Quiz Samples for Chapter 10 General Physics I Name:...

General Physics IQuiz Samples for Chapter 10

RotationApril 27, 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).

10-1 Rotational Variables

1. (a) (�) A rigid body is a body that can rotatewith all its parts locked together and withoutany change in its shape.

(b) (�) A fixed axis means that the rotationoccurs about an axis that does not move. Afixed axis is called the axis of rotation orthe rotation axis.

(c) (�) The angular position of this line is theangle of the line relative to a fixed direction,which we take as the zero angular position.

θ =s

r,

where s is the length of a circular arc thatextends from the zero angular position to thereference line and r is the radius of the circle.

(d) (�) The angular displacement ∆θ is givenby

∆θ = θ2 − θ1.

(e) (�) If a wheel turning at a constant ratecompletes 100 revolutions in 10 s its angularspeed is

ω =100× 2π

10 s= 20π rad/s.

(f) (�) If the angular velocity vector of aspinning body points out of the page then,when viewed from above the page, the body isspinning counterclockwise about an axis thatis perpendicular to the page.

(g) (�) The angular velocity ω is given by

ω = lim∆t→0

∆θ

∆t=dθ

dt= θ̇.

(h) (�) The angular acceleration α is given by

α = lim∆t→0

∆ω

∆t=dω

dt= ω̇ =

d2θ

dt2= θ̈.

(i) (�) The angular velocity of a rotating wheelincreases 2 rev/s every minute. The angularacceleration of this wheel is

α =2× 2π rad/s

60 s=

2π

30rad/s2.

10-2 Rotation with Constant AngularAcceleration

1. Constant angular acceleration (α = constant) is animportant special case of rotational motion. Theappropriate kinematic equations are

(a) (�) ω − ω0 = αt,

(b) (�) θ − θ0 = ω0t+1

2αt2,

(c) (�) ω2 = ω20 + 2α(θ − θ0),

(d) (�) θ − θ0 =1

2(ω0 + ω)t,

(e) (�) θ − θ0 = ωt− 1

2αt2,

where θ0 (ω0) is the initial value of θ (ω).

2. (�) A wheel rotates with a constant angularacceleration of α. During a certain time interval,its angular displacement is ∆θ. At the end of theinterval, its angular velocity is ω. Because thewheel rotates with a constant angular accelerationof α,

ω2 − ω20 = 2α∆θ.

Therefore, its angular velocity at the beginning ofthe interval is

ω0 =√ω2 − 2α∆θ.

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10-3 Relating the Linear and Angular Variables

1. We often need to relate the linear variables s, v,and a for a particular point in a rotating body tothe angular variables θ, ω, and α for that body.The two sets of variables are related by r, theperpendicular distance of the point from therotation axis.

(a) (�) s = rθ.

(b) (�)ds

dt=dθ

dtr.

(c) (�) v = rω.

(d) (�) If the motion is a uniform circularmotion, the period of revolution T for themotion of each point and for the rigid bodyitself is given by

T =2πr

v.

(e) (�) If the motion is a uniform circularmotion, the period of revolution T for themotion of each point and for the rigid bodyitself is given by

T =2π

ω.

(f) (�)

A particle (or point) moving in a circular pathhas a radial component of linear acceleration,

ar =v2

r= rω2,

directed radially inward, that is responsiblefor changes in the direction of the linearvelocity v.

(g) (�) The linear acceleration is tangent to thepath of the point in question. We call it thetangential component at of the linearacceleration of the point, and we write

at = αr.

2. The magnitude of the angular velocity is multipliedby a factor of m and the magnitude of the angularacceleration is multiplied by a factor of n.

(a) (�) If we set ω′ and α′ are the final angularfrequency and acceleration, then

ω′ = mω, α′ = nα.

(b) (�) The radial component of the accelerationis the centripetal acceleration:

a′r = rω′2 = m2ar.

(c) (�) The tangential component of theacceleration is

a′t = rα′ = nat.

(d) (�) The magnitude of the final acceleration is

a′ =√a′2r + a′2t =

√m4a2

r + n2a2t .

(e) (�) If n = m2, then

a′ =√a′2r + a′2t =

√m4a2

r + n2a2t = m2a.

(f) (�) A wheel starts from rest and spins with aconstant angular acceleration. As time goeson the acceleration vector for a point on therim increases in magnitude and becomes morenearly radial.

10-4 Kinetic Energy of Rotation

1. We want to express the kinetic energy of a rotatingobject in terms of its angular velocity.

(a) (�) The kinetic energy Krotation of rotationcan be computed as

Krotation =1

2

n∑i=1

miv2i .




(b) (�) The speed of each particle is

vi = riω.

(c) (�) The kinetic energy Krotation of rotationcan be computed as

Krotation =1

2

(n∑i=1

mir2i

)ω2.

(d) (�) I is the rotational inertia of the body:

I =n∑i=1

mir2i .

(e) (�) The kinetic energy Krotation of a rigidbody rotating about a fixed axis is given by

Krotation =1

2Iω2.

2. Four identical particles, each with mass m, arearranged in the xy plane as shown. They areconnected by light (massless) sticks to form a rigidbody.

(a) (�) The rotational inertia of this array aboutthe x axis is

Ix = 2ma2.

(b) (�) The rotational inertia of this array aboutthe y axis is

Iy = 6ma2.

10-5 Calculating the Rotational Inertia

1. (�) For a body with continuously distributedmass, the rotational inertia is defined as

I =

∫r2dm =

∫ρr2dV,

where r is the perpendicular distance from therotation axis to a differential mass element dm.Here, dm = ρdV , where ρ is the volume densityand dV is the differential volume elementcontaining the mass element dm.

2. (�) The parallel-axis theorem relates therotational inertia I of a body with mass M aboutan axis to that of the same body about a parallelaxis through the center of mass:

I = Icom +Mh2.

Here h is the perpendicular distance between thetwo axes, and Icom is the rotational inertia of thebody about the axis through the com.

3. Compute the following rotational inertia.

Axis Axis Axis

AxisAxis

Axis

Axis

AxisAxis

Hoop Solid cylinder Solid Sphere

Thin spherical shell Solid cylinder Slab

HoopThin rodAnnularcylinder

(a) (�) Hoop about central axis: I = MR2.

(b) (�) Annular cylinder (or ring) about centralaxis: I = 1

2M(R21 +R2

2).

(c) (�) Solid cylinder (or disk) about central axis:I = 1

2MR2.

(d) (�) Solid cylinder (or disk) about centraldiameter: I = 1

4MR2 + 112ML2.




(e) (�) Thin rod about axis through centerperpendicular to length: I = 1

12ML2.

(f) (�) Solid sphere about any diameter:I = 2

5MR2.

(g) (�) Thin spherical shell about any diameter:I = 2

3MR2.

(h) (�) Hoop about any diameter: I = 12MR2.

(i) (�) Slab about perpendicular axis throughcenter. I = 1

12M(a2 + b2).

4. Let us compute the rotational inertia for a uniformthin rod about an axis through its centerperpendicular to the length.

(a) (�) The linear mass density is

ρ =M

L.

(b) (�) The differential mass element is

dm = ρdx =M

Ldx = Mdt,

where x = Lt and t ∈ [−12 ,

12 ].

(c) (�) We carry out the integration to find that

I = M

∫ 12

− 12

x2dt

= ML2

∫ 12

− 12

t2dt

= 2ML2

∫ 12

0t2dt

= 2ML2 1

3

(1

2

)3

=1

12ML2.

5. Let us compute the rotational inertia for a uniforndisk of radius R and mass M about a symmetryaxis normal to the disk.

(a) (�) The areal mass density is

ρ =M

πR2.


dm = ρdA,

where the areal element is an annular regionwithin the interval [r, r + dr]:

dA = 2πrdr.

Here, r ∈ [0, R].


I =

∫ R

0r2dm

= 2πρ

∫ R

0r2rdr

=2M

R2

∫ R

0r3dr

=1

2MR2.

6. Let us compute the rotational inertia for a uniformsphere of radius R and mass M about an axispassing the center.

(a) (�) The volume mass density is

ρ =M

43πR

3.

(b) (�) The differential rotational inertia elementcan be chosen as a thin disc with thickness dzand radius r =

√R2 − z2.

dI =1

2r2dm,

where the volume element is a disk of radius rwith [z, z + dz]:

dV = πr2dz.

Thus

dI =1

2r2ρdV

=1

2r2 M

43πR

3πr2dz

=3M

8R3r4dz

=3M

8R3(R2 − z2)2dz ← z = Rt

=3M

8R2(1− t2)2dt, t ∈ [−1, 1]

Here, z ∈ [−R,R] and r =√R2 − z2.





I =

∫dI

=3M

8R2

∫ 1

−1(1− t2)2dt

=3M

4R2

∫ 1

0(1− t2)2dt

=3M

4R2

∫ 1

0(1− 2t2 + t4)dt

=3M

4R2

(1− 2

3+

1

5

)=

3M

4R2 × 8

15

=2

5MR2.

7. Let us compute the rotational inertia for a uniformspherical shell of radius R and mass M about anaxis passing the center.

(a) (�) The areal mass density is

ρ =M

4πR2.

(b) (�) The differential rotational inertia elementcan be chosen as a thin hoop-like within polarangle interval [θ, θ + dθ] for −π

2 ≤ θ ≤π2 .

Here, θ is equivalent to the latitude of theEarth.

dI = r2dm,

where

r = R cos θ

and the areal element is a disk of radius rwith polar angle [θ, θ + dθ]:

dA = 2πrds,

where ds is the arc length

ds = Rdθ.

Thus,

dI = r2dm

= r2ρdA

= r2 M

4πR22πrRdθ

= r3 M

2Rdθ

=MR2

2cos3 θdθ.

=MR2

2cos2 θ(cos θdθ)

=MR2

2(1− sin2 θ)d(sin θ)

=MR2

2(1− t2)dt. ← t = sin θ

Here, θ ∈ [−π2 ,

π2 ] and t ∈ [−1, 1].


I =

∫dI

=MR2

2

∫ 1

−1(1− t2)dt

= MR2

∫ 1

0(1− t2)dt

= MR2

(1− 1

3

)=

2

3MR2.

8. A and B are two solid cylinders made of aluminum.Their dimensions are shown.

RB = mRA, LB = nLA.

Let us compute the ratio of the rotational inertiaof B to that of A about the common axis X-X’.




(a) (�) The mass ratio can be found as

MB = m2nMA.

(b) (�) The ratio of the rotational inertia is

IB = IA ×MB

MA×(RBRA

)2

= IA × (m2n)×m2

= m4nIA.

9. A thin rod of length L has a density that increasesalong its length,

ρ(x) = ρ0x.

Let us compute the rotational inertia of the rodaround its less dense end.

(a) (�) The total mass is

M =

∫ L

0ρ(x)dx

= ρ0

∫ L

0xdx

=1

2ρ0L

2.

Thus

ρ0 =2M

L2.


dm = ρ0xdx =2M

L2xdx = 2Mtdt,

where x = Lt and t ∈ [0, 1].


I = 2M

∫ 1

0x2tdt

= 2ML2

∫ 1

0t3dt

=1

2ML2.

10. (�) When a thin uniform stick of mass M andlength L is pivoted about its midpoint, itsrotational inertia is ML2

12 . When pivoted about a

parallel axis through one end, its rotational inertiais

I ′ = Icom +Mh2

=1

12ML2 +M

(1

2L

)2

=

(1

12+

1

4

)ML2

=1

3ML2.

10-6 Torque

1. Torque is a turning or twisting action on a bodyabout a rotation axis due to a force F .

(a) (�) If a force is exerted at a point given bythe position vector relative to the axis, thenthe magnitude of the torque is

τ = rFt,

where r is the radius and Ft is the tangentialcomponent of the force, which isperpendicular to the radius.

(b) (�) The magnitude of the torque is

τ = r⊥F,

where r⊥ is the moment arm of F . r⊥ isthe perpendicular distance from the axis tothe extended line of F .

(c) (�) The magnitude of the torque is

τ = rF sinφ,

where φ is the angle between r and F .

(d) (�) The SI unit of torque is thenewton-meter, N·m.

(e) (�) A torque τ is positive if it tends to rotatea body at rest counterclockwise.

(f) (�) A torque τ is negative if it tends to rotatethe body clockwise.

2. (�) A force with a given magnitude is to be appliedto a wheel. The torque can be maximized byapplying the force at the rim, tangent to the rim.




3. (�) A force is applied to a billiard ball. In order tocalculate the torque created by the force, you alsoneed to know the location and orientation of theaxis of rotation of the ball.

10-7 Newton’s Second Law for Rotation

1. (�) The rotational analog of Newton’s second lawis

τnet = Iα,

where τnet is the net torque acting on a particle orrigid body, I is the rotational inertia of the particleor body about the rotation axis, and α is theresulting angular acceleration about that axis.

2. (�) τnet = Iα for an object rotating about a fixedaxis, where τnet is the net torque acting on it, I isits rotational inertia, and α is its angularacceleration. This expression follows directly fromNewton’s second law.

3. A uniform disk, a thin hoop, and a uniform solidsphere, all with the same mass and same outerradius, are each free to rotate about a fixed axisthrough its center. Assume the hoop is connectedto the rotation axis by light spokes. With theobjects starting from rest, identical forces aresimultaneously applied to the rims, as shown.

Disk Hoop Sphere

(a) (�) The rotation inertias for the three objectsare

Idisk =1

2MR2,

Ihoop = MR2,

Isphere =2

5MR2.

(b) (�) All of them are applied by the same nettorque,

τnet = RF.

(c) (�) The angular accelerations for the threeobjects are

αdisk =2F

MR,

αhoop =F

MR,

αsphere =5

2

F

MR.

(d) (�) The angular velocities for the threeobjects are

ωdisk =2F

MRt,

ωhoop =F

MRt,

ωsphere =5

2

F

MRt.

4. A block of mass M is attached to a cord that iswrapped around the rim of a flywheel of diameterD and hangs vertically, as shown. The rotationalinertia of the flywheel is I. Let us compute theacceleration of the block when the block is releasedand the cord unwinds.

(a) (�) The equation of motion for the flywheel is

Iα =D

2T,

where T is the tension.

(b) (�) The equation of motion for the block is

Ma = Mg − T,

where

a =D

2α.




(c) (�) Therefore, the acceleration of the block is

a =Mg

M +4I

D2

.

10-8 Work and Rotational Kinetic Energy

1. (�) The equation used for calculating work inrotational motion is

W =

∫ θf

θi

τdθ.

When τ is constant,

W = τ(θf − θi).

2. (�) The equation used for calculating power inrotational motion is

P =dW

dt= τω.

3. (�) The work–kinetic energy theorem used forrotating bodies is

∆K = Kf −Ki =1

2Iω2

f −1

2Iω2

i = W.

4. (�) A disk starts from rest and rotates about afixed axis, subject to a constant net torque. Thework done by the torque during the secondrevolution is the same as the work done during thefirst revolution.


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