pages 451-500

Radial acceleration at the surface of the Earth example 15. What is your radial acceleration due to the rotation of the earthif you are at the equator?

. At the equator, your distance from the Earth’s rotation axis isthe same as the radius of the spherical Earth, 6.4 × 106 m. Yourangular velocity is

ω =2π radians

1 day= 7.3× 10−5 s−1,

which gives an acceleration of

ar = ω2r

= 0.034 m/s2.

The angular velocity was a very small number, but the radius wasa very big number. Squaring a very small number, however, givesa very very small number, so the ω2 factor “wins,” and the finalresult is small.

If you’re standing on a bathroom scale, this small acceleration isprovided by the imbalance between the downward force of gravityand the slightly weaker upward normal force of the scale on yourfoot. The scale reading is therefore a little lower than it shouldbe.

Dynamics

If we want to connect all this kinematics to anything dynamical,we need to see how it relates to torque and angular momentum.Our strategy will be to tackle angular momentum first, since angu-lar momentum relates to motion, and to use the additive propertyof angular momentum: the angular momentum of a system of par-ticles equals the sum of the angular momenta of all the individualparticles. The angular momentum of one particle within our rigidlyrotating object, L = mv⊥r, can be rewritten as L = r p sin θ,where r and p are the magnitudes of the particle’s r and momen-tum vectors, and θ is the angle between these two vectors. (The rvector points outward perpendicularly from the axis to the parti-cle’s position in space.) In rigid-body rotation the angle θ is 90◦,so we have simply L = rp. Relating this to angular velocity, wehave L = rp = (r)(mv) = (r)(mωr) = mr2ω. The particle’s con-tribution to the total angular momentum is proportional to ω, witha proportionality constant mr2. We refer to mr2 as the particle’scontribution to the object’s total moment of inertia, I, where “mo-ment” is used in the sense of “important,” as in “momentous” — abigger value of I tells us the particle is more important for deter-mining the total angular momentum. The total moment of inertia

Section 15.7 Rigid-body rotation 451

ag / Analogies between rota-tional and linear quantities.

ah / Example 16

is

I =∑

mir2i , [definition of the moment of inertia;

for rigid-body rotation in a plane; r is the distance

from the axis, measured perpendicular to the axis]

The angular momentum of a rigidly rotating body is then

L = Iω. [angular momentum of

rigid-body rotation in a plane]

Since torque is defined as dL/dt, and a rigid body has a constantmoment of inertia, we have τ = dL/dt = I dω/dt = Iα,

τ = Iα, [relationship between torque and

angular acceleration for rigid-body rotation in a plane]

which is analogous to F = ma.

The complete system of analogies between linear motion andrigid-body rotation is given in figure ag.

A barbell example 16. The barbell shown in figure ah consists of two small, dense,massive balls at the ends of a very light rod. The balls havemasses of 2.0 kg and 1.0 kg, and the length of the rod is 3.0m. Find the moment of inertia of the rod (1) for rotation about itscenter of mass, and (2) for rotation about the center of the moremassive ball.

. (1) The ball’s center of mass lies 1/3 of the way from the greatermass to the lesser mass, i.e., 1.0 m from one and 2.0 m from theother. Since the balls are small, we approximate them as if theywere two pointlike particles. The moment of inertia is

I = (2.0 kg)(1.0 m)2 + (1.0 kg)(2.0 m)2

= 2.0 kg·m2 + 4.0 kg·m2

= 6.0 kg·m2

Perhaps counterintuitively, the less massive ball contributes farmore to the moment of inertia.

(2) The big ball theoretically contributes a little bit to the momentof inertia, since essentially none of its atoms are exactly at r=0.However, since the balls are said to be small and dense, we as-sume all the big ball’s atoms are so close to the axis that we canignore their small contributions to the total moment of inertia:

I = (1.0 kg)(3.0 m)2

= 9.0 kg·m2

This example shows that the moment of inertia depends on thechoice of axis. For example, it is easier to wiggle a pen about itscenter than about one end.

452 Chapter 15 Conservation of angular momentum

The parallel axis theorem example 17. Generalizing the previous example, suppose we pick any axisparallel to axis 1, but offset from it by a distance h. Part (2) ofthe previous example then corresponds to the special case ofh = −1.0 m (negative being to the left). What is the moment ofinertia about this new axis?

. The big ball’s distance from the new axis is (1.0 m)+h, and thesmall one’s is (2.0 m)-h. The new moment of inertia is

I = (2.0 kg)[(1.0 m)+h]2 + (1.0 kg)[(2.0 m)− h]2

= 6.0 kg·m2 + (4.0 kg·m)h − (4.0 kg·m)h + (3.0 kg)h2.

The constant term is the same as the moment of inertia about thecenter-of-mass axis, the first-order terms cancel out, and the thirdterm is just the total mass multiplied by h2. The interested readerwill have no difficulty in generalizing this to any set of particles(problem 27, p. 483), resulting in the parallel axis theorem: If anobject of total mass M rotates about a line at a distance h fromits center of mass, then its moment of inertia equals Icm + Mh2,where Icm is the moment of inertia for rotation about a parallel linethrough the center of mass.

Scaling of the moment of inertia example 18. (1) Suppose two objects have the same mass and the sameshape, but one is less dense, and larger by a factor k . How dotheir moments of inertia compare?(2) What if the densities are equal rather than the masses?

. (1) This is like increasing all the distances between atoms by afactor k . All the r ’s become greater by this factor, so the momentof inertia is increased by a factor of k2.(2) This introduces an increase in mass by a factor of k3, so themoment of inertia of the bigger object is greater by a factor ofk5.

Iterated integrals

In various places in this book, starting with subsection 15.7.5,we’ll come across integrals stuck inside other integrals. These areknown as iterated integrals, or double integrals, triple integrals, etc.Similar concepts crop up all the time even when you’re not doingcalculus, so let’s start by imagining such an example. Suppose youwant to count how many squares there are on a chess board, and youdon’t know how to multiply eight times eight. You could start fromthe upper left, count eight squares across, then continue with thesecond row, and so on, until you how counted every square, givingthe result of 64. In slightly more formal mathematical language,we could write the following recipe: for each row, r, from 1 to 8,consider the columns, c, from 1 to 8, and add one to the count for


each one of them. Using the sigma notation, this becomes

8∑r=1

8∑c=1

1.

If you’re familiar with computer programming, then you can thinkof this as a sum that could be calculated using a loop nested insideanother loop. To evaluate the result (again, assuming we don’t knowhow to multiply, so we have to use brute force), we can first evaluatethe inside sum, which equals 8, giving

8∑r=1

8.

Notice how the “dummy” variable c has disappeared. Finally we dothe outside sum, over r, and find the result of 64.

Now imagine doing the same thing with the pixels on a TVscreen. The electron beam sweeps across the screen, painting thepixels in each row, one at a time. This is really no different than theexample of the chess board, but because the pixels are so small, younormally think of the image on a TV screen as continuous ratherthan discrete. This is the idea of an integral in calculus. Supposewe want to find the area of a rectangle of width a and height b, andwe don’t know that we can just multiply to get the area ab. Thebrute force way to do this is to break up the rectangle into a grid ofinfinitesimally small squares, each having width dx and height dy,and therefore the infinitesimal area dA = dx dy. For convenience,we’ll imagine that the rectangle’s lower left corner is at the origin.Then the area is given by this integral:

area =

∫ b

y=0

∫ a

x=0dA

=

∫ b

y=0

∫ a

x=0dx dy

Notice how the leftmost integral sign, over y, and the rightmost dif-ferential, dy, act like bookends, or the pieces of bread on a sandwich.Inside them, we have the integral sign that runs over x, and the dif-ferential dx that matches it on the right. Finally, on the innermostlayer, we’d normally have the thing we’re integrating, but here’s it’s1, so I’ve omitted it. Writing the lower limits of the integrals withx = and y = helps to keep it straight which integral goes with which


differential. The result is

area =

∫ b

y=0

∫ a

x=0dA

=

∫ b

y=0

∫ a

x=0dx dy

=

∫ b

y=0

(∫ a

x=0dx

)dy

=

∫ b

y=0ady

= a

∫ b

y=0dy

= ab.

Area of a triangle example 19. Find the area of a 45-45-90 right triangle having legs a.

. Let the triangle’s hypotenuse run from the origin to the point(a, a), and let its legs run from the origin to (0, a), and then to(a, a). In other words, the triangle sits on top of its hypotenuse.Then the integral can be set up the same way as the one before,but for a particular value of y , values of x only run from 0 (on they axis) to y (on the hypotenuse). We then have

area =∫ a

y=0

∫ y

x=0dA

=∫ a

y=0

∫ y

x=0dx dy

=∫ a

y=0

(∫ y

x=0dx)

dy

=∫ a

y=0y dy

=12

a2

Note that in this example, because the upper end of the x valuesdepends on the value of y , it makes a difference which order wedo the integrals in. The x integral has to be on the inside, and wehave to do it first.

Volume of a cube example 20. Find the volume of a cube with sides of length a.

. This is a three-dimensional example, so we’ll have integralsnested three deep, and the thing we’re integrating is the volumedV = dx dy dz.


volume =∫ a

z=0

∫ a

y=0

∫ a

x=0dx dy dz

=∫ a

z=0

∫ a

y=0a dy dz

= a∫ a

z=0

∫ a

y=0dy dz

= a∫ a

z=0a dz

= a3

Area of a circle example 21. Find the area of a circle.

. To make it easy, let’s find the area of a semicircle and thendouble it. Let the circle’s radius be r , and let it be centered on theorigin and bounded below by the x axis. Then the curved edgeis given by the equation r2 = x2 + y2, or y =

√r2 − x2. Since

the y integral’s limit depends on x , the x integral has to be on theoutside. The area is

area =∫ r

x=−r

∫ √r2−x2

y=0dy dx

=∫ r

x=−r

√r2 − x2 dx

= r∫ r

x=−r

√1− (x/r )2 dx .

Substituting u = x/r ,

area = r2∫ 1

u=−1

√1− u2 du

The definite integral equals π, as you can find using a trig sub-stitution or simply by looking it up in a table, and the result is, asexpected, πr2/2 for the area of the semicircle.

Finding moments of inertia by integration

When calculating the moment of inertia of an ordinary-sized ob-ject with perhaps 1026 atoms, it would be impossible to do an actualsum over atoms, even with the world’s fastest supercomputer. Cal-culus, however, offers a tool, the integral, for breaking a sum downto infinitely many small parts. If we don’t worry about the exis-tence of atoms, then we can use an integral to compute a moment


of inertia as if the object was smooth and continuous throughout,rather than granular at the atomic level. Of course this granular-ity typically has a negligible effect on the result unless the objectis itself an individual molecule. This subsection consists of threeexamples of how to do such a computation, at three distinct levelsof mathematical complication.

Moment of inertia of a thin rod

What is the moment of inertia of a thin rod of mass M andlength L about a line perpendicular to the rod and passing throughits center? We generalize the discrete sum

I =∑

mir2i

to a continuous one,

I =

∫r2 dm

=

∫ L/2

−L/2x2 M

Ldx [r = |x|, so r2 = x2]

=1

12ML2

In this example the object was one-dimensional, which madethe math simple. The next example shows a strategy that can beused to simplify the math for objects that are three-dimensional,but possess some kind of symmetry.

Moment of inertia of a disk

What is the moment of inertia of a disk of radius b, thickness t,and mass M , for rotation about its central axis?

We break the disk down into concentric circular rings of thick-ness dr. Since all the mass in a given circular slice has essentiallythe same value of r (ranging only from r to r + dr), the slice’s con-tribution to the total moment of inertia is simply r2 dm. We thenhave

I =

∫r2 dm

=

∫r2ρdV ,

where V = πb2t is the total volume, ρ = M/V = M/πb2t is thedensity, and the volume of one slice can be calculated as the volumeenclosed by its outer surface minus the volume enclosed by its innersurface, dV = π(r + dr)2t− πr2t = 2πtr dr.

I =

∫ b

0r2 M

πb2t2πt r dr

=1

2Mb2.


In the most general case where there is no symmetry about therotation axis, we must use iterated integrals, as discussed in subsec-tion 15.7.4. The example of the disk possessed two types of symme-try with respect to the rotation axis: (1) the disk is the same whenrotated through any angle about the axis, and (2) all slices perpen-dicular to the axis are the same. These two symmetries reduced thenumber of layers of integrals from three to one. The following ex-ample possesses only one symmetry, of type (2), and we simply setit up as a triple integral. You may not have seen multiple integralsyet in a math course. If so, just skim this example.

Moment of inertia of a cube

What is the moment of inertia of a cube of side b, for rotationabout an axis that passes through its center and is parallel to fourof its faces? Let the origin be at the center of the cube, and let xbe the rotation axis.

I =

∫r2 dm

= ρ

∫r2 dV

= ρ

∫ b/2

−b/2

∫ b/2

−b/2

∫ b/2

−b/2

(y2 + z2

)dx dy dz

= ρb

∫ b/2

−b/2

∫ b/2

−b/2

(y2 + z2

)dy dz

The fact that the last step is a trivial integral results from the sym-metry of the problem. The integrand of the remaining double in-tegral breaks down into two terms, each of which depends on onlyone of the variables, so we break it into two integrals,

I = ρb

∫ b/2

−b/2

∫ b/2

−b/2y2 dy dz + ρb

∫ b/2

−b/2

∫ b/2

−b/2z2 dy dz

which we know have identical results. We therefore only need toevaluate one of them and double the result:

I = 2ρb

∫ b/2

−b/2

∫ b/2

−b/2z2 dy d z

= 2ρb2∫ b/2

−b/2z2 dz

=1

6ρb5

=1

6Mb2

Figure ai shows the moments of inertia of some shapes, whichwere evaluated with techniques like these.


aj / Example 23.

ai / Moments of inertia of somegeometric shapes.

The hammer throw example 22. In the men’s Olympic hammer throw, a steel ball of radius 6.1 cmis swung on the end of a wire of length 1.22 m. What fraction ofthe ball’s angular momentum comes from its rotation, as opposedto its motion through space?

. It’s always important to solve problems symbolically first, andplug in numbers only at the end, so let the radius of the ball be b,and the length of the wire `. If the time the ball takes to go oncearound the circle is T , then this is also the time it takes to revolveonce around its own axis. Its speed is v = 2π`/T , so its angularmomentum due to its motion through space is mv` = 2πm`2/T .Its angular momentum due to its rotation around its own cen-ter is (4π/5)mb2/T . The ratio of these two angular momenta is(2/5)(b/`)2 = 1.0×10−3. The angular momentum due to the ball’sspin is extremely small.

Toppling a rod example 23. A rod of length b and mass m stands upright. We want to strikethe rod at the bottom, causing it to fall and land flat. Find themomentum, p, that should be delivered, in terms of m, b, andg. Can this really be done without having the rod scrape on thefloor?

. This is a nice example of a question that can very nearly beanswered based only on units. Since the three variables, m, b,and g, all have different units, they can’t be added or subtracted.The only way to combine them mathematically is by multiplicationor division. Multiplying one of them by itself is exponentiation, soin general we expect that the answer must be of the form

p = Amjbkg l ,

where A, j , k , and l are unitless constants. The result has to haveunits of kg·m/s. To get kilograms to the first power, we need

j = 1,


meters to the first power requires

k + l = 1,

and seconds to the power −1 implies

l = 1/2.

We find j = 1, k = 1/2, and l = 1/2, so the solution must be of theform

p = Am√

bg.

Note that no physics was required!

Consideration of units, however, won’t help us to find the unit-less constant A. Let t be the time the rod takes to fall, so that(1/2)gt2 = b/2. If the rod is going to land exactly on its side,then the number of revolutions it completes while in the air mustbe 1/4, or 3/4, or 5/4, . . . , but all the possibilities greater than 1/4would cause the head of the rod to collide with the floor prema-turely. The rod must therefore rotate at a rate that would causeit to complete a full rotation in a time T = 4t , and it has angularmomentum L = (π/6)mb2/T .

The momentum lost by the object striking the rod is p, and byconservation of momentum, this is the amount of momentum, inthe horizontal direction, that the rod acquires. In other words,the rod will fly forward a little. However, this has no effect onthe solution to the problem. More importantly, the object strikingthe rod loses angular momentum bp/2, which is also transferredto the rod. Equating this to the expression above for L, we findp = (π/12)m

√bg.

Finally, we need to know whether this can really be done withouthaving the foot of the rod scrape on the floor. The figure showsthat the answer is no for this rod of finite width, but it appearsthat the answer would be yes for a sufficiently thin rod. This isanalyzed further in homework problem 46 on page 486.


15.8 Angular momentum in three dimensionsConservation of angular momentum produces some surprising phe-nomena when extended to three dimensions. Try the following ex-periment, for example. Take off your shoe, and toss it in to the air,making it spin along its long (toe-to-heel) axis. You should observea nice steady pattern of rotation. The same happens when you spinthe shoe about its shortest (top-to-bottom) axis. But somethingunexpected happens when you spin it about its third (left-to-right)axis, which is intermediate in length between the other two. Insteadof a steady pattern of rotation, you will observe something morecomplicated, with the shoe changing its orientation with respect tothe rotation axis.

Rigid-body kinematics in three dimensions

How do we generalize rigid-body kinematics to three dimensions?When we wanted to generalize the kinematics of a moving particleto three dimensions, we made the numbers r, v, and a into vectorsr, v, and a. This worked because these quantities all obeyed thesame laws of vector addition. For instance, one of the laws of vectoraddition is that, just like addition of numbers, vector addition givesthe same result regardless of the order of the two quantities beingadded. Thus you can step sideways 1 m to the right and thenstep forward 1 m, and the end result is the same as if you steppedforward first and then to the side. In order words, it didn’t matterwhether you took ∆r1 + ∆r2 or ∆r2 + ∆r1. In math this is calledthe commutative property of addition.

ak / Performing the rotations inone order gives one result, 3,while reversing the order gives adifferent result, 5.

Angular motion, unfortunately doesn’t have this property, asshown in figure ak. Doing a rotation about the x axis and then

Section 15.8 Angular momentum in three dimensions 461

al / The right-hand rule forassociating a vector with adirection of rotation.

about y gives one result, while doing them in the opposite ordergives a different result. These operations don’t “commute,” i.e., itmakes a difference what order you do them in.

This means that there is in general no possible way to constructa ∆θ vector. However, if you try doing the operations shown infigure ak using small rotation, say about 10 degrees instead of 90,you’ll find that the result is nearly the same regardless of whatorder you use; small rotations are very nearly commutative. Notonly that, but the result of the two 10-degree rotations is about thesame as a single, somewhat larger, rotation about an axis that liessymmetrically at between the x and y axes at 45 degree angles toeach one. This is exactly what we would expect if the two smallrotations did act like vectors whose directions were along the axisof rotation. We therefore define a dθ vector whose magnitude isthe amount of rotation in units of radians, and whose direction isalong the axis of rotation. Actually this definition is ambiguous,because there it could point in either direction along the axis. Wetherefore use a right-hand rule as shown in figure al to define thedirection of the dθ vector, and the ω vector, ω = dθ/ dt, based onit. Aliens on planet Tammyfaye may decide to define it using theirleft hands rather than their right, but as long as they keep theirscientific literature separate from ours, there is no problem. Whenentering a physics exam, always be sure to write a large warning noteon your left hand in magic marker so that you won’t be tempted touse it for the right-hand rule while keeping your pen in your right.

self-check DUse the right-hand rule to determine the directions of the ω vectors ineach rotation shown in figures ak/1 through ak/5. . Answer, p. 563

Because the vector relationships among dθ, ω, and α are strictlyanalogous to the ones involving dr, v, and a (with the proviso thatwe avoid describing large rotations using ∆θ vectors), any operationin r-v-a vector kinematics has an exact analog in θ-ω-α kinematics.

Result of successive 10-degree rotations example 24. What is the result of two successive (positive) 10-degree rota-tions about the x and y axes? That is, what single rotation about asingle axis would be equivalent to executing these in succession?

. The result is only going to be approximate, since 10 degreesis not an infinitesimally small angle, and we are not told in whatorder the rotations occur. To some approximation, however, wecan add the ∆θ vectors in exactly the same way we would add ∆rvectors, so we have

∆θ ≈ ∆θ1 + ∆θ2

≈ (10 degrees)x + (10 degrees)y.

This is a vector with a magnitude of√

(10 deg)2 + (10 deg)2 =


14 deg, and it points along an axis midway between the x and yaxes.

Angular momentum in three dimensions

The vector cross product

In order to expand our system of three-dimensional kinematics toinclude dynamics, we will have to generalize equations like vt = ωr,τ = rF sin θrF , and L = rp sin θrp, each of which involves threequantities that we have either already defined as vectors or that wewant to redefine as vectors. Although the first one appears to differfrom the others in its form, it could just as well be rewritten asvt = ωr sin θωr, since θωr = 90◦, and sin θωr = 1.

It thus appears that we have discovered something general aboutthe physically useful way to relate three vectors in a multiplicativeway: the magnitude of the result always seems to be proportional tothe product of the magnitudes of the two vectors being “multiplied,”and also to the sine of the angle between them.

Is this pattern just an accident? Actually the sine factor hasa very important physical property: it goes to zero when the twovectors are parallel. This is a Good Thing. The generalization ofangular momentum into a three-dimensional vector, for example, ispresumably going to describe not just the clockwise or counterclock-wise nature of the motion but also from which direction we wouldhave to view the motion so that it was clockwise or counterclock-wise. (A clock’s hands go counterclockwise as seen from behind theclock, and don’t rotate at all as seen from above or to the side.) Nowsuppose a particle is moving directly away from the origin, so thatits r and p vectors are parallel. It is not going around the originfrom any point of view, so its angular momentum vector had betterbe zero.

Thinking in a slightly more abstract way, we would expect theangular momentum vector to point perpendicular to the plane ofmotion, just as the angular velocity vector points perpendicular tothe plane of motion. The plane of motion is the plane containingboth r and p, if we place the two vectors tail-to-tail. But if r andp are parallel and are placed tail-to-tail, then there are infinitelymany planes containing them both. To pick one of these planes inpreference to the others would violate the symmetry of space, sincethey should all be equally good. Thus the zero-if-parallel propertyis a necessary consequence of the underlying symmetry of the lawsof physics.

The following definition of a kind of vector multiplication is con-sistent with everything we’ve seen so far, and on p. 473 we’ll provethat the definition is unique, i.e., if we believe in the symmetry ofspace, it is essentially the only way of defining the multiplication of


am / The right-hand rule forthe direction of the vector crossproduct.

an / The magnitude of thecross product is the area of theshaded parallelogram.

ao / A cyclic change in the x ,y , and z subscripts.

two vectors to produce a third vector:

Definition of the vector cross product:The cross product A × B of two vectors A and B is defined asfollows:(1) Its magnitude is defined by |A×B| = |A||B| sin θAB, where θABis the angle between A and B when they are placed tail-to-tail.(2) Its direction is along the line perpendicular to both A and B.Of the two such directions, it is the one that obeys the right-handrule shown in figure am.

The name “cross product” refers to the symbol, and distin-guishes it from the dot product, which acts on two vectors butproduces a scalar.

Although the vector cross-product has nearly all the propertiesof numerical multiplication, e.g., A× (B + C) = A×B + A×C, itlacks the usual property of commutativity. Try applying the right-hand rule to find the direction of the vector cross product B × Ausing the two vectors shown in the figure. This requires startingwith a flattened hand with the four fingers pointing along B, andthen curling the hand so that the fingers point along A. The onlypossible way to do this is to point your thumb toward the floor, inthe opposite direction. Thus for the vector cross product we have

A×B = −B×A,

a property known as anticommutativity. The vector cross productis also not associative, i.e., A× (B×C) is usually not the same as(A×B)×C.

A geometric interpretation of the cross product, an, is that ifboth A and B are vectors with units of distance, then the mag-nitude of their cross product can be interpreted as the area of theparallelogram they form when placed tail-to-tail.

A useful expression for the components of the vector cross prod-uct in terms of the components of the two vectors being multipliedis as follows:

(A×B)x = AyBz −ByAz(A×B)y = AzBx −BzAx(A×B)z = AxBy −BxAy

I’ll prove later that these expressions are equivalent to the previ-ous definition of the cross product. Although they may appearformidable, they have a simple structure: the subscripts on the rightare the other two besides the one on the left, and each equation isrelated to the preceding one by a cyclic change in the subscripts,


ap / The position and momentumvectors of an atom in the spinningtop.

aq / The right-hand rule forthe atom’s contribution to theangular momentum.

ao. If the subscripts were not treated in some completely symmet-ric manner like this, then the definition would provide some way todistinguish one axis from another, which would violate the symme-try of space.

self-check EShow that the component equations are consistent with the rule A×B =−B× A. . Answer, p. 563

Angular momentum in three dimensions

In terms of the vector cross product, we have:

v = ω × r

L = r× p

τ = r× F

But wait, how do we know these equations are even correct?For instance, how do we know that the quantity defined by r × pis in fact conserved? Well, just as we saw on page 375 that thedot product is unique (i.e., can only be defined in one way whileobserving rotational invariance), the cross product is also unique,as proved on page 473. If r×p was not conserved, then there couldnot be any generally conserved quantity that would reduce to ourold definition of angular momentum in the special case of planerotation. This doesn’t prove conservation of angular momentum— only experiments can prove that — but it does prove that ifangular momentum is conserved in three dimensions, there is onlyone possible way to generalize from two dimensions to three.

Angular momentum of a spinning top example 25As an illustration, we consider the angular momentum of a spin-ning top. Figures ap and aq show the use of the vector crossproduct to determine the contribution of a representative atom tothe total angular momentum. Since every other atom’s angularmomentum vector will be in the same direction, this will also bethe direction of the total angular momentum of the top. This hap-pens to be rigid-body rotation, and perhaps not surprisingly, theangular momentum vector is along the same direction as the an-gular velocity vector.

Three important points are illustrated by this example: (1)When we do the full three-dimensional treatment of angular mo-mentum, the “axis” from which we measure the position vectors isjust an arbitrarily chosen point. If this had not been rigid-bodyrotation, we would not even have been able to identify a single lineabout which every atom circled. (2) Starting from figure ap, we hadto rearrange the vectors to get them tail-to-tail before applying theright-hand rule. If we had attempted to apply the right-hand ruleto figure ap, the direction of the result would have been exactly theopposite of the correct answer. (3) The equation L = r× p cannot


ar / A top is supported at itstip by a pinhead. (More practicaldevices to demonstrate thiswould use a double bearing.)

as / The torque made by gravityis in the horizontal plane.

at / The ∆L vector is in thesame direction as the torque, outof the page.

be applied all at once to an entire system of particles. The totalmomentum of the top is zero, which would give an erroneous resultof zero angular momentum (never mind the fact that r is not welldefined for the top as a whole).

Doing the right-hand rule like this requires some practice. Iurge you to make models like aq out of rolled up pieces of paper andto practice with the model in various orientations until it becomesnatural.

Precession example 26Figure ar shows a counterintuitive example of the concepts we’vebeen discussing. One expects the torque due to gravity to causethe top to flop down. Instead, the top remains spinning in the hor-izontal plane, but its axis of rotation starts moving in the directionshown by the shaded arrow. This phenomenon is called preces-sion. Figure as shows that the torque due to gravity is out of thepage. (Actually we should add up all the torques on all the atomsin the top, but the qualitative result is the same.) Since torqueis the rate of change of angular momentum, τ = dL/dt , the ∆Lvector must be in the same direction as the torque (division bya positive scalar doesn’t change the direction of the vector). Asshown in at, this causes the angular momentum vector to twist inspace without changing its magnitude.

For similar reasons, the Earth’s axis precesses once every 26,000years (although not through a great circle, since the angle betweenthe axis and the force isn’t 90 degrees as in figure ar). This pre-cession is due to a torque exerted by the moon. If the Earth wasa perfect sphere, there could be no precession effect due to sym-metry. However, the Earth’s own rotation causes it to be slightlyflattened (oblate) relative to a perfect sphere, giving it “love han-dles” on which the moon’s gravity can act. The moon’s gravity onthe nearer side of the equatorial bulge is stronger, so the torques donot cancel out perfectly. Presently the earth’s axis very nearly linesup with the star Polaris, but in 12,000 years, the pole star will beVega instead.

The frisbee example 27The flow of the air over a flying frisbee generates lift, and the liftat the front and back of the frisbee isn’t necessarily balanced. Ifyou throw a frisbee without rotating it, as if you were shooting abasketball with two hands, you’ll find that it pitches, i.e., its nosegoes either up or down. When I do this with my frisbee, it goesnose down, which apparently means that the lift at the back ofthe disc is greater than the lift at the front. The two torques areunbalanced, resulting in a total torque that points to the left.

The way you actually throw a frisbee is with one hand, putting alot of spin on it. If you throw backhand, which is how most peo-ple first learn to do it, the angular momentum vector points down


au / Example 28.

(assuming you’re right-handed). On my frisbee, the aerodynamictorque to the left would therefore tend to make the angular mo-mentum vector precess in the clockwise direction as seen by thethrower. This would cause the disc to roll to the right, and there-fore follow a curved trajectory. Some specialized discs, used inthe sport of disc golf, are actually designed intentionally to showthis behavior; they’re known as “understable” discs. However, thetypical frisbee that most people play with is designed to be stable:as the disc rolls to one side, the airflow around it is altered in waythat tends to bring the disc back into level flight. Such a disc willtherefore tend to fly in a straight line, provided that it is thrownwith enough angular momentum.

Finding a cross product by components example 28. What is the torque produced by a force given by x + 2y + 3z (inunits of Newtons) acting on a point whose radius vector is 4x + 5y(in meters)?

. It’s helpful to make a table of the components as shown in thefigure. The results are

τx = ryFz − Fy rz = 15 N·mτy = rzFx − Fzrx =−12 N·mτz = rxFy − Fx ry = 3 N·m

Torque and angular momentum example 29In this example, we prove explicitly the consistency of the equa-

tions involving torque and angular momentum that we provedabove based purely on symmetry. Starting from the definition oftorque, we have

τ =dLdt

=ddt

∑i

ri × pi

=∑

i

ddt

(ri × pi ).

The derivative of a cross product can be evaluated in the sameway as the derivative of an ordinary scalar product:

τ =∑

i

[(dri

dt× pi

)+(

ri ×dpi

dt

)]The first term is zero for each particle, since the velocity vector isparallel to the momentum vector. The derivative appearing in thesecond term is the force acting on the particle, so

τ =∑

i

ri × Fi ,

which is the relationship we set out to prove.


Rigid-body dynamics in three dimensions

The student who is not madly in love with mathematics maywish to skip the rest of this section after absorbing the statementthat, for a typical, asymmetric object, the angular momentum vectorand the angular velocity vector need not be parallel. That is, onlyfor a body that possesses symmetry about the rotation axis is it truethat L = Iω (the rotational equivalent of p = mv) for some scalarI.

Let’s evaluate the angular momentum of a rigidly rotating sys-tem of particles:

L =∑i

ri × pi

=∑i

miri × vi

=∑i

miri × (ω × ri)

An important mathematical skill is to know when to give up andback off. This is a complicated expression, and there is no reasonto expect it to simplify and, for example, take the form of a scalarmultiplied by ω. Instead we examine its general characteristics. Ifwe expanded it using the equation that gives the components of avector cross product, every term would have one of the ω compo-nents raised to the first power, multiplied by a bunch of other stuff.The most general possible form for the result is

Lx = Ixxωx + Ixyωy + Ixzωz

Ly = Iyxωx + Iyyωy + Iyzωz

Lz = Izxωx + Izyωy + Izzωz,

which you may recognize as a case of matrix multiplication. Themoment of inertia is not a scalar, and not a three-component vector.It is a matrix specified by nine numbers, called its matrix elements.

The elements of the moment of inertia matrix will depend on ourchoice of a coordinate system. In general, there will be some specialcoordinate system, in which the matrix has a simple diagonal form:

Lx = Ixxωx

Ly = Iyyωy

Lz = Izzωz.

The three special axes that cause this simplification are calledthe principal axes of the object, and the corresponding coordinatesystem is the principal axis system. For symmetric shapes such asa rectangular box or an ellipsoid, the principal axes lie along theintersections of the three symmetry planes, but even an asymmetricbody has principal axes.


We can also generalize the plane-rotation equationK = (1/2)Iω2

to three dimensions as follows:

K =∑i

1

2miv

2i

=1

2

∑i

mi(ω × ri) · (ω × ri)

We want an equation involving the moment of inertia, and this hassome evident similarities to the sum we originally wrote down forthe moment of inertia. To massage it into the right shape, we needthe vector identity (A×B)·C = (B×C)·A, which we state withoutproof. We then write

K =1

2

∑i

mi [ri × (ω × ri)] · ω

=1

2ω ·∑i

miri × (ω × ri)

=1

2L · ω

As a reward for all this hard work, let’s analyze the problem ofthe spinning shoe that I posed at the beginning of the chapter. Thethree rotation axes referred to there are approximately the principalaxes of the shoe. While the shoe is in the air, no external torques areacting on it, so its angular momentum vector must be constant inmagnitude and direction. Its kinetic energy is also constant. That’sin the room’s frame of reference, however. The principal axis frameis attached to the shoe, and tumbles madly along with it. In theprincipal axis frame, the kinetic energy and the magnitude of theangular momentum stay constant, but the actual direction of theangular momentum need not stay fixed (as you saw in the caseof rotation that was initially about the intermediate-length axis).Constant |L| gives

L2x + L2

y + L2z = constant.

In the principal axis frame, it’s easy to solve for the componentsof ω in terms of the components of L, so we eliminate ω from theexpression 2K = L · ω, giving

1

IxxL2x +

1

IyyL2y +

1

IzzL2z = constant #2.

The first equation is the equation of a sphere in the three di-mensional space occupied by the angular momentum vector, whilethe second one is the equation of an ellipsoid. The top figure cor-responds to the case of rotation about the shortest axis, which hasthe greatest moment of inertia element. The intersection of the two


av / Visualizing surfaces ofconstant energy and angularmomentum in Lx -Ly -Lz space.

aw / The Explorer I satellite.

surfaces consists only of the two points at the front and back of thesphere. The angular momentum is confined to one of these points,and can’t change its direction, i.e., its orientation with respect to theprincipal axis system, which is another way of saying that the shoecan’t change its orientation with respect to the angular momentumvector. In the bottom figure, the shoe is rotating about the longestaxis. Now the angular momentum vector is trapped at one of thetwo points on the right or left. In the case of rotation about theaxis with the intermediate moment of inertia element, however, theintersection of the sphere and the ellipsoid is not just a pair of iso-lated points but the curve shown with the dashed line. The relativeorientation of the shoe and the angular momentum vector can andwill change.

One application of the moment of inertia tensor is to video gamesthat simulate car racing or flying airplanes.

One more exotic example has to do with nuclear physics. Al-though you have probably visualized atomic nuclei as nothing morethan featureless points, or perhaps tiny spheres, they are often el-lipsoids with one long axis and two shorter, equal ones. Althougha spinning nucleus normally gets rid of its angular momentum viagamma ray emission within a period of time on the order of picosec-onds, it may happen that a deformed nucleus gets into a state inwhich has a large angular momentum is along its long axis, whichis a very stable mode of rotation. Such states can live for secondsor even years! (There is more to the story — this is the topic onwhich I wrote my Ph.D. thesis — but the basic insight applies eventhough the full treatment requires fancy quantum mechanics.)

Our analysis has so far assumed that the kinetic energy of ro-tation energy can’t be converted into other forms of energy such asheat, sound, or vibration. When this assumption fails, then rota-tion about the axis of least moment of inertia becomes unstable,and will eventually convert itself into rotation about the axis whosemoment of inertia is greatest. This happened to the U.S.’s first ar-tificial satellite, Explorer I, launched in 1958. Note the long, floppyantennas, which tended to dissipate kinetic energy into vibration. Ithad been designed to spin about its minimimum-moment-of-inertiaaxis, but almost immediately, as soon as it was in space, it beganspinning end over end. It was nevertheless able to carry out itsscience mission, which didn’t depend on being able to maintain astable orientation, and it discovered the Van Allen radiation belts.

15.9 ? Proof of Kepler’s elliptical orbit lawKepler determined purely empirically that the planets’ orbits wereellipses, without understanding the underlying reason in terms ofphysical law. Newton’s proof of this fact based on his laws of motion


ax / The r −ϕ representation of acurve.

ay / Proof that the two an-gles labeled ϕ are in fact equal:The definition of an ellipse is thatthe sum of the distances fromthe two foci stays constant. If wemove a small distance ` along theellipse, then one distance shrinksby an amount ` cosϕ1, while theother grows by ` cosϕ2. Theseare equal, so ϕ1 = ϕ2.

and law of gravity was considered his crowning achievement bothby him and by his contemporaries, because it showed that the samephysical laws could be used to analyze both the heavens and theearth. Newton’s proof was very lengthy, but by applying the morerecent concepts of conservation of energy and angular momentumwe can carry out the proof quite simply and succinctly, and withoutcalculus.

The basic idea of the proof is that we want to describe the shapeof the planet’s orbit with an equation, and then show that this equa-tion is exactly the one that represents an ellipse. Newton’s originalproof had to be very complicated because it was based directly onhis laws of motion, which include time as a variable. To make anystatement about the shape of the orbit, he had to eliminate timefrom his equations, leaving only space variables. But conservationlaws tell us that certain things don’t change over time, so they havealready had time eliminated from them.

There are many ways of representing a curve by an equation, ofwhich the most familiar is y = ax + b for a line in two dimensions.It would be perfectly possible to describe a planet’s orbit using anx − y equation like this, but remember that we are applying con-servation of angular momentum, and the space variables that occurin the equation for angular momentum are the distance from theaxis, r, and the angle between the velocity vector and the r vector,which we will call ϕ. The planet will have ϕ=90◦when it is movingperpendicular to the r vector, i.e., at the moments when it is at itssmallest or greatest distances from the sun. When ϕ is less than90◦the planet is approaching the sun, and when it is greater than90◦it is receding from it. Describing a curve with an r−ϕ equationis like telling a driver in a parking lot a certain rule for what direc-tion to steer based on the distance from a certain streetlight in themiddle of the lot.

The proof is broken into the three parts for easier digestion.The first part is a simple and intuitively reasonable geometrical factabout ellipses, whose proof we relegate to the caption of figure ay;you will not be missing much if you merely absorb the result withoutreading the proof.

(1) If we use one of the two foci of an ellipse as an axis fordefining the variables r and ϕ, then the angle between the tangentline and the line drawn to the other focus is the same as ϕ, i.e., thetwo angles labeled ϕ in figure ay are in fact equal.

The other two parts form the meat of our proof. We state theresults first and then prove them.

(2) A planet, moving under the influence of the sun’s gravitywith less than the energy required to escape, obeys an equation of

Section 15.9 ? Proof of Kepler’s elliptical orbit law 471

az / Proof of part (3).

the form

sinϕ =1√

−pr2 + qr,

where p and q are positive constants that depend on the planet’senergy and angular momentum.

(3) A curve is an ellipse if and only if its r−ϕ equation is of theform

sinϕ =1√

−pr2 + qr,

where p and q are positive constants that depend on the size andshape of the ellipse.

Proof of part (2)

The component of the planet’s velocity vector that is perpen-dicular to the r vector is v⊥ = v sinϕ, so conservation of angularmomentum tells us that L = mrv sinϕ is a constant. Since theplanet’s mass is a constant, this is the same as the condition

rv sinϕ = constant.

Conservation of energy gives

1

2mv2 − GMm

r= constant.

We solve the first equation for v and plug into the second equationto eliminate v. Straightforward algebra then leads to the equationclaimed above, with the constant p being positive because of ourassumption that the planet’s energy is insufficient to escape fromthe sun, i.e., its total energy is negative.

Proof of part (3)

We define the quantities α, d, and s as shown in the figure. Thelaw of cosines gives

d2 = r2 + s2 − 2rs cosα.

Using α = 180◦−2ϕ and the trigonometric identities cos(180◦−x) =− cosx and cos 2x = 1− 2 sin2 x, we can rewrite this as

d2 = r2 + s2 − 2rs(2 sin2 ϕ− 1

).

Straightforward algebra transforms this into

sin ϕ =

√(r + s)2 − d2

4rs.

Since r + s is constant, the top of the fraction is constant, and thedenominator can be rewritten as 4rs = 4r(constant − r), which isequivalent to the desired form.


15.10 Some theorems and proofsIn this section I prove three theorems stated earlier, and state afourth theorem whose proof is left as an exercise.

Uniqueness of the cross product

The vector cross product as we have defined it has the followingproperties:(1) It does not violate rotational invariance.(2) It has the property A× (B + C) = A×B + A×C.(3) It has the property A× (kB) = k(A×B), where k is a scalar.

Theorem: The definition we have given is the only possible methodof multiplying two vectors to make a third vector which has theseproperties, with the exception of trivial redefinitions which just in-volve multiplying all the results by the same constant or swappingthe names of the axes. (Specifically, using a left-hand rule ratherthan a right-hand rule corresponds to multiplying all the results by−1.)

Proof: We prove only the uniqueness of the definition, withoutexplicitly proving that it has properties (1) through (3).

Using properties (2) and (3), we can break down any vectormultiplication (Axx +Ayy +Azz)× (Bxx +Byy +Bzz) into termsinvolving cross products of unit vectors.

A “self-term” like x × x must either be zero or lie along thex axis, since any other direction would violate property (1). If itwas not zero, then (−x) × (−x) would have to lie in the oppositedirection to avoid breaking rotational invariance, but property (3)says that (−x)× (−x) is the same as x× x, which is a contradiction.Therefore the self-terms must be zero.

An “other-term” like x× y could conceivably have componentsin the x-y plane and along the z axis. If there was a nonzero compo-nent in the x-y plane, symmetry would require that it lie along thediagonal between the x and y axes, and similarly the in-the-planecomponent of (−x)× y would have to be along the other diagonal inthe x-y plane. Property (3), however, requires that (−x)× y equal−(x× y), which would be along the original diagonal. The only wayit can lie along both diagonals is if it is zero.

We now know that x × y must lie along the z axis. Since weare not interested in trivial differences in definitions, we can fixx × y = z, ignoring peurile possibilities such as x × y = 7z or theleft-handed definition x× y = −z. Given x× y = z, the symmetryof space requires that similar relations hold for y× z and z× x, withat most a difference in sign. A difference in sign could always beeliminated by swapping the names of some of the axes, so ignoringpossible trivial differences in definitions we can assume that thecyclically related set of relations x× y = z, y× z = x, and z× x = y

Section 15.10 Some theorems and proofs 473

holds. Since the arbitrary cross-product with which we started canbe broken down into these simpler ones, the cross product is uniquelydefined.

Choice of axis theorem

Theorem: Suppose a closed system of material particles conservesangular momentum in one frame of reference, with the axis takento be at the origin. Then conservation of angular momentum isunaffected if the origin is relocated or if we change to a frame ofreference that is in constant-velocity motion with respect to thefirst one. The theorem also holds in the case where the system isnot closed, but the total external force is zero.

Proof: In the original frame of reference, angular momentum isconserved, so we have dL/ dt=0. From example 29 on page 467,this derivative can be rewritten as

dL

dt=∑i

ri × Fi,

where Fi is the total force acting on particle i. In other words, we’readding up all the torques on all the particles.

By changing to the new frame of reference, we have changedthe position vector of each particle according to ri → ri + k − ut,where k is a constant vector that indicates the relative position ofthe new origin at t = 0, and u is the velocity of the new frame withrespect to the old one. The forces are all the same in the new frameof reference, however. In the new frame, the rate of change of theangular momentum is

dL

dt=∑i

(ri + k− ut)× Fi

=∑i

ri × Fi + (k− ut)×∑i

Fi.

The first term is the expression for the rate of change of the angu-lar momentum in the original frame of reference, which is zero byassumption. The second term vanishes by Newton’s third law; sincethe system is closed, every force Fi cancels with some force Fj . (Ifexternal forces act, but they add up to zero, then the sum can bebroken up into a sum of internal forces and a sum of external forces,each of which is zero.) The rate of change of the angular momentumis therefore zero in the new frame of reference.

Spin theorem

Theorem: An object’s angular momentum with respect to someoutside axis A can be found by adding up two parts:(1) The first part is the object’s angular momentum found by usingits own center of mass as the axis, i.e., the angular momentum the


object has because it is spinning.(2) The other part equals the angular momentum that the objectwould have with respect to the axis A if it had all its mass concen-trated at and moving with its center of mass.

Proof: Let the system’s center of mass be at rcm, and let particlei lie at position rcm + di. Then the total angular momentum is

L =∑i

(rcm + di)× pi

= rcm ×∑i

pi +∑i

di × pi,

which establishes the result claimed, since we can identify the firstterm with (2) and the second with (1).

Parallel axis theorem

Suppose an object has mass m, and moment of inertia Io for ro-tation about some axis A passing through its center of mass. Givena new axis B, parallel to A and lying at a distance h from it, theobject’s moment of inertia is given by Io +mh2.

The proof of this theorem is left as an exercise (problem 27,p. 483).

Section 15.10 Some theorems and proofs 475

SummarySelected vocabularyangular momen-tum . . . . . . . .

a measure of rotational motion; a conservedquantity for a closed system

axis . . . . . . . . An arbitrarily chosen point used in the defini-tion of angular momentum. Any object whosedirection changes relative to the axis is consid-ered to have angular momentum. No matterwhat axis is chosen, the angular momentum ofa closed system is conserved.

torque . . . . . . the rate of change of angular momentum; anumerical measure of a force’s ability to twiston an object

equilibrium . . . a state in which an object’s momentum andangular momentum are constant

stable equilibrium one in which a force always acts to bring theobject back to a certain point

unstable equilib-rium . . . . . . . .

one in which any deviation of the object fromits equilibrium position results in a force push-ing it even farther away

NotationL . . . . . . . . . . angular momentumt . . . . . . . . . . torqueT . . . . . . . . . the period the time required for a rigidly ro-

tating body to complete one rotationω . . . . . . . . . . the angular velocity, dθ/dtmoment of iner-tia, I . . . . . . .

the proportionality constant in the equationL = Iω

Summary

Angular momentum is a measure of rotational motion which isconserved for a closed system. This book only discusses angularmomentum for rotation of material objects in two dimensions. Notall rotation is rigid like that of a wheel or a spinning top. An exampleof nonrigid rotation is a cyclone, in which the inner parts take lesstime to complete a revolution than the outer parts. In order to definea measure of rotational motion general enough to include nonrigidrotation, we define the angular momentum of a system by dividingit up into small parts, and adding up all the angular momenta ofthe small parts, which we think of as tiny particles. We arbitrarilychoose some point in space, the axis, and we say that anythingthat changes its direction relative to that point possesses angularmomentum. The angular momentum of a single particle is

L = mv⊥r,

where v⊥ is the component of its velocity perpendicular to the linejoining it to the axis, and r is its distance from the axis. Positive and


negative signs of angular momentum are used to indicate clockwiseand counterclockwise rotation.

The choice of axis theorem states that any axis may be used fordefining angular momentum. If a system’s angular momentum isconstant for one choice of axis, then it is also constant for any otherchoice of axis.

The spin theorem states that an object’s angular momentumwith respect to some outside axis A can be found by adding up twoparts:

(1) The first part is the object’s angular momentum found byusing its own center of mass as the axis, i.e., the angular momentumthe object has because it is spinning.

(2) The other part equals the angular momentum that the ob-ject would have with respect to the axis A if it had all its massconcentrated at and moving with its center of mass.

Torque is the rate of change of angular momentum. The torquea force can produce is a measure of its ability to twist on an object.The relationship between force and torque is

|τ | = r|F⊥|,

where r is the distance from the axis to the point where the force isapplied, and F⊥ is the component of the force perpendicular to theline connecting the axis to the point of application. Statics problemscan be solved by setting the total force and total torque on an objectequal to zero and solving for the unknowns.

In the special case of a rigid body rotating in a single plane, wedefine

ω =dθ

dt[angular velocity]

and

α =dω

dt, [angular acceleration]

in terms of which we have

L = Iω

and

τ = Iα,

where the moment of inertia, I, is defined as

I =∑

mir2i ,

Summary 477

summing over all the atoms in the object (or using calculus to per-form a continuous sum, i.e. an integral). The relationship betweenthe angular quantities and the linear ones is

vt = ωr [tangential velocity of a point]

vr = 0 [radial velocity of a point]

at = αr. [radial acceleration of a point]

at a distance r from the axis]

ar = ω2r [radial acceleration of a point]

at a distance r from the axis]

In three dimensions, torque and angular momentum are vectors,and are expressed in terms of the vector cross product, which is theonly rotationally invariant way of defining a multiplication of twovectors that produces a third vector:

L = r× p

τ = r× F

In general, the cross product of vectors b and c has magnitude

|b× c| = |b| |c| sin θbc,

which can be interpreted geometrically as the area of the parallel-ogram formed by the two vectors when they are placed tail-to-tail.The direction of the cross product lies along the line which is per-pendicular to both vectors; of the two such directions, we choose theone that is right-handed, in the sense that if we point the fingers ofthe flattened right hand along b, then bend the knuckles to pointthe fingers along c, the thumb gives the direction of b× c. In termsof components, the cross product is

(b× c)x = bycz − cybz(b× c)y = bzcx − czbx(b× c)z = bxcy − cxby

The cross product has the disconcerting properties

a× b = −b× a [noncommutative]

and

a× (b× c) 6= (a× b)× c [nonassociative],

and there is no “cross-division.”

For rigid-body rotation in three dimensions, we define an angularvelocity vector ω, which lies along the axis of rotation and bears aright-hand relationship to it. Except in special cases, there is noscalar moment of inertia for which L = Iω; the moment of inertiamust be expressed as a matrix.


Problem 4.

ProblemsKey√

A computerized answer check is available online.∫A problem that requires calculus.

? A difficult problem.

1 A skilled motorcyclist can ride up a ramp, fly through the air,and land on another ramp. Why would it be useful for the rider tospeed up or slow down the back wheel while in the air?

2 An object thrown straight up in the air is momentarily at restwhen it reaches the top of its motion. Does that mean that it is inequilibrium at that point? Explain.

3 An object is observed to have constant angular momentum.Can you conclude that no torques are acting on it? Explain. [Basedon a problem by Serway and Faughn.]

4 The sun turns on its axis once every 26.0 days. Its mass is2.0 × 1030 kg and its radius is 7.0 × 108 m. Assume it is a rigidsphere of uniform density.(a) What is the sun’s angular momentum?

√

In a few billion years, astrophysicists predict that the sun will useup all its sources of nuclear energy, and will collapse into a ball ofexotic, dense matter known as a white dwarf. Assume that its radiusbecomes 5.8 × 106 m (similar to the size of the Earth.) Assume itdoes not lose any mass between now and then. (Don’t be fooledby the photo, which makes it look like nearly all of the star wasthrown off by the explosion. The visually prominent gas cloud isactually thinner than the best laboratory vacuum ever produced onearth. Certainly a little bit of mass is actually lost, but it is not atall unreasonable to make an approximation of zero loss of mass aswe are doing.)(b) What will its angular momentum be?(c) How long will it take to turn once on its axis?

√

5 (a) Alice says Cathy’s body has zero momentum, but Bobsays Cathy’s momentum is nonzero. Nobody is lying or making amistake. How is this possible? Give a concrete example.(b) Alice and Bob agree that Dong’s body has nonzero momentum,but disagree about Dong’s angular momentum, which Alice says iszero, and Bob says is nonzero. Explain.

6 Two objects have the same momentum vector. Assume thatthey are not spinning; they only have angular momentum due totheir motion through space. Can you conclude that their angularmomenta are the same? Explain. [Based on a problem by Serwayand Faughn.]

Problems 479

Problem 8.

Problem 9.

Problem 13.

7 You are trying to loosen a stuck bolt on your RV using a bigwrench that is 50 cm long. If you hang from the wrench, and yourmass is 55 kg, what is the maximum torque you can exert on thebolt?

√

8 The figure shows scale drawing of a pair of pliers being used tocrack a nut, with an appropriately reduced centimeter grid. Warn-ing: do not attempt this at home; it is bad manners. If the forcerequired to crack the nut is 300 N, estimate the force required of theperson’s hand. . Solution, p. 557

9 Make a rough estimate of the mechanical advantage of thelever shown in the figure. In other words, for a given amount offorce applied on the handle, how many times greater is the resultingforce on the cork?

10 A physical therapist wants her patient to rehabilitate hisinjured elbow by laying his arm flat on a table, and then lifting a2.1 kg mass by bending his elbow. In this situation, the weight is33 cm from his elbow. He calls her back, complaining that it hurtshim to grasp the weight. He asks if he can strap a bigger weightonto his arm, only 17 cm from his elbow. How much mass shouldshe tell him to use so that he will be exerting the same torque? (Heis raising his forearm itself, as well as the weight.)

√

11 Two horizontal tree branches on the same tree have equaldiameters, but one branch is twice as long as the other. Give aquantitative comparison of the torques where the branches join thetrunk. [Thanks to Bong Kang.]

12 A ball is connected by a string to a vertical post. The ball isset in horizontal motion so that it starts winding the string aroundthe post. Assume that the motion is confined to a horizontal plane,i.e., ignore gravity. Michelle and Astrid are trying to predict thefinal velocity of the ball when it reaches the post. Michelle saysthat according to conservation of angular momentum, the ball hasto speed up as it approaches the post. Astrid says that according toconservation of energy, the ball has to keep a constant speed. Whois right? [Hint: How is this different from the case where you whirla rock in a circle on a string and gradually reel in the string?]

13 A person of weight W stands on the ball of one foot. Findthe tension in the calf muscle and the force exerted by the shinboneson the bones of the foot, in terms of W , a, and b. For simplicity,assume that all the forces are at 90-degree angles to the foot, i.e.,neglect the angle between the foot and the floor.

√


Problem 14.

Problems 15 and 16.

Problem 17.

14 The rod in the figure is supported by the finger and thestring.(a) Find the tension, T , in the string, and the force, F , from thefinger, in terms of m, b,L, and g.

√

(b) Comment on the cases b = L and b = L/2.(c) Are any values of b unphysical?

15 A uniform ladder of mass m and length L leans against asmooth wall, making an angle θ with respect to the ground. The dirtexerts a normal force and a frictional force on the ladder, producinga force vector with magnitude F1 at an angle φ with respect to theground. Since the wall is smooth, it exerts only a normal force onthe ladder; let its magnitude be F2.(a) Explain why φ must be greater than θ. No math is needed.(b) Choose any numerical values you like for m and L, and showthat the ladder can be in equilibrium (zero torque and zero totalforce vector) for θ = 45.00◦ and φ = 63.43◦.

16 Continuing problem 15, find an equation for φ in terms ofθ, and show that m and L do not enter into the equation. Do notassume any numerical values for any of the variables. You will needthe trig identity sin(a− b) = sin a cos b− sin b cos a. (As a numericalcheck on your result, you may wish to check that the angles givenin part b of the previous problem satisfy your equation.)

√?

17 (a) Find the minimum horizontal force which, applied atthe axle, will pull a wheel over a step. Invent algebra symbols forwhatever quantities you find to be relevant, and give your answerin symbolic form. [Hints: There are four forces on the wheel atfirst, but only three when it lifts off. Normal forces are alwaysperpendicular to the surface of contact. Note that the corner of thestep cannot be perfectly sharp, so the surface of contact for thisforce really coincides with the surface of the wheel.](b) Under what circumstances does your result become infinite?Give a physical interpretation.

18 In the 1950’s, serious articles began appearing in magazineslike Life predicting that world domination would be achieved by thenation that could put nuclear bombs in orbiting space stations, fromwhich they could be dropped at will. In fact it can be quite difficultto get an orbiting object to come down. Let the object have energyE = KE +PE and angular momentum L. Assume that the energyis negative, i.e., the object is moving at less than escape velocity.Show that it can never reach a radius less than

rmin =GMm

2E

(−1 +

√1 +

2EL2

G2M2m3

).

[Note that both factors are negative, giving a positive result.]

Problems 481

Problem 19.

19 You wish to determine the mass of a ship in a bottle withouttaking it out. Show that this can be done with the setup shown inthe figure, with a scale supporting the bottle at one end, providedthat it is possible to take readings with the ship slid to severaldifferent locations. Note that you can’t determine the position ofthe ship’s center of mass just by looking at it, and likewise for thebottle. In particular, you can’t just say, “position the ship right ontop of the fulcrum” or “position it right on top of the balance.” ?

20 Two atoms will interact via electrical forces between theirprotons and electrons. One fairly good approximation to the poten-tial energy is the Lennard-Jones potential,

PE(r) = k

[(ar

)12− 2

(ar

)6]

,

where r is the center-to-center distance between the atoms.

Show that (a) there is an equilibrium point at r = a, (b) the equi-librium is stable, and (c) the energy required to bring the atomsfrom their equilibrium separation to infinity is k. [Hints: The firsttwo parts can be done more easily by setting a = 1, since the valueof a only changes the distance scale. One way to do part b is bygraphing.]

21 Suppose that we lived in a universe in which Newton’s lawof gravity gave forces proportional to r−7 rather than r−2. Which,if any, of Kepler’s laws would still be true? Which would be com-pletely false? Which would be different, but in a way that could becalculated with straightforward algebra?

22 Show that a sphere of radius R that is rolling without slippinghas angular momentum and momentum in the ratio L/p = (2/5)R.

23 Suppose a bowling ball is initially thrown so that it has noangular momentum at all, i.e., it is initially just sliding down thelane. Eventually kinetic friction will get it spinning fast enough sothat it is rolling without slipping. Show that the final velocity of theball equals 5/7 of its initial velocity. [Hint: You’ll need the result ofproblem 22.]

24 Penguins are playful animals. Tux the Penguin invents a newgame using a natural circular depression in the ice. He waddles attop speed toward the crater, aiming off to the side, and then hopsinto the air and lands on his belly just inside its lip. He then belly-surfs, moving in a circle around the rim. The ice is frictionless, sohis speed is constant. Is Tux’s angular momentum zero, or nonzero?What about the total torque acting on him? Take the center of thecrater to be the axis. Explain your answers.


Problem 28.

Problem 29.

Problem 30.

Problem 32

25 A massless rod of length ` has weights, each of mass m, at-tached to its ends. The rod is initially put in a horizontal position,and laid on an off-center fulcrum located at a distance b from therod’s center. The rod will topple. (a) Calculate the total gravita-tional torque on the rod directly, by adding the two torques. (b)Verify that this gives the same result as would have been obtainedby taking the entire gravitational force as acting at the center ofmass.

26 Use analogies to find the equivalents of the following equationsfor rotation in a plane:

KE = p2/2m

∆x = vo∆t+ (1/2)a∆t2

W = F∆x

Example: v = ∆x/∆t→ ω = ∆θ/∆t

27 Prove the parallel axis theorem stated on page 475.

28 The box shown in the figure is being accelerated by pullingon it with the rope.(a) Assume the floor is frictionless. What is the maximum forcethat can be applied without causing the box to tip over?

. Hint, p. 542√

(b) Repeat part a, but now let the coefficient of friction be µ.√

(c) What happens to your answer to part b when the box is suffi-ciently tall? How do you interpret this?

29 (a) The bar of mass m is attached at the wall with a hinge,and is supported on the right by a massless cable. Find the tension,T , in the cable in terms of the angle θ.

√

(b) Interpreting your answer to part a, what would be the best angleto use if we wanted to minimize the strain on the cable?(c) Again interpreting your answer to part a, for what angles doesthe result misbehave mathematically? Interpet this physically.

30 (a) The two identical rods are attached to one another witha hinge, and are supported by the two massless cables. Find theangle α in terms of the angle β, and show that the result is a purelygeometric one, independent of the other variables involved.

√

(b) Using your answer to part a, sketch the configurations for β → 0,β = 45◦, and β = 90◦. Do your results make sense intuitively?

31 (a) Find the angular velocities of the earth’s rotation and ofthe earth’s motion around the sun.

√

(b) Which motion involves the greater acceleration?

32 Give a numerical comparison of the two molecules’ momentsof inertia for rotation in the plane of the page about their centersof mass.

√

Problems 483

Problem 39

33 Find the angular momentum of a particle whose position isr = 3x− y+ z (in meters) and whose momentum is p = −2x+ y+ z(in kg·m/s).

√

34 Find a vector that is perpendicular to both of the followingtwo vectors:

x + 2y + 3z

4x + 5y + 6z

√

35 Prove property (3) of the vector cross product from thetheorem on page 473.

36 Prove the anticommutative property of the vector cross prod-uct, A×B = −B×A, using the expressions for the components ofthe cross product.

37 Find three vectors with which you can demonstrate that thevector cross product need not be associative, i.e., that A× (B×C)need not be the same as (A×B)×C.

38 Which of the following expressions make sense, and which arenonsense? For those that make sense, indicate whether the result isa vector or a scalar.(a) (A×B)×C(b) (A×B) ·C(c) (A ·B)×C

39 (a) As suggested in the figure, find the area of the infinites-imal region expressed in polar coordinates as lying between r andr + dr and between θ and θ + dθ.

√

(b) Generalize this to find the infinitesimal element of volume incylindrical coordinates (r, θ, z), where the Cartesian z axis is per-pendicular to the directions measured by r and θ.

√

(c) Find the moment of inertia for rotation about its axis of a conewhose mass is M , whose height is h, and whose base has a radiusb.

√

40 Find the moment of inertia of a solid rectangular box of massM and uniform density, whose sides are of length a, b, and c, forrotation about an axis through its center parallel to the edges oflength a.

√


41 The nucleus 168Er (erbium-168) contains 68 protons (whichis what makes it a nucleus of the element erbium) and 100 neutrons.It has an ellipsoidal shape like an American football, with one longaxis and two short axes that are of equal diameter. Because thisis a subatomic system, consisting of only 168 particles, its behaviorshows some clear quantum-mechanical properties. It can only havecertain energy levels, and it makes quantum leaps between theselevels. Also, its angular momentum can only have certain values,which are all multiples of 2.109× 10−34 kg ·m2/s. The table showssome of the observed angular momenta and energies of 168Er, in SIunits (kg ·m2/s and joules).L× 1034 E × 1014

0 02.109 1.27864.218 4.23116.327 8.79198.437 14.873110.546 22.379812.655 31.13514.764 41.20616.873 52.223

(a) These data can be described to a good approximation as a rigidend-over-end rotation. Estimate a single best-fit value for the mo-ment of inertia from the data, and check how well the data agreewith the assumption of rigid-body rotation. . Hint, p. 542

√

(b) Check whether this moment of inertia is on the right order ofmagnitude. The moment of inertia depends on both the size andthe shape of the nucleus. For the sake of this rough check, ignorethe fact that the nucleus is not quite spherical. To estimate its size,use the fact that a neutron or proton has a volume of about 1 fm3

(one cubic femtometer, where 1 fm = 10−15 m), and assume theyare closely packed in the nucleus.

42 (a) Prove the identity a × (b × c) = b(a · c) − c(a · b)by expanding the product in terms of its components. Note thatbecause the x, y, and z components are treated symmetrically inthe definitions of the vector cross product, it is only necessary tocarry out the proof for the x component of the result.(b) Applying this to the angular momentum of a rigidly rotatingbody, L =

∫r× (ω× r) dm, show that the diagonal elements of the

moment of inertia tensor can be expressed as, e.g., Ixx =∫

(y2 +z2) dm.(c) Find the diagonal elements of the moment of inertia matrix ofan ellipsoid with axes of lengths a, b, and c, in the principal-axisframe, and with the axis at the center.

√?

Problems 485

Problem 44.

43 When we talk about rigid-body rotation, the concept of aperfectly rigid body can only be an idealization. In reality, anyobject will compress, expand, or deform to some extent when sub-jected to the strain of rotation. However, if we let it settle down fora while, perhaps it will reach a new equilibrium. As an example,suppose we fill a centrifuge tube with some compressible substancelike shaving cream or Wonder Bread. We can model the contents ofthe tube as a one-dimensional line of mass, extending from r = 0 tor = `. Once the rotation starts, we expect that the contents will bemost compressed near the “floor” of the tube at r = `; this is bothbecause the inward force required for circular motion increases withr for a fixed ω, and because the part at the floor has the greatestamount of material pressing “down” (actually outward) on it. Thelinear density dm/dr, in units of kg/m, should therefore increase asa function of r. Suppose that we have dm/dr = µer/`, where µ is aconstant. Find the moment of inertia.

√

44 Two bars of length L are connected with a hinge and placedon a frictionless cylinder of radius r. (a) Show that the angle θ shownin the figure is related to the unitless ratio r/L by the equation

r

L=

cos2 θ

2 tan θ.

(b) Discuss the physical behavior of this equation for very large andvery small values of r/L. ?

45 Let two sides of a triangle be given by the vectors A andB, with their tails at the origin, and let mass m be uniformly dis-tributed on the interior of the triangle. (a) Show that the distanceof the triangle’s center of mass from the intersection of sides A andB is given by 1

3 |A + B|.(b) Consider the quadrilateral with mass 2m, and vertices at theorigin, A, B, and A + B. Show that its moment of inertia, forrotation about an axis perpendicular to it and passing through itscenter of mass, is m

6 (A2 +B2).(c) Show that the moment of inertia for rotation about an axis per-pendicular to the plane of the original triangle, and passing throughits center of mass, is m

18(A2 +B2−A ·B). Hint: Combine the resultsof parts a and b with the result of problem 27. ?

46 In example 23 on page 459, prove that if the rod is sufficientlythin, it can be toppled without scraping on the floor.

. Solution, p. 557 ?


47 A yo-yo of total mass m consists of two solid cylinders ofradius R, connected by a small spindle of negligible mass and radiusr. The top of the string is held motionless while the string unrollsfrom the spindle. Show that the acceleration of the yo-yo is g/(1 +R2/2r2). [Hint: The acceleration and the tension in the string areunknown. Use τ = ∆L/∆t and F = ma to determine these twounknowns.] ?

48 We have n identical books of width w, and we wish to stackthem at the edge of a table so that they extend the maximum pos-sible distance Ln beyond the edge. Surprisingly, it is possible tohave values of Ln that are greater than w, even with fairly small n.For large n, however, Ln begins to grow very slowly. Our goal is tofind Ln for a given n. We adopt the restriction that only one bookis ever used at a given height.2 (a) Use proof by induction to findLn, expressing your result as a sum. (b) Find a sufficiently tightlower bound on this sum, as a closed-form expression, to prove that1,202,604 books suffice for L > 7w. ?

49 A certain function f takes two vectors as inputs and givesan output that is also a vector. The function can be defined in sucha way that it is rotationally invariant, and it is also well definedregardless of the units of the vectors. It takes on the followingvalues for the following inputs:

f(x, y) = −z

f(2x, y) = −8z

f(x, 2y) = −2z

Prove that the given information uniquely determines f , and givean explicit expression for it. ?

50 (a) Find the moment of inertia of a uniform square of massm and with sides of length b, for rotation in its own plane, aboutone of its corners.

√

(b) The square is balanced on one corner on a frictionless surface.An infinitesimal perturbation causes it to topple. Find its angularvelocity at the moment when its side slaps the surface.

√?

2When this restriction is lifted, the calculation of Ln becomes a much moredifficult problem, which was partially solved in 2009 by Paterson, Peres, Thorup,Winkler, and Zwick.

Problems 487

Problem 51.

51 The figure shows a microscopic view of the innermost tracksof a music CD. The pits represent the pattern of ones and zeroesthat encode the musical waveform. Because the laser that reads thedata has to sweep over a fixed amount of data per unit time, thedisc spins at a decreasing angular velocity as the music is playedfrom the inside out. The linear velocity v, not the angular velocity,is constant. Each track is separated from its neighbors on eitherside by a fixed distance p, called the pitch. Although the tracksare actually concentric circles, we will idealize them in this problemas a type of spiral, called an Archimedean spiral, whose turns haveconstant spacing, p, along any radial line. Our goal is to find theangular acceleration of this idealized CD, in terms of the constantsv and p, and the radius r at which the laser is positioned.(a) Use geometrical reasoning to constrain the dependence of theresult on p.(b) Use units to further constrain the result up to a unitless multi-plicative constant.(c) Find the full result. [Hint: Find a differential equation involvingr and its time derivative, and then solve this equation by separatingvariables.]

√

(d) Consider the signs of the variables in your answer to part c,and show that your equation still makes sense when the direction ofrotation is reversed.(e) Similarly, check that your result makes sense regardless of whetherwe view the CD player from the front or the back. (Clockwise seenfrom one side is counterclockwise from the other.) ?

52 Neutron stars are the collapsed remnants of dead stars. Theyrotate quickly, and their rotation can be measured extremely accu-rately by radio astronomers. Some of them rotate at such a pre-dictable rate that they can be used to count time about as accu-rately as the best atomic clocks. They do decelerate slowly, butthis deceleration can be taken into account. One of the best-studiedstars of this type3 was observed continuously over a 10-year period.As of the benchmark date April 5, 2001, it was found to have

ω = 1.091313551502333× 103 s−1

and

α = −1.085991× 10−14 s−2,

where the error bars in the final digit of each number are about±1. Astronomers often use the Julian year as their unit of time,where one Julian year is defined to be exactly 3.15576× 107 s. Findthe number of revolutions that this pulsar made over a period of 10Julian years, starting from the benchmark date.

3Verbiest et al., Astrophysical Journal 679 (675) 2008


Problems 57 and 58.

√

53 A disk, initially rotating at 120 radians per second, is sloweddown with a constant angular acceleration of magnitude 4.0 s−2.How many revolutions does the disk make before it comes to rest?[Problem by B. Shotwell.]

√

54 A bell rings at the Tilden Park merry go round in Berkeley,California, and the carousel begins to move with an angular accel-eration of 1.0 × 10−2 s−2. How much time does it take to performits first revolution?

√

55 A gasoline-powered car has a heavy wheel called a flywheel,whose main function is to add inertia to the motion of the engineso that it keeps spinning smoothly between power strokes of thecylinders. Suppose that a certain car’s flywheel is spinning withangular velocity ω0, but the car is then turned off, so that the engineand flywheel start to slow down as a result of friction. Assume thatthe angular acceleration is constant. After the flywheel has made Nrevolutions, it comes to rest. What is the magnitude of the angularacceleration? [Problem by B. Shotwell.]

√

56 A rigid body rotates about a line according to θ = At3 −Bt(valid for both negative and positive t).(a) What is the angular velocity as a function of time?

√

(b) What is the angular acceleration as a function of time?√

(c) There are two times when the angular velocity is zero. What isthe positive time for which this is true? Call this t+.

√

(d) What is the average angular velocity over the time interval from0 to t+? [Problem by B. Shotwell.]

√

57 A bug stands on a horizontal turntable at distance r fromthe center. The coefficient of static friction between the bug and theturntable is µs. The turntable spins at constant angular frequencyω.(a) Is the bug more likely to slip at small values of r, or large values?(b) If the bug walks along a radius, what is the value of r at whichit looses its footing? [Problem by B. Shotwell.]

√

58 A bug stands on a horizontal turntable at distance r fromthe center. The coefficient of static friction between the bug and theturntable is µs. Starting from rest, the turntable begins rotatingwith angular acceleration α. What is the magnitude of the angularfrequency at which the bug starts to slide? [Problem by B. Shotwell.]√

?

Problems 489

Problem 59.

Problem 60.

Problem 62.

59 The figure shows a tabletop experiment that can be used todetermine an unknown moment of inertia. A rotating platform ofradius R has a string wrapped around it. The string is threadedover a pulley and down to a hanging weight of mass m. The massis released from rest, and its downward acceleration a (a > 0) ismeasured. Find the total moment of inertia I of the platform plusthe object sitting on top of it. (The moment of inertia of the objectitself can then be found by subtracting the value for the emptyplatform.)

√

60 The uniform cube has unit weight and sides of unit length.One corner is attached to a universal joint, i.e., a frictionless bearingthat allows any type of rotation. If the cube is in equilibrium, findthe magnitudes of the forces a, b, and c.

√

61 In this problem we investigate the notion of division by avector.(a) Given a nonzero vector a and a scalar b, suppose we wish to finda vector u that is the solution of a · u = b. Show that the solutionis not unique, and give a geometrical description of the solution set.(b) Do the same thing for the equation a× u = c.(c) Show that the simultaneous solution of these two equations ex-ists and is unique.

Remark: This is one motivation for constructing the number system called thequaternions. For a certain period around 1900, quaternions were more popularthan the system of vectors and scalars more commonly used today. They stillhave some important advantages over the scalar-vector system for certain appli-cations, such as avoiding a phenomenon known as gimbal lock in controlling theorientation of bodies such as spacecraft. ?

62 The figure shows a slab of mass M rolling freely down aninclined plane inclined at an angle θ to the horizontal. The slab ison top of a set of rollers, each of radius r, that roll without slippingat their top and bottom surfaces. The rollers may for example becylinders, or spheres such as ball bearings. Each roller’s center ofmass coincides with its geometrical center. The sum of the massesof the rollers is m, and the sum of their moments of inertia (eachabout its own center) is I. Find the acceleration of the slab, andverify that your expression has the correct behavior in interestinglimiting cases.

√?

63 Vector A = (3.0x − 4.0y) meters, and vector B = (5.0x +12.0y) meters. Find the following: (a) The magnitude of vectorA− 2B.

√

(b) The dot product A ·B.√

(c) The cross product A × B (expressing the result in terms of itscomponents).

√

(d) The value of (A + B) · (A−B).√

(e) The angle between the two vectors. √[problem by B. Shotwell]


64 A disk starts from rest and rotates about a fixed axis, subjectto a constant torque. The work done by the torque during the firstrevolution is W . What is the work done by the torque during thesecond revolution?

√[problem by B. Shotwell]

65 Show that when a thin, uniform ring rotates about a diameter,the moment of inertia is half as big as for rotation about the axis ofsymmetry. . Solution, p. 558

Problems 491

Exercise 15: TorqueEquipment:

• rulers with holes in them

• spring scales (two per group)

While one person holds the pencil which forms the axle for the ruler, the other members of thegroup pull on the scale and take readings. In each case, calculate the total torque on the ruler,and find out whether it equals zero to roughly within the accuracy of the experiment. Finishthe calculations for each part before moving on to the next one.


Vibrations and resonance

493

The vibrations of this electric bassstring are converted to electricalvibrations, then to sound vibra-tions, and finally to vibrations ofour eardrums.

Chapter 16

Vibrations

Dandelion. Cello. Read those two words, and your brain instantlyconjures a stream of associations, the most prominent of which haveto do with vibrations. Our mental category of “dandelion-ness” isstrongly linked to the color of light waves that vibrate about half amillion billion times a second: yellow. The velvety throb of a cellohas as its most obvious characteristic a relatively low musical pitch— the note you are spontaneously imagining right now might beone whose sound vibrations repeat at a rate of a hundred times asecond.

Evolution has designed our two most important senses aroundthe assumption that not only will our environment be drenched withinformation-bearing vibrations, but in addition those vibrations willoften be repetitive, so that we can judge colors and pitches by therate of repetition. Granting that we do sometimes encounter non-repeating waves such as the consonant “sh,” which has no recogniz-able pitch, why was Nature’s assumption of repetition neverthelessso right in general?

Repeating phenomena occur throughout nature, from the orbitsof electrons in atoms to the reappearance of Halley’s Comet every 75years. Ancient cultures tended to attribute repetitious phenomena

495

a / If we try to draw a non-repeating orbit for Halley’sComet, it will inevitably end upcrossing itself.

b / A spring has an equilib-rium length, 1, and can bestretched, 2, or compressed, 3. Amass attached to the spring canbe set into motion initially, 4, andwill then vibrate, 4-13.

like the seasons to the cyclical nature of time itself, but we nowhave a less mystical explanation. Suppose that instead of Halley’sComet’s true, repeating elliptical orbit that closes seamlessly uponitself with each revolution, we decide to take a pen and draw awhimsical alternative path that never repeats. We will not be able todraw for very long without having the path cross itself. But at sucha crossing point, the comet has returned to a place it visited oncebefore, and since its potential energy is the same as it was on thelast visit, conservation of energy proves that it must again have thesame kinetic energy and therefore the same speed. Not only that,but the comet’s direction of motion cannot be randomly chosen,because angular momentum must be conserved as well. Althoughthis falls short of being an ironclad proof that the comet’s orbit mustrepeat, it no longer seems surprising that it does.

Conservation laws, then, provide us with a good reason whyrepetitive motion is so prevalent in the universe. But it goes deeperthan that. Up to this point in your study of physics, I have beenindoctrinating you with a mechanistic vision of the universe as agiant piece of clockwork. Breaking the clockwork down into smallerand smaller bits, we end up at the atomic level, where the electronscircling the nucleus resemble — well, little clocks! From this pointof view, particles of matter are the fundamental building blocksof everything, and vibrations and waves are just a couple of thetricks that groups of particles can do. But at the beginning ofthe 20th century, the tables were turned. A chain of discoveriesinitiated by Albert Einstein led to the realization that the so-calledsubatomic “particles” were in fact waves. In this new world-view,it is vibrations and waves that are fundamental, and the formationof matter is just one of the tricks that waves can do.

16.1 Period, frequency, and amplitude

Figure b shows our most basic example of a vibration. With noforces on it, the spring assumes its equilibrium length, b/1. It canbe stretched, 2, or compressed, 3. We attach the spring to a wallon the left and to a mass on the right. If we now hit the mass witha hammer, 4, it oscillates as shown in the series of snapshots, 4-13.If we assume that the mass slides back and forth without frictionand that the motion is one-dimensional, then conservation of energyproves that the motion must be repetitive. When the block comesback to its initial position again, 7, its potential energy is the sameagain, so it must have the same kinetic energy again. The motionis in the opposite direction, however. Finally, at 10, it returns to itsinitial position with the same kinetic energy and the same directionof motion. The motion has gone through one complete cycle, andwill now repeat forever in the absence of friction.

The usual physics terminology for motion that repeats itself over

496 Chapter 16 Vibrations

c / Position-versus-time graphsfor half a period and a full period.

d / The locomotive’s wheelsspin at a frequency of f cyclesper second, which can alsobe described as ω radians persecond. The mechanical link-ages allow the linear vibration ofthe steam engine’s pistons, atfrequency f , to drive the wheels.

e / Example 1.

and over is periodic motion, and the time required for one repetitionis called the period, T . (The symbol P is not used because of thepossible confusion with momentum.) One complete repetition of themotion is called a cycle.

We are used to referring to short-period sound vibrations as“high” in pitch, and it sounds odd to have to say that high pitcheshave low periods. It is therefore more common to discuss the rapid-ity of a vibration in terms of the number of vibrations per second,a quantity called the frequency, f . Since the period is the numberof seconds per cycle and the frequency is the number of cycles persecond, they are reciprocals of each other,

f = 1/T .

The forms of various equations turn out to be simpler whenthey are expressed not in terms of f but in terms of ω = 2πf .It’s not a coincidence that this relationship looks the same as theone relating angular velocity and frequency in circular motion. Inmachines, mechanical linkages are used to convert back and forthbetween vibrational motion and circular motion. For example, a carengine’s pistons oscillate in their cylinders at a frequency f , drivingthe crankshaft at the same frequency f . Either of these motions canbe described using ω instead of f , even though only in the case ofthe crankshaft’s rotational motion does it make sense to interpretω as the number of radians per second. When the motion is notrotational, we usually refer to ω as the angular frequency, and weoften use the word “frequency” to mean either f or ω, relying oncontext to make the meaning clear.

A carnival game example 1In the carnival game shown in figure e, the rube is supposed topush the bowling ball on the track just hard enough so that it goesover the hump and into the valley, but does not come back outagain. If the only types of energy involved are kinetic and poten-tial, this is impossible. Suppose you expect the ball to come backto a point such as the one shown with the dashed outline, thenstop and turn around. It would already have passed through thispoint once before, going to the left on its way into the valley. Itwas moving then, so conservation of energy tells us that it can-not be at rest when it comes back to the same point. The motionthat the customer hopes for is physically impossible. There isa physically possible periodic motion in which the ball rolls backand forth, staying confined within the valley, but there is no wayto get the ball into that motion beginning from the place where westart. There is a way to beat the game, though. If you put enoughspin on the ball, you can create enough kinetic friction so that asignificant amount of heat is generated. Conservation of energythen allows the ball to be at rest when it comes back to a point

Section 16.1 Period, frequency, and amplitude 497

f / 1. The amplitude of thevibrations of the mass on a springcould be defined in two differentways. It would have units ofdistance. 2. The amplitude of aswinging pendulum would morenaturally be defined as an angle.

like the outlined one, because kinetic energy has been convertedinto heat.

Period and frequency of a fly’s wing-beats example 2A Victorian parlor trick was to listen to the pitch of a fly’s buzz, re-produce the musical note on the piano, and announce how manytimes the fly’s wings had flapped in one second. If the fly’s wingsflap, say, 200 times in one second, then the frequency of theirmotion is f = 200/1 s = 200 s−1. The period is one 200th of asecond, T = 1/f = (1/200) s = 0.005 s.

Units of inverse second, s−1, are awkward in speech, so an abbre-viation has been created. One Hertz, named in honor of a pioneerof radio technology, is one cycle per second. In abbreviated form,1 Hz = 1 s−1. This is the familiar unit used for the frequencies onthe radio dial.

Frequency of a radio station example 3. KKJZ’s frequency is 88.1 MHz. What does this mean, and whatperiod does this correspond to?

. The metric prefix M- is mega-, i.e., millions. The radio wavesemitted by KKJZ’s transmitting antenna vibrate 88.1 million timesper second. This corresponds to a period of

T = 1/f = 1.14× 10−8 s.

This example shows a second reason why we normally speak interms of frequency rather than period: it would be painful to haveto refer to such small time intervals routinely. I could abbreviateby telling people that KKJZ’s period was 11.4 nanoseconds, butmost people are more familiar with the big metric prefixes thanwith the small ones.

Units of frequency are also commonly used to specify the speedsof computers. The idea is that all the little circuits on a computerchip are synchronized by the very fast ticks of an electronic clock, sothat the circuits can all cooperate on a task without getting aheador behind. Adding two numbers might require, say, 30 clock cycles.Microcomputers these days operate at clock frequencies of about agigahertz.

We have discussed how to measure how fast something vibrates,but not how big the vibrations are. The general term for this isamplitude, A. The definition of amplitude depends on the systembeing discussed, and two people discussing the same system maynot even use the same definition. In the example of the block on theend of the spring, f/1, the amplitude will be measured in distanceunits such as cm. One could work in terms of the distance traveledby the block from the extreme left to the extreme right, but itwould be somewhat more common in physics to use the distancefrom the center to one extreme. The former is usually referred to as


g / Sinusoidal and non-sinusoidalvibrations.

the peak-to-peak amplitude, since the extremes of the motion looklike mountain peaks or upside-down mountain peaks on a graph ofposition versus time.

In other situations we would not even use the same units for am-plitude. The amplitude of a child on a swing, or a pendulum, f/2,would most conveniently be measured as an angle, not a distance,since her feet will move a greater distance than her head. The elec-trical vibrations in a radio receiver would be measured in electricalunits such as volts or amperes.

16.2 Simple harmonic motionWhy are sine-wave vibrations so common?

If we actually construct the mass-on-a-spring system discussedin the previous section and measure its motion accurately, we willfind that its x−t graph is nearly a perfect sine-wave shape, as shownin figure g/1. (We call it a “sine wave” or “sinusoidal” even if it isa cosine, or a sine or cosine shifted by some arbitrary horizontalamount.) It may not be surprising that it is a wiggle of this generalsort, but why is it a specific mathematically perfect shape? Why isit not a sawtooth shape like 2 or some other shape like 3? The mys-tery deepens as we find that a vast number of apparently unrelatedvibrating systems show the same mathematical feature. A tuningfork, a sapling pulled to one side and released, a car bouncing onits shock absorbers, all these systems will exhibit sine-wave motionunder one condition: the amplitude of the motion must be small.

It is not hard to see intuitively why extremes of amplitude wouldact differently. For example, a car that is bouncing lightly on itsshock absorbers may behave smoothly, but if we try to double theamplitude of the vibrations the bottom of the car may begin hittingthe ground, g/4. (Although we are assuming for simplicity in thischapter that energy is never dissipated, this is clearly not a veryrealistic assumption in this example. Each time the car hits theground it will convert quite a bit of its potential and kinetic en-ergy into heat and sound, so the vibrations would actually die outquite quickly, rather than repeating for many cycles as shown in thefigure.)

The key to understanding how an object vibrates is to know howthe force on the object depends on the object’s position. If an objectis vibrating to the right and left, then it must have a leftward forceon it when it is on the right side, and a rightward force when it is onthe left side. In one dimension, we can represent the direction of theforce using a positive or negative sign, and since the force changesfrom positive to negative there must be a point in the middle wherethe force is zero. This is the equilibrium point, where the objectwould stay at rest if it was released at rest. For convenience of

Section 16.2 Simple harmonic motion 499

h / The force exerted by anideal spring, which behavesexactly according to Hooke’s law.

notation throughout this chapter, we will define the origin of ourcoordinate system so that x equals zero at equilibrium.

The simplest example is the mass on a spring, for which the forceon the mass is given by Hooke’s law,

F = −kx.

We can visualize the behavior of this force using a graph of F versusx, as shown in figure h. The graph is a line, and the spring constant,k, is equal to minus its slope. A stiffer spring has a larger value ofk and a steeper slope. Hooke’s law is only an approximation, butit works very well for most springs in real life, as long as the springisn’t compressed or stretched so much that it is permanently bentor damaged.

The following important theorem relates the motion graph tothe force graph.

Theorem: A linear force graph makes a sinusoidal motiongraph.

If the total force on a vibrating object depends only on theobject’s position, and is related to the objects displacementfrom equilibrium by an equation of the form F = −kx, thenthe object’s motion displays a sinusoidal graph with frequencyω =

√k/m.

Proof: By Newton’s second law, −kx = ma, so we need a functionx(t) that satisfies the equation d2 x/ dt2 = −cx, where for conve-nience we write c for k/m. This type of equation is called a differ-ential equation, because it relates a function to its own derivative(in this case the second derivative).

Just to make things easier to think about, suppose that we hap-pen to have an oscillator with c = 1. Then our goal is to finda function whose second derivative is equal to minus the originalfunction. We know of two such functions, the sine and the cosine.These two solutions can be combined to make anything of the formP sin t+Q cos t, where P and Q are constants, and the result will stillbe a solution. Using trig identities, such an expression can alwaysbe rewritten as A cos(t+ δ).

Now what about the more general case where c need not equal1? The role of c in d2 x/dt2 = −cx is to set the time scale. Forexample, suppose we produce a fake video of an object oscillatingaccording to A cos(t + δ), which violates Newton’s second law be-cause c doesn’t equal 1, so the acceleration is too small. We canalways make the video physically accurate by speeding it up. Thissuggests generalizing the solution to A cos(ωt + δ). Plugging in tothe differential equation, we find that ω =

√k/m, and T = 2π/ω

brings us to the claimed result.

We’ve proved that anything of this form is a solution, but we


pages 451-500

Documents

Transcript of pages 451-500