version 001 – rolling & angular momentum – …...version 001 – rolling & angular...

Version 001 – Rolling & Angular Momentum – ramadoss – (1234) 1

This print-out should have 76 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.One version of this assignment with keys is

on my website.

AP B 1998 MC 6001 10.0 points

A 2 kg object moves in a circle of radius 4 mat a constant speed of 3 m/s. A net force of4.5 N acts on the object.What is the magnitude of the angular mo-

mentum ‖~L‖ of the object with respect to anaxis perpendicular to the circle and throughits center?

1. ‖~L‖ = 13.5kg ·m2

s2

2. ‖~L‖ = 18N ·mkg

3. ‖~L‖ = 12m2

s

4. ‖~L‖ = 24kg ·m2

scorrect

5. ‖~L‖ = 9N ·mkg

Explanation:The angular momentum is

‖~L‖ = mv r = (2 kg) (3 m/s) (4 m)

= 24 kg ·m2/s .

AP M 1998 MC 32 33002 (part 1 of 2) 10.0 points

A wheel with rotational inertia I is mountedon a fixed, frictionless axle. The angularspeed ω of the wheel is increased from zero toωf in a time interval T .What is the average net torque τ on the

wheel during this time interval?

1. τ =ωf

T

2. τ =I ω2

f

T

3. τ =I ωf

Tcorrect

4. τ =ωf

T 2

5. τ =I ωf

T 2

Explanation:The change of angular momentum of the

wheel is∆L = I ωf ,

so the average net torque on the wheel duringthis time interval is

τ =L

T=

I ωf

T.

003 (part 2 of 2) 10.0 pointsWhat is the average power input to the wheelduring this time interval?

1. P =I2 ωf

2T 2

2. P =I2 ω2

f

2T 2

3. P =I ω2

f

2Tcorrect

4. P =I ωf

2T

5. P =I ω2

f

2T 2

Explanation:The change in the energy of the wheel is

1

2I ω2

f in the time interval T , so the average

power input to the wheel is

1

2I ω2

f

1

T=

I ω2f

2T.

AP M 1998 MC 6004 10.0 points

A wheel of mass M and radius R rolls on alevel surface without slipping.If the angular velocity of the wheel about

its center is ω, what is its linear momentumrelative to the surface?

1. p = M ω2R


2. p = 0

3. p =M ω2R2

2

4. p = M ωR2

5. p = M ωR correct

Explanation:First, we note that the wheel is rotating

about its center at an angular velocity of ω,so the velocity difference between the centerof the wheel and the lowest point (the point ofthe wheel that touches the surface) is ωR inmagnitude in the horizontal direction. Sincethe wheel rolls without slipping (which meansthe velocity of the lowest point is 0) the ve-locity v of the center of the wheel is ωR.Obviously, the center of the wheel is also thecenter of the mass, so the linear momentum isp = M ωR.

Atwood Machine 07005 (part 1 of 3) 10.0 points

Consider a massless, frictionless pulley at-tached to the ceiling. A massless, inextensiblestring is attached to the masses Mb and Ma,whereMb > Ma . The tensions Ty, Tz , Tx, andthe gravitational constant g are magnitudes.

ℓ

R

ω

Mb

Ma

Ty

Tx

Tz

What is true about the tensions Ty and Tx?

1. Ty > Mb g and Tx > Ma g

2. Ty < Mb g and Tx > Ma g correct

3. Ty = Mb g and Tx < Ma g

4. Ty = Mb g and Tx > Ma g

5. Ty < Mb g and Tx < Ma g

6. Ty > Mb g and Tx = Ma g

7. Ty = Mb g and Tx = Ma g

8. Ty > Mb g and Tx < Ma g

9. Ty < Mb g and Tx = Ma g

Explanation:The magnitude of the tension in the string

is the same at all points along the string:

T ≡ Ty = Tx .

Consider the free body diagrams

Mb Ma

Ty

Tx

Mbg

Mag

a a

Since the larger mass will move down andthe smaller mass up, we can take motiondownward as positive for Mb and motion up-ward as positive for Ma. Applying Newton’ssecond law to Ma and Mb, respectively,

∑

F = Tx −Ma g = Ma a (1)

Tx = Ma (g + a) > Ma g ,

and for mass Mb,

∑

F = Mb g − Ty = Mb a (2)

Ty = Mb (g − a) < Mb g .

006 (part 2 of 3) 10.0 pointsWhich relationship about Tz is true?


1. Tz = Mb g +Ma g

2. Tz < Mb g +Ma g correct

3. Tz > Mb g +Ma g

Explanation:The forces are in equilibrium for the pulley,

so

Tz = Ty + Tx

= Mb (g − a) +Ma (g + a)

= Mb g +Ma g − (Mb −Ma) a

< Mb g +Ma g

007 (part 3 of 3) 10.0 pointsWhat is the magnitude of the acceleration ofthe center of mass of the system?

1.Mb +Ma

Mb −Mag

2.Mb +Ma

Ma −Mb

g

3.

(

Mb +Ma

Mb −Ma

)2

g

4.

(

Mb −Ma

Mb +Ma

)2

g correct

5.Ma −Mb

Mb +Mag

6.Mb −Ma

Mb +Mag

Explanation:Since Ty = Tx, adding equations 1 and 2

yields

Mb g −Ma g = Ma a+Mb a

a =Mb −Ma

Ma +Mb

g .

Assume that the CMmoves downward (andthat aCM > 0):

(Mb +Ma) aCM = Mb a−Ma a

aCM =Mb −Ma

Mb +Maa

=

(

Mb −Ma

Mb +Ma

)2

g > 0 ,

so the assumption was correct.

keywords:

Bohr Model of Hydrogen 02008 (part 1 of 2) 10.0 points

In the Bohr’s model of the hydrogen atom,the electron moves in a circular orbit of radius5.125× 10−11 m around the proton. Assumethat the orbital angular momentum of theelectron is equal to h.Calculate the orbital speed of the electron.1. 14872000.02. 13070800.03. 12470200.04. 12940600.05. 17351900.06. 13603700.07. 14193000.08. 16531600.09. 16185800.010. 12650300.0

Correct answer: 1.4193× 107 m/s.

Explanation:The angular momentum of the electron in theground state of the hydrogen atom (this thecase here)in the Bohr’s model is h, therefore:

L = mv r = h

and solving for v,

v =h

mr

=(6.62607× 10−34 J s)

(9.10939× 10−31 kg)(5.125× 10−11 m)

= 1.4193× 107 m/s

009 (part 2 of 2) 10.0 pointsCalculate the angular frequency of the elec-tron’s motion1. 3.57614e+172. 2.97224e+173. 3.24846e+174. 2.63805e+17


5. 4.16706e+176. 3.17823e+177. 3.31538e+178. 3.02953e+179. 2.24748e+1710. 2.76936e+17

Correct answer: 2.76936× 1017 s−1.

Explanation:The angular frequency is given by

ω =v

r

=(1.4193× 107 m/s)

(5.125× 10−11 m)

= 2.76936× 1017 s−1

Bullet Rotates a Rod 01010 (part 1 of 2) 10.0 points

A wooden block of massM hangs from a rigidrod of length ℓ having negligible mass. Therod is pivoted at its upper end. A bullet ofmass m traveling horizontally and normal tothe rod with speed v hits the block and getsembedded in it.

Mv

m

ℓ

What is the angular momentum L of theblock-bullet system, with respect to the pivotpoint immediately after the collision?

1. L =

(

M m

M +m

)

v ℓ

2. L = mv ℓ correct

3. L = (M −m) v ℓ

4. L = (m+M) v ℓ

5. L = M v ℓ

Explanation:

If∑

~τext = 0, then∑

~L = 0 .

The net angular momentum of the systemconserves, and

Li = Lf = L = mv ℓ .

011 (part 2 of 2) 10.0 points

What is the fractionKf

Ki(the final kinetic

energy compared to the initial kinetic energy)in the collision?

1.Kf

Ki=

M

M −m

2.Kf

Ki=

M

M +m

3.Kf

Ki=

2m

m+M

4.Kf

Ki=

m

m+Mcorrect

5.Kf

Ki=

m

M −m

Explanation:By conservation of the angular momentum

Li = Lf = L

mv ℓ = (m+M) vf ℓ

vf = v

(

m

m+M

)

Ki =1

2mv2 and Kf =

1

2I ω2

f where

I = (M +m) ℓ2 and ωf =vfℓ, so

Kf =1

2(M +m) v2f and

Kf

Ki=

1

2

m2

M +mv2

1

2mv2

=m

M +m.

Bullet Hits a Cube 02012 10.0 points

Assume: A bullet of mass m and cube ofmass M undergo an inelastic collision, wherem ≪ M .


Note: The moment of inertia of this cube(with edges of length 2 a and mass M) about

an axis along one of its edges is8M a2

3.

A solid cube is resting on a horizontal sur-face. The cube is constrained to rotate aboutan axis at its bottom right edge (due to asmall obstacle on the table). A bullet withspeed vmin is shot at the left-hand face at a

height of4

3a. The bullet gets embedded in

the cube.

2 a

M

m~vmin

4

3a

ω

M gFind the minimum value of vmin required

to tip the cube so that it falls its right-handface.

1. vmin =m

M

√

2 g a(√

2− 1)

2. vmin =m

M

√

3 g a(√

3− 1)

3. vmin =m

M

√

5 g a(√

2− 1)

4. vmin =m

M

√

3 g a(√

5− 1)

5. vmin =M

m

√

3 g a(√

2− 1)

correct

Explanation:Basic Concepts:

∑

~L = const

∆U +∆K = 0

For the cube to tip over the center of mass(CM) must rise so that it is over the axis ofrotation AB. To do this the CM must beraised a distance of a

(√2− 1

)

.

From conservation of energy

M g a(√

2− 1)

=1

2Icube ω

2 . (1)

From conservation of angular momentum

4 a

3mv =

(

8M a2

3

)

ω

ω =( mv

2M a

)

. (2)

Thus, substituting Eq. 2 into 1, we have

1

2

(

8M a2

3

)(

m2 v2

4M2 a2

)

= M g a(√

2− 1)

Solving for v yields

vmin =M

m

√

3 g a(√

2− 1)

.

Child on a Merrygoround013 10.0 points

A child is standing on the edge of a merry-go-round that is rotating with frequency f. Thechild then walks towards the center of themerry-go-round.For the system consisting of the child plus

the merry-go-round, what remains constantas the child walks towards the center? (ne-glect friction in the bearing)

1. only mechanical energy

2. only angular momentum correct

3. neither mechanical energy nor angularmomentum

4. mechanical energy and angular momen-tum

Explanation:The forces external to the system are not

exerting any torque, so the angular momen-tum is conserved. On the other hand, the fric-tion force acting on the child is doing work, be-cause she is moving towards the center (whichis the direction of that force).

Circular Trajectory014 10.0 points

A particle of mass m moves in a circle ofradius R at a constant speed v. Assume: The


motion begins from the point Q, which hascoordinates (R, 0).Determine the angular momentum of the

particle about point P , which has coordinates(−R, 0) as a function of time.

1. ~L = mvR

[

sin

(

v t

R+ π

)

+ 1

]

k

2. ~L = mvR

[

cos

(

v t

R+ π

)

+ 1

)

k

3. ~L = mv2

R

[

sin

(

v t

R+ π

)

+ 1

]

k

4. ~L = mvR

[

sin

(

v t

R

)

+ 1

]

k

5. None of these

6. ~L = mv2

R

[

cos

(

v t

R+ π

)

+ 1

]

k

7. ~L = mvR

[

cos

(

v t

R

)

+ 1

]

k correct

8. ~L = mv2

R

[

cos

(

v t

R

)

+ 1

]

k

9. ~L = mv2

R

[

sin

(

v t

R

)

+ 1

]

k

Explanation:Basic Concept:

~L = ~r ×m~v .

Solution: The angular displacement of theparticle around the circle is

θ = ω t =v t

R.

The vector from the center of the circle to themass is then

R cos θ ı+R sin θ .

The vector from the point P to the point ofmass is

~r = R ı+R cos θ ı+R sin θ

~r = R

{[

1 + cos

(

v t

R

)]

ı+ sin

(

v t

R

)

}

.

The velocity is

~v =d~r

dt

= −v sin

(

v t

R

)

ı+ v cos

(

v t

R

)

.

So

~L = ~r ×m~v

= mvR {[1 + cos(ω t)] ı+ sin(ω t) }× [− sin(ω t) ı+ cos(ω t) ]

= mvR {[1 + cos(ω t)] [cos(ω t)]

−[sin(ωt)] [− sin(ω t)] } k

= mvR

[

cos

(

v t

R

)

+ 1

]

k .

Collision With a Cylinder015 10.0 points

A solid cylinder of mass M = 38 kg, radiusR = 0.17 m and uniform density is pivoted ona frictionless axle coaxial with its symmetryaxis. A particle of massm = 2.4 kg and initialvelocity v0 = 3.6 m/s (perpendicular to thecylinder’s axis) flies too close to the cylinder’sedge, collides with the cylinder and sticks toit.

Before the collision, the cylinder was not ro-tating. What is the magnitude of its angularvelocity after the collision?1. 2.339642. 2.374933. 5.91444. 13.04355. 9.736846. 11.21257. 14.79598. 5.284099. 25.757610. 19.3237

Correct answer: 2.37493 rad/s.


Explanation:Basic Concept: Conservation of Angu-

lar Momentum,

Lparticlez + Lcylinder

z = const.

The axle allows the cylinder to rotate withoutfriction around a fixed axis but it keeps thisaxis fixed. Let the z coordinate axis run alongthis axis of rotation; then the axle may exertarbitrary torques in x and y directions butτz ≡ 0. Consequently, the z componenent ofthe angular momentum must be conserved,Lz = const, hence when the particle collideswith the cylinder

Lbeforez,part + Lbefore

z,cyl = Lz,net = Lafterz,part + Lafter

z,cyl.

Before the collision, the cylinder did notrotate hence

Lbeforez,cyl = 0

while the particle had angular momentum

~Lbeforepart = ~r × ~P0 = ~r ×m~v0.

Both the radius-vector ~r and the velocity ~v0of the particle lie in the xy plane (⊥ to thez axis), and according to the picture, at themoment of collision the radius vector has mag-nitude |~r| = R equal to the cylinder’s radiusand direction perpendicular to the particle’svelocity. Hence, its angular momentum isparallel to the z axis and has magnitude

|~Lbeforepart | = Lbefore

z,part = Rmv0 .

Altogether, before the collision

Lbeforez,net = Rmv0

and therefore, after the collision we shouldalso have

Lafterz,net = Rmv0 . (1)

After the collision, the cylinder and theparticle rotate as a single rigid body of netmoment of inertia

Inet = Icyl + Ipart = 12MR2 + mR2. (2)

relatve to cylinder’s axis. For a rigid rotationlike this, the angular momentum points in zdirection and its magnitude is

Lz,net = ωInet . (3)

Combining eqs. (1), (2) and (3) together,we immeditaly obtain

ωafter =Lafterz,net

Inet=

Rmv012MR2 +mR2

=v0R

× m12M +m

= 2.37493 rad/s.

Component of Net Torque016 10.0 points

A force ~F = (2.1 N) ı + (2.9 N) is appliedto an object that is pivoted about a fixedaxis aligned along the z coordinate axis. Theforce is applied at the point ~R = (4.1 m) ı +(3.7 m) .Find the z-component of the net torque.1. 1.722. 3.573. 4.124. -0.075. -0.656. 2.767. 3.748. 5.289. 4.9610. 1.22

Correct answer: 4.12 Nm.


~τ = ~r × ~F

Solution: From the definition of torque

~τ = ~R× ~F

= [Rx ı+Ry ]× [Fx ı+ Fy ]

= [(4.1 m) ı+ (3.7 m) ]× [(2.1 N) ı+ (2.9 N) ]

= [(4.1 m) (2.9 N)− (3.7 m) (2.1 N)]k

= (4.12 Nm) k

In this case, the net torque only has a z-component.


Component of Net Torque 03017 10.0 points

A force ~F = (1.7 N) ı+ (3.4 N) is applied toan object that is pivoted about a fixed axisaligned along the z coordinate axis. The forceis applied at the point~r = (3.7 m) ı+(4.9 m) .Find the z-component of the net torque.1. 5.542. 1.113. 2.554. 6.565. 4.256. -0.697. 4.328. 3.429. 7.3810. 3.38



~τ = ~r × ~F

Solution: From the definition of torque

~τ = ~r × ~F

= [x ı+ y ]× [Fx ı+ Fy ]

= [(3.7 m) ı+ (4.9 m) ]× [(1.7 N) ı+ (3.4 N) ]

= [(3.7 m) (3.4 N)− (4.9 m) (1.7 N)]k

= (4.25 Nm) k

In this case, the net torque only has a z-component.

Conical Pendulum 04018 10.0 points

A small metallic bob is suspended from theceiling by a thread of negligible mass. Theball is then set in motion in a horizontal circleso that the thread describes a cone.

v

9.8 m/s2

3.5m

3 kg

24◦

Calculate the magnitude of the angular mo-mentum of the bob about a vertical axisthrough the supporting point. The acceler-ation of gravity is 9.8 m/s2 .1. 56.23372. 4.522143. 5.673374. 11.01345. 10.64386. 5.960387. 1.724128. 10.11069. 2.4002410. 6.54738

Correct answer: 10.6438 kg ·m2/s.

Explanation:

Let : ℓ = 3.5 m ,

θ = 24◦ ,

g = 9.8 m/s2 , and

m = 3 kg .

Consider the free body diagram.


T

mg

θ

Newton’s second law in the vertical andhorizontal projections, respectively, gives

T cos θ −mg = 0

T cos θ = mg and

T sin θ −mω2 ℓ sin θ = 0

T = mω2 ℓ ,

where the radius of the orbit is R = ℓ sin θ.Dividing,

cos θ =mg

mω2 ℓ

ω =

√

g

ℓ cos θ.

~r and ~v are perpendicular, where r =ℓ sin θ , so the angular momentum L = m~r×~vwill be

L = mω (ℓ sin θ)2

= m sin2 θ

√

g ℓ3

cos θ

= (3 kg) sin2 24◦

×√

(9.8 m/s2) (3.5 m)3

cos 24◦

= 10.6438 kg ·m2/s .

Cylinder rolls019 10.0 points

The coefficient of static friction between acertain cylinder and a horizontal floor is 0.2.If the rotational inertia of the cylinder aboutits symmetry axis is given by I = (1/2)MR2,then the maximum acceleration the cylindercan have without slipping is:

1. 0.4 g correct

2. 1 g

3. 0.2 g

4. 0.1 g

5. 0.05 g

Explanation:If the cylinder is not slipping, we know

that the angular acceleration is related to thelinear acceleration by a = αR.Let us consider the relevant force to be

applied at the center of the cylinder.Then, the only horizontal forces applied to

the cylinder are the accelerating force andthe frictional force. Taking the center of thecylinder to be the axis of rotation, we real-ize that the applied force exerts no torque(since ~r = 0) , and the rotational analogue ofNewton’s second law reads:

Iα = τnet(

1

2MR2

)

α = fsR

1

2MR2

( a

R

)

= µsMgR

a = 2µsg

= 2(0.2)(g)

= 0.4 g

Decelerated Grinding Wheel020 (part 1 of 2) 10.0 points

The motor driving a grinding wheel with arotational inertia of 0.84 kgm2 is switchedoff when the wheel has a rotational speed of33 rad/s. After 9.7 s, the wheel has sloweddown to 26.4 rad/s.What is the absolute value of the constant

torque exerted by friction to slow the wheeldown?1. 0.4622. 0.24363. 0.174872


4. 0.4153855. 0.3607276. 0.2546947. 0.5715468. 0.4819. 0.33049210. 0.453818


Explanation:We have

τ ∆t = ∆L = ∆(I ω) ,

so that

|τ | = I |ω1 − ω0|∆t1

=(0.84 kgm2) (33 rad/s− 26.4 rad/s)

9.7 s= 0.571546 Nm .

021 (part 2 of 2) 10.0 pointsIf this torque remains constant, how long afterthe motor is switched off will the wheel cometo rest?1. 55.02. 39.03. 48.54. 41.55. 42.56. 49.07. 33.58. 50.09. 31.010. 38.5

Correct answer: 48.5 s.

Explanation:When the wheel comes to rest, its angular

speed is ω2 = 0; hence

∆t2 =

∣

∣

∣

∣

I (ω0 − ω2)

τ

∣

∣

∣

∣

=I ω0

|τ |

=(0.84 kgm2) (33 rad/s)

(0.571546 Nm)= 48.5 s .

Disk and Hoop Race022 (part 1 of 2) 10.0 points

A uniform solid disk and a uniform hoop areplaced side by side at the top of an incline ofheight h.If they are released from rest and roll with-

out slipping, determine their speeds whenthey reach the bottom. (d = disk, h = hoop)

1. vd =√

h g , vh =√

2h g

2. vd =

√

2

3h g , vh =

√

3h g

3. vd =√

2h g , vh =√

h g

4. vd =

√

1

2h g , vh =

√

1

3h g

5. vd =

√

4

3h g , vh =

√

h g correct

Explanation:Because they roll without slipping, v = r ω

and the total kinetic energy is K =1

2mv2 +

1

2I ω2, so

for the disk, since I =1

2mr2 ,

K =1

2mv2d +

1

4mv2d =

3

4mv2d .

3

4mv2d = mhg ,

vd =

√

4

3h g .

Similarly, for the hoop, since I = mr2 con-servation of energy gives us

vh =√

h g .

023 (part 2 of 2) 10.0 pointsWhich object reaches the bottom first?

1. the hoop


2. both at the same time

3. the disk correct

Explanation:From the previous result one can see that

for any fallen height h, the disk has a velocitygreater than the velocity of the hoop, so thedisk will reach the bottom first (its velocityhas been greater than hoop’s all the time).

keywords:

Disk and Hoop Race 02024 (part 1 of 2) 10.0 points

A uniform solid disk and a uniform hoop areplaced side by side at the top of an incline ofheight h.If they are released from rest and roll with-

out slipping, determine their speeds whenthey reach the bottom. (d = disk, h = hoop)

1. vd =

√

2

3h g , vh =

√

3h g

2. vd =√

h g , vh =√

2h g

3. vd =√

2h g , vh =

√

1

2h g

4. vd =√

3h g , vh =

√

1

3h g

5. vd =√

3h g , vh =√

2h g

6. vd =

√

1

2h g , vh =

√

1

3h g

7. vd =√

2h g , vh =√

h g

8. vd =

√

1

3h g , vh =

√

h g

9. vd =

√

4

3h g , vh =

√

h g correct

10. vd =√

3h g , vh =√

h g

Explanation:Because they roll without slipping, v = r ω

and the total kinetic energy is K =1

2mv2 +

1

2I ω2, so

for the disk, since I =1

2mr2 ,

K =1

2mv2d +

1

4mv2d =

3

4mv2d .

3

4mv2d = mhg ,

vd =

√

4

3h g .

Similarly, for the hoop, since I = mr2 ,conservation of energy gives us

vh =√

h g .

025 (part 2 of 2) 10.0 pointsWhat is the ratio of their accelerations as theyroll down the incline?

1.adiskahoop

=

√

3

2

2.adiskahoop

=1

3

3.adiskahoop

=

√

4

3

4.adiskahoop

=√2

5.adiskahoop

=3

2

6.adiskahoop

=1

2

7.adiskahoop

= 3

8.adiskahoop

=4

3correct

9.adiskahoop

=√3

10.adiskahoop

= 2

Explanation:Both disk and hoop started from rest, so

v2 = 2 a s .


When the two roll down the same height hon the incline, they have covered the samedistance s, so

adiskahoop

=2 adisk s

2 ahoop s=

(vdisk)2

(vhoop)2=

4

3.

keywords:

Door Handle Torque026 10.0 points

A door is opened in the usual way, in thedirection indicated.

B

A

CD

In which direction is the net torque vector~τ due to the force applied to the door handle?The force is toward you, away from the door.

1. A correct

2. B

3. Insufficient information is given.

4. C

5. D

Explanation:

~τ = r×Fr ~τ

F~r points from the hinge to the spot where

~F acts, so by the right hand rule, ~r× ~F pointsin the direction A, along the hinge line.

Figure Skater027 (part 1 of 2) 10.0 points

A figure skater rotating on one spot with botharms and one leg extended has moment of

inertia Ii. She then pulls in her arms and theextended leg, reducing her moment of inertiato 0.75 Ii.What is the ratio of her final to initial ki-

netic energy?

1. 2

2. 3/4

3. 1

4. 8/3

5. 1/2

6. 9/16

7. 3/8

8. 4/3 correct

9. 16/9

Explanation:The angular momentum was conserved, so

Li = Lf

Since L = I ω and KEr =1

2I ω2,

KEr =L2

2 I

So,

KEf

KEi=

L2

f

2 If

L2

i

2 Ii

=IiIf

=Ii

0.75 Ii=

4

3

028 (part 2 of 2) 10.0 pointsConsider the following statements for the fig-ure skater:I. Angular momentum was conserved.II. Mechanical energy was conserved.III. The kinetic energy changed because of

energy dissipation due to friction.IV. Her rotation rate changed in response

to a torque exerted by pulling in her arms andleg.


Which is the correct combination of state-ments?

1. I and II

2. I, II, III

3. I, II, IV

4. II

5. I correct

Explanation:(I) The angular momentum was conserved

since there was no external torque acting onthe skater.(II) From part 1, we found that the kinetic

energy increased. and the potential energydidn’t change, so the total mechanical energyincreased. (II) is wrong.(III) We found that the total kinetic energy

increased, and therefore, we can infer thatthere was no energy dissipation.(IV) The force pulling her arms in would be

perpendicular to her rotation, and therefore,exert no torque.Ergo, the only correct statement is (I).

Figure Skater Spin 01029 10.0 points

A figure skater on ice spins on one foot. Shepulls in her arms and her rotational speedincreases.Choose the best statement below:

1. Her angular speed increases because sheis undergoing uniformly accelerated angularmotion.

2. Her angular speed increases because airfriction is reduced as her arms come in.

3. Her angular speed increases because herangular momentum increases.

4. Her angular speed increases because bypulling in her arms she creates a net torque inthe direction of rotation.

5. Her angular speed increases because herangular momentum is the same but her mo-ment of inertia decreases. correct

6. Her angular speed increases because herpotential energy increases as her arms comein.

Explanation:The initial angular momentum of the figure

skater is Ii ωi. After she pulls in her arms, theangular momentum of her is If ωf . Note thatIf < Ii because her arms now rotate closer tothe rotation axis and reduce the moment ofinertia. Since the net external torque is zero,angular momentum remains unchanged, andso Ii ωi = If ωf = L. Therefore, ωf > ωi.

Grindstone Energy030 10.0 points

A constant torque of 26.9 N ·m is appliedto a grindstone whose moment of inertia is0.161 kg ·m2.Using energy principles, and neglecting fric-

tion, find the angular speed after the grind-stone has made 17.2 rev.1. 184.9872. 201.3083. 139.2244. 195.5615. 164.2066. 128.7837. 190.0348. 179.5779. 160.42410. 172.888


Explanation:

Given : τ = 26.9 N ·m ,

I = 0.161 kg ·m2 , and

N = 17.2 rev .

Wnet = τ ∆θ =1

2I w2

f − 1

2I w2

i

Σ~τext =d~L

dt


If the torque is constant, then we can writefor the angle of rotation of the grindstone:

τ = Iα,

where α is the angular acceleration. There-fore,

α =τ

I=

d2θ

dt2,

andθ = N 2π,

where N is the number of revolutions. Thus,

ω2 = 2αθ,

ω =√2αθ

=

√

2τ

I2 πN

=

√

4 π (26.9 N ·m)(17.2 rev)

0.161 kg ·m2

= 190.034 rad/s .

Horizontal Rotation031 (part 1 of 3) 10.0 points

A mass m is attached to a cord passingthrough a small hole in a frictionless hori-zontal surface. The mass is initially orbitingwith speed v0 in a circle of radius r0. The cordis then slowly pulled from below, decreasingthe radius of the circle to r.What is the speed of the mass when the

radius is r?

1. v = v0r

r0

2. None of these

3. v = v0r0

r0 + r

4. v = v0r0r

correct

5. v = v0r0

r0 − r

6. v = v0r0 − r

r0 + r

7. v = 2 v0r0r

8. v = v0r0 − r

r0

9. v = v0r0 + r

r0

10. v = v0r0 + r

r0 − r

Explanation:Basic Concepts

~τ = ~r × ~F

d∑

~L

dt= ~τext

Solution

τ = ‖~r × ~F‖

= ‖~r‖‖~F‖ sin(0) = 0

Angular moment is conserved, so

Lf = Li

mr v = mr0 v0

Therefore,

v =r0 v0r

032 (part 2 of 3) 10.0 pointsFind the tension in the cord as a function ofr.

1. T =2m (r0 v0)

2

r3

2. None of these

3. T =m (r0 v0)

2

r3correct

4. T =m (r0 v0)

2

2 r3

5. T =m (r v0)

2

r30

6. T =m (r0 v0)

2

(r + r0)3

Explanation:

T =mv2

r=

m (r0 v0)2

r3


033 (part 3 of 3) 10.0 pointsHow much work W is done in moving m fromr0 to r?Note: The tension depends on r.

1. W =1

3mv20

(

r2

r20− 1

)

2. W =2

3mv20

(

r20r2

− 2

)

3. W =2

3mv20

(

r2

r20− 2

)

4. W =1

3mv20

(

r20r2

− 1

)

5. W =1

2mv20

(

(r0 + r)2

r2− 1

)

6. W =1

2mv20

(

r20r2

− 1

)

correct

7. None of these

8. W =1

2mv20

(

r2

r20− 1

)

Explanation:The work is done by the centripetal force in

the negative direction. Method 1:

W =

∫

~F d~ℓ

= −∫

T dr′

= −∫ r

r0

m (r0 v0)2

(r′)3dr′

=m (r0 v0)

2

2

(

1

r2− 1

r20

)

=1

2mv20

(

r20r2

− 1

)

Method 2:

W = ∆K =1

2mv2 − 1

2mv20

=1

2mv20

(

r20r2

− 1

)

Horizontal Circle 02034 (part 1 of 2) 10.0 points

A ball is rotating in a horizontal circle at theend of a string of length 3.7 m at an angularvelocity of 8.5 rad/s. The string is graduallyshortened to 3.2 m without any force beingexerted in the direction of the ball’s motion.Find the new angular velocity of the ball.1. 11.8082. 12.19913. 11.36384. 12.70075. 14.37586. 13.90627. 16.20688. 13.76579. 12.016310. 15.7756


Explanation:Since there is no force exerted in the di-

rection of the ball’s motion, the total angularmomentum of the ball is conserved. We have

I1 ω1 = I0 ω0

ml21 ω1 = ml20 ω0

ω1 =

(

l0l1

)2

ω0

=

(

3.7 m

3.2 m

)2

(8.5 rad/s)

= 11.3638 rad/s .

035 (part 2 of 2) 10.0 pointsFind its new linear speed.1. 48.31412. 37.07173. 38.94424. 32.51245. 28.64956. 39.35887. 34.68758. 33.04429. 38.545810. 36.3641

Correct answer: 36.3641 m/s.


Explanation:The new linear speed is

v1 = l1 ω1

= (3.2 m)(11.3638 rad/s)

= 36.3641 m/s .

Ice Skater pulls arms in036 10.0 points

An ice skater with rotational inertia I0 is spin-ning with angular velocity ω0. She pulls herarms in, decreasing her rotational inertia toI0/3. Her angular velocity becomes:

1. 3ω0 correct

2. ω0

3. ω0/3

4.√3ω0

5. ω0/√3

Explanation:Angular momentum is given by L = Iω,

and is conserved on ice, so:

I0ω0 = Iω

=

(

1

3I0

)

ω

ω = 3ω0

Impulse on a Cue Ball037 10.0 points

A cue stick strikes a cue ball and delivers ahorizontal impulse in such a way that the ballrolls without slipping as it starts to move.At what height above the ball’s center (if

the radius of the ball is 0.88 cm) was the blowstruck?1. 0.62. 0.3523. 0.524. 0.445. 0.686. 0.36

7. 0.488. 0.729. 0.32810. 0.56

Correct answer: 0.352 cm.


L = Iω

p = mv

ph = Icω0 (1)

p = M v0 (2)

If the ball rolls without slipping,

v0 = Rω0

So,

h =Icω0

p=

Icω0

M v0

=Ic

M R=

2

5R

=2

5(0.88 cm) = 0.352 cm

KE Comparison 01038 10.0 points

A object of mass M and radius R has arotational inertia of ICM = kM R2, where kis a constant.If the object rolls without slipping, what is

the ratio of its rotational kinetic energy to itslinear kinetic energy?

1.Kr

KCM

= kM

2.Kr

KCM

=k

k + 1

3.Kr

KCM

=k + 1

k

4.Kr

KCM

= k correct

5.Kr

KCM

= k R2


6.Kr

KCM

= k + 1

Explanation:

Kr =1

2ICM ω2 and KCM =

1

2M v2CM .

Because there is no slipping, vCM = Rω, and

Kr =1

2ICM ω2 =

1

2kM R2 ω2

=1

2kM v2CM = kKCM

Kr

KCM

= k .

KE Ratio039 10.0 points

A ball with Icm = κM R2, mass M and ra-dius R rolls along a horizontal surface withoutslipping.

What is the ratioKr

Kcmof rotational to

center-of-mass kinetic energy?

1.1

κ

2.κ

R

3.κ

R2

4.1

κ2

5. κ correct

6. 1; they have the same magnitude.

7. κ2

8.√κ

Explanation:

Kcm =1

2M v2cm, and for rolling without

slipping vcm = ω R , so the rotational kineticenergy is

Kr =1

2Icm ω2 =

1

2

(

κM R2)

(vcmR

)2

= κ

(

1

2M v2cm

)

= κKcm

Kr

Kcm= κ .

Kinetic Energy of a Rolling Wheel040 10.0 points

If a steel, thin-shelled wheel of radius r andmass M is moving along the road at 2 m/s,what is its total kinetic energy?

1. 1M

2. 4M correct

3. 2M

4. 5M

5. 3M

Explanation:The moment of inertia of a thin ring or

hollow cylinder about its axis is I = M r2

and the translational speed of the wheel isv = ω r .The wheel is rotating and translating, so its

total kinetic energy is

K = Krot +Ktrans

=1

2I ω2 +

1

2M v2

=1

2

(

M r2)

(v

r

)2

+1

2M v2

= M v2 = 4M .

Maximize Torque041 10.0 points

A force is to be applied to a wheel. The torquecan be maximized by:


1. applying the force near the axle, parallelto the tangent to the wheel

2. applying the force near the rim, radiallyoutward from the axle

3. applying the force at the rim, tangent tothe rim correct

4. applying the force at the rim, at 45◦ tothe tangent

5. applying the force near the axle, radiallyoutward from the axle

Explanation:Let’s look at how torque is defined via cross

product, where we take ~F to be the magnitudeof the force causing the torque, and we take~r to be the vector pointing from the center ofrotation of the object to the point at which ~Fis applied:

~τ = ~r × ~F

|~τ | = |~r|∣

∣

∣

~F∣

∣

∣sin θ

Where θ is the angle between the vectors ~rand ~F .So, to maximize the force, we want ~F to be

applied as far from the center of the wheel aspossible, and we want it to be perpendicularto the wheel. Thus, it should be near the rim,and tangent to the wheel.

Length Ratio042 10.0 points

A non-uniform horizontal beam of length ℓand massM has its center of mass at distancex from the left end, as shown. It is supportedby two ropes; the left-end rope makes an angleof θ with the vertical and has tension T1, andthe right-hand rope makes an angle of φ withthe vertical and has tension T2.

xcm

ℓb

T1

θ

T2

φ

If sin θ = cosφ =3

5and cos θ = sinφ =

4

5,

and the rod is stationary, what is the ratiox

ℓ?

1.3

8

2.1

2

3.1

4

4.5

8

5.3

7

6.3

5

7.9

25correct

8.16

25

9.1

5

10.2

5Explanation:Consider the free-body diagram.

xcm

ℓb

m~g

T1

θ

T2

φ

∑

F = 0 for the rod to be in static equi-

librium, so horizontally,

−T1 sin θ + T2 sinφ = 0

−T1

(

3

5

)

+ T2

(

4

5

)

= 0

T1 =4

3T2

and vertically,

T1 cos θ + T2 cosφ−mg = 0(

4

3T2

)(

4

5

)

+ T2

(

3

5

)

= mg


16T2 + 9T2 = 15mg

T2 =3

5mg .

Taking torques about the left end of therod,

∑

~τ = −mg x+ T2 ℓ cosφ = 0

x

ℓ=

T2 cosφ

mg=

(

3

5mg

)(

3

5

)

mg=

9

25.

Momentum of a Particle043 10.0 points

A particle whose mass is 2 kg moves in xy-plane with a constant speed of 7 m/s in thepositive x-direction along y = 6 m.Find the magnitude of its angular mo-

mentum relative to the point (x0, y0), wherex0 = 2 m and y0 = 10 m.1. 192.02. 32.03. 54.04. 198.05. 16.06. 42.07. 56.08. 96.09. 81.010. 126.0

Correct answer: 56 kgm2/s.

Explanation:Apply

~L ≡ ~r ×~p = m~r × ~v

~L =[

(y − y0) + (x− x0) ı]

×mv ı

= −mv (y − y0) k

Then the magnitude of the angular momen-tum is

‖~L‖ = |mv (y − y0)|= |(2 kg)(7 m/s)(6 m− 10 m)|= 56 kgm2/s

Momentum of Airplane044 10.0 points

An airplane of mass 998 kg flies level to theground at a constant speed of 175 m/s relativeto the Earth. An observer on the groundalong the path of the plane sees the plane adistance 5979.53 m away at an angle abovethe horizontal of 62.419 ◦.What is the magnitude of the airplane’s

angular momentum relative to a ground ob-server directly below the airplane?1. 1362270000.02. 943110000.03. 1344800000.04. 1921150000.05. 925645000.06. 1414660000.07. 558880000.08. 716065000.09. 1746500000.010. 1467060000.0

Correct answer: 9.25645× 108 kgm2/s.


~L = ~r× ~p

L = rmv sin(θ)

= (5979.53 m)(998 kg)(175 m/s) sin(62.419 ◦)

= 9.25645× 108 kgm2/s

Momentum of Moon045 10.0 points

There is a moon orbiting an Earth-like planet.The mass of the moon is 8.65 × 1022 kg, thecenter-to-center separation of the planet andthe moon is 7.04× 105 km, the orbital periodof the moon is 28.5 days, and the radius of themoon is 1420 km.What is the angular momentum of the

moon about the planet?1. 1.09391e+352. 1.72597e+333. 9.66136e+334. 1.28744e+33


5. 6.77481e+346. 6.38583e+347. 2.11845e+358. 2.16226e+349. 3.13956e+3410. 1.79894e+34

Correct answer: 1.09391× 1035 kgm2/s.

Explanation:The angular speed ω in radians per unit

time, for a complete circle is

ω =2 π

T.

Angular momentum L is

L = mv r = mr2 ω =2 πmr2

T= 2 π (8.65× 1022 kg)

×

[

(7.04× 105 km)1000 m

1 km

]2

(28.5 days)

(

24 hr

1 day

)

= 1.09391× 1035 kgm2/s .

Projectile Hits a Disk046 (part 1 of 2) 10.0 points

A particle of mass m and speed v0 collideswith and sticks to the edge of a uniform diskof mass M = 2m and radius R. That is, thedistance between the trajectory of the particleand the center of the disk is R.If the disk is initially at rest and is pivoted

about a frictionless axle through its centerO perpendicular to the plane of the paper,the angular velocity of the disk plus particlesystem after the collision in terms of v0 andR?

O

M

R

m

1.3 v04R

2.v08R

3.v04R

4.v0R

5.v02R

correct

Explanation:By conservation of angular momentum

~Li = ~Lf . The angular momentum before thecollision is

Li = mv0R .

The moment of inertia of the uniform disk is

I =1

2M R2 =

1

2(2m)R2 = mR2 .

The angular momentum after the collision isthe sum of the angular momentum due to themass and that due to the rotating disk:

Lf = mvR + i ω

= mR2 ω + I ω = 2mR2 ω ,

where ω is the angular velocity of the systemafter the collision. For the system of m plusdisk, there is no external torque acting on thesystem, and angular momentum is constantbefore, during and after the collision, i.e.,Li = Lf , so

ω =v02R

.

047 (part 2 of 2) 10.0 pointsWhat is the loss of kinetic energy due to thecollision process?

1.1

12mv20

2.3

8mv20

3.1

8mv20


4.3

16mv20

5.1

4mv20 correct

Explanation:The kinetic energy of the system before

collision is

Ki =1

2mv20

and after the collision

Kf =1

2I ω2

=1

2

(

1

2M R2 +mR2

)

I ω2

=1

2(2mR2) I ω2

= mR2 ω2 = mR2( v02R

)2

=1

4mv20 ,

where v = ωR, I = mR2 and ω =v02R

. The

loss in kinetic energy is

|∆K| =∣

∣Kf −Ki

∣

∣

=

∣

∣

∣

∣

1

4mv20 −

1

2mv20

∣

∣

∣

∣

=1

4mv20 .

Projectile Sticks to a Rod048 (part 1 of 2) 10.0 points

A projectile of mass m = 1.72 kg moves tothe right with speed v0 = 13.5 m/s. The pro-jectile strikes and sticks to the end of a sta-tionary rod of mass M = 5.46 kg and lengthd = 2.62 m that is pivoted about a frictionlessaxle through its center.

dO O

v0 ωm

M

Find the angular speed of the system rightafter the collision.1. 7.21137

2. 7.816673. 7.38064. 9.743255. 7.916446. 4.776637. 14.3868. 5.007129. 1.8634410. 5.60425


Explanation:The initial angular momentum of the pro-

jectile about the pivot O is Li =d

2mv0. With

the projectile stuck to the end of the rod, therotational inertia of the projectile and the rodcombined about O is,

I = m

(

d

2

)2

+1

12M d2

=

[

m

4+

M

12

]

d2 (1)

=

[

(1.72 kg)

4+

(5.46 kg)

12

]

(2.62 m)2

= 6.07499 kgm2 .

Using conservation of angular momentum, wehave

Li = If ω (2)

= ~r ×~p

=d

2mv0 (3)

Therefore, combining Eqs. (1), (2), and (3),we have

ω =

d

2mv0

md2

4+

M d2

12

· 1212

=6mv0

(3m+M) d

=6 (1.72 kg) (13.5 m/s)

[3 (1.72 kg) + (5.46 kg)] (2.62 m)

= 5.00712 rad/s .


049 (part 2 of 2) 10.0 pointsDetermine the ratio of the kinetic energy lostto the initial kinetic energy.1. 0.4909262. 0.8285813. 0.7400474. 0.5509125. 0.5141246. 0.6567137. 0.4101948. 0.6360599. 0.56963110. 0.706629

Correct answer: 0.514124.

Explanation:The initial energy of the system is

E0 =1

2mv20

=1

2(1.72 kg) (13.5 m/s)2

= 156.735 J .

The final energy of the system is

E =1

2I ω2

=1

2

[(

3m+M

12

)

d2]{

6mv0[3m+M ] d

}2

=1

2mv20

[

3m

3m+M

]

=1

2(1.72 kg) (13.5 m/s)2

×[

3 (1.72 kg)

3 (1.72 kg) + (5.46 kg)

]

= 76.1537 J .

Therefore, the fractional loss of the energy is

f =E0 −E

E0

=

1

2mv20

[

1− 3m

3m+M

]

1

2mv20

=M

3m+M

=(5.46 kg)

3 (1.72 kg) + (5.46 kg)= 0.514124 .

Qualitative String on Rope050 10.0 points

A particle, held by a string whose other endis attached to a fixed point C, moves counter-clockwise in a circle on a horizontal friction-less surface. If the string is cut, the angularmomentum of the particle about the point C:

1. decreases

2. does not change correct

3. changes direction but not magnitude

4. none of these

5. increases

Explanation:So long as no external work is done on

the particle, the angular momentum of theparticle must be constant.One might wonder how the angular mo-

mentum could possibly remain constant when~L = ~r×~p, and the momentum is staying con-stant while~r is changing. But, remember thatthe cross product depends on the relative ori-entation of ~p and ~r.Let’s set up a coordinate system, so that

the x and y directions are in the plane of thesurface, and the z direction is perpendicularto the surface. Let’s also assume that thestring is cut at a moment when the particle’sinstantaneous velocity is purely in the +xdirection, and the +y direction is the directionfrom the particle to the center of the circle.Then, we know that the particle’s momen-

tum forever has no y component and a con-stant x component. Meanwhile, we have

~L = ~r ×~p

⇒∣

∣

∣

~L∣

∣

∣= rp sin θ

= p(r sin θ)

= pry


(The direction of the angular momentumvector is in the z direction.) Since all motionof the particle is only in the x direction, thismeans that the y position of the particle isconstant, and therefore, we conclude that theangular momentum of the particle is constantas well.

Race Down a Plane051 (part 1 of 3) 10.0 points

A thin cylindrical shell and a solid cylinderhave the same mass and radius. The two arereleased side by side and roll down, withoutslipping, from the top of an inclined planethat is 3 m above the ground.Find the final linear velocity of the thin

cylindrical shell.The acceleration of gravity is9.8 m/s2 .1. 5.422182. 5.93973. 5.686834. 6.260995. 4.643276. 4.949757. 5.772358. 6.929659. 4.3150910. 5.85662


Explanation:

Let : H = 3 m , and

g = 9.8 m/s2 .

H

S

h

12S

Because there is no slipping, v1 = ω1R andthe rotational inertia of the cylindrical shell isI1 = mR2 . Thus, from conservation of energy

Krot +Ktrans = H1

2I1 ω

21 +

1

2m (ω1R)2 = mgH

1

2(mR2)

(v1R

)2

+1

2mv21 = mgH , so

v1 =√

g H

=√

(9.8 m/s2) (3 m)

= 5.42218 m/s .

052 (part 2 of 3) 10.0 pointsFind the final linear velocity of the solid cylin-der.1. 6.260992. 5.715483. 5.482094. 5.112085. 6.155766. 7.752857. 6.953188. 7.495789. 6.5665810. 8.00167


Explanation:Because there is no slipping, v2 = ω2R and

the rotational inertia of the solid cylinder is

I2 =1

2mR2 , so

Krot +Ktrans = H1

2I2 ω

22 +

1

2m (ω2R)2 = mgH

1

2

(

1

2mR2

)

ω22 +

1

2mv22 = mgH

v2 =

√

4

3g H

v2 =

√

4

3(9.8 m/s2) (3 m)

= 6.26099 m/s .

053 (part 3 of 3) 10.0 pointsWhen the first object reaches the bottom,what is the height above the ground of theother object?1. 1.05


2. 0.353. 0.754. 0.555. 1.26. 0.67. 0.4758. 0.49. 0.67510. 1.15

Correct answer: 0.75 m.

Explanation:Because v2 > v1, object 2 reaches the bot-

tom first. Since acceleration is constant, then

v =vf + vi

2=

vf2

Since vi = 0

Thusvf2

=∆s

∆tand the final velocity of the

second object is

v2 =2 s2t

.

In that same time t the first object travelsa distance s1 but its velocity is v′1 < v (thespeed at the buttom), so let

v′1 =2 s1t

.

Dividing these two equations, one obtains

s2s1

=v2v′1

=

√

43g H

√

g (H − h)

where h is the height of object 1 whenobject 2 reaches the bottom of the incline.From similar triangle in the figure

H

H − h=

s2s1

=

√

4

3g H

√

g (H − h)

H

H − h=

√

4H

3 (H − h)

√

H

H − h=

√

4

3

H

H − h=

4

3

3H = 4H − 4h

h =1

4H

=1

4(3 m)

= 0.75 m .

keywords:

Rigid Body Torque054 10.0 points

A force F = F0

(

ı+ + k)

acts on a rigid

body at a point r = r0 (ı− ) away from theaxis of rotation.What is the resulting torque on the body?

1. r0 F0

(

k − ı− )

.

2. r0 F0

(

k − ı− 2 )

.

3. r0 F0

(

2 k − ı− )

. correct

4. r0 F0

(

2 − ı− k)

.

5. r0 F0

(

k − 2 ı− )

.

6. Zero.

7. r0 F0

(

2 ı− k − )

.

8. r0 F0

(

ı+ + k)

.

9. r0 F0

(

k − 2 ı− 2 )

.

Explanation:

~τ = r×F

= F0 r0

(

ı× + ı× k − × ı− × k)

= F0 r0

(

k − + k − ı)

= F0 r0

(

2 k − − ı)

.


Ring and Disk Race055 10.0 points

A solid disk (thin cylindrical object with uni-form density) and a ring (all of its mass lo-cated at the radius of the ring) roll down anincline without slipping. Let mring , mdisk

be the inertial masses and rring and rdisk theradii.The ring is slower than the disk only if

1. mring = mdisk and rring = rdisk.

2. mring > mdisk .

3. The ring is always slower regardless of therelative values of m and r. correct

4. rring > rdisk .

Explanation:Say the two objects start at the top of an

incline of vertical height h. Since v = r ω(noslipping) conservation of energy gives

mgh =1

2mv2 +

1

2I ω2

1

2mv2 +

1

2Iv2

r2=

v2

2

(

m+I

r2

)

.

Iring = mr2 and Idisk =1

2mr2 so I =

Cmr2 and

v2

2(m+ Cm) = mg h

mv2

2(1 + C) = mg h ,

v =

√

2 g h

1 + C .

Therefore, the speed is independent of bothmass and radius and depends only on theshape of the object. Since Cring > Cdisk andthe constant C appears in the denominator,the ring is always slower.

keywords:

Rolling Basketball

056 10.0 pointsA 150 g basketball has a 26.5 cm diameterand may be approximated as a thin sphericalshell.

150 g

µ=0.394.

97m

40◦

Starting from rest, how long will it takea basketball to roll without slipping 4.97 mdown an incline that makes an angle of 40◦

with the horizontal? The moment of inertia ofa thin spherical shell of radius R and mass m

is I =2

3mR2, the acceleration due to gravity

is 9.8 m/s2 , and the coefficient of friction is0.39 .1. 1.331492. 1.733453. 1.62174. 1.693535. 1.538676. 1.386587. 2.164118. 1.210989. 2.0120410. 1.92138


Explanation:

Let : ℓ = 4.97 m ,

θ = 40◦ , and

R = 0.1325 m .


P

mg sinθ

N

mg

θ

The moment of inertia of the ball aboutthe point of contact between the ball and theinclined plane is

IP = Icm +md2

=2

3mR2 +mR2

=5

3mR2 .

The net torque about the point of contactbetween the ball and the inclined plane is

mgR sin θ = IP α =5

3mR2 α

a =mgR2 sin θ5

3mR2

=3

5g sin θ .

Because the sphere starts from rest, its cen-ter of mass moves a distance

ℓ =1

2a t2

t =

√

2 ℓ

a=

√

2 (4.97 m)

3.77959 m/s2

= 1.6217 s .

Revolving Mass on a Spring057 10.0 points

A 56 kg mass is on a spring of length 62 cmand stiffness constant 27 N/cm. It is spun ata speed of 0.4 1/s in the absence of gravity.How far will the spring stretch?1. 0.206433

2. 0.2318033. 0.2397624. 0.1883165. 0.1775346. 0.1465817. 0.1159878. 0.1007579. 0.13310410. 0.23528

Correct answer: 0.206433 cm.

Explanation:The centripetal force is

Fc =mv2

r=

m(rω)2

r= mrω2

Since the spring has an unstretched lengthL, and the rotation will cause it to stretch adistance x, the radius of rotation is L+ x and

Fc = m(L + x)ω2 = mLω2 +mxω2

The unbalanced force on the mass is suppliedby the spring, so

Fc = Fsp

mLω2 +mxω2 = kx

mLω2 = kx−mxω2

x =mLω2

k −mω2

The terms in the denominator are not com-patible:

N

cm− kg

(1

s

)2

=kgm/s2

cm− kg

s2

so a conversion is necessary:

kgm/s2

cm

100 cm

1m− kg

s2=

kg

s2− kg

s2

System Angular Momentum058 10.0 points

Suppose a physics instructor seated him-self at rest on a low-friction “piano stool” andheld a spinning bike wheel with its angular ve-locity vector initially horizontal. Suppose theinstructor then turned the bike wheel so that


its angular velocity vector pointed vertically

downward. When this turn was completed,the instructor would be spinning rapidly onthe stool.What would be the final, total angular mo-

mentum of the system of instructor plus bikewheel, projected along the vertical axis, theonly axis along which the system can spin?

1. Zero, as it was initially. correct

2. The same, nonzero value it had initially.

3. Equal but opposite to what the wheel hadinitially.

4. Twice its initial value.

Explanation:The system originally had zero angular mo-

mentum about a vertical axis, so no matterwhat happens within the system of person,bike wheel and stool, as long as no externaltorque is involved, the angular momentumalong this axis, say z, is always zero. Whenthe bike wheel’s angular momentum vectoris twisted downward by an internal torque,the rest of the system develops an angularmomentum vector of equal magnitude in theopposite direction to maintain a total angularmomentum about the vertical axis of zero.There is, of course, a third law for torques;

when the man exerts a torque to twist thebike wheel, it exerts an equal and oppositetorque on him and the stool.

Supernova Explosion059 10.0 points

A star of radius 3.8 × 105 km rotates aboutits axis with a period of 23 days. The starundergoes a supernova explosion, whereby itscore collapses into a neutron star of radius7.2 km.Estimate the period of the neutron star

(assume the mass remains constant).1. 0.001154822. 0.005286493. 0.00276024. 0.00185225. 0.0571298

6. 6.6564e-057. 0.008485598. 0.000713419. 0.27556210. 0.000535146


Explanation:

Given : T = 23 days = 1.9872× 106 s ,

ri = 3.8× 105 km , and

rf = 7.2 km ,

We will assume that during the collapse ofthe star,(1) no torque acts on it,(2) it remains spherical,(3) its mass remains constant.

Since I is proportional to r2, and ω =2 π

T,

conservation of angular momentum gives

Tf = Ti

(

rfri

)2

= (1.9872× 106 s)

(

7.2 km

3.8× 105 km

)2

= 0.00071341 s .

Yo Yo on table with friction060 10.0 points

A yo-yo, arranged as shown, rests on a surface.

When a force ~F is applied to the string asshown, the yo-yo:

1. moves to the left and rotates clockwise

2. moves to the right and does not rotate

3. moves to the right and rotates clockwisecorrect


4. moves to the right and rotates counter-clockwise

5. moves to the left and rotates counter-clockwise

Explanation:To solve this problem, the first thing we

note is that there are two horizontal forcesacting on the yo-yo: static friction and theapplied force. Let the radius of the yo-yobe R, and the distance from the center ofthe yo-yo to the point of application of ~Fto be r, and the moment of inertial of theyo-yo to be I. In order to determine thenet direction of motion, we need to determinethe relative sizes of these two forces. First,analyze the torques at the point of contactbetween the ground and the yo-yo, which willbe instantaneously at rest if the yo-yo is notmoving. In this reference frame, we have(I +mR2)α = (R − r)F , where we used theparallel axis theorem to translate the centerof rotation to the hub. We can see fromthe direction of the torques that α must becounterclockwise.Now, analyze the problem with the center

of rotation being the center of the yo-yo. Wefind that the frictional force must act to theleft, because otherwise the angular accelera-tion would be in the wrong direction, as bothtorques would be to the left, and we found theangular acceleration already.In this frame,

Iα = FsR − FrI(R− r)F

I +mR2= FsR − Fr

FsR = F

(

I(R− r)

I +mR2+ r

)

= F

(

IR− Ir + Ir +mR2r

I +mR2

)

Fs = F

(

I +mR2(

rR

)

I +mR2

)

So, we can see from this expression that, ifr < R, then the numerator is less than the

denominator, and therefore Fs < F , and thenet force on the yo-yo is to the right.

Yo Yo N 07061 (part 1 of 4) 10.0 points

Assume: Drag (friction) is negligible.Given: g = 9.81 m/s2 . The density of this

large Yo-Yo like solid is uniform throughout.The Yo-Yo like solid has a mass of 2.9 kg .

ℓ 2 ℓ ℓ

Front ViewA cord is wrapped around the stem of the

Yo-Yo like solid and attached to the ceiling.The radius of the stem is 5 m and the radiusof the disk is 8 m.

h 8 m

5 m 2.9 kg

ω

Cross sectional Side ViewCalculate the moment of inertia about the

center of mass (axis of rotation).1. 26.85982. 55.69723. 88.98294. 66.30625. 64.7046. 23.63667. 82.72628. 54.72459. 61.337810. 76.9152

Correct answer: 76.9152 kgm2.

Explanation:


Let : g = 9.81 m/s2 ,

m = 2.9 kg ,

r1 = 5 m ,

r2 = 8 m , and

∆h = 22 m . see, Part 4

The areal density of the Yo-Yo is

σ =A1 +A2

M

=π (r21 + r22)

M.

Therefore the masses of the disks are

M1 = σ A1 , so

M1 =r21

r21 + r22M , and

M2 = σ A2 , so

M2 =r22

r21 + r22M .

And the moment of inertia are

I1 =1

2M1 r

21

=1

2

r41r21 + r22

M ,

I2 =1

2M2 r

22

=1

2

r42r21 + r22

M , and

I = I1 + I2

=1

2

r41 + r42r21 + r22

M

=1

2

(5 m)4 + (8 m)4

(5 m)2 + (8 m)2

× (2.9 kg)

= 76.9152 kgm2 .

062 (part 2 of 4) 10.0 pointsWhat is the vertical acceleration of the centerof mass of the Yo-Yo?1. 3.633332. 5.75889

3. 5.075824. 6.182685. 5.609286. 3.069277. 4.555178. 2.607019. 4.7600610. 5.39652

Correct answer: 4.76006 m/s2.

Explanation:Basic Concepts: General rotational kine-

matics, energy conservation. In particular

Krot =1

2I ω2 ,

and the moment of inertia for a uniform diskis

I =1

2M R2 .

Solution: On the center of mass the totalforce and torque is∑

F : T −M g = −M a , so

T = M (g − a) , (1)∑

τ : T r1 = I α

=1

2

r41 + r42r21 + r22

Ma

r1, so

T =r41 + r42

2 r21 (r21 + r22)

M a . (2)

Setting Eq. 1 equal to 2, we have

M (g − a) =1

2

r41 + r42r21 (r

21 + r22)

M a

g =1

2

2 r21 (r22 + r22) + r41 + r42r21 (r

21 + r22)

a

g =1

2

3 r41 + 2 r21 r22 + r42

r21 (r21 + r22)

a

a =2 r21 [r

21 + r22]

2 r41 + [r21 + r22]2g (3)

=2 (5 m)2 [(5 m)2 + (8 m)2]

2 (5 m)4 + [(5 m)2 + (8 m)2]2

× (9.81 m/s2)

= 4.76006 m/s2 .


063 (part 3 of 4) 10.0 pointsWhat is magnitude of the torque the cordexerts on the center of mass of the Yo-Yo?1. 43.55442. 59.84873. 76.69374. 56.81015. 68.05876. 40.89767. 40.06048. 73.22419. 41.586310. 42.6757


Explanation:Substituting a from Eq. 3 into Eq. 2, we the

tension is

T =r41 + r42

2 r41 + [r21 + r22]2M g (4)

=(5 m)4 + (8 m)4

2 (5 m)4 + [(5 m)2 + (8 m)2]2

× (2.9 kg) (9.81 m/s2)

= 14.6448 N .

Therefore, the torque is

τ = r1 T

=r1 [r

41 + r42]

2 r41 + [r21 + r22]2M g (5)

=(5 m) [(5 m)4 + (8 m)4]

2 (5 m)4 + [(5 m)2 + (8 m)2]2

× (2.9 kg) (9.81 m/s2)

= 73.2241 Nm .

064 (part 4 of 4) 10.0 pointsThe Yo-Yo is released from rest at height h.Find the velocity v of the center of mass of

the disk at the height ∆h = 22 m .1. 13.09052. 14.47213. 10.79964. 12.35315. 12.64386. 16.4087

7. 4.696958. 19.47239. 12.101910. 15.1156


Explanation:By energy conservation we know that the

initial energy equals the final energy. In thiscase energy conservation implies

M g h = Klinear +Krot .

where Klinear =1

2M v2 . We also know that

Krot is equal to1

2I ω2 where I =

1

2

r41 + r42r21 + r22

M

for our uniform disk, and we know ω =v

R.

This implies

Krot =1

2I ω2

=1

4

r41 + r42r21 (r

21 + r22)

M v2 (6)

Thus

M g h = Klinear +Krot

=1

2M v2 +

1

4

r41 + r42r21 [r

21 + r22]

M v2

=1

4

2 r41 + 2 r21 r22 + r41 + r42

r21 [r21 + r22]

M v2

=3 r41 + 2 r21 r

22 + r42

4 r21 [r21 + r22]

M v2 (7)

Therefore, the velocity is

v =

√

4 r21 [r21 + r22]

3 r41 + 2 r21 r22 + r42

g h (8)

=

√

4 (5 m)2 [(5 m)2 + (8 m)2]

3 (5 m)4 + 2 (5 m)2 (8 m)2 + (8 m)4

×√

(9.81 m/s2) (22 m)

= 14.4721 m/s .

Yo Yo MC 05


065 (part 1 of 2) 10.0 pointsYou are playing with a Yo-Yo which you candescribe as a uniform disk with mass m andradius r. The string of the Yo-Yo has a lengthL. The Yo-Yo rolls down vertically.

hr

m

ω

Determine the tension T of the string ata height ∆h below the original position h asindicated in the sketch.

1. T =1

2mg

2. T =1

4mg

3. T =5

2mg

4. T =7

4mg

5. T =1

3mg correct

6. T =r

Lmg

7. T =L− h

rmg

8. T = 3mg

9. T = mg

10. T =2

5mg

Explanation:Apply “F = ma” to the center of mass of

the disk,

Fnet =∑

F : mg − T = ma . (1)

Apply “τ = I α” to the Yo-Yo (where thetorque is calculated about the center),

T r = I α

=1

2mr2 α

=1

2mr a , so (2)

T =1

2ma . (3)

In the 3rd step we used a = α r. SubstitutingEq. 3 into 1 leads to,

mg − 1

2ma = ma , so

a =2

3g , and

T =1

2ma =

1

3mg .

066 (part 2 of 2) 10.0 pointsAfter the Yo-Yo has been elastically reflectedat length L, it moves upward.Determine the speed of its center when it is

at a height ∆h below the original position h.

1. v =√

3 g∆h

2. v =

√

1

2g∆h

3. v =√

g∆h

4. v =√

2 g∆h

5. v =

√

1

8g∆h

6. v =√

g∆h− g L

7. v =

√

r

Lg∆h

8. v =

√

4

3g∆h correct

9. v =

√

3

2g∆h

10. v =

√

5

3g∆h

Explanation:The application of work-energy equation

gives

mg∆h =1

2mv2 +

1

2I ω2 .


But

1

2I ω2 =

1

2

(

1

2mr2

)

ω2 =1

4mv2 .

So

mg∆h =

(

1

2+

1

4

)

mv2

=⇒ v =

√

4

3g∆h .

Yo Yo MC 06067 (part 1 of 5) 10.0 points

A yo-yo consists of two identical, uniformdisks whose total mass is M and radius Rconnected between the disks by a shaft ofradius r and negligible mass. See the figurebelow.

h R

r M

ω

The combined mass of the three disks (theyo-yo) is M . The yo-yo rolls down an inex-tensible, thin string of length h + ℓ, whereℓ .Neglect the mass and moment of inertia of

the shaft. The string is initially wound aroundthe shaft and unwinds as the yo-yo rolls downthe string.What is the moment of inertia, I, of the

yo-yo about the axis through the centers ofmass of the two disks?

1. I =1

2M (R2 + r2)

2. I =1

2M (r +R)2

3. I =1

4M (R− r)2

4. I =1

2M r2

5. I = M (r +R)2

6. I = M R2

7. I = M r2

8. I =1

4M (r +R)2

9. I = M (R2 + r2)

10. I =1

2M R2 correct

Explanation:Basic Concepts: ~Fnet = m~a

~τnet = I αNo-slip condition (magnitudes):

s = r θ , v = r ω , a = r α .

Solution: For a disk, the moment of inertiais

Idisk =1

2M R2

Since there are two disks, each with massM

2, the moment of inertia of the yo-yo is

I = 21

2

(

M

2

)

R2 =1

2M R2.

Remember: We are neglecting the mass andmoment of inertia of the connecting spindle.

068 (part 2 of 5) 10.0 pointsWhat is the ratio of the magnitude of thegravitational acceleration g to the magnitudeof the acceleration a of the center of mass ofthe yo-yo?

1.∣

∣

∣

g

a

∣

∣

∣=

1

2

√

R

r

2.∣

∣

∣

g

a

∣

∣

∣= 1 +

√

R

r

3.∣

∣

∣

g

a

∣

∣

∣= 1 +

1

2

R

r

4.∣

∣

∣

g

a

∣

∣

∣=

R

r

5.∣

∣

∣

g

a

∣

∣

∣= 1 +

1

2

√

R

r

6.∣

∣

∣

g

a

∣

∣

∣= 1 +

1

2

(

R

r

)2

correct

7.∣

∣

∣

g

a

∣

∣

∣= 1 +

(

R

r

)2


8.∣

∣

∣

g

a

∣

∣

∣=

1

2

R

r

9.∣

∣

∣

g

a

∣

∣

∣= 1 +

R

r

10.∣

∣

∣

g

a

∣

∣

∣=

(

R

r

)2

Explanation:We will take the upward direction as the

positive y direction and clockwise as the pos-itive sense of rotation.

T

θ

W = M g

y

Looking at our free body diagram above,we see that the net force on the yo-yo is

Fnet = T −W = M a .

Similarly, the net torque on the yo-yo aboutthe center of mass is

τnet = T r sin(90◦)

= T r

= I α ,

where the sign of τ is given by the right-handrule, in the sense that clockwise is positivehere. To relate the rotational and transla-tional motion we have the fact that the stringdoes not slip as it unrolls from the yo-yo.This is expressed in the no-slip condition,a = −r α . Substituting for α, we have

T = −I a

r2.

(If we looked the yo-yo from the other side sothe tension acted on the left side of the yo-yo,the sense of rotation would be reversed; in thiscase, τ and α would both have the oppositesign.) We can now substitute the tension Tinto our equation for the acceleration.

T = M a+W = M (a+ g) , so

M(a+ g) = −I a

r2

g = − I a

M r2− a

= −[

1 +I

M r2

]

a , so

∣

∣

∣

g

a

∣

∣

∣= 1 +

I

M r2.

The ratio of the moment of inertia, I, to M r2

is

I

M r2=

1

2

M R2

M r2

=1

2

(

R

r

)2

.

So, the ratio of accelerations is

∣

∣

∣

g

a

∣

∣

∣= 1 +

I

M r2

= 1 +1

2

(

R

r

)2

.

069 (part 3 of 5) 10.0 pointsAssume: The yo-yo started from rest.What is the ratio of the final kinetic energy

Kf to the rotational kinetic energyKrot aboutthe axis through the center of mass when theyo-yo has fallen the entire length of the string?

1.Kf

Krot= 1 + 2

( r

R

)2

correct

2.Kf

Krot= 1 + 2

√

r

R

3.Kf

Krot= 1 +

( r

R

)2

4.Kf

Krot=( r

R

)2

5.Kf

Krot=

√

r

R

6.Kf

Krot= 1 +

r

R

7.Kf

Krot= 1 +

√

r

R

8.Kf

Krot=

r

R


9.Kf

Krot= 1 + 2

r

R

Explanation:Here, we will use the principle of conserva-

tion of energy. There are no dissipative forces.This means that the total energy of the yo-yoplus string system is conserved:

Kf + Uf = Ki + Ui .

Since the yo-yo starts from rest, Ki = 0. If wemeasure the potential energy from the start-ing height, Ui = 0. The height the yo-yo fallsis equal to the length, ℓ, of string unwoundfrom the yo-yo. Putting this together, we get

Kf −mg ℓ = 0 , so

Kf = mg ℓ .

But, this is the total kinetic energy of bothtranslation and rotation! We need to relatethe two type of motion to each other. As be-fore, this comes through a no-slip condition;this time it is v = −r ω. The ratio of trans-lational kinetic energy to rotational kineticenergy is then

Ktr

Krot=

1

2M v2

1

2I ω2

=M v2

I(

−v

r

)2

=M r2

1

2M R2

Ktr = 2( r

R

)2

Krot .

This means that

Kf = Krot

[

1 + 2( r

R

)2]

, so

Kf

Krot= 1 + 2

( r

R

)2

.

070 (part 4 of 5) 10.0 pointsIf E is the total mechanical energy, ~p is thelinear momentum of the center of mass and~L is the angular momentum relative to therotational axis through the center of mass,mark the correct statement.

1. E and ~L are conserved, but not ~p.

2. Neither E, ~p, or ~L is conserved.

3. E, ~p, and ~L are all conserved.

4. E and ~p are conserved but not ~L.

5. ~p is conserved, but not E and ~L.

6. ~p and ~L are conserved, but not E.

7. E is conserved but not ~p and ~L. correct

8. ~L is conserved, but not E and ~p.

Explanation:• ~p of the yo-yo is NOT conserved. There

is a net force acting on the yo-yo. You can seethis explicitly because the initial momentumis zero and the final momentum is not zero.• ~L of the yo-yo is NOT conserved. There

is a net torque acting on the yo-yo. Again,you could also see this because the initial an-gular momentum is zero and the final angularmomentum is nonzero.• E is conserved. Although there is an ex-

ternal force acting on the string-yo-yo-system,it does not do any work to change the totalenergy of the system. This force is exerted onthe string by the person holding the end of thestring fixed. The displacement is zero, how-ever, so no work is done. If the person hadinstead pulled the end of the string up whilethe yo-yo fell, then the person would do workon the system and increase its energy. Theamount of energy added would be equal to theamount of work done in lifting the string.

071 (part 5 of 5) 10.0 pointsWhat is the SI unit of torque τ?

1. [τ ] = kgm/s2

2. [τ ] = kgm2/s2 correct

3. [τ ] = Nkg


4. [τ ] = Nm/s

5. [τ ] = kgm/s

6. [τ ] = N/m

7. [τ ] = kg s/m

8. [τ ] = kg s2/m2

Explanation:The SI unit of torque is Nm. By writing

Newtons in terms of fundamental quantities:

[F ] = N = kgm/s2 .

(compare, for example, F = ma). Therefore

[τ ] = Nm = kgm2/s2 .


Given: M , g, and R.A string is wound around a uniform disk of

radius R and mass M . The disk is releasedfrom rest at a height h with the string verticaland its tip end tied to a fixed support as infigure.

h R

M

ω

Derive an expression for the tension, T ,on the string at height h + ∆h, as the diskdescends.

1. T =1

2M gR +M g∆h

2. T = M gR

∆h

3. T =1

3M g correct

4. T = M g(R −∆h)

R

5. T = M g(R2 −∆h2)

R∆h

6. T = M g

7. T = 2M g∆h

R

8. T = M g∆h

R

9. T = 2M g

10. T =2

3M g

Explanation:Basic Concepts: General rotational kine-

matics, energy conservation. In particular

Krot =1

2I ω2 ,

and the moment of inertia for a uniform diskis

I =1

2M R2 .

Solution: On the center of mass the totalforce and torque is

∑

F : T −M g = −M a , so

T = M (g − a) , (1)∑

τ : T R = I α

=1

2M R2

( a

R

)

, so

2T = M a . (2)

Adding Eq. 1 and 2, we have

3T = M g .

Therefore

T =1

3M g . (3)

073 (part 2 of 2) 10.0 pointsWhat is the equation for the vertical accelera-tion of the center of mass in terms of the givenparameters?

1. a =2 (R2 − h2)

Rhg

2. a = 4 g


3. a =2 (R− h)

Rg

4. a = 2 gR

h

5. a =3

4g

6. a =2

3g correct

7. a = g

8. a =4M

3g

9. a =1

4g

Explanation:To find a, we substitute our value of T in

Eq. 3 into the Eq. 2 giving

21

3M g = M a , so

a =2

3g . (4)

Alternate Solution: Using the pointwhere the string touches the disk as the pivotand the parallel axis theorem

Iedge = Icm +M R2

=1

2M R2 +M R2

=3

2M R2 .

The torque equation at this pivot point is

∑

τ : M gR = Iedga α

=3

2M R2

( a

R

)

, so

a =2

3g .


A string is wrapped around the stem of a Yo-Yo which has a mass M and a moment ofinertia I about its center axis. The radius ofthe stem is r1 and the radius of the disk is

r2. Denote the tension of the string T and thedescending acceleration a.

hr2

r1 M

ω

The correct torque equation is given by

1. T r1 =a I

r1. correct

2. Tr1 + r2

2=

2 a I

r2.

3. Tr1 + r2

2=

2 a I

r1.

4. T r1 =2 a I

r1 + r2.

5. Tr1 + r2

2=

2 a I

r2 − r1.

6. T r2 =2 a I

r1 + r2.

7. T r2 =a I

r2.

8. Tr1 + r2

2=

2 a I

r1 + r2.

9. Tr2 − r1

2=

2 a I

r1 + r2.

Explanation:The torque created by the tension is

τ = T r1 = I α = Ia

r1=

a I

r1.

075 (part 2 of 3) 10.0 pointsThe correct vertical force equation is given by

1. T −M g = M a .

2. M g − T = M a . correct

3. M g + T = M a .

Explanation:The net force on the Yo-Yo is

Fnet =∑

F : M g − T = M a .


076 (part 3 of 3) 10.0 pointsRelease the Yo-Yo from rest. It reaches acenter of mass speed v when it falls for aheight ∆h. Using the same notation as in theprevious questions, the correct work-energyequation is given by

1.M g∆h =1

2M v2 +

1

2I

(

v

r1

)2

. correct

2.1

2M v2 =

1

2Iv2 .

3. M g∆h =1

2I

(

v

r1

)2

.

4. M g∆h =1

2I

(

v

r2

)2

.

5. M g∆h =1

2M v2 +

1

2I

(

v

r2

)2

.

6. M g∆h =1

2M v2 .

7.1

2M v2 =

1

2I

(

v

r1

)2

.

8.1

2M v2 =

1

2I

(

v

r2

)2

.

Explanation:The work done by the gravity isW = M gh.

The combined (translational and rotational)kinetic energy is

K =1

2M v2+

1

2I ω2 =

1

2M v2+

1

2I

(

v

r1

)2

.

Thus the work energy theorem gives

M g∆h =1

2M v2 +

1

2I

(

v

r1

)2

.

version 001 – rolling & angular momentum – …...version 001 – rolling & angular...

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