Review Review -- Nov. 30th 2004Nov. 30th 2004

Chapters: 10, 11, 12, 13, 15, 16Chapters: 10, 11, 12, 13, 15, 16

Review of rotational variables (scalar notation)Review of rotational variables (scalar notation)

Angular position:Angular position: ( )in radianssr

θ =

Angular displacement:Angular displacement: 2 1θ θ θ∆ = −

Average angular velocity:Average angular velocity: 2 1

2 1avg t t t

θ θ θω − ∆= =

− ∆

Instantaneous angular velocity:Instantaneous angular velocity:0

Limt

dt dtθ θω

∆ →

∆= =

∆

Average angular acceleration:Average angular acceleration: 2 1

2 1avg t t t

ω ω ωα − ∆= =

− ∆

Instantaneous angular acceleration:Instantaneous angular acceleration:0

Limt

dt dtω ωα

∆ →

∆= =

∆

Relationships between linear and angular variablesRelationships between linear and angular variables

Position:Position: ( )in radianss rθ θ=

Time period for rotationTime period for rotation::circumference 2 2

velocityrT

vπ π

ω= = =

Tangential acceleration:Tangential acceleration: ta rα=

Centripetal acceleration:Centripetal acceleration:2

2r

va rr

ω= =

dv r rdtθ ω= =Velocity:Velocity:

Kinetic energy of rotationKinetic energy of rotation2

i iI m r=∑212K Iω=

2 2I r dm r dVρ= =∫ ∫Therefore, for a continuous rigid object:Therefore, for a continuous rigid object:

h

rotationaxis

c.o.m. axis

Icom

••If moment of inertia is known about If moment of inertia is known about an axis though the center of mass an axis though the center of mass ((c.o.m.c.o.m.), then the moment of inertia ), then the moment of inertia about any parallel axis is:about any parallel axis is:

Parallel axis theoremParallel axis theorem

2comI I Mh= +

••It is essential that these axes are It is essential that these axes are parallel; as you can see from table parallel; as you can see from table 1010--2, the moments of inertia can be 2, the moments of inertia can be different for different axes.different for different axes.

Some rotational inertiaSome rotational inertia

TorqueTorque

( )( )

( )( )

sin

sin

tr F rF

r F r F

τ φ

φ ⊥

= =

= =

••There are two ways to compute torque:There are two ways to compute torque:

••The direction of the force vector is The direction of the force vector is called the called the line of actionline of action, and , and rr⊥⊥ is called is called the the moment armmoment arm..

••The first equation shows that the The first equation shows that the torque is equivalently given by the torque is equivalently given by the component of force tangential to the component of force tangential to the line joining the axis and the point where line joining the axis and the point where the force acts.the force acts.••In this case, In this case, rr is the moment arm of is the moment arm of FFtt..

Summarizing relations for translational and Summarizing relations for translational and rotational motionrotational motion

••Note: work obtained by multiplying torque by an angle Note: work obtained by multiplying torque by an angle -- a a dimensionless quantity. Thus, torque and work have the same dimensionless quantity. Thus, torque and work have the same dimensions, but you see that they are quite different.dimensions, but you see that they are quite different.

Rolling motion as rotation and translationRolling motion as rotation and translation

s Rθ=The wheel moves with speed ds/dt

comdv Rdtθ ω⇒ = =

The kinetic energy of rolling2 21

2

2 2 21 12 2

2 21 12 2

P P com

com

com com r t

K I I I MR

K I MR

K I Mv K K

ω

ω ω

ω

= = +

= +

= + = +

Torque and angular momentumTorque and angular momentum( )definitionr Fτ = ×

••Here, Here, pp is the linear momentum is the linear momentum mv mv of of the object.the object.

sinl mvr

rp rmv

r p r mv

φ

⊥ ⊥

⊥ ⊥

=

= =

= =

••SI unit is Kg.mSI unit is Kg.m22/s./s.

( ) is defined as:l l r p m r v= × = ×Angular momentum

••Torque was discussed in the previous chapter; cross products Torque was discussed in the previous chapter; cross products are discussed in chapter 3 (section 3are discussed in chapter 3 (section 3--7) and at the end of this 7) and at the end of this presentation; torque also discussed in this chapter (section 7).presentation; torque also discussed in this chapter (section 7).

Angular momentum of a rigid body about a fixed axisAngular momentum of a rigid body about a fixed axisWe are interested in the component of We are interested in the component of angular momentum parallel to the axis of angular momentum parallel to the axis of rotation:rotation:

( )

1 1

2

n n

z iz i i ii i

L l m v r vr dm

r r dm r dm Iω ω ω

⊥ ⊥= =

⊥ ⊥ ⊥

= = =

= = =

∑ ∑ ∫

∫ ∫In fact:In fact: L Iω=

Conservation of angular momentumConservation of angular momentum

If the net external torque acting on a system is zero, the If the net external torque acting on a system is zero, the angular momentum of the system remains constant, no angular momentum of the system remains constant, no matter what changes take place within the system.matter what changes take place within the system.

It follows from Newton's second law that:It follows from Newton's second law that:

What happens to kinetic energy?What happens to kinetic energy?2 2

2 21 1 12 2 22

i i i if f f f i i i

f f f

I I IK I I I KI I Iωω ω

⎛ ⎞= = = =⎜ ⎟⎜ ⎟

⎝ ⎠••Thus, if you increase Thus, if you increase ωω by reducing by reducing II, you end , you end up increasing up increasing KK..••Therefore, you must be doing some work.Therefore, you must be doing some work.••This is a very unusual form of work that you do This is a very unusual form of work that you do when you move mass radially in a rotating frame. when you move mass radially in a rotating frame. ••The frame is accelerating, so Newton's laws do The frame is accelerating, so Newton's laws do not hold in this frame not hold in this frame

a constantL =

i fL L=

i i f fI Iω ω=

f i

i f

II

ωω

=

EquilibriumEquilibriumA system of objects is said to be in equilibrium if:A system of objects is said to be in equilibrium if:

1.1. The linear momentum The linear momentum PP of its center of mass is constant.of its center of mass is constant.2.2. Its angular momentum Its angular momentum LL about its center of mass, or about about its center of mass, or about

any other point, is also constant.any other point, is also constant.

If, in addition, If, in addition, LL and and PP are zero, the system is said to are zero, the system is said to be in be in static equilibriumstatic equilibrium..

0net netdPF Fdt

= ⇒ =

0net netdLdt

τ τ= ⇒ =

1.1. The vector sum of all the external forces that act on a The vector sum of all the external forces that act on a body must be zero.body must be zero.

2.2. The vector sum of all the external torques that act on a The vector sum of all the external torques that act on a body, body, measured about any axismeasured about any axis, must also be zero., must also be zero.

The requirements of equilibriumThe requirements of equilibrium

1.1. The vector sum of all the external forces that act on a The vector sum of all the external forces that act on a body must be zero.body must be zero.

2.2. The vector sum of all the external torques that act on a The vector sum of all the external torques that act on a body, body, measured about any axismeasured about any axis, must also be zero., must also be zero.

, ,

, ,

, ,

0 0

0 0

0 0

net x net x

net y net y

net z net z

F

F

F

τ

τ

τ

= =

= =

= =

Balance oftorques

Balance offorces

3.3. The linear momentum The linear momentum PP of the body must be zero.of the body must be zero.

One more requirement for static equilibrium:One more requirement for static equilibrium:

ElasticityElasticity

••A A stressstress, a force per unit area, produces a , a force per unit area, produces a strainstrain, or dimensionless , or dimensionless unit deformation.unit deformation.••These various stresses and strains are related via a These various stresses and strains are related via a modulus of modulus of elasticityelasticity

Hydraulic stressTensile stress Shear stress

••All of these deformations have the following in common:All of these deformations have the following in common:

stress = modulus × strain

Tension and compressionTension and compression••The figure left shows a graph of stress The figure left shows a graph of stress versus strain for a steel specimen.versus strain for a steel specimen.••Stress = force per unit area (Stress = force per unit area (FF//AA))••Strain = extension (Strain = extension (∆∆LL) / length () / length (LL))••For a substantial range of applied For a substantial range of applied stress, the stressstress, the stress--strain relation is strain relation is linear.linear.••Over this soOver this so--called called elasticelastic region, the, region, the, the specimen recovers its original the specimen recovers its original dimensions when the stress is removed.dimensions when the stress is removed.••In this region, we can write:In this region, we can write:

F LEA L

∆=

This equation, This equation, stress = stress = EE × strain× strain, is known as , is known as Hooke's Hooke's lawlaw, and , and the modulus the modulus EE is called is called Young's modulusYoung's modulus. . The dimensions of The dimensions of EE are are the same as stress, the same as stress, i.e.i.e. force per unit area.force per unit area.

Vp BV∆

=

••BB is called the is called the bulk modulusbulk modulus..••VV is the volume of the specimen, and is the volume of the specimen, and ∆∆VVits change in volume under a its change in volume under a hydrostatic hydrostatic pressure pressure pp..

Shear stressShear stress

F xGA L

∆=

••GG is called the is called the shear modulusshear modulus..

Hydraulic stressHydraulic stress

Newton's law Newton's law of gravitationof gravitation

11 2 21 22 6.67 10 N m /kgm mF G G

r−= = × ⋅

A uniform spherical shell of matter attracts a particle A uniform spherical shell of matter attracts a particle that is outside the shell as if the shell's mass were that is outside the shell as if the shell's mass were concentrated at its center.concentrated at its center.

A uniform spherical shell of matter exerts no net A uniform spherical shell of matter exerts no net gravitational force on a particle located inside itgravitational force on a particle located inside it

Shell theoremsShell theorems

Gravitational potential energyGravitational potential energy

••But, close to the Earth's surface,But, close to the Earth's surface,

••Further away from Earth, we must choose a Further away from Earth, we must choose a reference point against which we measure reference point against which we measure potential energy.potential energy.

r1

( )1 22 1 1 2

1 1g

GMW GMm m r rr r r r

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠

21 2

GM GM grr r

≈ ≈

. . gi e U W mgh∆ = − =

The natural place to chose as a reference point is The natural place to chose as a reference point is rr == ∞∞, since , since UUmust be zero there, must be zero there, i.e.i.e. we set we set rr11 == ∞∞ as our reference point. as our reference point.

1 2g

GmM GmM GmMU Wr r r

= − = − = −

1.1. THE LAW OF ORBITS: All planets move in elliptical THE LAW OF ORBITS: All planets move in elliptical orbits, with the sun at one focus.orbits, with the sun at one focus.

Planets and satellites: Planets and satellites: Kepler's Kepler's lawslaws

2.2. THE LAW OF AREAS: A line that connects a planet THE LAW OF AREAS: A line that connects a planet to the sun sweeps out equal areas in the plane of to the sun sweeps out equal areas in the plane of the planet's orbit in equal times; that is, the rate the planet's orbit in equal times; that is, the rate dAdA//dtdt at which it sweeps out area at which it sweeps out area AA is constant.is constant.

2 21 12 2

dA dr rdt dt

θ ω≈ =

3.3. THE LAW OF PERIODS: The square of the period THE LAW OF PERIODS: The square of the period of any planet is proportional to the cube of the of any planet is proportional to the cube of the semimajor semimajor axis of the orbit.axis of the orbit.

( ) ( )2 22 3

2

2 2T r

GMπ πω

= =

Ener

gy

Satellites: Orbits and EnergySatellites: Orbits and Energy

( )2

2

GMm vmr r

⎛ ⎞= ⎜ ⎟

⎝ ⎠

••Again, we'll do the math for a Again, we'll do the math for a circular orbit, but it holds quite circular orbit, but it holds quite generally for all elliptical orbits.generally for all elliptical orbits.••Applying Applying FF = = mama::

2

2

totalGMm GMmE K U

r r

GMm Kr

= + = −

= − = −

212 2 2

GMm UK mvr

= = = −

••Thus,Thus,

2GMmE

a= −

Simple Harmonic MotionSimple Harmonic Motion••The simplest possible version of harmonic motion is called The simplest possible version of harmonic motion is called Simple Simple Harmonic Motion (SHM)Harmonic Motion (SHM)..••This term implies that the periodic motion is a This term implies that the periodic motion is a sinusoidalsinusoidal function function of time,of time,

••The positive constantThe positive constant xxmm is called the is called the amplitudeamplitude..••The quantity The quantity ((ωωtt + + φφ)) is called the is called the phasephase of the motion.of the motion.••The constant The constant φφ is called the is called the phase constantphase constant or or phase anglephase angle..••The constant The constant ωω is called the is called the angular frequency angular frequency of the motion.of the motion.••TT is the period of the oscillations, and is the period of the oscillations, and ff is the frequency.is the frequency.

2 2 fTπω π= =

The velocity and acceleration of SHMThe velocity and acceleration of SHM

••The positive quantity The positive quantity ωωxxmm is called the velocity amplitude is called the velocity amplitude vvmm..

( )

( )

( )( ) cos

( ) sin

m

m

dx t dv t x tdt dt

v t x t

ω φ

ω ω φ

⎡ ⎤= = +⎣ ⎦

= − +

Velocity:Velocity:

( )

( )2

2

( )( ) sin

( ) cos

( ) ( )

m

m

dv t da t x tdt dt

a t x t

a t x t

ω ω φ

ω ω φ

ω

⎡ ⎤= = − +⎣ ⎦

= − +

= −

Acceleration:Acceleration:

In SHM, the acceleration is proportional to the In SHM, the acceleration is proportional to the displacement but opposite in sign; the two quantities displacement but opposite in sign; the two quantities are related by the square of the angular frequencyare related by the square of the angular frequency

The force law for SHMThe force law for SHM2 2( ) ( )F ma m x m xω ω= = − = −

••Note: SHM occurs in situations where the force is proportional tNote: SHM occurs in situations where the force is proportional to o the displacement, and the proportionality constant the displacement, and the proportionality constant ((−−mmωω22)) is is negative, negative, i.e.i.e. F kx= −••This is very familiar This is very familiar -- it is it is Hooke's Hooke's law.law.

SHM is the motion executed by a particle of mass SHM is the motion executed by a particle of mass mmsubjected to a force that is proportional to the subjected to a force that is proportional to the displacement of the particle but of opposite sign.displacement of the particle but of opposite sign.

2k mTm k

ω π= =

Mechanical energy:Mechanical energy: 212 mE U K kx= + =

xxmm is the maximum displacement or amplitudeis the maximum displacement or amplitude

Waves I Waves I -- wavelength and frequencywavelength and frequency

2k πλ

= kk is the is the angular angular wavenumberwavenumber..

Tran

sver

se w

ave

Tran

sver

se w

ave

2Tπω = ωω is the is the angular frequencyangular frequency..

12

fT

ωπ

= =

v fk Tω λ λ= ± = ± = ±

frequencyfrequency

velocityvelocity

∓

Review Review -- traveling waves on a stringtraveling waves on a string

v τµ

=

••The tension in the string is The tension in the string is ττ..••The mass of the element The mass of the element dmdm isis µµdldl, where , where µµ is the mass per unit is the mass per unit length of the string.length of the string.

VelocityVelocity

2 2 212 cos ( )kinetic m

dKP v y kx tdt

µ ω ω= = −

2 2 212 cos ( )elastic m

dUP v y kx tdt

µ ω ω= = −

Energy transfer ratesEnergy transfer rates

2 2 212

2 2 2 21 1 12 2 2

2 cos ( )

2

avg m

m m

P v y kx t

v y v y

µ ω ω

µ ω µ ω

= × −

= × × =

( , ) sin( )my x t y kx tω= −