Quiz Samples for Chapter 11 General Physics I May 5, 2020...

General Physics IQuiz Samples for Chapter 11

Rolling, Torque, and Angular MomentumMay 5, 2020

Name: Department: Student ID #:

Notice

+2 (−1) points per correct (incorrect) answer.

No penalty for an unanswered question.

Fill the blank ( ) with (8) if the statement iscorrect (incorrect).

Textbook: Walker, Halliday, Resnick, Principlesof Physics, Tenth Edition, John Wiley & Sons(2014).

11-1 Rolling as Translation and RotationCombined

1. For a wheel of radius R rolling smoothly,

Pure rotation Pure translation

Rolling motion

(a) () The linear speed of the wheel’s center ofmass is determined by

vcom = ωR,

where ω is the angular speed of the wheelabout its center.

(b) () The wheel may also be viewed as rotatinginstantaneously about the point P of the“road” that is in contact with the wheel. Theangular speed of the wheel about this point isthe same as the angular speed of the wheelabout its center.

(c) () If it is seen on the ground, then the speedof the contact point of the wheel on theground is 0.

(d) () If it is seen on the ground, then the speedof the top of the wheel is 2vcom.

2. A wheel of radius R rolls without sliding on ahorizontal surface as shown. Starting from rest, thewheel moves with constant angular acceleration α.

(a) () During time t, the wheel rolls

θ =1

2αt2. (1)

(b) () When a wheel of radius r rotates θwithout sliding, it moves forward by adistance rθ. Therefore, the wheel movesforward by a distance

s =1

2Rαt2. (2)

3. () Two wheels of radii R1 and R2, respectively,roll side-by-side without sliding, at the sameinstantaneous speed.

R2 = aR1,

where a is a constant.

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(a) () The angular velocity of the wheel Ri is ωi

for i = 1 and 2. Then,

ω2 =ω1

a.

(b) () The angular acceleration of the wheel Ri

is αi for i = 1 and 2. Then,

α2 =α1

a.

11-2 Forces and Kinetic Energy of Rolling

1. Consider a smoothly rolling wheel of radius R,mass M , and rotational inertia Icom about the axispassing the center of mass.

(a) () The rotational kinetic energy is

Krotation =1

2Icomω

2,

where ω is the angular velocity of the wheel.

(b) () The translational kinetic energy is

Ktranslation =1

2Mv2com,

where vcom is the velocity of the center ofmass of the wheel.

(c) () The total kinetic energy is

K = Krotation +Ktranslation

=1

2Icomω

2 +1

2Mv2com.

(d) () The translational kinetic energy dependson the choice of the refrerence frame.

(e) () The rotational kinetic energy is invariantas long as the frame of reference is inertial.

(f) () If the wheel is being accelerated but isstill rolling smoothly, the acceleration of thecenter of mass, acom is related to the angularacceleration about the center with

acom = |acom| = αR.

(g) ()

If the wheel rolls smoothly down a ramp ofangle θ, its acceleration along an x axisextending up the ramp is

acom,x = − g sin θ

1 +IcomMR2

.

2. A thin-walled hollow tube of radiun R and mass Mrolls without sliding along the floor.

(a) () The translational kinetic energy is

Ktranslation =1

2Mv2com.

(b) () The rotational kinetic energy is

Krotation =1

2Icomω

2 =1

2M(Rω)2,

whereIcom = MR2.

(c) () Because vcom = Rω,

Ktranslation = Krotation.

(d) () In the frame where the center of mass isat rest,

Ktranslation = 0.

In this frame, the rotational kinetic energy isinvariant:

Krotation =1

2M(Rω)2.

3. The speed of a rear-drive car is increasing on ahorizontal road.

(a) () A torque is applied to the axle of the reartires by the engine to increase the speed.

(b) () The axle of the front tires does notreceive any torque by the engine and the onlytorque applied to the front tires is thefrictional force on the ground.




(c) () The rear tires’ action to the ground isbackward and the reaction of the ground tothe rear tire is forward.

(d) () The direction of the frictional force on thetires is forward for the rear tires.

(e) () The direction of the frictional force on thetires is backward for the front tires.

4. A solid sphere and a solid cylinder of equal massM and radius R are simultaneously released fromrest at point A on the same inclined plane rollingdown the incline smoothly to point B. The heightdifference between A and B is h.

(a) () At a point P between A and B

vcom = Rω.

Here, vcom is the speed of the center of massand ω is the angular velocity in each case.

(b) () The rotational inertias are

Isphere, com =2

5MR2,

Icylinder, com =1

2MR2.

(c) () At a point P , the total kinetic energies are

Ksphere =1

2Isphere, comω

2s +

1

2Mv2sphere, com

=1

2× 2

5MR2ω2

s +1

2MR2ω2

s

=7

10MR2ω2

s ,

Kcylinder =1

2Icylinder, comω

2c +

1

2Mv2cylinder, com

=1

2× 1

2MR2ω2

c +1

2MR2ω2

c

=3

4MR2ω2

c ,

where ωs (ωc) is the angular velocity of thesphere (cylinder) at the point P .

(d) () At point B,

vsphere, com =

√10

7gh,

vcylinder, com =

√4

3gh.

(e) () The sphere reaches the bottom firstbecause it has the smaller inertia.

11-3 The Yo-Yo

1. Consider a yo-yo of mass M , which travelsvertically up or down a string. The unstrechablestring, of assumed negligible thickness, is woundaround an axle of radius R0. The rotational inertiaabout the center of mass is Icom.

(a) () It can be treated as a wheel rolling alongan inclined plane at angle θ = 90.

(b) () Effectively, the gravitational force Fg isapplied to the center of mass and Fg does notcontribute to the torque.

(c) () As long as the string attached to theyo-yo is vertical, the tension T exerts a torquethat is the net torque:

τnet = R0T = Icomα.

(d) () The acceleration of the center of mass isdetermined by

acom = R0α = g − T

M.

(e) () The solutions for α, acom, and T are

acom =g

1 +IcomMR2

0

,

α =acomR0

,

T =Mg

1 +MR2

0

Icom

.

11-4 Torque Revisited

1. In three dimensions, torque is a vector quantitydefined relative to a fixed point (usually an origin);it is

τ = r × F ,

where F is a force applied to a particle and r is aposition vector locating the particle relative to thefixed point.




(a) () The magnitude of τ is given by

τ = rF sinφ = rF⊥ = r⊥F,

where φ is the angle between F and r, F⊥ isthe component of F perpendicular to r, andr⊥ is the moment arm of F .

(b) () The direction of τ is given by theright-hand rule for cross products.

(c) () The fundamental dimensions of torque are

[M ][L]2[T ]−2.

(d) () Possible units of torque are

kg ·m2/s2.

11-5 Angular Momentum

1. The angular momentum ` of a particle with linearmomentum p, mass m, and linear velocity v is avector quantity defined relative to a fixed point(usually an origin) as

` = r × p = mr × v.

(a) () The magnitude of ` is given by

` = rmv sinφ = rp⊥ = rmv⊥ = r⊥p = r⊥mv,

where φ is the angle between p and r, p⊥ andv⊥ are the components of p and vperpendicular to r, and r⊥ is theperpendicular distance the fixed point and theextension of p.

(b) () The direction of ` is given by theright-hand rule: Position your right hand sothat the fingers are in the direction of r. Thenrotate them around the palm to be in thedirection of p. Your outstretched thumb givesthe direction of `.

(c) () N·s is a unit of linear momentum.

(d) () kg·m2/s is a unit of angular momentum.

(e) () A particle moves along the x axis. Inorder to calculate the angular momentum ofthe particle, you need to know the pointabout which the angular momentum is to becalculated.

2. () The angular momentum vector of Earth, dueto its daily rotation, is directed north.

3. Consider two objects 1 and 2. The mass, position,and velocity of each particle are

m1 = m,

m2 = 2m,

r1 = 2di,

r2 = dj,

v1 = 3vj,

v2 = −2vi.

The magnitude of their total angular momentum(about the origin O) is:

(a) () The angular momentum of particle 1 is

`1 = m1r1 × v1= m2di× 3vj

= 6mvdk.

(b) () The angular momentum of particle 2 is

`2 = m2r2 × v2= 2mdj× (−2v)i

= −4mvd(−k).

(c) () The total angular momentum L is

L = `1 + `2 = 10mvdk.

11-6 Newton’s Second Law in Angular Form

1. Newton’s second law for a particle can be writtenin angular form as

τnet = ˙ =d`

dt.

where τnet is the net torque acting on the particleand ` is the angular momentum of the particle.

(a) () The definition of the angular momentumof a particle is

` = mr × v,

where r is the position vector of the particleand v is the velocity of the particle.




(b) () Differentiating each side with respect totime t yields

d`

dt= mv × v +mr × a.

(c) () v × v = 0 and ma = Fnet. Therefore,

d`

dt= r × Fnet = τnet.

11-7 Angular Momentum of a Rigid Body

1. The angular momentum of a system of particles isthe vector sum of the angular momenta of theindividual particles:

L =

n∑i=1

ì.

(a) () The time rate of change of this angularmomentum is equal to the net external torqueon the system (the vector sum of the externaltorques):

τnet =dL

dt.

(b) () For a rigid body rotating about a fixedaxis, the component of its angular momentumparallel to the rotation axis is

L = Iω.

(c) () The definition of the angular momentumof a particle is

L =n∑

i=1

ì =n∑

i=1

miri × vi,

where ì, ri, vi are the angular momentum,position, and velocity vectors of the ithparticle.

(d) () Differentiating each side with respect totime t yields

dL

dt=

n∑i=1

mivi × vi +

n∑i=1

miri × ai.

(e) () vi × vi = 0 and miri × a = τi. Therefore,

dL

dt=

n∑i=1

τi = τnet.

(f) () The net external torque τnet acting on asystem of particles is equal to the time rate ofchange of the system’s total angularmomentum L.

2. We are interested in the component of a totalangular momentum L that is parallel to therotation axis, where the z axis.

(a) () That z component of the angularmomentum of the ith particle is

ìz = ì sin θ = (ri sin θ)(∆mivi) = r⊥i∆mivi.

(b) () The z component of the angularmomentum for the rotating rigid body as awhole is found by adding up the contributionsof all the mass elements that make up thebody.

(c) () Thus, because v = ωr⊥, we may write

Lz =

n∑i=1

ìz

=

n∑i=1

∆mivir⊥i

=

n∑i=1

∆mi(ωr⊥i)r⊥i

= ω

n∑i=1

∆mir2⊥i.

(d) () The quantity

n∑i=1

∆mir2⊥i is the rotational

inertia I of the body about the fixed axis.Thus

L = Iω.

3. A uniform rod of length L and mass M rests onfrictionless ice along the x axis on the xy plane andits center of mass is at the origin. Forces that areequal in magnitude and opposite in direction aresimultaneously applied to its ends for a tinyinterval ∆t as shown.




(a) () The net force during the interval is 0.

(b) () The impulse during the interval is 0.

(c) () The center of mass is at rest at any time.

(d) () The total linear momentum is 0.

(e) () During the time interval [0,∆t], the nettorque is

τnet = LF k,

where k is the unit normal vector out of thexy plane.

(f) () The mangnitude of the angularmomentum of the rod after ∆t is

`rod = τnet∆t = LF∆t.

(g) () The angular momentum of the rod afterthe hit can also be expressed as

`rod = LF∆t = Icomω,

where

Icom =1

12ML2

and the angular velocity ω is

ω =12F∆t

ML.

4. A pulley of rotational inertia I and radius R is freeto rotate on a horizontal fixed axis through itscenter. A string passes over the pulley. Mass m1 isattached to one end and mass m2 (< m1) isattached to the other. The portion of the string

attached to m1 has tension T1 and the portionattached to m2 has tension T2.

Let us compute the magnitude of the total externaltorque, about the pulley center, acting on themasses and pulley, considered as a system.

(a) () The equations of motion are

τnet = RT1 −RT2 = Iα,

m1a1 = m1g − T1,m2a2 = T2 −m2g,

where a1 and a2 are the accelerations of m1

and m2, respectively. The directions are downfor a1 and up for a2.

(b) () The constraint equations are

a1 = a2 = Rα.

(c) () T1 − T2 can be expressed as

T1 − T2 =I

Rα,

T1 − T2 = (m1 −m2)g − (m1 +m2)Rα.

(d) () The angular acceleration α is

α =(m1 −m2)gR

I + (m1 +m2)R2.

(e) () The tensions are given by

T1 =m1g(I + 2m2R

2)

I + (m1 +m2)R2,

T2 =m2g(I + 2m1R

2)

I + (m1 +m2)R2.




11-8 Conservation of Angular Momentum

1. Law of conservation of angular momentum:

(a) () The angular momentum L of a systemremains constant if the net external torqueacting on the system is zero:

L = constant.

This states that every component is conserved.

(b) () If the component of the net externaltorque on a system along a certain axis iszero, then the component of the angularmomentum of the system along that axiscannot change, no matter what changes takeplace within the system.

(c) () This is true even if the rotational inertiavaries:

Iiωi = Ifωf .

Here the subscripts refer to the values of therotational inertia I and angular speed ωbefore and after the redistribution of mass.

(d) () As a result,

ωf =Iiωi

If.

If Ii < If , then the angular velocity decreases.If Ii > If , then the angular velocity increases.

(e) () An ice skater with rotational inertia I0 isspinning with angular speed ω0. As she pullsher arms in, the mass is concentrated at theaxis, thereby decreasing her rotational inertiato I = I0/n. By the law of conservation ofangular momentum, her angular velocity isthen

ω = nω0.

2. A uniform sphere of radius R and mass M rotatesabout a diameter with the angular momentum ofmagnitude L. Under the action of internal forces,the sphere collapses to a uniform sphere of radiusR/2.

(a) () The angular momentum is conserved afterthe collapse.

(b) () The rotational inertia of the initial stateabout its own axis passing the center of massis

Ii =2

5MR2.

(c) () The angular velocity of the initial state is

ωi =L

Ii=

5L

2MR2.

(d) () The rotational inertia of the final stateabout its own axis passing the center of massis

If =2

5M

(R

2

)2

=1

10MR2.

(e) () The angular velocity of the final state is

ωf =L

If=

10L

MR2.

(f) () The ratio ωf/ωi is

ωf

ωi=IiIf

= 4.

3. A playground merry-go-round (1) has a radius Rand a rotational inertia I. When themerry-go-round is at rest, a child (2) with mass mruns with speed v along a line tangent to the rimand jumps on. We use subscripts 1 and 2 for themerry-go-round and child, respectively.

(a) () The initial value for the angularmomentum of the merry-go-round is

L1i = I1ω1i = 0.

Here, ω1i = 0 is the initial angular velocity ofthe merry-go-round.

(b) () The initial value for the angularmomentum of the child is

L2i = I2ω2i = mRv.

The rotational inertia of the child is

I2 = mR2.

(c) () The initial value for the total angularmomentum is

Ltotal = L1i + L2i = mRv.

The angular momentum is conserved. Thusthe final-state total angular momentum is thesame as mRv.




(d) () The rotational inertia of the compositesystem consisting of the merry-go-round andthe child is

Itot = I1 + I2 = I +mR2.

(e) () The final angular velocity ωtot can becomputed as

Itotωtot = Ltot → ωtot =Ltot

Itot=

mRv

I +mR2.

11-9 Precession of a Gyroscope

1. A spinning gyroscope can precess about a verticalaxis through its support at the rate

Ω =Mgr

Iω,

where M is the gyroscope’s mass, r is the momentarm, I is the rotational inertia, and ω is the spinrate. We assume the spin rate is so rapid that theangular momentum due to precession is negligiblerelative to L. We also assume the shaft ishorizontal when precession begins. The magnitudeof L is given by

L = Iω.

(a) () According to Newton’s second law for theangular momentum, the incremental changedL in the angular momentum of the gyroscopein an incremental time interval dt; that is,

dL = τdt.

() For a rapidly spinning gyroscope, themagnitude of L is fixed by

L = Iω.

Thus the torque can change only the directionof L, not its magnitude.

(b) () The only way that L can be changed inthe direction of without the magnitude Lbeing changed is for to rotate around the zaxis. The differential change is along thedirection of the torque.

dL = τdt = Mgrdt.

(c) () As L changes by an incremental amountin an incremental time interval dt, the shaftand precess around the z axis throughincremental angle dφ.

dφ =dL

L=Mgrdt

Iω.

(d) () We define

Ω =dφ

dt.

Then we find that

Ω =Mgr

Iω.

This result is valid under the assumption thatthe spin rate ω is rapid.

(e) () Ω decreases as ω is increased.

(f) () There would be no precession if thegravitational force Mg did not act on thegyroscope.

(g) () Because I is a function of M , masscancels from

Ω =Mgr

Iω.

Thus Ω is independent of the mass.


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