and is generally described by a
stress-strain curve. Let us first define these concepts:
Definition 4.2 (Stress) The internal force that neighboring particles of a
solid object exert on each other. It is denoted by ~σ.
In particular, the stress between two particles is measured as the force
that a particle applies on the other particle across a small surface of the
object, and it can have any direction relative to said surface. If the distorting
force (of magnitude F ) causes a deformation, then stress is given by:
σ ≡ F
A(4.9)
in magnitude, where A is the so-called cross-sectional area, i.e. the area
of the object’s cross section that is perpendicular to the direction of the force
applied. Here, σ is in Newtons per unit area, so it has the dimensions of
pressure.
Definition 4.3 (Strain) A measure of deformation representing the dis-
placement between two particles in a solid object, relative to its initial state.
It is denoted by ε.
In other words, while stress is a measure of internal forces between pairs
of particles, strain measures the proportional change in position of the ob-
ject. Obviously, the deformation (weakly) increases with the stress. If ` is
length of the object when in a relaxed state, and ∆` is the displacement it
suffers under the distorting force (both relative to the same fixed point of
reference), then strain is defined by:8
ε ≡ ∆`
`(4.10)
Therefore, ε is a measure of proportional deformation, and it is dimen-
sionless.8 By convention, ∆` is positive (negative) if the object is being stretched (com-
pressed).
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The stress-strain curve (Figure 4.4) is then the function ε(σ), i.e. it is
the graph of the strain ε of an object under various levels of stress σ.9 The
curve can be used to determine the properties of the materials of the object,
including the object’s modulus of elasticity (E):
Definition 4.4 (Modulus of Elasticity) The modulus of elasticity is de-
fined by:
E(σ) ≡ σ
ε(σ)
That is, it is the slope of the stress-strain curve.
Therefore, the modulus of elasticity of an object is a measure of the ob-
ject’s resistance to being deformed elastically (i.e. non-permanently) when
stress is applied to it. A stiffer object will have a higher elasticity modulus.
For many objects, if the deformation force is small enough, the elasticity
modulus is constant in the strain. That is, the stress-strain curve is linear for
low levels of stress. For these so-called linear-elasticity objects, the elasticity
modulus is called Young’s modulus. Let us look at these specific cases, and
derive Young’s modulus.
Example 4.7 (Linear Elasticity) Consider a body that behaves according
to Hooke’s Law (Principle 1.3), e.g. a spring. When the body is in a relaxed
state, we normalize its position to x = 0 and let ` be the object’s length. Sup-
pose a linear force of magnitude F is applied on the object, thereby stretching
it until extended to some position x = ∆`.
We make the following observations:
• By Hooke’s Law (i.e. F = −kx), the displacement is proportional to
the force applied: if the force is twice as strong, the displacement is
doubled. Thus, ∆` ∝ F .
9 By convention, strain is plotted as the independent variable (on the x axis), al-though we really should think of the stress-strain curve as showing strain levels undervarious stress forces, and not the other way around.
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• Clearly, if the object is twice as long, then the amount of displacement
is doubled as well, for a given force. Thus, ∆` ∝ `.
• Finally, the amount of displacement is inversely proportional to the
cross-sectional area. If the force is applied over a certain area A, then
increasing the area by a factor of two will decrease the displacement
by the same factor. Thus, ∆` ∝ 1A .10
Thus, we have found that the displacement ∆` is proportional to the
force, proportional to the length, and inversely proportional to the cross-
sectional area over which the force is applied, or ∆` ∝ F`A .
Using equations (4.9)-(4.10), we have obtained ε ∝ σ. The constant of
proportionality is the elasticity modulus, given by:
Y ≡ F/A
∆`/`=σ
ε
The elasticity modulus in this case is denoted Y for Young’s modulus.11
A large number of materials behave in this manner, including most met-
als and some materials such as rubber. When the strain is small, the elas-
ticity modulus is constant. If released (e.g. a spring is stretched and then
released), the object eventually returns to its original position. In particu-
lar, since F = Y A` ∆` (be definition), then the released object will describe
a harmonic oscillation with oscillation constant:
k =Y A
`(4.11)
so that its position will be x = xmax cos(ωt+ϕ), with angular frequency
10 For instance, consider a rod of cross-sectional area A and length `. If two identicalsuch rods are placed together in parallel, then they each counteract F with half theforce, showing ∆` ∝ `. If instead a single rod of cross-sectional area 2A is used, thedisplacement is 1
2∆` for the same F , showing ∆` ∝ 1/A.
11 For example, consider a rod of radius r = 0.5 cm, cross-sectional area A = 8×10−5
m2, length ` = 1 m, and mass M = 500 kg. Suppose a force of F = 5, 000 N isapplied. If the rod is made of steel (Y = 20 × 1010 N/m2), then the rod will beextended by ∆` ≈ 0.3 mm. If the rod is made of nylon (Y = 0.36× 1010 N/m2), theextension is ∆` ≈ 17 mm. Indeed, ropes are much more elastic than metal rods.
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ω =√k/m =
√Y A`m , period of oscillation T = 2π/ω = 2π
√`mY A , and fre-
quency 1/T Hz.
However, for most objects, the linear relationship disappears after a cer-
tain level of stress is applied on the object. At this level (point A in Figure
4.4), the elasticity modulus achieves its so-called elasticity limit. Past this
point, and for higher levels of stress, Hooke’s law ceases to hold, the elas-
ticity modulus is no longer constant, and the object becomes permanently
deformed. Namely, if the force were to be released past point A, the object
would not return to its original form. At some maximum strain (point D),
the object breaks.
Strain (ε)
Stress (σ)
AB
C
D
Figure 4.4: The typical shape of the stress-straincurve.
Different materials will exhibit different behaviors between points A and
D. The object’s elasticity may become negative past some certain elasticity
(a yield point), represented by point B. Typically, the object also reaches a
maximum stress (point C) before it breaks, and it may even plateau around
it (that is, there might be a range of strains whereby the object experiences
deformation without any further increase in stress). The object’s state as it
goes through this plateau is called the plastic flow state.
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Notice that the stress decreases as the object’s strain increases just before
the breaking point (segment from C to D). This is because, just before
breaking, the object’s atoms get “squeezed out” of place, and the surface
area decreases (with equal force), making the stress lower.
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Chapter 5
Waves, Fluids, and
Oscillations
5.1 Waves and the Doppler shift
In this section, we study the emission and perception of waves. In par-
ticular, we will focus on the so-called Doppler effect.
The Doppler effect (or Doppler shift) is the change that occurs in the
frequency of a wave for an observer who is in motion relative to the source
of the wave. We experience the Doppler shift when the sound source moves
relative to our position, for instance when an ambulance or a police car drive
by in the street.
In particular, the relationship between the observed (or perceived) fre-
quency f and the originally emitted frequency f0 is given by:
f =
(c+ vrc+ vs
)f0 (5.1)
where c is the velocity of waves in the medium of consideration, vr is
the speed of the receiver, and vs is the speed of the source.1 If the source is
approaching (vr > vs), then f > f0, so the sound is perceived with higher
1 For sound in air, c = 331.5 + 0.6tc m/sec, where tc is temperature in Celsius. Forlight in a vacuum, c = 299, 792, 458 m/sec.
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pitch. If the source is receding (vr < vs), then f < f0, so the pitch is lower.
If the speeds vs and vr are small relative to the speed of the wave, then
a Tylor approximation yields:
∆f ≡ f − f0 =∆v
cf0 (5.2)
where ∆v ≡ vr − vs is the relative velocity of the receiver. Thus, the
change in frequency is proportional to the difference in speeds.
By definition (recall Definitions 1.6-1.7), the frequency of oscillation re-
lates to the period via T = 1f . Thus, in T seconds, the wave (or, rather, a par-
ticle moving in the field of consideration) moves by a distance of λ ≡ cT = c/
f m. This is called wavelength:
Definition 5.1 (Wavelength) The distance over which a period wave re-
peats its pattern (i.e. the distance between successive crests), defined by:
λ ≡ c
f
where f is the wave’s frequency, and c is the speed of the wave.
Therefore, by definition, waves with higher frequency have lower wave-
length, and the factor of proportionality between the two is given by c (e.g.
the speed of sound for acoustic waves, or the speed of light for electromag-
netic waves).
Suppose that source and receiver are separated by distance ~r. Let ~vs be
the velocity of the source, and θ = ∠~r~vs. The x component of the velocity
(often called the radial component) is vx = vs cos θ. If the receiver perceives
frequency f , then:
f = f0
(1 +
vsc
cos θ)
(5.3)
by equation (5.2). In terms of λ = c/f , equation (5.3) reads:
λ = λ0
(1− vs
ccos θ
)(5.4)
Thus:
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• If θ = π/2 (i.e 90◦), then cos θ = 0, so f = f0 and λ = λ0. In words,
when the source and the receiver keep their distance constant, the
frequency and wavelength arrive unaltered to the receiver.
• If θ < π/2, then cos θ > 0, so f > f0 and λ < λ0. In words, when the
source and the receiver are approaching, the receiver reads a higher
frequency and a lower wavelength.
• If θ > π/2, then cos θ < 0, so f < f0 and λ > λ0. In words, when
the source and the receiver are receding, the receiver reads a lower
frequency and a higher wavelength.
The Doppler shift, DS, can thus be just calculated as the change in
wavelength, λλ0
, from equation (5.4):
DS ≡ 1− vsc
cos θ
Let’s now examine different types of waves, and how the Doppler effect
is manifested:
Sound waves The Doppler effect is most popularly present in sounds
waves (e.g. sirens). Suppose that the wave source exhibits a circular motion
and the receiver is on the same plane of the circumference, so the source
oscillates between being close and far from the receiver. Since the circular
motion can be described by a sinusoidal wave (as in Figure 0.1), then the
frequency f takes the shape of such a wave: as the source approaches the
receiver, the frequency increases, and vice versa.
Let fmax be the frequency when the source is the closest. Then, the
velocity of the source can be deduced using equation (5.1). Moreover, the
period T (the time elapsed between two crests) satisfies, as we know, 2πRT =
vs, from which the radius of the orbit R can be deduced.
Electromagnetic waves Electromagnetic waves reflect electromagnetic
radiation. They include waves involving the motion of light particles, so c in
equation (5.1) stands here for the speed of light. Typically, the speed of the
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transmitter (vs) is way smaller than the speed of light (c), so approximation
(5.2) applies.
Most frequencies are invisible to the human eye and, for those that are
visible, the frequency determines the color at which the human eye per-
ceives light. From lowest to highest frequency (that is, highest to lowest
wavelength), the electromagnetic waves are:2
1. Long radio waves, with f < 106 Hz and λ > 103 m.
2. Radio waves, with f ∈ [106, 108] Hz and λ ∈ [1, 103] m. Among these:
• AM waves, with f ∈ [106, 107] Hz and λ ∈ [102, 103] m.
• FM waves, with f ∈ [107, 108] Hz and λ ∈ [1, 102] m.
3. Microwaves, with f ∈ [108, 1011] Hz and λ ∈ [10−3, 1] m.
4. Infrared waves, with f ∈ [1011, 1012] Hz and λ ∈ [10−3, 10−6] m.
5. Visible light waves, with f ∈ [430 × 1012, 770 × 1012] Hz and λ ∈[390×10−9, 700×10−9] m. These are the frequencies and wavelengths
to which a typical human eye can respond. Within these:
• Red, with f ∈ [400, 484]× 1012 Hz and λ ∈ [620, 750]× 10−9 m.
• Orange, with f ∈ [484, 508] × 1012 Hz and λ ∈ [590, 620] × 10−9
m.
• Yellow, with f ∈ [508, 526] × 1012 Hz and λ ∈ [570, 590] × 10−9
m.
• Green, with f ∈ [526, 606]× 1012 Hz and λ ∈ [495, 570]× 10−9 m.
• Blue, with f ∈ [606, 668]× 1012 Hz and λ ∈ [450, 495]× 10−9 m.
• Violet, with f ∈ [668, 789]×1012 Hz and λ ∈ [380, 450]×10−9 m.
6. Ultraviolet waves, with f ∈ [1013, 1016] Hz and λ ∈ [10−8, 10−7] m.
7. X-rays, with f ∈ [1016, 1020] Hz and λ ∈ [10−12, 10−8] m.
2 Frequencies and wavelength given in the list are crudely approximated.
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8. Gamma rays, with f > 1020 Hz and λ < 10−12 m.
In astronomy, the frequency of light cannot be measured, and only the
wavelength λ is observable. Changes in the wavelength can then be used
to infer if the objects (say, the Earth and a distant planet) are approaching
or receding. Using the classification for color used above, astronomers have
adopted the following terminology:
Definition 5.2 (Redshift and blueshift) Redshift occurs when the elec-
tromagnetic radiation of an object is increased in wavelength (i.e. decreased
in frequency). When the opposite occurs, we call it a blueshift.
This terminology is used even when light is not visible by the human
eye. For example, for the relative motion of planets:
• When astronomers observe an increase in the wavelength emitted by
a distant planet as it arrives on Earth, we say that a redshift has
occurred, and thus the planet is receding from the Earth.
• A blueshift in the wavelength will thus denote that the planet is ap-
proaching the Earth.
The detection of wavelengths has permitted astronomers to make many
predictions about the behavior of the cosmos. For example, when explor-
ing a star, astronomers can determine which elements are present in the
composition of the star by looking at the so-called spectral lines of their
electromagnetic spectrum.3 By determining the color shift in the spectrum,
3 Spectral lines of the spectrum are discontinuities present in an the otherwise con-tinuous electromagnetic spectrum of light. They are due to either the emission orthe absorption of light within narrow frequency ranges. In the visible light spectrum,absorption lines thus appear as black lines, and emission lines appear as white lines.Each atoms and molecule has its own unique pattern of spectral lines along the spec-trum, so these can be used in practice to identify which elements are present in thecomposition of a star. For instance, the helium atom was found to be emitted by theradiation of our Sun by studying the Sun’s spectrum.
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the velocity at which the stars are moving relative to each other can be
inferred (using equation (5.4)).4
Throughout its life-cycle, a star survives by burning its nuclear fuel in its
core, a process that is sufficiently powerful to counteract the effects of the
star’s own gravity. When this fuel is exhausted, nuclear fusion gives way to
gravity, and the star implodes into a nova. At this point, depending on the
mass of the original star, the dying star can become one of three objects:
1. A white dwarf, for stars whose mass is below 1.4 the mass of our Sun.
2. A neutron star, which are denser and smaller than white dwarfs, for
stars whose mass is between 1.4 and 3 times that of the Sun.
3. A black hole (for masses at least 3 times larger than that of our Sun)
is an object of such enormous density that it produces a gravitational
field so strong that no light or matter can escape its pull within its
vicinity.
Let’s study these with some detail:
Example 5.1 (Binary-Star Systems) Today, astronomers know that many
star systems in the universe are binary star systems, that is, systems com-
posed of two stars orbiting each other in a spiral-like motion. This can
be observed by looking at the spectra of the stars and observing continuous
red- and blueshifts in the spectral lines, an unequivocal manifestation that a
Doppler effect is occurring and that, therefore, stars are oscillating relative
to each other and to Earth. Moreover, for each one of those stars, the radial
velocity, the radius of the orbit, and the orbital period, can all be determined.
Consider a binary system. Star 1 has mass m1, and describes a circular
orbit about the center of mass of the system, at radius r2. Star 2 has mass
m2, goes around the center of mass, and r2 is the radius of its orbit. By
definition of the center of mass (Definition 1.27), m1r1 = m2r2. Suppose
the observer is on the plane of the two orbits.
4 More specifically, knowing the shift in the spectrum, λ/λ0, one can infer the radialvelocity, v cos θ, but the actual velocity v remains unknown unless the angle θ (andthus the direction of motion of the stars) can be calculated.
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Because the orbits are circular, the period of motion is:
T = 2π
√(r1 + r2)3
G(m1 +m2)
Now, we want to make the Doppler shift measurement of both stars. The
Doppler shift (in wavelength) of star i = 1, 2 is DSi = 1 − vic cos θi. Thus,
by measuring the Doppler shift, once can know the radial velocity of each
star. Using the period, one can infer r1 + r2 and m1 + m2, and using that
m1r1 = m2r2, one can solve for m1, m2, r1, and r2.
This shows that measuring the Doppler effect is extremely informative
of the properties of the binary star system (including the positions, masses,
and velocities of each one of its component stars).
Example 5.2 (X-ray binaries) An X-ray binary is a class of binary stars
that are visible only in X-rays. They are composed of matter falling from
one of the component stars (the so-called donor), to the other component
(the so-called accretor). The latter is a very compact star, usually a neutron
star or a black hole.
When a neutron star accretor is sufficiently close to the donor, mass
from the latter is transferred to the former. But because the donor and
the accretor are orbiting each other, the motion of the transferring particle
describes a spiral motion as it falls into the accretor’s own orbit. The spiral
trajectory is usually called an accretor disk. As a result of the rotation of
the neutron star, the poles of the star heat up and radiate light, which from
Earth appear as pulsations (in a similar way that a lighthouse would). Such
types of neutron stars are called pulsars.
A piece of matter with mass m spiraling toward the accretor will arrive
and release kinetic energy T = 12mv
2, where v is the particle’s velocity at
arrival.5 By the conservation of energy, kinetic energy and gravitational
5 To illustrate the amount of energy released by neutron stars, for m as little as 10grams, the kinetic energy that would be released from the impact is as large as anatomic bomb explosion. The reason is that v is phenomenally high, given the stronggravitational pull of the neutron star. The transfer of matter from the donor to theaccretor has been calculated to be about dm
dt= 1014 kg/sec, corresponding to a power
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potential energy add up to zero, so 12mv
2 = mMGR , where M is here the mass
of the accretor and R is the radius of its orbit. Thus, the speed at which
matter reaches the neutron star is:
v =
√2MG
R
This is, no surprise, equal to the escape velocity of the circular orbit
(recall Result 3.7): the velocity at which matter will arrive from infinitely
far is the same as the one that would take to escape the neutron star’s
gravitational pull.
Example 5.3 (Black Holes) A black hole, the most extreme manifesta-
tion of a star’s death, is an object with no size and, thus, infinite density.
Its mass, M , is at least three times that of the Sun.
The event horizon is a sphere with radius REH around the black hole,
which limits the frontier below which objects cannot escape from the hole’s
gravitational pull. At the event horizon, therefore, objects need a speed of
v ≥ vesc =√
2MGREH
to get away from the hole. Since not even light escapes
this pull, and because nothing travels faster than light, we have c =√
2MGREH
.
Thus, the radius of the event horizon is:
REH =2MG
c2
For example, if the black hole is the mass of the Earth, REH = 1 cm. If
it is the mass of the Sun, REH = 3 km.
Since a black hole has no surface (because it has no size), it does not pul-
sate as neutron stars in binary systems do. To determine its mass, therefore,
one cannot use the Doppler shift, for no electromagnetic radiation is being
emitted. Instead, one can use the Doppler shift of the donor (if the black
hole is the accretor in the binary system) and, if one has an estimate of the
mass of the donor, one can infer the mass of the accretor.
of P = 2 × 1030 J/sec. When kinetic energy is turned into heat energy, the neutronstar can achieve temperatures of 107 K. These are enormous values. The temperatureis so high, in fact, that all radiation is exclusively X-ray radiation. Hence, the nameof the system.
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5.2 Fluid Statics
In this section, we study the behavior of fluids (e.g. water, air, or oil)
that are at rest. We will explore the properties of fluids in various states
and how they react to forces. This branch of physics is called fluid statics.
The dynamics of fluids will be explored in Section 5.3.
5.2.1 Pressure and Pascal’s Principle
Consider a fluid which is fully contained in a closed compartment. The
compartment has an opening of cross-sectional area A. Suppose that a force
F is applied everywhere on this area, and let the direction of the force be
perpendicular to the cross section of the fluid. (In short, let F be a normal
force to the fluid’s surface). Then, we will say that the liquid’s pressure is
being increased:
Definition 5.3 (Pressure) The force applied perpendicular to the surface
of an object per unit area over which the force is distributed. It is defined
by:
P ≡ F
A
where F is the magnitude of the normal force, and A is the area of the
surface on contact.
Pressure is measured in Pascals (Pa), which are force, in Newtons, per
unit area, in squared meters. Our first principle is due to Pascal himself:
Principle 5.1 (Pascal’s Principle) The pressure applied to an enclosed
and static fluid is transmitted undiminished to every point in the fluid and
to the walls in the container.
The principle says that a pressure change that occurs anywhere in the
container is transmitted throughout the fluid such that the same change
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occurs everywhere. When gravity is not at play, this implies that the pres-
sure is literally the same everywhere in the vessel.6 Moreover, gravity or
no gravity, the force by the liquid onto the (inner) walls of the vessel must
be everywhere perpendicular to the surface of the walls. Were this not the
case, the fluid would clearly not be in a motionless state.
Formally, for every small area of size ∆A, if a force ∆F is exerted
by the fluid onto the wall and this force is perpendicular to ∆A, then
lim∆A→0∆F∆A = P , where P is the liquid’s unique pressure as postulated
by Pascal.
Example 5.4 (Hydraulic jacks) A hydraulic jack is a device that is de-
signed to lift heavy loads with relative ease and which directly exploits Pas-
cal’s principle. Consider two closed vessels whose openings have surface
areas A1 and A2, and suppose that the two vessels are connected by a tube.
Fluid fills both vessels and the tube. Forces with magnitudes F1 and F2
are applied over the two areas, in a direction that is perpendicular to the
corresponding surface.
By Pascal’s principle (and ignoring the effects of gravity), it must be thatF1A1
= F2A2
if the fluid is at rest. Therefore, if A2 is substantially larger than A1
(or vice versa), so will F2 be much higher than F1. Indeed, F2 = A2A1F1 � F1,
i.e. a small force F1 translates into a very strong one in surface 2. For
instance, by simply placing a light object on a platform covering the small
surface (e.g. a foot), we may be able to lift a great heavy object (e.g. a car)
placed on a platform over the larger surface.
Suppose that, by pressing on the small platform, we displace the fluid
down by a distance d1. The total volume of fluid displaced is d1A1. The
distance d2 that the fluid on the larger vessel is displaced up must satisfy:
d2A2 = d1A1
for all the fluid that is lost in vessel 1 is shifted into vessel 2 through
the connecting tube. Using Pascal’s principle, this means that d2F2 = d1F1.
6 The effects of gravity will be introduced shortly. In that case, we will see that thepressure depends on the region of the vessel we consider.
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Thus, the work on the small surface is W1 = F1d1 = F2d2 = W2, thereby
guaranteeing that total energy is conserved.
Incidentally, this illustrates an impracticality in this method of heavy
lifting: in order to lift a heavy object in surface 2 by a certain distance
d2, we would need to push platform 1 down by a much grater distance, as
d1 = A2A1d2 � d2. In practice, this is resolved by a lever that shifts fluid into
vessel 2 every time it is pushed down, but lets it back into vessel 1 when
brought back up.
Gravity plays an important role in pressure. For example, in the deep
oceans, water pressure is much higher due to the effects of gravity. To
understand this, consider taking a portion of fluid. For simplicity, suppose
we take a rectangular cuboid (i.e. a “box”) at some location (x, y) (see
Figure 5.1).
y + ∆yx+ ∆x
∆mg
F2
F1
Figure 5.1: A cuboid filled with liquid.
The box has height y + ∆y and the area of the upper (and lower) face
is A. The whole box has a mass ∆m, and the liquid has density ρ, which is
possibly a function of y. Since the volume of a cuboid is V = A∆y, then by
Definition 1.20:
∆m = ρV = ρA∆y
A force of magnitude ∆mg pulls down on the fluid due to gravity. Ev-
erywhere in the fluid, the box is subject to normal forces which are per-
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pendicular to each one of the six faces of the box. Figure 5.1 depicts only
two such forces, on the upper face, with magnitude F2, and on the lower
face, with magnitude F1, which act upon the whole surface area (omitting
the horizontal forces is without loss of generality because all forces on the x
plane clearly cancel).
For the fluid element to be in static equilibrium, in must the case that:
F1 − F2 −∆mg = 0
Moreover, F1 = PyA1, where Py is the pressure on the lower surface
(at position y), and similarly for the upper surface (where the position is
y+ ∆y, so pressure is denoted Py+∆y). Since both areas are equal, we have
(P1 − P2)A = ∆mg = ρA∆yg, orPy+∆y−Py
∆y = −ρg. Taking the limit as
∆y → 0, we then obtain:
dP
dy= −ρg (5.5)
a formula for the rate of change in pressure. Therefore, as we go deeper
into the ocean (lower values of y), pressure goes up by ρg per unit displace-
ment. This phenomenon is called hydrostatic pressure:
Definition 5.4 (Hydrostatic pressure) The pressure on static fluids that
is due to the action of gravity. The change in hydrostatic pressure from a
displacement in some direction y is given by equation (5.5).
Liquids are virtually incompressible. By this, we mean that their density
ρ is constant in y. By equation (5.5), this means that the change in the
pressure is everywhere the same, exactly what Pascal’s principle postulates.
Gases, on the other hand, are very much compressible, and their density
ρ can be increased easily by reducing the volume they occupy. For example,
the air’s density, and hence changes in atmospheric pressure, are very much
a function of the altitude, with air being denser at lower altitudes.7
7 We will study gases in more detail in Chapter 6.
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In sum, while the change in pressure in liquids is everywhere the same
(equation 5.5), the same is not true for gases: Pascal’s principle applies only
to incompressible fluids.
Example 5.5 (Shooting at cans) Imagine filling a can to its very rim
with water, covering it with a lid, and hitting it with a hammer. The counter
force will be extremely strong and the can will resist the hit. However, recall
that P = F/A and, by Pascal’s principle, pressure propagates undiminished
on the whole fluid. Thus, if we shoot a bullet (whose surface area is extremely
small) through the can, pressure inside the liquid will build up enormously,
the can’s materials might give way, and the can itself might explode. If
the can is filled with air instead of water, however, none of this will occur.
Again, this is because air, unlike water, is compressible.
Henceforth, we will assume that liquids are strictly incompressible (so
that ρ is constant in y). Integrating equation (5.5) within some bounds
[P1, P2], corresponding to the pressures at some locations y1 and y2, we
obtain: ∫ P2
P1
dP = −ρg∫ y2
y1
dy ⇒ P2 − P1 = −ρg(y2 − y1)
We have just derived the so-called Pascal’s Law:
Result 5.1 (Pascal’s Law) For static and incompressible fluids, the change
in hydrostatic pressure ∆P is proportional to the height of the fluid above
the point of measurement, ∆y. In particular:
∆P = ρg∆y (5.6)
where ρ is the fluid’s density.
Of course, this is nothing but the mathematical analogous to the state-
ment we have made in words in Principle 5.1: a change in the pressure at
any point in an enclosed, static, and incompressible fluid is transmitted to
all points in the fluid.
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Unlike liquids, the rate of change of pressure in gases is not constant. For
example, as we go up in the air into the thinner atmosphere, pressure goes
down, but not linearly with our displacement (as density ρy is a function
of altitude y). At sea level (y = 0), pressure on Earth is approximately
1.03 kg/cm2 (that is, it weighs roughly 10.1 N/cm2). This results in an
atmospheric pressure of about 101 kilopascals, kPa. This unit is generally
called an atmosphere. In particular:
1 atm ≡ 101, 325 Pa
The atmosphere pressure may also be called barometric pressure. What
this means is that the air that fills a column of cross-sectional area 1 cm2
running from sea level to the very top of Earth’s atmosphere has a mass of
about 1.03 kg. With these correspondences, we can see the equivalence be-
tween barometric and hydrostatic pressure. In particular, given the density
of water:
10 meters of water ≡ 1 atm
Since the change in barometric pressure depends on the density, which
is itself a function of the position, measuring it in practice can be challeng-
ing. One way to do it is to use a barometer, which exploits the following
experiment:
Example 5.6 (Measuring barometric pressure) Consider filling a glass
with liquid (e.g. mercury). In the glass, we introduce a tube with a single
opening, with the opening submerged in the liquid (e.g. a straw whose upper
opening is blocked by our finger). Let y1 denote the vertical height corre-
sponding to the liquid’s surface on the glass, and y2 be the height the liquid
reaches inside the tube or straw, with h ≡ y2 − y1. Let P1 and P2 be the
corresponding pressures. Since the liquid inside the tube is not subject to air
pressure, we have P2 = 0.
Then, we can simply apply Pascal’s Law (equation 5.6) to find:
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P1 = ρgh
where ρ is the density of the liquid. Therefore, in practice, to know the
barometric pressure, all we have to do is take a liquid of a certain density
ρ, measure how far we can pull the liquid up in a tube before it breaks lose
due to gravity, and then use the measured distance h to compute barometric
pressure P by P = ρgh.8 This is exactly how barometers work.
Here are a couple of real-world examples in which pressure plays a role:
Submarines Submarines today can submerge up to 900 meters, where
the hydrostatic pressure is 90 atm. On every square meter of a submarine
at that depth there is a force equivalent to 900 tons of weight, and that force
is perpendicular to the surface of the submarine and distributed across the
entire outside of the vessel. Yet, submarines are able to maintain a steady
1 atm in their interior. If some air was to be sucked out of the interior of
the submarine, its shell would crush due to the enormous forces of water
pressure.
Divers Humans cannot breathe through a tube below water (i.e. snorkel)
below a depth of about 1 meter. To measure how deep we can snorkel, we
use a so-called manometer (see Figure 5.2). This device is used to measure
how much over-pressure human lungs can produce to counteract the water
pressure from outside.
In a manometer, air is blow on one end of the tube (the left end, in the
figure), which displaces the liquid (say, water) by a distance h. Letting the
pressure be P1 at y1 and P2 at y2, Pascal’s Law says P1 − P2 = ρhg, where
8 For example, using mercury (ρ = 13.6×103 kg/m3), it has been found that mercurygives way to gravity at typical heights of about h ≈ 0.76 m, so the barometric pressureis P = ρgh ≈ 1.03 × 105 Pa, i.e. exactly 1 atm. Water is much less denser thanmercury, so the tube would have to be much higher (about 13.6 times higher, makingh ≈ 10 meters) if we are to measure the barometric pressure. This is why mercury isoften the most practical choice for use in barometers and thermometers.
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h
y1
y2
Figure 5.2: A manometer filled with water. Blowingair into the left end generates a displacement equal toh on the right end.
ρ is the density of water. Since P2 = 1 atm (as the right end is open ended
and thus subject to air pressure9), and 1 atm = 101,325 Pa, we then have:
P1 = 101, 325 + ρgh
Thus, the manometer indicates how much over-pressure ρgh the human
lungs can generate over and above 1 atm. For human lungs, the resulting
distance of this experiment is about h = 1 meter, which means humans
cannot snorkel at more than 1 meter under water: human lungs can only
generate up to one-tenth of an atmosphere of over-pressure. At lower depths,
we must rely on pressured oxygen, which they breath through a can, to
restore pressure within their chest. Incidentally, human lungs can create
about the same amount of under-pressure (i.e. sucking instead of blowing
air in the manometer).
9 Because both ends are open, the liquid is subject to barometric pressure before theexperiment starts, so the water levels are equalized (i.e. h = 0).
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5.2.2 Archimedes’ Principle and Stability
Consider an object (e.g. a cylinder) that is partially submerged in a
fluid, for instance a liquid (Figure 5.3). The object has mass m, density ρ,
radius r, length `, and cross-sectional area A.10 The fluid has density ρf ,
and the object is submerged at depth h ≡ y2−y1, where y1 is the bottom of
the object and y2 is surface of the liquid. The pressure levels at these two
points are P1 and P2, respectively.
h
ry2
y1
mg
F1
F2
Figure 5.3: A partially submerged cylinder.
As usual, a force of magnitude mg pulls down on the object. Counter-
acting this force, a force of magnitude F1 pushes up on the object. Finally, a
force of magnitude F2, due to atmospheric (or barometric) pressure, pushes
down on the object. As discussed in the previous section, the direction of
these forces are all orthogonal to the surface of the cylinder (or else the
object, fluid, and air would not be in a static equilibrium).
From Pascal’s Law, we have P1 − P2 = ρfgh. For the object to be in a
static equilibrium, it must be that F1 − F2 −mg = 0, that is:
10 Recall that the volume of a cylinder is V = πr2`, while density is defined ρ = m/V .Therefore, m = πρr2`.
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Fb = mg
where Fb ≡ F1 − F2 is called the buoyant force.
Definition 5.5 (Buoyant force) The upward force Fb that is exerted by
a fluid and opposes the weight of a submerged object.11
Next, by definition of pressure (Definition 5.3), F = AP , so Fb = F1 −F2 = A(P1 − P2) = Aρfgh. Thus:
Fb = Aρfgh
Importantly, note that Ah is the volume of the displaced fluid, so ρfAh
is the mass of the displaced fluid (because density is the ratio of mass to
volume). Multiply mass by g and we obtain weight.12 Thus, Aρfgh is the
weight of the displaced fluid.
We have just derived the physical law of buoyancy, popularly known as
Archimedes’ Principle after its discoverer, Archimedes of Syracuse (287 BC
– 212 BC):
Principle 5.2 (Archimedes’ Principle) The buoyant force on a (totally
or partially) submerged body has the same magnitude as the weight of the
fluid that the body displaces.
For instance, take an object of weight W1 ≡ V ρg, where V is the volume
and ρ is the density (so V ρ is the mass) of the object. If we immerse
the object in water, the weight becomes Wimmersed ≡ V ρg − Fb. Since,
by Archimedes’ Principle, the buoyant force Fb equals the weight of the
displaced fluid, then Fb = V ρwaterg, so the weight loss is:
Wloss ≡W1 −Wimmersed = V ρwaterg
11 If the object is partially submerged, the buoyant force is net of the barometricforce. If it is totally submerged, air pressure is obviously not acting upon the object.12 Recall that a body’s weight is its mass times the acceleration on the body (Defi-nitions 1.14-1.15). Since the body is in a stable equilibrium, the acceleration is onlydue to gravity.
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In other words, W1Wloss
= ρρwater
. Knowing the density of water (ρwater =
0.9998 g/cm3 at 0 ◦C) and computing weight losses for the object in question
(e.g. by placing a fully submerged scale at the bottom of the container) thus
allows us to recover the original object’s density.13
Example 5.7 (Icebergs) Consider a block of ice floating on water (e.g. an
iceberg). The density of ice is ρice = 0.92 g/cm3 at 0 ◦C, slightly lower than
water. Since the block is floating, mg = Fb, where m = V ρice and V is the
total volume of the block of ice. By Archimedes’ Principle, Fb = Vuwρwaterg,
where Vuw is the volume of the portion of the block of ice that is underwater.
In sum, mg = Fb implies:
VuwV
=ρice
ρwater= 0.92
That is, 92% of the block of ice is underwater. Hence, when we see an
iceberg floating in the ocean, we can be certain that we are only seeing 8%
of it.
Archimedes’ Principle can also be used to derive conditions for floating.
Return to the cylinder example (Figure 5.3). For the object to float, we
have established that the buoyant force must overcome gravity, so Fb = mg.
The buoyant force is, by Archimedes’ Principle, Fb = Ahρfg, where ρf is
the density of the fluid. Therefore, if the object itself has density ρ, we have
Ahρfg = A`ρg or, simplifying:
`
h=ρfρ
Now, if the object floats, then ` ≥ h (recall Figure 5.3). Therefore, a
necessary condition for floating is:
ρf > ρ
13 Legend has it that Archimedes’ himself used this approach when commanded toexamine whether certain objects were made of gold, knowing that the density of goldis ρgold = 19.32 g/c3.
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In words, for an object to float in a fluid, it is necessary that the object
itself is less dense than the fluid. Otherwise, the object sinks. Remarkably,
this condition is completely independent of the dimensions of the object. The
only thing that matters is the object’s density. For example, a gold pebble
will always sink in water no matter how small, and a piece of wood will
always float in water no matter how large, because ρgold > ρwater > ρwood.
5.2.3 Stability
Archimedes’ Principle is also crucial for understanding the conditions
under which floating objects (e.g. ships) will remain stable.
Consider an object floating in liquid and let C denote its center of mass
(see Figure 5.4). The center of mass is subject to a gravitational acceleration
mg. If the object’s distribution of mass is such that the center of mass does
not coincide with the point where the buoyant force acts (point B in the
figure), there will be a torque (relative to any reference point on the object),
and the object will rotate (clockwise, in our example).
B
C
mg
FbWater line
Figure 5.4: A floating object with center of mass C.
As we learned in Example 4.3, the object will be stable when B and C
are directly aligned on a vertical line. If, furthermore, the center of mass
is below the center of the buoyant force (as in the figure), the torque is
restoring, so any tilting of the object (due to an external torque) will bring
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it back to equilibrium (through the internal restoring torque). The lower
is the center of mass relative to the object’s point of buoyancy, the more
stability the object will possess. This is why ships are designed to have very
low centers of mass, as close as possible to their keel.
Example 5.8 (Balloons) Similar intuitions can be applied to gaseous flu-
ids. Consider a balloon that is filled with gas. The ballon has mass m =
mgas + mrest, which includes the mass of the gas inside the balloon, mgas,
and the mass of the rest of materials, mrest (including rubber and string).
Let ρgas be the density of the gas inside, and ρair be the air’s density outside.
For the balloon to rise in the air, it must be that Fb > mg. Here,
the buoyant force Fb is, by Archimedes’ Principle, the weight of the fluid
(air) that is displaced by the ballon, namely V ρairg, where V is the balloon’s
volume. Thus, a necessary condition for rise is Fb = V ρairg > mg. Using
mgas = V ρgas, the necessary condition reads:
ρair > ρgas +mrest
V
Thus, the only way in which the balloon can rise is that the gas inside
has sufficiently low density.
For example, if we fill the balloon with air (so ρgas = ρair), the balloon
will never rise, as the weight of its materials (rubber and string) always
bring it down. But if we fill it up with helium, which is substantially less
dense than air, the balloon will rise provided (i) its materials are not too
heavy; (ii) the balloon is sufficiently large in volume.
Example 5.9 (Objects in a box) Consider a sealed compartment in outer
space in static equilibrium, i.e. with no external forces or torques (see Figure
5.5). Inside the compartment there is an apple (object A) and a helium-filled
ballon (object B). Suppose the box is accelerated with magnitude a in some
direction (in this case, horizontally). The acceleration generates a percep-
tion of gravity for everything that is inside the box, of the same magnitude
but opposite direction as the acceleration, denoted by gp.
Then:
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B
A
a
gp
x1 x2
B
A
a
gp
x1 x2
Figure 5.5: An apple (A) and a ballon (B) inside asealed box in outer space. Left: Vacuum. Right: Boxfilled with air.
• If there is vacuum in the box (Figure 5.5, left panel), then both apple
and balloon would move in the direction in which gravity is perceived,
i.e. opposite to the acceleration.
• If the box is filled with air (Figure 5.5, right panel), accelerating the
box generates the perception of gravity as well. Crucially, the air inside
the box is being accelerated, just as the apple and the balloon are. The
air pushing on the box creates a pressure differential between points x1
and x2. Just as air pressure is higher on the Earth’s surface because
of gravity, pressure will be greater on the left wall of the box than on
the right, or P1 > P2 (where P1 is the pressure on any point of the
box’s surface at position y1, and similarly for P2). This means that
the ballon will “rise” relative to the perceived gravity (i.e. move in the
direction of a) whereas the apple will “fall” relative to the perceived
gravity (i.e. move leftward).
Now suppose that the same box is placed on the Earth’s surface, where
there is an actual gravitational force g. To overcome gravity, we suspend
the apple from the ceiling of the box with a string, and attach the helium
balloon to another string which is attached to the floor of the box (Figure
5.6). Again, the box is accelerated in the horizontal direction, generating a
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perception of gravity for everything inside the box.
• If there is vacuum in the box (Figure 5.6, left panel), both the apple and
the balloon will arc to the left to some position A′ and B′, respectively,
because of the direction of the perceived gravity gp.
• If the box is filled with air (Figure 5.6, right panel), accelerating the
box horizontally generates the perception of gravity gp in the horizontal
direction. The apple will arc again toward the left to some position A′.
However, now the acceleration generates a pressure differential between
x1 and x2. In particular, P1 > P2 and, because of the low density of
helium, the balloon will go forward! The latter is counterintuitive, but
completely consistent with Archimedes’ Principle.
B
A
a
gp
g
x1 x2
A′
B′
B
A
a
gp
g
x1 x2
A′
B′
Figure 5.6: An apple and a ballon inside a sealed boxon Earth. The apple (object A) hangs from a string;the ballon (object B) floats on a string. Left: Vacuum.Right: Box filled with air.
5.3 Fluid Dynamics
In the previous section, we have analyzed the behavior of objects and
fluids when the latter are at rest. Now, we study situations in which fluids
are in motion.
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Consider an incompressible fluid (e.g. a liquid) running inside a vessel
(e.g. a pipe), as in Figure 5.7. Let A2 denote the cross-sectional area of the
liquid on one end of the pipe, and P2 be the pressure across that surface.
The liquid is moving at speed v2. On the other end of the pipe, a given cross
section of area A1 experiences pressure P1 and velocity v1.
A1
A2
v1dt
y1
P1
v2dt
y2
P2
Figure 5.7: Liquid running through a pipe.
If the fluid were completely static (i.e. v1 = v2 = 0), then by Pascal’s
Principle we would have P1 − P2 = ρgh > 0, where h ≡ y2 − y1 > 0. Thus,
if the liquid was static, the pressure at y1 would be higher than at y2. Since
ρ = m/V , then we can write:
P1 − P2 =mgh
V
We identify in mgh the gravitational potential energy of the system.
Thus, P1 − P2 is expressed in energy per unit volume.
Let’s now set the liquid in motion. In this process, three actors are at
play: the kinetic energy of motion (per unit volume), the gravitational po-
tential energy (per unit volume), and pressure. What is key is that the total
energy is conserved, and therefore the sum of the three must be constant.
Intuitively, as we move the liquid from one place to another, we trade-off
speed for either height h or pressure P .
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At any location y:
• Kinetic energy is T = 12mv
2 or, per unit volume, 12ρv
2.
• Gravitational potential energy is Vg = mgy or, per unit volume, ρgy.
Therefore, the conservation of total energy requires that:
1
2ρv2 + ρgy + Py = constant (5.7)
where Py is the pressure at point y. This is called Bernoulli’s equation,
after Daniel Bernoulli (1700 – 1782). In words:
Principle 5.3 (Bernoulli’s Principle) An increase in the speed of a fluid
implies a decrease in pressure or a decrease in the fluid’s potential energy,
as per equation (5.7).
Indeed, from equation (5.7) it is clear that an increase in speed must
come at the expense of a lower potential energy or a lower pressure (or both,
to some extent). Else, total energy would not be conserved. In particular,
specializing equation (5.7) to our two cross sections in Figure 5.7:
1
2ρ(v2
2 − v21) + ρgh+ ∆P = 0 (5.8)
where ∆P ≡ P2−P1 denotes the pressure change. Let’s apply Bernoulli’s
Principle to a few special cases of equation (5.8):
• First, consider the special case when h = 0, so that the liquid flows
through a pipe along a straight line at height y (Figure 5.8). Since the
liquid is incompressible, the same amount of matter must flow through
disk 1 in a certain unit of time as through disk 2, so A1v1 = A2v2.
Therefore, in this example, since A1 < A2, then v1 > v2.
By Bernoulli’s Principle, i.e. equation (5.8) with h = 0, we have:
1
2ρv2
1 + P1 =1
2ρv2
2 + P2
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yA1 A2
P1 P2v1 v2
Figure 5.8: Special case 1: h = 0.
and since v1 > v2, then it must be that P1 < P2. Thus, surprisingly, in
the region of the pipe where the speed is fastest is where the liquid’s
pressure is lowest.
• Next, consider Figure 5.9. A tube is introduced into a container filled
with liquid. This device is called a siphon.
Take the open end and suck in the air so the tube becomes completely
filled with the liquid (as reflected in the figure by the shaded areas).
The liquid exits the tube at velocity v2, and it empties the container
at velocity v2. Because the cross-sectional area of the tube is a lot
smaller than the surface area in the container, v2 ≈ 0, i.e. the water
line descends very slowly.14
Because both ends of the tube are open, the pressure at both ends
is the same and equal to P1 = P2 = 1 atm (as there is barometric
pressure). Therefore, here we have a special case in which h > 0 but
∆P ≡ P2 − P1 = 0.
Using equation (5.8) for this special case thus gives 12ρv
21+ρgy1 = ρgy2,
that is 12v
21 = gh. Thus:
v1 =√
2gh
14 Again, here we use that A1v1 = A2v2 because the liquid is incompressible. In otherwords, A1
A2= v2
v1, but A1
A2≈ 0 because the tube’s opening is very small relative to the
surface area in the container.
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h
y2
y1
d
y3
v1 =√
2gh
v2 ≈ 0
Figure 5.9: Special case 2: ∆P = 0.
is the speed at which the liquid is running out. We have seen this
formula before: this is exactly the same speed that would be reached
at y1 by an object in free fall from y2 (recall e.g. Example 2.16).15
In the first special case, the change in the liquid’s speed is only at the
expense of a pressure change. In the second special case, instead, since the
pressure does not change due to the vast differences in the cross-sectional
areas, all the speed is coming from gravitational potential energy being
converted into the kinetic energy involved in the liquid’s motion.
15 This provides an ingenious method for stealing gasoline from a car: introduce atube in the car’s gas tank, suck the air out of the tube, and as long as the dry end ofthe tube is below the tank (so that h > 0), the gasoline will automatically flow withno further effort from our part.
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5.4 Oscillations
Oscillatory motion has appeared in many examples in the previous chap-
ters. In this section, we first review simple harmonic oscillators, and then
explore some other topics on oscillations.
5.4.1 Simple Harmonic Oscillators
In very general terms, consider an object of mass m with center of mass
C that rotates about some point P (e.g. Figure 4.2). The torque about
point P is ~τP = ~rP × ~F , and its magnitude is:
τP = bmg sin θ = −IP θ
where ~F is the gravitational force vector, θ ≡ ∠~rP ~F , and b ≡ |~rP | is the
distance between P and C. The second equality is because the motion is
rotational (recall Result 3.6), where IP is the moment of inertia about point
P .
In a stable equilibrium, τP = 0. Using the Small Angle approximation,
sin θ ≈ θ around θ = 0, so stability implies:
θ +
(bmg
IP
)θ = 0
This is a simple harmonic oscillator in θ. Therefore, the solution is:
θ = θmax cos(ωt+ ϕ)
where θmax is the amplitude, ω is the angular frequency (a constant),16
and ϕ is the phase angle. As we have seen before (e.g. Example 1.9), the
16 Again, the angular frequency ω should not be confused with the angular velocity,ω ≡ θ.
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solution to this DE gives the angular frequency and period of motion:
ω =
√bmg
IPradians/sec (5.9a)
T =2π
ωsec (5.9b)
In the previous chapters, we have seen that many objects behave in this
manner, e.g. pendulums, springs, or disks. For each case, the moment of
inertia IP is different. Let’s examine each one in turn:
• Rod The moment of inertia of a rod of length L is:
IP =1
12mL2 +mb2
where the first term is the moment of inertia about the center of mass,
and the second term follows from the Parallel Axis Theorem (Result
3.1). If the mass of the rod is evenly distributed, then C is located
in the middle of the rod, so b = 12L. Therefore, by equations (5.9a)-
(5.9b), the angular frequency and period of the rod are:
ω =
√3g
2Land T = 2π
√2L
3g
• Pendulum The moment of inertia of a pendulum is:
IP = m`2 +mb2
by the Parallel Axis Theorem. A pendulum is a bob attached to a
massless string of length `. Since the string is massless, the center of
mass is on the rod, so ` = b, and thus IP = 2m`2. Therefore, the
angular frequency and period of the pendulum are:
ω =
√g
`and T = 2π
√`
g
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in agreement with our results from Example 1.10.
• Ring The moment of inertia of a ring (i.e. a hula-hoop) of radius r is:
IP = mr2 +mb2
by the Parallel Axis Theorem. A hula-hoop has center of mass at the
center of the circle, so b = r, and thus IP = 2mr2. Therefore, the
angular frequency and period of the pendulum are:
ω =
√g
2rand T = 2π
√2r
g
exactly as we found in Example 3.8.
• Disk The moment of inertia of a disk of radius R is:
IP =1
2mR2 +mb2
by the Parallel Axis Theorem. A disk has center of mass at the center
of the circle, so b = R, and thus IP = 32mR
2. Therefore, the angular
frequency and period of the pendulum are:
ω =
√2g
3Rand T = 2π
√3R
2g
Comparing our results across the different objects, we observe that a rod
(of length L), a pendulum (of length `), a ring (of radius r), and a disk (of
radius R), will all have the same period of motion if 23L = ` = 2r = 3
2R.
Simple harmonic oscillations are not limited to these four objects, how-
ever. Let us see one such example:
Example 5.10 (Oscillating liquid) Consider the manometer (Figure 5.2).
If air is blown into one end of the tube and then released, the liquid will os-
cillate.
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Let the liquid have mass m and density ρ. The cross-sectional area of
the tube is A. The length of the liquid (i.e. of the segment of the tube that is
filled with liquid) is L. By definition of density m = V ρ, where V = AL is
the volume of the liquid. Let y be the total displacement of the liquid (with
y = 0 when the liquid is at rest17).
Since the cross-sectional area of the tube is everywhere the same, the
velocity of the liquid (when released) is the same everywhere along the tube
at any given time, equal to v = y. Assuming, for simplicity, that the liquid’s
motion inside the tube generates no energy loss18, we can now invoke the
Conservation of Mechanical Energy. First, the kinetic energy of the system
is
T =1
2m(y)2
where recall m = ALρ. For the gravitational potential energy, first normalize
V to V = 0 when y = 0 (i.e. when the liquid is at its stable equilibrium).
When the liquid has been displaced by a distance y, the accumulated mass
is ∆m = ρ∆V , where the increase in volume is ∆V = Ay. Thus, the
gravitational potential energy is V = (∆m)gy = Aρgy2. Thus, the total
energy is:
E = T + V =1
2ALρ(y)2 +Aρgy2
Thus, the change in total energy is E = ALρyy + 2Aρgyy. By the
conservation of total energy, E = 0, and thus:
y +
(2g
L
)y = 0
a simple harmonic oscillator. Therefore, as we know, the solution will
be y = ymax cos(ωt + ϕ), where ω =√
2gL is the angular frequency, ymax is
the amplitude, and ϕ the phase angle. The period of oscillation is T = 2π/
ω = 2π√
L2g .
17 That is, in the notation of Figure 5.2, y1 ≡ 0.18 In reality, some energy loss might occur from the friction between the liquid andthe inner walls of the tube, transforming some (though a small amount of) potentialenergy into heat.
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Compared to our examples before, therefore, an oscillating liquid has the
same period as a pendulum of length ` if L = 2`, that is, if the length of the
liquid in the tube is twice the length of the pendulum.
Example 5.11 (Torsional Pendulum) The torsional pendulum (Figure
5.10) is another example of a simple harmonic oscillator. A torsional pen-
dulum is a rod (or disk) hanging from a wire or rope and placed horizontally.
The disk is then offset over some angle, and it oscillates back and forth.
P
Figure 5.10: The torsional pendulum.
Let P be the center of the disk. The torque relative to P is restoring
and obeys a rotational version of Hooke’s law: just like the force in a simple
spring is proportional to the spring’s linear displacement, the rotational force
in the torsional pendulum is proportional to the pendulum’s angular position.
In magnitude:
τP = −κθ
where θ is the angle, and κ is the torsional spring constant. Because the
motion is rotational, τP = IPα, where α ≡ θ is the angular acceleration.
Thus, in the stable equilibrium (τP = 0):
θ +
(κ
IP
)θ = 0
which is, again, a simple harmonic oscillator. The solution is θ =
θmax cos(ωt + ϕ), with angular frequency ω =√
κIP
and period of rotation
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T = 2π√
IPκ .19
5.4.2 Forced Oscillations: Resonance and Damping
So far, we have analyzed situations in which objects are allowed to os-
cillate freely with a certain frequency after an external force or a torque
displaces them from their stable equilibrium. In this section, we examine
situations in which certain frequencies are forced upon the system.
Consider an otherwise simple harmonic oscillator (e.g. a spring, Figure
5.11). By Hooke’s Law, there is a restoring internal force −kx at each
position x. Now, consider adding an external force F (t), which we may call
a driving force. For simplicity, we add this force in a sinusoidal fashion (that
is, the added force is itself oscillating), so that:
F (t) ≡ F0 cos(ωt)
Here, the forced amplitude F0 and angular frequency ω are our choice.
Newton’s Second Law now says that ma = −kx + F0 cos(ωt). Using
a = x, we have:
x+
(k
m
)x =
F0
mcos(ωt) (5.10)
Since the right-hand side is non-zero, this DE is no longer a simple
harmonic oscillator. If it were, we know the process would be described
by x = xmax cos(ω0t + ϕ), with angular frequency ω0 =√
km , which in
this context we may call the natural frequency. Under the driving force
F , however, we have a second-order nonhomogenous ODE with constant
coefficients, so we must use Method IV (Remark 0.5) to find its general
19 The constant κ will depend on the cross-sectional area A and the length ` of the wireor rope holding the disk. Recall that when objects are composed of linear-elasticitymaterials and obey linear oscillations, equation (4.11) says that the oscillation con-stant is given by k = Y A
`, where Y is Young’s modulus (whose value depends on the
exact material that the wire or rope is made of). Intuitively, thicker (higher A) andshorter (lower `) ropes would oscillate less when the object is stretched. Here, therope is not being stretched but twisted, but the same intuition applies: κ is increasingin A and decreasing in ` (indeed it is easier to twist longer and thinner ropes).
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x > 0x = 0
−kx F0 cos(ωt)
Figure 5.11: Spring motion as a harmonic oscilla-tor. Above: Spring in a relaxed position (x = 0); Be-low : Stretched spring, in position x, with counteract-ing force F0 cos(ωt).
solution.
Here, we will just find a particular integral of x. Since the non-homogenous
part is F0m cos(ωt), we propose a guess with a similar functional form:
x = A cos(ωt)
where A is the amplitude of the oscillation. Then, xp = −Aω sin(ωt),
and xp = −Aω2 cos(ωt). Substituting into (5.10), we get:
−Aω2 cos(ωt) + ω20A cos(ωt) =
F0
mcos(ωt)
where we have used ω20 ≡ k
m (and ω0 denotes the angular frequency
of the process when it oscillates freely). Matching coefficients, we get the
amplitude:
A =F0
m(ω20 − ω2)
(5.11)
This means that the process:
x =F0
m(ω20 − ω2)
cos(ωt) (5.12)
is a particular integral of DE (5.10). Under certain initial conditions, the
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system will oscillate with an amplitude A (given by equation (5.11)), and
angular frequency ω. Under general initial conditions, the system might go
through a period in which amplitude and frequency are different than A and
ω, but eventually these oscillations will die out in favor of the ones that are
forced upon the system, and the process will follow the law of motion given
in equation (5.12).
• The initial phase, if it exists, is called the transient response to F (t).
• The phase in which the system has converged to the forced oscillation
is called the steady-state.
Equation (5.12) says that the object oscillates at the same frequency as
the driving force F (t), but with an amplitude A which depends on (i) the
frequency ω of the driving force, and (ii) the frequency ω0 of the natural
motion of the oscillator. Figure 5.12 plots the amplitude A as a function of
ω.20
First, note that:
A
> 0 if ω < ω0
< 0 if ω > ω0
In words, if the forced frequency is lower than the natural frequency,
then the amplitude is positive and the motion is sinusoidal. But if the forced
frequency is higher than the natural frequency, the amplitude becomes neg-
ative, which means that the process is still sinusoidal but becomes out of
phase by 180◦ (recall Figure 1.5). Thus, as frequencies increase above ω0,
there is a phase shift of 180◦.
There are three limiting cases of interest:
• If ω → 0, then A→ F0k .
20 The angular frequency ω is in radians per second, but it can be converted intoHertz via the identity ω = 2πf , where the frequency f is in Hz, as f = 1/T (whereT is the period of oscillation, in seconds).
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ω0 Imposed frequency (ω)
F0
k
Amplitude
Figure 5.12: The amplitude A = F0
m(ω20−ω2)
as a
function of the imposed (or driving) frequency, ω, inradians per second. The natural frequency is ω0. Note:To convert to Hz, one can use that ω = 2πf , where fis frequency in Hz.
That is, if the system is driven to very low frequencies (compared to its
natural frequency ω0), then by equation (5.11) the amplitude of motion
is proportional to the amplitude of the driving force by a factor of 1k .
• If ω → +∞, then A→ 0.
That is, if the system is driven to very high frequencies, then by equa-
tion (5.11) the amplitude decays to zero.
• If ω → ω0, then A→ +∞.
That is, if the system’s frequency is forced to being the natural one
(i.e. if the two are synchronized), the amplitude explodes to infinity,
and we would see an enormous displacement.
The latter case is pathological: when the system is forced to oscillate at
the same frequency as its natural one, but is driven by an external force with
a non-zero amplitude, the system’s amplitude increases to infinity. In Figure
5.12, this appears as an asymptote. This phenomenon is called resonance,
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and the frequency at which it occurs (here the natural frequency ω0) is called
the resonant frequency.
Definition 5.6 (Resonance) A phenomenon in which an external force
drives a system to oscillate with greater amplitude at specific frequencies.
Definition 5.7 (Resonant frequency) Frequency (or frequencies) at which
the response amplitude is a relative maximum. Also called normal mode or
natural frequency.
In practice, the frequency will obviously be very high at frequencies close
or at the resonant frequency, but certainly it will not be infinite. The reason
is that there is always a frictional force, or damping, which will limit the
oscillations to finite amplitudes.
Definition 5.8 (Damping) A frictional force within or upon an oscilla-
tory system that reduces, restricts, or prevents its oscillations.
Thus, in fact, amplitudes look more like Figure 5.13 than Figure 5.12.
Figure 5.13 shows examples for the amplitude (in absolute value) as a func-
tion of the driving frequency, for more or less levels of damping. This graph
is sometimes called the resonance curve.
Definition 5.9 (Resonance curve) The graph of the (absolute value of)
amplitude on the driving frequency, when friction (i.e. damping) is at play.
In principle, the frictional term of damping might depend on the oscil-
lator’s displacement in a complex way. In many cases, however, damping
is often a (near) proportional function of the object’s speed. In this case,
damping is defined as a frictional force
Fdamping = −bx
for some factor of proportionality b, where the minus sign indicates that it
is always a restoring force.
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Figure 5.13: The resonance curve: amplitude A(in absolute value) as a function of the driving fre-quency, in practice, with more or less friction (damp-ing). For frequencies higher than the resonant fre-quency (marked f0 Hz in the plot), there is a phaseshift of 180◦.
If this is the case, then Newton’s Second Law for the oscillator reads
ma = −kx− bx, and thus:
x+ γx+ ω20x =
F0
mcos(ωt)
where we have defined γ ≡ bm and, once again, ω0 ≡
√km is the reso-
nant frequency, and the driving force is assumed to be sinusoidal: F (t) =
F0 cos(ωt). Here, γ therefore measures the strength of damping.
We have now a nonhomogenous second-order ODE with constant coeffi-
cients. We conjecture a solution of the form:
x = Aeiωt
where i =√−1. By Euler’s formula (Result 0.1), stating that eiωt =
cos(ωt) + i sin(ωt), this conjectured solution is sinusoidal, with a real part
cos(ωt) and, potentially, an imaginary part.
For this conjecture, x = Aiωeiωt and x = A(iω)2eiωt. Moreover, by
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Euler’s formula, we can write the driving-force component F0m cos(ωt) as
F0m e
iωt. Substituting these into the ODE, we get:
A(iω)2eiωt + γAiωeiωt + ω20Ae
iωt =F0
meiωt
Note that (iω)2 = −ω2. Since eiωt 6= 0, we can divide through by eiωt to
obtain:
A(− ω2 + γiω + ω2
0
)=F0
m(5.13)
If there was no driving force (F0 = 0), equation (5.13) would imply
ω2 − iγω − ω20 = 0, the characteristic equation of the system. In that case,
the roots are:
ω1 =1
2
(√4ω2
0 − γ2 + iγ
)and ω2 =
1
2
(−√
4ω20 − γ2 + iγ
)(5.14)
Thus, the general solution with no forced oscillations is x = A1eiω1t +
A2eiω2t, for some arbitrary amplitudes A1 and A2.
With a driving force (F0 > 0), equation (5.13) implies a driving ampli-
tude of:
A =F0
m(ω20 − ω2 + iγω)
and therefore the displacement is:
x =F0
m(ω20 − ω2 + iγω)
cos(ωt)
The resulting amplitude is similar to the one we obtained under no damp-
ing (equation (5.11)), except for the term iγω. The resonant frequency is
ω = ω0, as before. However, unlike before, for this frequency the amplitude
does not shoot to infinity.
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To see this last point formally, call:
R ≡ 1
m(ω20 − ω2 + iγω)
so that x = RF0 cos(ωt). Recall from Euler’s formula (Result 0.1) that any
complex number can be written in an exponential form. Here, we can write
R = p + qi for some real numbers (p, q) and, by Euler’s formula (equation
(0.7)):
R = ρeiθ
where ρ ≡√p2 + q2 and some angle θ. Therefore, x = ρF0e
iθ cos(ωt) or,
using again Euler’s formula:
x = ρF0 cos(ωt+ θ)
This way of writing the displacement shows that:
• The oscillation is not in phase with the driving force (whose frequency
is ω), but it is shifted by some extra amount, θ. The θ angle solves
1/R = (1/ρ)e−iθ, where 1/R = m(ω20 − ω2 + iγω), and therefore:
tan θ = −γ(
ω
ω20 − ω2
)< 0
where we have used that tan(−θ) = − tan θ. Notice we get θ < 0,
meaning that the oscillator “lags behind” the driving force for any
value of the frequency of the driving force, ω.
• The amplitude of the oscillator is proportional, but not equal, to that
of the driving force (which is F0), with some factor of proportionality
ρ between the two.
So what happens as we approach the resonant frequency? For this,
consider computing R2 (so as to have everywhere positive values for the
amplitude):21
21 To compute R2, we use the rule that the square of a complex number equals theproduct of the number with its complex conjugate (equation (0.5)).
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R2 =1
m2(ω20 − ω2 + iγω)(ω2
0 − ω2 − iγω)
=1
m2[(ω20 − ω2)2 + γ2ω2]
(5.15)
Therefore, as ω → ω0, we have R2 → 1m2γ2ω2 , or:
R→ 1
mγω
Thus, for as long as there is damping (γ 6= 0), the amplitude does not
shoot to infinity as we approach the resonant frequency ω0. Moreover, as
ω → ω0, we have:
tan θ = −γ(
ω
ω20 − ω2
)→ −∞
for any value of γ > 0. Since tan θ = sin θcos θ = −∞, then it must be that
sin θ = −1 and cos θ = 0, that is, θ = −π2 (i.e. 90◦). In words, as we
approach the resonant frequency, the oscillator approaches a state of out of
phase of 90◦ relative to the phase of the driving force (see Figure ??).
Again, γ controls the strength of damping: when frictional forces are
stronger (higher γ), the maximum amplitude R is smaller. Conversely, the
resonance is sharper as damping is made smaller. In a plot of equation 5.15,
i.e. of ρ2 (the resulting amplitude) on ω (the driving frequency), we see that
γ controls the width of resonant frequencies, with lower γ meaning more
width.
Another measure of width that is used in practice is:
Q ≡ ω0
γ
so that narrower resonances correspond to higher values for Q.
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ω0
Driving frequency (ω)
-180◦
-90◦
0◦
θ
Figure 5.14: The phase angle θ with damping, as afunction of the driving frequency, ω.
5.4.3 Coupled Oscillators
Systems need not have only one resonance frequency, as we have assumed
so far. More complex systems may exhibit multiple resonance frequencies.
Systems in which different oscillating objects are linked are called coupled
oscillators. For instance, a double spring (i.e. a spring attached at the end
of another spring) is a system of two coupled oscillators which has two reso-
nance frequencies (see Example 5.13 below). In this case, the amplitude (in
absolute value) seen in Figure 5.13 would exhibit two peaks. More generally,
a system with n coupled oscillators will have n resonant frequencies, so the
amplitude curve will exhibit n peaks (see Example 5.14 below).
Let us examine these examples in detail.
Example 5.12 (Two coupled pendulums) Consider two pendulums of
equal length ` and equal mass m, attached to each other with a spring (Figure
5.15). Let x1 and x2 denote their respective displacements.
If there was no spring, each pendulum would oscillate independently as a
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Classical Mechanics Pau Roldan-Blanco
x1 x2
Figure 5.15: Two coupled pendulums.
simple harmonic oscillator, with natural frequency ω0 =√
g` (recall Example
1.10), obeying:
xi + ω20xi = 0
for i = 1, 2. Consider adding back the spring. The spring force is, by
Hooke’s law, proportional to the distance between the two pendulums. Letting
the spring constant be k, Newton’s law then says ma1 = −mω20x1−k(x1−x2)
for pendulum 1, and ma2 = −mω20x2 − k(x2 − x1) for pendulum 2. Thus:
mx1 +mω20x1 − k(x1 − x2) = 0
mx2 +mω20x2 − k(x2 − x1) = 0
a system of two ODEs. This system will have two solutions for the
common frequency of the pendulums, let’s call them (ωs, ωf ) for slow and
fast (i.e. ωs < ωf ). To find them:22
22 Alternative method: Conjecture solutions of the form x1 = Aeiωt and x2 =Beiωt, i.e. the pendulums will (eventually) oscillate at the same frequency ω, butwith potentially different amplitudes (A,B). Plugging into the ODEs, we obtain(ω2 − ω2
0 − km
)A = − k
mB and
(ω2 − ω2
0 − km
)B = − k
mA. Multiplying the two to-
gether, we will readily get (ω2s , ω
2f ) = (ω2
0 , ω20 + 2k/m). Plugging the solutions back,
we will get A = B for ωs, and A = −B for ωf . That is, the pendulums movesymmetrically when ω = ωs, and antisymmetrically when ω = ωf .
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• First, add the two together to obtain:
¨x+ ω20x = 0
a simple harmonic oscillator in x ≡ x1 + x2. That is, one solution
is that the two pendulums oscillate together symmetrically as a single
system with frequency ωs = ω0. When one pendulum moves right, the
other moves right as well.
• Second, subtract the second equation from the first to obtain:
¨x+
(ω2
0 −2k
m
)x = 0
an oscillator for x ≡ x1−x2. Thus, the second solution is that the two
pendulums oscillate antisymmetrically with common frequency ω2f =
ω20 + 2k
m , that is:
ωf =
√ω2
0 +2k
m
That is, when one pendulum moves right, the other moves left, and
vice versa.
The frequencies ωf and ωs < ωf are the resonant frequencies (or normal
modes) of the system. When the pendulum oscillates at the slow resonant fre-
quency, then the pendulums oscillate together symmetrically, i.e. when one
moves right, the other does so as well (Figure 5.16, left panel). As a result,
the spring between the two pendulums is neither stretched nor compressed.
But if we increase the frequency and reach ωf , then the two pendulums will
start moving antisymmetrically: when one moves to the right, the other one
moves to the left, and vice versa (Figure 5.16, right panel). As a result, the
spring between the two pendulums becomes stretched when the pendulums are
far apart, and compressed when they are close together. In both cases, how-
ever, the two pendulums have the same frequency and, therefore, the same
period of motion.
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x1 x2 x1 x2
Figure 5.16: Left: Oscillation at the lower resonantfrequency, ω = ωs. Right: Oscillation at the higherresonant frequency, ω = ωf .
A very similar result will be obtained in the following example, which
considers coupled springs as opposed to coupled pendulums.
Example 5.13 (Two coupled springs) Consider two coupled springs (Fig-
ure 5.17): two bobs, of masses m1 and m2, are attached to the ends of
springs. The springs constants are k1, k2, and k3, respectively for each
spring. Let x1 and x2 denote, respectively, the displacement of the first and
second bobs away from the stable equilibrium.
x2x1
−k1x1 −k2x2 k3x2
Figure 5.17: Two coupled springs.
For the first mass, Newton’s Second Law says m1a = F1, where F1 is the
magnitude of the force acting on mass 1. This is composed of the force on
mass 1 from moving mass 1 (equal to −k1x1− k2x1), and the force on mass
1 from moving mass 2 (equal to k2x2). Thus, Newton’s law on mass 1 says:
mx1 = −(k1 + k2)x1 + k2x2 (5.16)
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Similarly, for mass 2, we have:
mx2 = −(k2 + k3)x2 + k2x1 (5.17)
We can now find two solutions to this system of ODEs:
• Adding equations (5.16)-(5.17), we obtain m(x1 + x2) = −k1x1 +
−k3x2. Note that k2 has disappeared, because the influence of the
middle section on either mass cancels out with its effect on the other
mass.
Assuming k1 = k3 = k and m1 = m2 = m for simplicity, we can write:
¨x+
(k
m
)x = 0
a simple harmonic oscillator in x ≡ x1+x2, with solution x = As cos(ωst+
ϕs), where As is the amplitude, ωs =√
km is the angular frequency,
and ϕs is the phase.
• Consider now, instead, subtracting equation (5.17) from equation (5.17).
Then, we obtain m(x1 − x2) = −(k1 + 2k2)x1 − k3x2.
Assuming, once again, that k1 = k3 = k and m1 = m2 = m, we can
write:
¨x+
(k + 2k2
m
)x = 0
a simple harmonic oscillator in x ≡ x1 − x2. Thus, the solution is
x = Af cos(ωf t+ϕf ), where Af is the amplitude, ωf =√
k+2k2m is the
angular frequency, and ϕf is the phase.
Thus, we have found two solutions, each of which oscillates at a fixed
frequency. These are the resonant frequencies for the system. The frequen-
cies ωs and ωf are the resonant frequencies of this system. Comparing the
two solutions, notice that ωs =√
km <
√k+2k2m = ωf . That is, the second
solution oscillates faster than the first one.
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In particular, we have that the displacement for each mass is:
x1 =x+ x
2=
1
2
(As cos(ωst+ ϕs) +Af cos(ωf t+ ϕf )
)x2 =
x− x2
=1
2
(As cos(ωst+ ϕs)−Af cos(ωf t+ ϕf )
)the first equalities by definition, and the second ones by our results above.
We observe:
• Symmetric oscillation mode:
If we excite the masses so that Af = 0, then x1 = x2, so that both
masses will oscillate at the same frequency ωs. In this case, both
masses move right and left together, in unison. In practice, the masses
will have this motion if we start them off at the same positions, x1(0) =
x2(0). This is called the symmetric oscillation mode.
• Antisymmetric oscillation mode:
If we excite the masses so that As = 0, then x1 = x2, so that both
masses will oscillate at the same frequency ωf . However, in this case,
the masses are out of phase: when one moves to the left, the other one
moves to the right. In practice, the masses will have this motion if we
start them off at the opposite positions, x1(0) = −x2(0). This is called
the antisymmetric oscillation mode.
Example 5.14 (n coupled oscillators) More generally, consider a sys-
tem with n masses. All the masses are potentially linked through harmonic
oscillators. Let kij denote the spring constant between masses i and j.
Newton’s Second Law now says:
m1x1 =k11x1 + k12x2 + · · ·+ k1nxn
...
mnxn =kn1x1 + kn2x2 + · · ·+ knnxn
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To solve this system of linear ODEs, recall from Section 0.7 that we may
guess xi = cieiωt for each i = 1, . . . , n, where i =
√−1 and (c1, . . . , cn) are
constants to be found. Plugging the guess into the system of ODEs, we find
the following collection of characteristic equations:
−ω2c1 =k11
m1c1 +
k12
m1c2 + · · ·+ k1n
m1cn
...
−ω2cn =kn1
mnc1 +
kn2
mnc2 + · · ·+ knn
mncn
or, in matrix notation:
−ω2~c = M~c
where ~c ≡ (c1, . . . , cn)>, and the matrix M ≡ (Mij) has entries Mij =kijmi
. Note the last equation can be written as (M + ω21n)~c = ~0, where 1n is
the n× n identity matrix. Since we are looking for a vector ~c 6= ~0, then we
must solve:
Det(M + ω21n) = 0
Letting λ ≡ −ω2, this means that we are looking for the n eigenvalues
of the matrix M (recall Definition 0.13). This will give us n solutions
for ω =√−λ (which are potentially complex), each with a corresponding
eigenvector ~c. The collection of solutions (ω1, . . . , ωn) then compose the
resonant frequencies of the system.
Coupled oscillators therefore may have a large number of resonance fre-
quencies. For example, a string (e.g. a violin string) can be thought of a
system that is composed of a continuum of coupled oscillators. Indeed, when
plucked, violin strings will oscillate transversally (i.e. in the y direction) and
exhibit a very large number of resonances.23
23 In contrast, the example of the spring (Example 5.13) was one of longitudinal
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Musical instruments in general have countably many resonant frequen-
cies, fn for n = 1, 2, 3, . . . , called harmonics. For a given harmonic fn, the
vibrating string oscillates about (n−1) nodes, which are points on the string
that remain at rest (i.e. not vibrating) and which are equidistant to each
other. For example, the third harmonic has nodes at 13L and 2
3L os the
string, where L is the length of the string.
Moreover, the harmonic frequencies are proportional to the number of
nodes: fn = nf , where f is called the fundamental harmonic.24
oscillations, because the spring oscillates only in the x direction. Of course, nothingprevents us from making the spring oscillate transversally as well. These more generalcases were considered in Example 5.14.24 For example, for musical instruments, f ≈ v
2LHz, where v = 340 m/sec is the
speed of sound, and L is the length of the instrument. For example, flutes have holesso that, by putting our finger on one hole, we effectively decrease the length L of theinstrument and increase the frequency. As a result, the sound pitch becomes higher.
235
Chapter 6
Heat, Temperature, and
Thermodynamics
Heat is a form of energy transfer between bodies. It is a physical phe-
nomenon whereby energy is transmitted between objects in thermal contact.
When the temperature of an object changes, its so-called thermometric prop-
erties change. For instance:
• Objects tend to expand as they heat up, and shrink as they cool down.
• Contained gases increase in pressure as their temperature raises.
• If a hot object is put in contact with a cold one, the first object cools
off and shrinks, while the second one heats up and expands, due to
the flow of thermal energy between one object and the other.
In this last example, when the process stops, the objects reach a so-called
thermal equilibrium.
Definition 6.1 (Thermal Equilibrium) Two or more objects are in ther-
mal equilibrium if there exists a zero net thermal energy flow between them,
so their temperatures are the same. An object is in thermal equilibrium with
itself if the temperature within it is spatially and temporally uniform.
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This chapter studies heat and temperature changes. We start off with un-
derstanding how solids and liquids expand and contract due to heat. Then,
we study how heat can change the pressure in gases. Finally, we will review
the basic laws of thermodynamics.
6.1 Thermal Expansion of Solids and Liquids
Solid and liquid bodies expand or contract when their temperature changes.
This is in fact how temperature scales were developed. Swedish astronomer
Anders Celsius (1701–1744) invented a linear scale whereby the temperature
of 0◦C was attributed to the length of a rod when dipped in melting ice,
whereas a temperature of 100◦C was given when the rod’s length changed
due to it being subject to boiling water. According to his system, the change
in the rod’s length is linear in the temperature change. Dutch-German-
Polish physicist Daniel Gabriel Fahrenheit (1686–1736), who is also credited
for inventing the mercury thermometer, used instead the human body as
the reference for his (also linear) scale, mistakenly attributing 100◦F to the
customary heat of the human body (which is typically closer to 97◦F).1
There is no upper limit to how hot systems can get. But if there ex-
ists a system which cannot transfer thermal energy to any another system
that it is in thermal contact with, then such system must be at the lowest
possible temperature. In Celsius, this minimal temperature is -273.15◦C, or
-459.67◦F. Inspired by the observation that there is in fact a lowest tempera-
ture, the physicist William Thomson, 1st Baron Kelvin (1824–1907) created
yet another temperature scale, with 0◦K given to the lowest possible tem-
perature. In physics, Kelvin is now the standard scale for temperatures.2
To understand the thermal expansion of solids and liquids formally, con-
sider a rod of length L going through a temperature change of ∆T . Then,
1 The conversion between the two scales is TF = 95TC + 32, where TF (respectively,
TC) denotes Fahrenheit (respectively, Celsius) temperature.2 The increments in Kelvin are the same as those in Celsius, i.e. TK = TC + 273.15.
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the change in length is:
∆L = αL∆T (6.1)
where α is the so-called linear expansion coefficient, in units per ◦C,
whose exact value depends on the material. Typically, bodies expand with
heat, so α > 0.3 For instance, for copper, α = 17 × 10−6/◦C; for steel,
α = 12 × 10−6/◦C. So copper expands with temperature more easily than
steel does.4
While the length expands linearly to the temperature change, in terms
of volume there is of course a cubic expansion. To see this, consider for
simplicity a perfect cube with sides of equal length, L. Its volume is V = L3.
Consider changing the temperature by ∆T . Then, the new volume is:
V + ∆V = (L+ ∆L)3
= V
(1 +
∆L
L
)3
≈ V + 3∆LL2
= V + 3αV∆T
where the third line uses the Binomial theorem,5 and the fourth line uses
equation (6.1). In sum:
∆V = βV∆T (6.2)
where β ≡ 3α. Here, β is the so-called cubical expansion coefficient.
Comparing (6.1) and (6.2), the only difference between the rate of expansion
3 A rare exception is water in the range T ∈ [0, 4]◦C. Indeed, just before hitting itsfreezing temperature, cooling down water will cause it to expend instead of contract.4 For example, if there is a temperature change of ∆T = 50◦C between Summer
and Winter, rail tracks (made of steel) will expand by ∆L = 0.6 meters. For thisreason, rail tracks typically have small gaps every few meters, allowing the materialto expand without there being a risk of bulging. Incidentally, this causes distinctiveclicking sounds as the train wheels go over these gaps.5 That is, if x� 1, then (1 + x)n ≈ 1 + nx.
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in length and that in volume is the factor of proportionality relative to the
temperature change, with this factor being larger (and exactly three times
larger in the simple case of perfect cube-like shapes) for volume than for
length.
Example 6.1 (Mercury thermometers) Consider a glass tube that is
attached to a spherical chamber containing a volume V of mercury. The
tube has radius r. The change in volume when the system is heated up by
∆T degrees is ∆V = βHgV∆T , where βHg ≈ 18 × 10−5/◦C is the cubic
expansion coefficient of mercury.6 If the mercury is rising up the tube, then
the change in the volume of mercury can also be written as ∆V = πr2h,
where h is the height by which the mercury rises. Thus:
h =βHgV∆T
πr2
is the height by which the mercury will rise. For example, if V = 1 cm3,
r = 1 mm, and ∆T = 1◦C, then h ≈ 5.7 mm. Thus, for every 5.7 mm rise
in the mercury inside the tube, we can be certain that the temperature has
gone up by 1◦C.
6.2 The Ideal Gas Law
We have seen that temperature changes in liquids and solids lead to their
expansion or compression. With gases, temperature changes lead to changes
in pressure (Definition 5.3).
In Sections 5.2 and 5.3, we argued that the fact that liquids are incom-
pressible means that the change in pressure is everywhere the same (see
equation 5.5). Gases, on the other hand, are compressible, so Pascal’s prin-
ciple does not apply.
It is an experimental fact, however, that there is a simple mathematical
relationship between the pressure of a gas (P , in Pa), its temperature (T ,
6 Here we are considering that the tube itself does not expand as it is heated. Inreality, these tubes are typically made of pyrex, a synthetic glass with a very lowexpansion coefficient.
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in Kelvin), and the volume (V , in m3) that it occupies. This relationship,
which holds approximately true for most gases, is called the Ideal Gas Law :7
Result 6.1 (Ideal Gas Law I) The state of gases can be described, with
a good approximation, with the following identity:
PV = nRT
where n is the number of moles, and R = 8.3 Joules/◦K is the so-called
universal gas constant. If the gas in question obeys this law exactly, then
the gas is called an ideal gas.8
In the definition, we have used the term “mole”. A mole is a standard
unit of measure and it is equal to NA ≡ 6.02×1023 (the so-called Avogadro’s
number). In particular, a mole is defined as the amount of substance that
contains as many particles as there are atoms in 12 grams of the isotope
carbon-12. That is, one mole of carbon weighs, by definition, 12 grams.
According to this measure, one mole of helium then weighs 4 grams, and
one mole of oxygen-2 (the gas form of oxygen) weighs 32 grams.
These weights are the atomic masses (A), which are calculated as the
sum of the number of protons (Z), which are positively charged, and the
number of neutrons (N), with neutral charge, that compose the atom:
A = N + Z
Atoms also contain electrons, with negative charge and in equal number
to protons, but electrons are nearly weightless, so their contribution to the
atomic mass is negligible. The mass of neutrons (mn) and protons (mz)
happens to be nearly identical and equal to 1.66× 10−27 kg, so the atomic
(or molecular, depending on the case) mass, denoted by m, is simply:
7 In what is coming, we will say for simplicity that gases are collections of molecules,i.e. of groups of two or more atoms. While this is true for some gases (for instanceoxygen or carbon dioxide), it is not true for the so-called atomic (or noble) gases,which are composed of individual atoms (these are helium, neon, argon, krypton,xenon, radon, and oganesson).8 For example, oxygen is very close to being an ideal gas.
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m = Nmn + Zmz ≈ A 1.66× 10−27 kg
This means that atoms and molecules weights are directly proportional
to the number of protons and neutrons they contain, which the factor of
proportionality being constant across the elements.
Example 6.2 (Gas Volumes) Consider an environment with 1 atm (i.e.
P = 101, 325 Pa, or atmospheric pressure) and n = 1 mole at room tem-
perature (i.e. T = 293◦K). By the ideal gas law, the volume of the gas is
then:
V =nRT
P=
8.3× 293
101, 325≈ 1, 000 cm3
or about 24 liters. Notice this is independent of the type of gas, whether
it is helium, or oxygen, or carbon dioxide, etc.
A surprising aspect of the Ideal Gas Law is that the mass of the particles
(atoms or molecules) of the gas does not show up at all in the equation.
What this implies is that gases whose particles are heavier will exhibit lower
particle velocities.
To see this, consider two gases with different particle masses. The two
gases are contained inside a container, and they have the same number of
moles n, the same volume V , the same temperature T , and therefore, by the
Ideal Gas Law, the same pressure P . The molecules of the two gases move
around and bounce off the inner walls of the container, describing elastic
collisions. If the mass of a molecule is m, and the (average) speed of the
molecule is v, the total momentum transfer from the collision is (proportional
to) mv. In units per second, the transfer is proportional to mv2. Moreover:
mv2 ∝ F ∝ P
the first sign by definition of momentum, the second one by definition of
pressure. However, by the Ideal Gas Law, P is constant in m. Therefore,
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we must conclude that the product mv2 is constant in m, for each given
temperature.
And this is indeed what one observes experimentally. For example,
comparing helium (He) to oxygen (O2), with a mass ratio of 1-to-8, then
mHev2He = mO2v
2O2
implies that the average velocity of oxygen particles at a
given temperature is√
8 ≈ 2.83 times smaller than that of helium particles.9
More generally, therefore, heavier gases exhibit lower particle velocities.
A second law that ideal gases must obey is as follows:
Result 6.2 (Ideal Gas Law II) The state of gases can be described, with
a good approximation, with the following identity:
PV = NkT
where N is the number of molecules in the gas, k is the so-called Boltz-
mann constant.
Comparing this law with our first Ideal Gas Law (Result 6.1), we see
that Nk = nR. Since n = NNA
(in words, the number of gas molecules is
Avogadro’s number times the number of moles), it follows that Boltzmann’s
constant is given by:
k =R
NA≈ 1.38× 10−23 J/K
Finally, we can also write these laws in terms of density, ρ. First, the
mass of the gas m can be expressed as the molar mass M (in grams per
mole) times the number of moles n, so n = mM . Therefore, we can write
PV = nRT as PV = mMRT . By the definition of density (Definition 1.20),
ρ = m/V . Plugging it in, we get:
P = ρR
MT
9 Indeed, oxygen particles at room temperature move at about 480 m/s, while heliumparticles at room temperature move at around 480×
√8 ≈ 1357.64 m/s.
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where RM is a specific gas constant, which is only a function of the sub-
stance. This says that, keeping pressure constant, increasing the tempera-
ture of a gas always decreases its density by the same proportion.
6.3 Phase Transitions
Substances can typically exist in three forms (gaseous, solid, and liq-
uid), and change abruptly from one form to another. These changes are
called phase transitions, and they depend on the substance in question, the
temperature, and the pressure. To understand these transitions, we usually
make use of phase diagrams, which depict the state of the substance for
different levels of pressure and temperature.
Temprature →
Pre
ssu
re→
Gas
Solid
Liquid
Figure 6.1: A typical phase diagram.
An example is given in Figure 6.1. In the figure, we see that if we take a
gas at a given temperature and increase its pressure (for example by reducing
the volume), the gas will eventually turn into a solid, if the temperature was
low enough to begin with, or even (though maybe temporarily) into a liquid,
if the temperature was higher. Similarly, if we fix the pressure (at 1atm, for
example), increasing the temperature usually will turn solids into liquids,
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Classical Mechanics Pau Roldan-Blanco
and then into gases. For low pressures, the liquid phase might be altogether
bypassed.
These phase transitions are discontinuous (i.e. abrupt), and they occur
as temperature or pressure surpass certain critical points, depicted in the
figure by the lines describing the different colored areas, and called phase
boundaries. In the figure, the dashed line represents the set of critical boiling
points, where the substance suddenly becomes gaseous if temperature is
increased for a given pressure level. The dotted line represents the set of
critical melting points, where the substance transitions from solid to liquid
under a temperature increase, for a given pressure. The intersection of the
phase boundaries (which is indicated in the figure with a black dot) is called
the triple point, and it marks the temperature and pressure conditions at
which the three different phases can coexist in a stable equilibrium.10
Figure 6.2: The phase diagram for water.
10 For example, for water the triple point is at T = 273.16 K (about 0.01◦C) andP = 611.657 Pa (about 0.006 atm). See Figure 6.2.
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6.4 Thermodynamics
In Sections 5.2-5.3, we studied the static and dynamic behavior of fluids,
mostly in their liquid state. We can now extend what we learned for liquids
to other states of substances.
Using Pascal’s principle, we learned that the condition for there to be
a hydrostatic equilibrium in liquids is given by equation (5.5), which we
reproduce here:
dP
dy= −ρg
We argued that, since liquids are incompressible, density ρ is constant
in (hydrostatic) pressure, P . Thus, the equation can be integrated out very
easily (as we did) to get a linear relationship between pressure and position
(Result 5.1): ∆P = ρg∆y. In words, changes in altitude translate linearly
into pressure changes.
For gases, however, the density does depend on pressure. For simplicity,
assume an isothermal atmosphere, that is an atmosphere where temperature
is everywhere the same (i.e. T is constant in y). Suppose our gas has N
molecules, each with mass m, and it is contained in a volume V . The
density of the gas is, therefore, ρ = NmV . By the Ideal Gas Law, N
V = PkT ,
and therefore ρ = PmkT .
Plugging this density into equation (5.5) we get that dPdy = −Pm
kT g, or:
dP
P= − 1
H0dy
where we have denoted H0 ≡ kTmg . In words, the rate of change in pres-
sure is proportional to the absolute change in altitude, with a constant of
proportionality 1/H0.11 Integrating out between P0 (pressure at sea level,
y = 0), and Ph (pressure at some altitude of h meters above sea level), we
get:
11 Again, this object is truly constant if temperature is everywhere the same, i.e. if Tis constant in y. This is not a terrible approximation at low enough altitudes in thetroposphere. In this case, if T = 273 K (i.e. room temperature), and our gas is air(for which the approximate atomic mass is approximately 29), then H0 ≈ 8, 000 m.
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∫ Ph
P0
dP
P= − 1
H0
∫ h
0dy
Thus, lnPh − lnP0 = − hH0
. Solving:
Ph = P0e−h/H0
This equation then allows us to compute air pressure at various altitudes
h to a good approximation.12
12 For example: for Mount Everest, h = 8.9 km, and we can calculate that pressure isonly a third of an atmosphere. At such low pressure, water boils at 72◦C as opposedto the 100◦ C at sea level, where P0 = 1 atm. At h = 30 km altitude, pressure is onlyone forty-fifth of an atmosphere, and water boils at only 20◦C!
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Chapter 7
Lagrangian Mechanics
In the previous chapters, we argued that the main goal of classical me-
chanics is to describe the trajectories (or orbits) of systems from their equa-
tions of motion. For this, we argued, we need to know three objects: (i)
the masses of the particles; (ii) a set of forces (or, by the Potential Energy
principle, the potential energy function V ); and (iii) initial conditions for
position and velocity.
Henceforth, we will tackle this problem from a slightly different perspec-
tive: given an initial and a final condition for the system, what is the set
of actions on the particle that minimize the difference between potential
and kinetic energies? We shall call this principle the Principle of Least Ac-
tion. The mathematics that operationalize it go by the name of Lagrangian
mechanics.
Formulating the problem in this way is convenient for two reasons: (i)
we can use it to obtain the equations of motion for the system, and thus it
encodes all its descriptive properties (e.g. mass and potential energy); (ii) it
encompasses the description of not only classical mechanics, but also most
other major theories in physics (e.g. Maxwell’s theory of electrodynam-
ics, Einstein’s theory of relativity, and the Standard Model of elementary
particles).
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7.1 The Euler-Lagrange Equation
We begin with some definitions:
Definition 7.1 (Lagrangian Equation) The Lagrangian equation of a
system of i = 1, . . . , N particles with trajectory ~r is:
L(~r, ~r) ≡ T − V (~r) =1
2
N∑i=1
mi|~ri|2 −N∑i=1
Vi(~r) (7.1)
where T is the kinetic energy and V is the potential energy.
Note we write the Lagrangian explicitly as a function of the position
(which influences the potential energy) and the velocity (which influences
the kinetic energy).
Definition 7.2 (Action) The action of a system between two instants of
time t0 and t1 > t0 is the integral of the Lagrangian between these points in
time, or:
A ≡∫ t1
t0
L(~r(t), ~r(t)
)dt (7.2)
Principle 7.1 (Principle of Least Action) The Principle of Least Ac-
tion consists of choosing the trajectory ~r(t) that minimizes A, i.e. for which
the action is stationary to first order.
Because the Principle of Least Action requires a minimization over a
set of trajectories, and trajectories are functions of (continuous) time, the
problem involves the minimization of a function of functions, namely a func-
tional. The optimization of functionals involves a branch of mathematics
called variational calculus. The minimization problem is written as:
δA = 0
to signify that we must find the least among all the possible variations
over the whole trajectory space.
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The equation that allows us to find the stationary action is the Euler-
Lagrange equation:
Definition 7.3 (Euler-Lagrange Equation) The Euler-Lagrange equa-
tion is a second-order partial differential equation (PDE) whose solutions
are the functions for which a given functional is stationary. In this case,
the functional is (7.2), and the equation reads:
d
dt
∂L∂~r− ∂L∂~r
= ~0 (7.3)
where L is the Lagrangian, equation (7.1).
Note that (7.3) is really a system of second-order difference equations,
one for each dimension of space and each particle.1 Then, we have the
following result:
Result 7.1 The stationary action (i.e. the solution to δA = 0) is given by
the function ~r(t) that solves equation (7.3).
We offer two proofs of the result:
Proof 1. First, a heuristic proof. Discretize the time space into T ≡{∆, 2∆, 3∆, . . . }. For each n ∈ T, approximate x(t) and x(t) with x(t) ≈xn+xn+1
2 and x(t) ≈ xn+1−xn∆t , and similarly for the y and z space dimensions.
In vector notation, ~r(t) ≈ ~rn ≡(xn+xn+1
2 , yn+yn+1
2 , zn+zn+1
2
)and ~r(t) ≈
~rn ≡(xn+1−xn
∆t , yn+1−yn∆t , zn+1−zn
∆t
).
We can approximate the action as follows:
A ≈ A ≡+∞∑n=1
L(~rn, ~rn
)∆t
Then, for some k ∈ N, consider taking the derivative along some dimen-
sion (say, x). This yields:
1 Recall (from Definition 0.15) that we use the same symbol, ∂, to denote both apartial derivative and a vector of partials (or gradient).
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∂A
∂xn
∣∣∣∣∣n=k
=∂
∂xn
[L(xn+1 − xn
∆t,xn + xn+1
2
)∆t+ L
(xn − xn−1
∆t,xn−1 + xn
2
)∆t
] ∣∣∣∣∣n=k
=1
∆t
(− ∂L∂xn
∣∣∣∣∣n=k+1
+∂L∂xn
∣∣∣∣∣n=k
)+
1
2
(∂L∂xn
∣∣∣∣∣n=k
+∂L∂xn
∣∣∣∣∣n=k+1
)
Taking the continuous time limit of the right-hand side yields
∂A
∂x= − d
dt
∂L∂x
+∂L∂x
Thus, ∂A∂x = 0 if, and only if, − d
dt∂L∂x + ∂L
∂x = 0. Repeating the argument for
the y and z coordinates proves the result. �
Proof 2. The second proof uses a more formal variational argument.
Suppose ~r(t) is a the true trajectory of the system between two states ~r0 =
~r(t0) and ~r1 = ~r(t1) and two points in time, t0 and t1. Let ~ε(t) be a small
perturbation that is zero at the endpoints, i.e. ~ε(t0) = ~ε(t1) = ~0. To a first
order, the change in the action functional, δA, can be written:
δA =
∫ t1
t0
L(~r +~ε, ~r + ~ε
)dt =
∫ t1
t0
(~ε · ∂L
∂~r+ ~ε · ∂L
∂~r
)dt
where the second equality follows from a first-order expansion of the
Lagrangian. Using integration by parts, we find:
δA =
(~ε · ∂L
∂~r
) ∣∣∣∣∣t1
t0
+
∫ t1
t0
(~ε · ∂L
∂~r−~ε · d
dt
∂L∂~r
)dt
Since ~ε(t0) = ~ε(t1) = ~0, then the first term vanishes, and we have:
δA =
∫ t1
t0
~ε ·(∂L∂~r− d
dt
∂L∂~r
)dt
Thus, δA = 0 if, and only if, ∂L∂~r −
ddt∂L∂~r
= 0. �
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Remark 7.1 (Obtaining Newton’s Laws) To see the power of the Prin-
ciple of Least Action, we now demonstrate that one can derive Newton’s
Second Law directly from the Euler-Lagrange Equation, (7.3). Using the
Lagrangian (7.1), note:
d
dt
∂L∂~r
=d
dt
(∂L∂~r1
, . . . ,∂L∂~rN
)=
(m1
d
dt~r1, . . . ,mN
d
dt~rN
)and
∂L∂~r
= −(∂V1(~r)
∂~r1, . . . ,
∂VN (~r)
∂~rN
)=(~F1(~r), . . . , ~FN (~r)
)where the second equality uses the definition of potential energy. There-
fore, the Euler-Lagrange equation implies:
~Fi = mid2~ridt2
for all i = 1, . . . , N . This is exactly Newton’s Second Law of Motion.
Equivalently, using the definition of momentum (Definition 1.25), the Euler-
Lagrange equation gives:
d~pidt
= mid2~ridt2
where ~pi = middt~ri
Example 7.1 (Spring Motion Revisited) Consider again the single-particle
case of Example 2.10. The Lagrangian is:2
L(~r, ~r) = T − V =1
2
(m|~r|2 − k|~r|2
)The Euler-Lagrange equation says d
dt∂L∂~r− ∂L
∂~r = ~0, so:
m~r = −k~r ⇒ ~r = −ω2~r
2 For a derivation of the formula for kinetic and potential energy in this case, seeExample 2.10.
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where ω =√
km . This is exactly what we found using Newton’s Second
Law in Equation (2.7).
7.2 Non-Inertial Reference Frames
Recall that Newton’s Laws work only within the realm of inertial refer-
ence frames (Definition 1.12), that is, reference frames in which forces add
up to zero and bodies are not being accelerated. An important advantage
of using the Lagrangian formulation over Newton’s original Laws is that it
allows us to operate easily with other reference frames. This will be very
useful when we study non-inertial reference frames, i.e. reference frames
that undergo acceleration relative to inertial frames. This will be the case,
for instance, when we study Einstein’s relativity.
Example 7.2 (Reference Frames I) Consider two reference frames, A
and B. For simplicity, suppose there is only one particle and one dimension,
x. Observers in A (an inertial reference frame) are standing at rest, and
use the coordinate x to locate the particle. Observers in B (a non-inertial
reference frame) are subject to acceleration. Let f(t) describe the location
of observers in B relative to those in A at time t. Thus, observers in B use
the coordinate X = x − f(t) at time t to locate the particle. That is, when
observers in A locate the particle at x(t), those at B locate it at X(t) +f(t).
Observers in A and B define the Lagrangians:
AA =
∫ t1
t0
(1
2mx(t)2 − V (x(t))
)dt
AB =
∫ t1
t0
(1
2m(X(t) + f(t)
)2 − V (X(t)))
dt
respectively. Both observers then find the stationary action by means of
the Euler-Lagrange equations. For observers in B, this gives:
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mX(t) = −
(dV(X(t)
)dX(t)
+mf(t)
)
while, for observers in A, mX(t) = −dV (X)dX . Thus, observers in the
non-inertial reference frame perceive an additional force of mf(t) relative to
those in frame A as a result of the underlying acceleration f(t) that these
observers are undergoing.
Example 7.3 (Reference Frames II) Consider a single particle, now in
two dimensions. Consider again an inertial reference frame A, using coordi-
nates (x, y), and a non-inertial reference frame B, using coordinates (X,Y ).
Now, observers in B are moving with rotation, in uniform circular motion,
relative to observers in A, at rest. (For instance, observers in B are on
a carousel, while observers in A are not). Time dependence is omitted for
brevity.
Using what we know about uniform circular motion, we can then relate
the two coordinate systems as follows:
x = X cos(ωt) + Y sin(ωt) (7.4a)
y = −X sin(ωt) + Y cos(ωt) (7.4b)
For simplicity, suppose that observers in A observe that the particle
moves with no forces acting on it (no potential energy). Thus, their La-
grangian is simply given by kinetic energy:
LA =1
2m(x2 + y2
)Thus, the stationary action for these observers solves the Euler-Lagrange
equation:
m~r = ~0
In sum, according to these observers, the particle possesses no accel-
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eration. What about observers in the non-inertial frame? Differentiating
(7.4a)-(7.4b) gives:
x = X cos(ωt)− ωX sin(ωt) + Y sin(ωt) + ωY cos(ωt)
y = − X sin(ωt)− ωX cos(ωt) + Y cos(ωt)− ωY sin(ωt)
Some algebra then shows:
x2 + y2 = X2 + Y 2 + ω2(X2 + Y 2) + 2ω(XY − Y X)
Therefore, the Lagrangian for observers in frame B is:
LB =1
2m(X2 + Y 2
)︸ ︷︷ ︸
(i)
−[−mω
2
2
(X2 + Y 2
)]︸ ︷︷ ︸
(ii)
+mω(XY − Y X
)︸ ︷︷ ︸
(iii)
We can identify three terms in the Lagrangian:
(i) This term is just the kinetic energy according to the observers in B.
(ii) This term is the potential energy according to observers in B:
V (X,Y ) = −mω2
2
(X2 + Y 2
)That is, even though observers at rest see the particle as having no
force, those in rotational motion relative to them perceive a potential
energy on the particle. By Example 2.10, the potential energy has
parameter k = mω2. Thus, ω =√k/m is the particle’s angular fre-
quency. Using the Potential Energy Principle, we have:
~F (~r) = mω2~r
Since this force is positive, it works as a centrifugal force.3
3 A centrifugal force is a centripetal force (recall Example 1.2) but in the opposite
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(iii) This term is less familiar. It is called the Coriolis force, and it depends
not only on the position of the particle but also on its velocity.
Finally, we can work out the Euler-Lagrange equation to find the sta-
tionary action for observers in reference frame B. This gives:
m
(X
Y
)= mω2
(X
Y
)+
(−2mω 0
0 2mω
)(X
Y
)Again, this shows that, even though the particle has no acceleration from
the perspective of observers in the inertial reference frame A, from those in
the non-inertial reference frame B it is perceived as having both centrifugal
and Coriolis forces.
7.3 Generalized Coordinates
Not only can we accommodate different reference frames, but we can
also use the Principle of Least Action to describe the laws of physics in
generalized coordinates, that is, in non-Cartesian coordinates. This will be
useful when we describe systems that move in non-Euclidean spaces, such
as spherical surfaces.
Let us thus see how to set up the equations of classical mechanics in
a general way that applies to any coordinate system. For each particle
i = 1, . . . , N , the set of generalized coordinates is denoted:
q(t) ≡(q1(t), . . . , qN (t)
)For instance:
• In a Cartesian coordinate system with three spatial dimensions, q(t) ={(xi(t), yi(t), zi(t)
)}Ni=1
, where (xi, zi, zi) are the Cartesian coordinates
of particle i.
direction.
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• In a polar coordinate system (Definition 1.11), q(t) = (R(t),θ(t)),
where R = (R1, . . . , RN ) are the radii (or distances from the pole),
and θ = (θ1, . . . , θN ) are angles from the polar direction.
More generally, q is a point in the configuration space of the system,
whatever this may be.4
The position of each particle, ri, is potentially a function of that of all
other particles:
ri = ri(q(t), t)
Finally, the generalized velocities are then defined by:
q ≡ d
dtq(t) =
(q1(t), . . . , qN (t)
)The velocity vector over the i-th particle, vi, is then:
vi ≡ ri ≡d
dtri =
N∑j=1
∂ri(q(t), t)
∂qjqj +
∂ri(q(t), t)
∂t
Though the equations of motion in a generalized system may be quite
complicated, the Principle of Least Action always applies. Conveniently, the
system (whether a wave, a field, or otherwise) can always be characterized
by a Lagrangian, which summarizes the equations of motion.
Therefore, we must look for the trajectory for which:
δA = δ
∫ t1
t0
L(q, q, t)dt = 0
Again, the stationary action solves the Euler-Lagrange equation. Writ-
ten in generalized coordinates:
d
dt
∂L∂q
=∂L∂q
4 Note we do not use the vector notation ~q to emphasize that, in a generalizedcoordinate system, the position of a particle may not be a vector in the Euclideansense of the term.
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where L is short for L(q, q, t). This Euler-Lagrange equation encom-
passes all of classical physics in its most general form. For a given initial
condition (q, q, t), it describes the motion of any particle within the isolated
system.
Often, this equation is expressed in terms of momentum. First, we define
momentum in generalized coordinates:
Definition 7.4 (Generalized Momentum Conjugate) The generalized
momentum conjugate to q is defined by:
p ≡ ∂L∂q
In Cartesian coordinates, where we denote position by ~r, it is clear that∂L∂~ri
= ~pi for all particles i = 1, . . . , N (recall Remark 7.1), where ~pi ≡mi~ri is momentum. The previous definition simply generalizes this idea to
generalized coordinates.
Using this definition, therefore, the Euler-Lagrange equation for gener-
alized coordinate systems is often expressed as follows:
d
dtp =
∂L∂q
Let’s see some examples in the context of non-Cartesian coordinate sys-
tems:
Example 7.4 (Polar Coordinates) Consider a polar coordinate system.
Recall (Definition 1.11) that particle location in such a system is character-
ized by a distance R to the pole (the radius), and an angle θ from the polar
direction. Thus, our set of (generalized) coordinates is:
q(t) = (R(t),θ(t))
where R = (R1, . . . , RN ) and θ = (θ1, . . . , θN ) are the radii and angles
for each particle relative to the pole. For simplicity, consider N = 1 (ex-
tending it to N ≥ 2 is straightforward). Moreover, assume that potential
energy is zero.
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First, let’s work out the Lagrangian equation. To transform Cartesian
coordinates (x, y), i.e. linear motion, into polar coordinates (R, θ), i.e. ro-
tational motion, we use:
x = R cos θ
y = R sin θ
Differentiating with respect to time, we get:
x = R cos θ −Rθ sin θ
y = R sin θ +Rθ cos θ
Some algebra shows: x2 + y2 = R2 +R2θ2. Thus, the Lagrangian is:
L(q, q) =1
2m(R2 +R2θ2
)(7.5)
where recall q ≡ (R, θ). Let’s now compute the generalized momenta.
The generalized momentum conjugate to R, denoted here by pR, is:
pR ≡∂L∂R
= mR
Thus, using the Euler-Lagrange equation, the equation of motion on the
R coordinate is ddtpR = ∂L
∂R = mRθ2. Equivalently, using pR = mR (by
definition), we have mR = mRθ2, or simply:
R = Rθ2
Interestingly, we have derived the equation of motion on the R coordinate
with no mention of the object’s mass or momentum.
Similarly, the generalized momentum conjugate to θ, denoted here by pθ,
is:
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pθ ≡∂L∂θ
= mR2θ
What this gives us is the momentum (mass times velocity) along the
angular coordinate. Accordingly, we often call this the system’s angular mo-
mentum (recall Definition 3.4). The equation of motion on the θ coordinate
comes from the Euler-Lagrange equation:
d
dtpθ =
∂L∂θ
= 0
Therefore, we have found, using the Euler-Lagrange equation, that an-
gular momentum is always conserved (recall Result 3.4). For example, a
pegtop remains upright when spinning because its angular momentum is be-
ing conserved, until friction eventually slows it down and it topples over.
Another implication of the conservation of angular momentum is that
the angular velocity of the particle is higher as the particle gets closer to
the origin. To see this, use pθ = mR2θ to write conservation of angular
momentum as ddt
(R2θ
)= 2Rθ +R2θ = 0, or:
θ = −2R
R2θ
Therefore, smaller R implies that θ must be higher for angular acceler-
ation to be conserved. To sum up our results, the generalized momentum
conjugate to q = (R, θ) is:
p ≡ (pR, pθ) =∂L∂q
=
(∂L∂R
,∂L∂θ
)=(mR,mR2θ
)and the equations of motion read p ≡ (pR, pθ) = (mRθ2, 0) or, written
in terms of coordinates:
q =(R, θ
)=
(Rθ2,−2R
R2θ
)
In this example, both the R coordinate and its velocity R appeared
in the Lagrangian (equation 7.5). However, only the angular velocity θ,
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and not the θ coordinate itself, showed up. As a result, the partial of the
Lagrangian with respect to θ was zero, meaning that the momentum along
this coordinate is conserved (Result 3.4).
More generally, when a coordinate has the property that shifting its
value does not change the Lagrangian, we call it a cyclic coordinate.
Definition 7.5 (Cyclic coordinates) For a generalized coordinate sys-
tem q, a coordinate q ∈ q is called cyclic if:
∂L∂q
= 0
A direct implication (and alternative definition) is, therefore:
Result 7.2 (Conservation of momentum) The momentum conjugate of
cyclic coordinates is always conserved. Indeed, using the Euler-Lagrange
equation, if q ∈ q is a cyclic coordinate, then:
d
dtpq ≡
d
dt
∂L∂q
=∂L∂q
= 0
Example 7.5 (Cartesian coordinates, zero forces) The standard exam-
ple of a cyclic coordinate is the angular momentum in polar coordinates
(previous example). Let’s now see an example of conservation of linear
momentum.
Again, consider a single particle, now in coordinates (x, y, z). Assuming
that potential energy is zero (all forces are zero), the Lagrangian is:
L =1
2m(x2 + y2 + z2
)Therefore, ∂L
∂x = ∂L∂y = ∂L
∂z = 0, so all coordinates of the system are cyclic,
and all components of momentum are conserved. Of course, this would not
be true if potential energy was not zero.
Example 7.6 (Cartesian coordinates, non-zero forces) Let’s thus rein-
troduce potential energy. Consider two particles i = 1, 2 moving in a one-
dimensional Cartesian coordinate system (a line), with coordinate x. Say
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m1 = m2 = m for simplicity, and suppose that the potential energy depends
on the distance x1−x2 between them (where xi is the position of particle i).
The Lagrangian for this case is:
L =m
2
(x2
1 + x22
)− V
(x1 − x2
)Since ∂L
∂x1, ∂L∂x2
6= 0, then x is non-cyclic for both particles, and momen-
tum is not conserved for either particle.
Remark 7.2 (Introduction to Symmetries) Importantly, the previous
example can be shown to conserve momentum if we introduce a slight change
of coordinates. This will introduce the topic of the next section: symmetries.
Define new coordinates (x+, x−) by:
x+ =x1 + x2
2x− =
x1 − x2
2
Some simple algebra steps show that the kinetic energy in the new coor-
dinate system is T = m(x2
+ + x2−). The Lagrangian is:
L = m(x2
+ + x2−)− V (2x−)
Importantly, potential energy (and thus the Lagrangian) is only a func-
tion of x−, and not x+. Hence, x+ is a cyclic coordinate, and the conjugate
momentum to x+ (call it p+) is conserved, i.e. p+ = 0. Computing p+ by
Definition 7.4, we have p+ = ∂L∂x+
= 2mx+. Using x+ = x1+x22 , then we
obtain:
p+ = m(x1 + x2)
That is, p+ is just the total momentum (i.e. the sum of momenta of the
system). Thus, though there are no cyclic coordinates, total momentum is
conserved.
Example 7.7 (Double Pendulum) This example clearly shows the power
of the Lagrangian method in non-Cartesian coordinates. In Example 1.10
we explored the motion of a simple pendulum, that is, a harmonic oscillator
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with a single massive body swinging from a massless string. Now we explore
the double pendulum, that is, a simple pendulum that is attached to the end
of the body of another simple pendulum (see Figure 7.1).
Figure 7.1: The double pendulum.
Let us start with Cartesian coordinates (x, y), with x (rep. y) being
the horizontal (resp. vertical) direction, and for simplicity suppose that the
string of the upper pendulum is attached to the origin (x, y) = (0, 0). Index-
ing each pendulum by i = 1, 2, where i = 1 is the upper one, let θi be the
angle of the string about the vertical axis, li be the length of the string, and
mi be the mass of the object.
We start by decomposing the tension vector into its x and y components.
For the upper pendulum, the position of the body in Cartesian coordinates
can be written:
x1 = l1 sin θ1
y1 = − l1 cos θ1
respectively. Similarly, the positions of the second pendulum’s body are:
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x2 = x1 + l2 sin θ2
y2 = y1 − l2 cos θ2
By looking at the components of tension, we have now moved to polar
coordinates. Thus, henceforth we work in the generalized coordinate system
q = (l, θ), where l is the radius (length of the pendulum) and θ is the angle.
Differentiating with respect to time, we obtain the components of the
velocity vector for each of the two objects:
~v1 =
(x1
y1
)=
(l1θ1 cos θ1
l1θ1 sin θ1
)
~v2 =
(x2
y2
)=
(l1θ1 cos θ1 + l2θ2 cos θ2
l1θ1 sin θ1 + l2θ2 sin θ2
)
where θi is thus the angular velocity of the i-th body.
We now want to describe the equations of motion of the system. One way
to do this is to use Newton’s Laws directly (as we did for the simple pendu-
lum, Example 1.10). This would involve invoking Newton’s Laws of Motion,
which would depend on position, velocity and acceleration of particles in the
Cartesian system (x, y).
A much faster and convenient way is to use the Lagrangian method in
the polar coordinate system, (l, θ). Let’s then write down the Lagrangian.
First, kinetic energy is:
T =1
2m1v
21 +
1
2m2v
22 =
1
2m1
(x2
1 + y21
)+
1
2m2
(x2
2 + y22
)=
1
2m1l
21θ
21 +
1
2m2
(l21θ
21 + l22θ
22 + 2l1l2θ1θ2 cos(θ1 − θ2)
)where we have used equation (0.3). To know the potential energy V , we
make use of the Potential Energy Principle to back out V from the forces
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at work. The gravitational force acting on each body is mig, for pendulum
i = 1, 2. Thus, Vi =∫migdy, so:
V = m1gy1 +m2gy2 = −m1gl1 cos θ1 −m2g(l1 cos θ1 + l2 cos θ2
)=− (m1 +m2)gl1 cos θ1 −m2gl2 cos θ2
Therefore, the Lagrangian is:
L = T − V =1
2(m1 +m2)l21θ
21 +
1
2m2l
22θ
22 +m2l1l2θ1θ2 cos(θ1 − θ2)
+ (m1 +m2)gl1 cos θ1 +m2gl2 cos θ2
The momentum conjugate to the θi coordinate for each i = 1, 2, denoted
pθi, is:
pθ1 ≡∂L∂θ1
= (m1 +m2)l21θ1 +m2l1l2θ2 cos(θ1 − θ2)
pθ2 ≡∂L∂θ2
= m2l22θ2 +m2l1l2θ1 cos(θ1 − θ2)
respectively. The equations of motion of the system can be obtained from
the Euler-Lagrange equations:
d
dtpθi =
∂L∂θi
, for each i = 1, 2 (7.6)
Let’s compute each component. First, using (pθ1 , pθ2) from above, we
have:
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Classical Mechanics Pau Roldan-Blanco
d
dtpθ1 = (m1 +m2)l21θ1 +m2l1l2θ2 cos(θ1 − θ2)
−m2l1l2θ1θ2 sin(θ1 − θ2) +m2l1l2θ22 sin(θ1 − θ2)
d
dtpθ2 = m2l
22θ2 +m2l1l2θ1 cos(θ1 − θ2)
−m2l1l2θ21 sin(θ1 − θ2) +m2l1l2θ1θ2 sin(θ1 − θ2)
On the other hand, note:
∂L∂θ1
= −m2l1l2θ1θ2 sin(θ1 − θ2)− (m1 +m2)gl1 sin θ1
∂L∂θ2
= m2l1l2θ1θ2 sin(θ1 − θ2)−m2gl2 sin θ2
Then, some algebra shows that (7.6) simplifies to:
0 = (m1 +m2)l1θ1 +m2l2θ2 cos(θ1 − θ2)
+m2l2θ22 sin(θ1 − θ2) + (m1 +m2)g sin θ1 (7.7a)
0 = l2θ2 + l1θ1 cos(θ1 − θ2)− l1θ21 sin(θ1 − θ2) + g sin θ2 (7.7b)
This is a system of coupled second-order non-linear ODEs describing the
angular acceleration of each pendulum as functions of the angular velocities
and the angles themselves. It is impossible to solve these equations by hand
(even if we invoke the Small-Angle Approximation method, Definition 1.16),
so we will need a non-linear solver such as Mathematica.
The double pendulum is one of the most famous examples of a chaotic
system (Definition 0.4): slightly different initial conditions give rise to dras-
tically different trajectories. Because we are working here with generalized
(polar) coordinates, the initial conditions are initial positions (radii) and
angular velocities for both bodies.5
5 For an interactive simulation illustrating the chaotic behavior of the double pen-
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Classical Mechanics Pau Roldan-Blanco
7.4 Symmetry and Conservation Laws
In the previous section, we have introduced Lagrangian mechanics in gen-
eralized coordinates. As we have argued, this is the most general description
that one can give for the trajectory of a particle. We now introduce the con-
cept of symmetries. A symmetry is a coordinate transformation that leaves
the physical properties of the system unchanged, no matter where the sys-
tem is located in the configuration space. Or, in the language of Lagrangian
mechanics:
Definition 7.6 (Symmetry) A symmetry of a physical system is a coor-
dinate transformation that leaves the Lagrangian unchanged.
Since the Lagrangian of the system describes all that there is to know
about the system, a symmetry is a transformation that does not alter its
properties.
Example 7.2 showed a case in which a re-labeling (i.e. a transformation)
of the coordinate system did not change the Lagrangian. As another exam-
ple, rotating the orbit described by the motion of a particle is a symmetry
because it is a transformation that does not alter the equations of motion
of the object. Let’s now see more examples:
Example 7.8 (Symmetries, Example I)
Example 7.9 (Symmetries, Example II)
dulum, visit http://bestofallpossibleurls.com/double-pendulum.html.
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Part II
ELECTRICITY
AND MAGNETISM
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