PHY 303K - Florin - Exam IV - Fall 2013

Version 022 – Test 4 – florin – (57850) 1

This print-out should have 20 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 pointsAmechanical lift is used to raise a heavy load.The lift consists of a horizontal platform oflength L drawn by vertical cables on eitherend of the platform. The load is placed onthe platform and drawn slowly upward at aconstant velocity so that acceleration is notimportant. The platform has a mass m. Theload has mass 4m and is placed a distance L/4from the right end of the lift. The accelerationdue to gravity is g. What is the tension in theright vertical cable?

1. 1/3mg

2. 7/2mg correct

3. 5/4mg

4. mg

5. 2/3mg

6. 3/2mg

7. 7/4mg

8. 3mg

9. 4/5mg

10. 2/7mg

Explanation:The net torque about any axis must be zero.

Since we want the tension in the right cable,compute the net torque about the left end:

∑

τL = −3

4L(4mg)−1

2L(mg)+L(TR)+0(TL) = 0.

Therefore TR = 7/2mg.

002 10.0 pointsA sizable quantity of soil is washed down

the Mississippi River and deposited in theGulf of Mexico each year. Thus, there is anet movement of mass southward towards theequator.What effect does this tend to have on the

length of a day?

1. Impossible to determine

2. shorten the day

3. lengthen the day correct

4. no change

Explanation:Soil washed down the river is deposited at

a greater distance from the Earth’s rotationalaxis. Just as a man on a turntable slowsdown when one of the masses is extended, theEarth slows down in its rotational motion,extending the length of the day. The amountof slowing, of course, is exceedingly small.(Interestingly, the construction of many damsin the Northern hemisphere has the oppositeeffect; shortening our days!)

003 10.0 pointsA length of light nylon cord is wound arounda uniform cylinder of radius 0.735 m and mass1.36 kg. The cylinder is mounted on a fric-tionless axle and is initially at rest. The cordis pulled from the cylinder with a constantforce of magnitude 5.43 N.How fast will the spool be rotating after the

string has been pulled for ∆t = 3.63 s?The moment of inertia of the cylinder is

I =1

2mr2.

1. 145.9042. 51.24943. 31.66954. 189.2245. 113.856. 44.21997. 24.90118. 11.30129. 39.437610. 26.338


Correct answer: 39.4376 rad/s.

Explanation:

Let : F = 5.43 N ,

m = 1.36 kg ,

r = 0.735 m , and

t = 3.63 s .

The angular momentum after 3.63 s is

L = τ∆t

= Fr ·∆t

= (5.43 N)(0.735 m)(3.63 s)

= 14.4875 kg ·m2/s .

The moment of inertia of the cylinder is

ω =

(

1

2mr2

)

=1

2(1.36 kg) (0.735 m)2

= 0.367353 kg ·m2 .

Therefore, the rotational velocity will be

ω = L/I

= (14.4875 kg ·m2/s)/(0.367353 kg ·m2)

= 39.4376 rad/s .

004 10.0 pointsA small puck moves in a circle on a frictionlessairtable. The circular motion is enforced bystring tied to the puck and going through atiny hole in the middle of the table. Initially,the puck moves in a circle of radius R1 atspeed v1. But later the string is pulled downthrough the hole forcing the puck to move in

a smaller circle of radius R2 =1

2R1 .

R

v

What is the new speed of the puck?

1. v2 = 2 v1 correct

2. v2 = 0

3. v2 =1√2v1

4. v2 = v1

5. v2 =1

2v1

6. v2 =1

2√2v1

7. v2 = 2√2 v1

8. v2 =1

4v1

9. v2 = 4 v1

10. v2 =√2 v1

Explanation:Let the hole in the airtable be the origin

of our coordinate system. Because the holeis tiny, the string always pulls the puck inthe radial direction. Consequently, the stringtension force ~T has zero torque (about theorigin). The other two forces on the puck —

the weight ~W and the normal force ~N of thetable — cancel each other and each other’storques. Altogether, we have zero net torque,so the angular momentum of the puck mustbe conserved:

~L = ~R×m~v = const.

When the puck moves in a circle, the directionof the angular momentum is vertically up, andits magnitude is L = mvR . This is true bothbefore and after the string being pulled down,so

L = mv1R1 = mv2R2

v2 =R1

R2v1 = 2 v1 .

005 10.0 points


A rod has a pivot at one end and is free torotate without friction at the other end, asshown. A force ~F is applied to the free endat an angle θ to the rod creating a torque ~τabout the pivot.

L

m

F

θ

If instead the same force is applied perpen-dicular to the rod, at what distance d from thepivot should it be applied in order to producethe same net torque ~τ about the pivot?

1. d = L/ tan θ

2. d =√5 L

3. d = L cos θ

4. d = L/ sin θ

5. d = L sin θ correct

6. d = L/ cos θ

7. d = L

8. d = L tan θ

9. d =√2 L

10. d = L/2

Explanation:The force generates a torque of

τ = F L sin θ ,

so the distance is L sin θ .

006 10.0 pointsHarry and Beth cycle at the same speed, i.e.their centers of mass move with the samevelocity. The bike tires all rotate withoutslipping, but the tires on Harry’s bike have alarger radius than those on Beth’s bike.

Which tires have the greater rotationalspeed?

1. It depends on which tires have moremass.

2. It depends on the angular acceleration ofthe wheels.

3. Beth’s tires correct

4. Harry’s tires

5. It depends on which tires have more an-gular momentum.

6. The rotational speeds are the same.

7. It depends on the which tires have a largermoment of inertia.

8. It depends on the center of mass speed.

Explanation:v = r ω. Tires with a smaller radius needs

a larger rotational speed to obtain the samelinear speed.

007 10.0 pointsA uniform meter-stick with length L pivotsat point O. The meter stick can rotate freelyabout O and is released from the horizontalposition at t = 0.

OL

Determine the angular acceleration of themeter stick at the moment it is released.

1.5g

6L

2.3g

4L

3.5g

4L


4.g

2L

5.g

4L

6.g

3L

7.7g

4L

8.g

L

9.3g

2Lcorrect

10.2g

LExplanation:The equation of motion τ = Iα gives:

mgL

2= Iα =

1

3mL2α,

So

α =3g

2L.

008 10.0 pointsA flywheel with a very low friction bearingtakes 1.7 h to stop after the motor poweris turned off. The flywheel was originallyrotating at ω = 3.14159 rad/s.Assuming the angular deceleration is con-

stant, how many revolutions does the flywheelmake before it stops? Remember that thenumber of revolutions is given by the ∆θ/2π.1. 2310.02. 2484.03. 2250.04. 1617.05. 2205.06. 2829.07. 3192.08. 3960.09. 1530.010. 2760.0

Correct answer: 1530 rev.

Explanation:

Let : t = 1.7 h .

The angular acceleration of the flywheel is

α =−ω0

t,

so the angle through which it rotates beforestopping is given by

∆θ = ω0 t+1

2α t2 = ω0 t+

1

2

(−ω0

t

)

t2

=1

2ω0 t

=1

2(3.14159 rad/s) (1.7 h)

3600 s

1 h

1 rev

2 π

= 1530 rev .

009 10.0 points

Consider an isolated system with 100quanta of energy distributed between twoblocks in contact. Block 1 has 300 quantummechanical harmonic oscillators and Block 2has 200 oscillators. The figure shows plots ofentropy of block 1 S1 and entropy of block 2S2 versus quanta q. Note that the entropy ofblock 2 is plotted with it’s x-axis reversed sothat m2 has a positive upward slope in thefigure.Initially, there are 90 quanta in block one

and 10 quanta in block 2, but the system isnot in equilibrium. At equilibrium, there are60 quanta in block 1 and 40 quanta in block2. The initial slopes of the entropy curvesare indicated in the figure as m1 and m2,respectively. After the system reaches equi-librium, what will be the relationship betweenthe slopes of the two curves.

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1. more information is needed

2. m1 =1

2m2

3. m1 = m2 = 0

4. m1 = m2 = ∞

5. m1 = m2 correct

6. m1 < m2

7. m1 > m2

8. the slopes do not change

Explanation:The slope of the entropy curve gives the

temperature of each block with a given quantaof energy in that block. When equilibrium isreached, the temperatures will be the sameso the slopes should be the same. However,the axis for block 2 is reversed so the correctanswer is m1 = −m2.

010 10.0 pointsUsing the Einstein model of a solid, whatis the change in entropy when adding twoquanta of energy to a system of 5 atoms thatalready has 5 quanta of energy stored in it?

1. ∆S = kB ln

(

21

7 · 5

)

2. ∆S = kB ln

(

16!

7! · 7

)

3. ∆S = kB ln

(

10

7

)

4. ∆S = kB ln

(

17

6

)

5. ∆S = kB ln (10) correct

6. ∆S = kB ln

(

10!

7!

)

7. ∆S = kB ln

(

21!

14! · 7!

)

8. ∆S = kB ln

(

20!

14! · 5!

)

9. ∆S = kB ln

(

20!

7! · 5!

)

10. ∆S = kB ln

(

21!

7! · 5!

)

Explanation:Entropy is defined as S = kB lnΩ where Ω

is the number of possible microstates, whichis the number of ways to arrange q quanta inn harmonic oscillators, i.e.

Ω =(n− 1 + q)!

(n− 1)! · q!

Here, n − 1 = 3 · 5 − 1 = 14 and initiallyqi = 5, but in the final state qf = 7. Thus,the change in entropy is

∆S = Sf − Si

= kb ln

(

(14 + 7)!

14! · 7!

)

− kb ln

(

(14 + 5)!

14! · 5!

)

= kb ln

(

(14+7)!14!·7!

)

(

(14+5)!14!·5!

)

= kb ln

(

21 · 207 · 6

)

= kb ln

(

21 · 107 · 3

)

= kb ln (10)

011 10.0 pointsTwo masses of 22 kg and 11 kg are suspendedby a pulley that has a radius of 15 cm anda mass of 6 kg. The cord has a negligiblemass and causes the pulley to rotate withoutslipping. The pulley rotates without friction

its moment of inertia is given by I =1

2MR2.

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h

15 cm

6 kg

α

22 kg

11 kg

Determine the angular acceleration α of thepulley after the masses are released and beforethey fall off of the pulley.The acceleration of gravity is 9.8 m/s2 .1. 42.23232. 36.33673. 29.44. 27.22225. 23.0386. 14.20297. 62.53248. 38.86329. 19.96310. 21.9103

Correct answer: 19.963 rad/s2.

Explanation:

Let : M = 6 kg ,

R = 15 cm ,

m1 = 22 kg , and

m2 = 11 kg

The tension in the cord attached to thefirst mass T1 is different than the tension inthe cord attached to the second mass T2. Wemust use both the Momentum Principle andthe Angular Momentum Principle to solve theproblem.The Momentum Principle for m1 gives:

m1 (−a) = T1 −m1g

T1 = m1g −m1a

where we have taken care to make the accel-eration negative since it will be positive for

m2

m2 (+a) = T2 −m2g

T2 = m2g +m2a

So far we have two equations and three un-known: T1, T2, and a. However, the AngularMomentum Principle gives us another equa-tion

Iα = τnet

= T1R − T2R

where again we have taken care to get thesigns of alpha and the torque correct. We cannow plug in the expressions for tension whilemaking the substitution that a = αR sincethere the cord does not slip.

Iα = (m1g −m1αR)R− (m2g +m2αR)R

= +(m1 −m2)gR− (m1R2 +m2R

2)α

α = [(m1 −m2)gR]/(I +m1R2 +m2R

2)

= 19.963 rad/s2 .

012 (part 1 of 2) 10.0 pointsA mass m is located at a distance R fromthe center of a planet of mass M . The initialspeed ofm is v0 and it’s velocity vector makesan angle θ with respect to the line joining themass and the center of the planet. You canassume that the planet remains stationary incourse of the motion of the projectile.Find the magnitude of the angular momen-

tum of the mass m relative to the center ofthe plant.

1. Mv0R cos θ

2. mv0R sin θ +Mv0R sin θ

3. mv0R sin θ −Mv0R sin θ

4. mv0R cos θ −Mv0R cos θ

5. 0

6. Mv0R sin θ

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7. mv0R cos θ

8. mv0R sin θ correct

Explanation:The magnitude of the angular momentum

vector,

|~Lm| = |~r ×~p| = mv0R sin θ.

~r is the position of the mass m w.r.t. centerof the planet and ~p is the momentum vectorof the mass.

013 (part 2 of 2) 10.0 pointsWhat is the magnitude of the net torque onthe mass m about the center of the planet?

1. Need more information

2. GmM cos θ/R

3. 0 correct

4. GmM cos θ/R2

5. GmM sin θ/R

6. GmM sin θ/R2

7. GmM/R

Explanation:The net torque on the mass m due to the

gravitational attraction of the planet,

~τ = ~r × ~Fgrav.

Since~r and ~Fgrav are anti-parallel at all times,~τ = 0 always.

014 10.0 pointsTwo objects share a total energyE = E1+E2.There are 10 ways to arrange an amount ofenergy E1 in the first object and 15 ways toarrange an amount of energy E2 in the secondobject. How many different ways are there toarrange the total energy E = E1 +E2 so thatthere is E1 in the first object and E2 in theother?

1. 10

2. cannot be determined

3. 15

4. 1× 1015

5. 225

6. 150 correct

7. 25

Explanation:The total number of microstates or num-

ber of ways of arranging energy in the systemis the product of the number of ways of ar-ranging the energy in respective objects, i.e.Ωtotal = Ω1Ω2

015 10.0 pointsA force

~F = (−3ı+ 6) N

acts at a distance from the point of rotationgiven by

~r = (8ı+ 4) m,

where ı and are unit vectors pointing in the+x and +y directions respectively. What isthe torque about the point of rotation createdby this force? (Note that k is a unit vectorpointing in the +z direction.)

1. (50ı+ 60) N ·m

2. (60k) N ·m correct

3. (14ı+ 50k) N ·m

4. (60) N ·m

5. (50ı− 36) N ·m

6. (−60+ 50k) N ·m

7. (−60k) N ·m

8. (36ı) N ·m

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9. (36k) N ·m

10. (14ı− 50k) N ·mExplanation:Torque is defined as ~r × ~F . Because both

vectors are in the xy plane, their cross productmust be in the ±z direction. So we only needto look at the z-component of the cross prod-uct. The z component of the cross productwill be

rxFy − ryFx = 60N ·m.

So the resultant vector is (60k) N ·m .

016 10.0 pointsA ball of putty with mass m falls verticallyonto the outer rim of a turntable of radiusR and moment of inertia I0 that is rotatingfreely with angular speed ωi about its verticalfixed symmetry axis, i.e. the turntable ishorizontal.What is the post-collision angular speed of

the turntable plus putty?

1. ωf =ωi

2 +mR4

I02. ωf =

ωi

3m+mR3

I0

3. ωf =ωi

1 +mR

I04. ωf =

ωi

2 +mR2

I0

5. ωf =ωi

2m+mR3

I06. ωf =

ωi

2 +mR

I0

7. ωf =ωi

1 +mR3

I08. ωf =

ωi

m+mR3

I0

9. ωf =ωi

1 +mR4

I010. ωf =

ωi

1 +mR2

I0

correct

Explanation:The final rotational inertia of the turntable-

plus-putty is

If = I0 + Iblob = I0 +mR2 .

Since there is no external torque on the systemof the putty plus the turntable, we know Lf =Li = I0ωi.

If ωf = I0 ω0

ωf =I0 ωi

If

ωf =I0 ωi

I0 +mR2

=ωi

1 +mR2

I0

.

017 10.0 pointsIn the figure, two objects of the same massm = 2.1 kg are connected by a massless rodof length d = 1.25 m. At a particular instantthey have velocity magnitudes that are v1 =32 m/s and v2 = 79 m/s, respectively. Thesystem is moving in outer space far from anyother objects. The x direction is to the right,y is up, and z is out of the page toward you.

d

b

b

m

m

v1

v2

What is the magnitude of the rotationalangular momentum Lrot of the system?1. 42.0

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2. 61.68753. 22.31254. 107.6255. 70.8756. 87.93757. 86.6258. 93.18759. 63.010. 59.0625

Correct answer: 61.6875 kgm2/s.

Explanation:The rotational angular momentum is the

angular momentum of the system about itscenter of mass. As the objects are of equalmass, the center of mass of the system ishalfway between them. Summing the angularmomentum of each mass,

~Lrot = ~r1,CM ×~p1 +~r2,CM ×~p2

= 〈0, d2, 0〉 × 〈mv1, 0, 0〉

+ 〈0,−d

2, 0〉 × 〈mv2, 0, 0〉

= 〈0, 0, (0.625 m)[(2.1 kg)(79 m/s)− (2.1 kg)(32 m/s)]〉= 〈0, 0, 61.6875 kgm2/s〉 .

The magnitude is thus 61.6875 kgm2/s .

018 10.0 pointsTwo rigid rods of length ℓ and mass M arerigidly attached as shown

pivot

ω

What is the magnitude of the angular mo-mentum L of this system when it is rotating atan angular velocity ω about an axis throughthe end of one rod, as indicated in the sketch?The rotational inertia of a rod about an axis

through one end is I =1

3M ℓ2, while the

rotational inertia about an axis through the

center of mass is ICM =1

12M ℓ2.

1. L =2

3M ℓ2 ω.

2. L =5

4M ℓ2 ω.

3. L =5

12M ℓ2 ω.

4. L =11

3M ℓ2 ω.

5. L =13

12M ℓ2 ω.

6. L =9

12M ℓ2 ω.

7. L = M ℓ2 ω.

8. L =11

12M ℓ2 ω.

9. L =1

3M ℓ2 ω.

10. L =17

12M ℓ2 ω. correct

Explanation:Using the parallel axis theorem,

I =1

3M ℓ2 +

1

12M ℓ2 +M ℓ2

=

(

4 + 1 + 12

12

)

M ℓ2 =17

12M ℓ2 .

Therefore, the angular momentum is

L = Iω =17

12M ℓ2 ω

019 10.0 pointsConsider a uniform ladder leaning against

a smooth wall and resting on a smooth floorat point P . There is a rope stretched horizon-tally, with one end tied to the bottom of theladder essentially at P and the other end tothe wall. The top of the ladder is at a heightis h up the wall and the base of the ladder isat a distance b from the wall.

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P

W1

W2

ℓ

d

T

θ

b

h

b

b

F

The weight of the ladder is W1 . Jill, with

a weight W2, is one-fourth the way

(

d =ℓ

4

)

up the ladder. The force which the wall exertson the ladder is F .What is the torque equation about P ?

1.b

2W2 + bW1 = F h

2. (W1 +W2)b

2= F h

3.h

4W2 +

h

2W1 = F b

4.b

4W2 +

b

2W1 = F h correct

5.h

2W2 + hW1 = F b

6. (W1 +W2)h

2= F b

Explanation:

Pivot

b

F

T

Nf

W2

W1 θ

Applying rotational equilibrium,

∑

τP= W2 d cos θ+W1

ℓ

2cos θ−F ℓ sin θ = 0 ,

where d is the distance of the person from the

bottom of the ladder, sin θ =h

ℓand cos θ =

b

ℓ.

2F ℓ sin θ = 2W2 d cos θ +W1 ℓ cos θ

2F h = 2W2d b

ℓ+W1 b

= 2W2b

4+W1 b

F h =b

4W2 +

b

2W1

020 10.0 pointsA sticky blob strikes and sticks to a free rod,which is initially at rest, as shown.

Let E be the mechanical energy of the sys-tem, ~P the linear momentum of the system,and ~L the angular momentum of the system.What is conserved?

1. ~P and E

2. ~L, ~P , and E

3. ~L and E

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4. ~L only

5. ~L and ~P correct

6. ~P only

Explanation:The mechanical energy of the system is not

conserved because this is not an elastic colli-sion, but the linear momentum and angularmomentum are always conserved in such freecollisions.

Version 068 – Test 4 – swinney – (58385) 1


001 10.0 points

A spool floating in space has radius r massm and moment of inertia about its centerI = β mr2. The spool is unwound by aconstant force F .

b

F

r

If initially the spool is motionless, at somelater time what is the ratio of translationalkinetic energy to rotational kinetic energy?

1.Ktrans

Krot= (1− β)2

2.Ktrans

Krot= (1 + β)2

3.Ktrans

Krot= β2

4.Ktrans

Krot=

1

β

5.Ktrans

Krot=

1− β

β

6.Ktrans

Krot= βcorrect

7.Ktrans

Krot= 1 + β

8.Ktrans

Krot=

1 + β

β

9.Ktrans

Krot= 1− β

10.Ktrans

Krot=

1

β2

Explanation:

From the momentum principle

a =F

m

and from the angular momentum principle

α =rF

I.

From kinematics for any time t

vtrans =F

mt (1)

and

ωrot =rF

It. (2)

Now equation (1) implies

Ktrans =1

2mv2trans =

1

2m

(

F

m

)2

t2, (3)

while equation (2) gives

Krot =1

2I ω2

trans =1

2I

(

rF

I

)2

t2, (4).

Finally, dividing equation (3) by equation (4),

Ktrans

Krot=

I/r2

m= β.

002 10.0 points

A door is opened by a pulling force in the xdirection.

y

xz

In which direction is the net torque vector~τ about the hinge due to the force applied tothe door handle?

1. y correct

2. Insufficient information is given.

3. −x

4. −z

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5. −y

6. z

7. x

Explanation:

~τ = r×Fr ~τ

F

~r points in the z direction from the hinge tothe spot where ~F acts, so by the right handrule, ~r× ~F points in the direction y, along thehinge line.

003 10.0 points

A crane of mass M supports a load m. Thecrane’s boom is length ℓ and the angle itmakes with the horizontal is θ. The distancebetween the front and rear wheels is d andthe center of mass of the crane plus boom islocated a distance d/3 from the rear wheel.The ground exerts Nr and Nf at the wheels.The acceleration of gravity is g .

M

ℓ

θ

NfdNr

m

What is the magnitude of the normal forceNf?

1. Nf = g

[(

ℓ

dsin θ + 1

)

m+1

3M

]

2. Nf = g

[(

ℓ

dcos θ + 1

)

m− 2

3M

]

3. Nf = g

[(

ℓ

dsin θ + 1

)

m− 2

3M

]

4. Nf = g

[(

ℓ

dsin θ + 1

)

m− 2

3cos θM

]

5. Nf = g

[(

ℓ

dcos θ − 1

)

m− 2

3M

]

6. Nf = g

[(

ℓ

dcos θ + 1

)

m+1

3M

]

cor-

rect

7. Nf = g

[(

ℓ

dsin θ − 1

)

m− 2

3M

]

8. Nf = g

[(

ℓ

dsin θ − 1

)

m+1

3M

]

9. Nf = g

[(

ℓ

dcos θ − 1

)

m+1

3M

]

10. Nf = g

[(

ℓ

dcos θ + 1

)

m+1

3sin θM

]

Explanation:

Since the crane and boom is in static equi-librium, the net torque and net force actingon it are both zero.The equation for net torque about the front

wheel is

Mg2

3d−mgℓ cos θ −Nrd = 0

−Nr =ℓ

dcos θmg − 2

3Mg

And the equation for net force is

Nf +Nr − (m+M) g = 0

Nf +Nr = (m+M) g

Adding these two equations eliminates Nr

in favor of Nf , i.e.,

Nf = (m+M) g +ℓ

dcos θmg − 2

3Mg

= g

[(

ℓ

dcos θ + 1

)

m+1

3M

]

004 10.0 points

A ball of radius r with a wire glued to onespot on its surface can be pulled along thefloor and will slide without any tendency toroll only if when the wire is horizontal it isalso a distance h above the floor.


r

~Fh

~acm

In terms of F , the magnitude of the appliedforce, fk, the magnitude of the force of kineticfriction, and the radius r of the ball, what ish?

1. h = r

(

F

fk

)

2. h = r

(

1− F

fk

)

3. h = r

(

1− fkF

)

correct

4. h = r

5. h = r (F − fk)

6. h = r

(

1 +fkF

)

7. h = 0 ; the wire must run along the floor.

8. h = r (fk − F )

9. h = r

(

fkF

)

10. h = r

(

1 +F

fk

)

Explanation:

Taking torques about the center of mass,for no rolling,

fk

rr-h

~F

0 = (r − h)F − r fk

h = r

(

1− fkF

)

.

005 10.0 points

The figure shows a system made of two pointmasses each with mass m that rotates at anangular speed of ω . The masses are connectedby massless flexible spokes of length r that canbe elongated or shortened.

x

y

r

m

m

ω

What is the the final kinetic energy Kf interms of the initial kinetic energy Ki if all thespokes lengths are decreased by a factor of 2?

1. Kf =1

4Ki

2. Kf =1

16Ki

3. Kf = 4Ki correct

4. None of these answers are correct.

5. Kf =1

2Ki

6. Kf = Ki

7. Kf =1

8Ki

8. Kf = 2Ki

9. Kf = 8Ki

10. Kf = 16Ki

Explanation:

Basic Concepts:

L = I ω =(

M r2)

ω

L2 = L1


First the new moment of inertia is

I ′ = 2m(

r′ 2)

= 2m(

1/2)2 r2)

= (1/4) 2m(

r2)

I ′ = (1/4) I,

Notice that rotational kinetic energy can

be written K =L2

2I, hence

K ′

K=

I

I ′since L

is constant. And so for the problem at handK ′

K= 4.

keywords:

006 10.0 points

An isolated system has 100 quanta of en-ergy distributed between two blocks in con-tact. Block 1 and Block 2 have 300 and 200quantum mechanical harmonic oscillators re-spectively and have q1 and q2 quanta respec-tively. An plot of a nonequilibrium state isshown.At equilibrium, what is the relationship be-

tween m1 =dS1

dq1and m2 =

dS2

dq2?

1. m1 = m2 = 0.

2. m1 6= 0 and m2 = 0.

3. m1 = m2. correct

4. m1 > m2 but neither are 0 or ∞.

5. m1 = m2 = ∞.

6. m1 = 0 and m2 6= 0.

7. m1 = ∞ and m2 is neither 0 nor ∞.

8. m1 < m2 but neither are 0 or ∞.

9. It is impossible to know.

10. m1 is neither 0 nor ∞ and m2 = ∞.

Explanation:

The slope of the entropy curve gives thetemperature of each block with a given quantaof energy in that block. When equilibrium isreached, the temperatures, i.e., (dS/dE)−1,will be the same so the slopes will be thesame. Thus the correct answer is m1 = m2.

007 10.0 points

Using the Einstein model of a solid, whatis the change in entropy when adding twoquanta of energy to a system of 5 atoms thatalready has 5 quanta of energy stored in it?(In an Einstein solid each atom correspondsto three independent oscillators.)

1. ∆S = kB ln

(

10!

7!

)

2. ∆S = kB ln

(

21

7 · 5

)

3. ∆S = kB ln

(

17

6

)

4. ∆S = kB ln

(

10

7

)

5. ∆S = kB ln

(

20!

7! · 5!

)

6. ∆S = kB ln

(

16!

7! · 7

)

7. ∆S = kB ln (10) correct

8. ∆S = kB ln

(

21!

14! · 7!

)

9. ∆S = kB ln

(

21!

7! · 5!

)

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10. ∆S = kB ln

(

20!

14! · 5!

)

Explanation:

Entropy is defined as S = kB lnΩ where Ωis the number of possible microstates, whichis the number of ways to arrange q quanta inn harmonic oscillators, i.e.

Ω =(n− 1 + q)!

(n− 1)! · q!Here, n − 1 = 3 · 5 − 1 = 14 and initially

qi = 5, but in the final state qf = 7. Thus,the change in entropy is

∆S = Sf − Si

= kb ln

(

(14 + 7)!

14! · 7!

)

− kb ln

(

(14 + 5)!

14! · 5!

)

= kb ln

(

(14+7)!14!·7!

)

(

(14+5)!14!·5!

)

= kb ln

(

21 · 207 · 6

)

= kb ln

(

21 · 107 · 3

)

= kb ln (10)

008 10.0 points

A person of weight WP stands motionlesson the end of a uniform diving board of weightWD that has two supports A and B which, ifthey exert vertical forces, exert forces SA andSB respectively. The normal force exerted onthe person is N .

A B

What is the free body diagram for the div-ing board?

1.

SB

WD WP

2.

SA

SB WD WP

3.More than one of these free body diagramsis possible.

4.

SA SB

WD WP

5.

SA

SB

WD

N

6.

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SA SB

WD

N

7.

SA

SB

WD WP

correct

8.

SA SB WD WP

N

9.

SA WD WP

N

10.

SB WD WP

N

Explanation:

For static equilibrium,∑

F = 0 and∑

τ = 0 .

SA SB WD WP

cannot have a net transitional force of zerosince all forces act downward.For the other three diagrams, only in

SA

SB

WD WP

could the net torque about the point B of thediving board be zero.

009 10.0 points

The centers of mass of three trucks parked ona hill are shown by the mark and each ofthe dotted lines are oriented vertically.

12

3

Which truck(s) will tip over?

1. Trucks 1 and 2

2. Trucks 2 and 3

3. Truck 2

4. Truck 3

5. None of the trucks

6. All three trucks

7. Truck 1 correct


8. Trucks 1 and 3

Explanation:

12

3

The center of gravity (CG) of Truck 1 is notabove its support base; the CGs of Trucks 2and 3 are above their support bases. There-fore, only Truck 1 will tip.

010 10.0 points

Two objects share a total energyE = E1+E2.There are 5 ways to arrange an amount ofenergy E1 in the first object and 10 ways toarrange an amount of energy E2 in the secondobject. How many different ways are there toarrange the total energy E = E1 +E2 so thatthere is E1 in the first object and E2 in theother?

1.14!

5!9!

2. 225

3. 15

4. 10

5. 1015

6.14!

4!10!

7. 50 correct

8. 5

9.15!

5!10!

10. 105

Explanation:

The total number of microstates or num-ber of ways of arranging energy in the system

is the product of the number of ways of ar-ranging the energy in respective objects, i.e.Ωtotal = Ω1Ω2

011 10.0 points

Two disks of identical mass but different radii(r and 2 r) are spinning on frictionless bear-ings at the same angular speed ω0 but in op-posite directions. The two disks are broughtslowly together. The resulting frictional forcebetween the surfaces eventually brings themto a common angular velocity.

2 r

ω0

r

ω0

What is the ratio of final kinetic energy Kf

to the initial kinetic energy Ki?

1.Kf

Ki=

2

5

2.Kf

Ki=

1

25

3.Kf

Ki=

4

25

4.Kf

Ki= 1

5.Kf

Ki=

9

25correct

6.Kf

Ki=

16

25

7.Kf

Ki=

1

5

8.Kf

Ki=

3

5

9.Kf

Ki= 2

10.Kf

Ki=

4

5

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Explanation:

Note: Since the disks are spinning in oppo-site directions, let ω1 = ω0 and ω2 = −ω0 .The inertia of the larger disk is

I1 =1

2m (2 r)2 = 2mr2 ,

and of the smaller disk

I2 =1

2mr2 .

Using conservation of angular momentum,

Ii ωi = If ωf

I1 ω0 − I2 ω0 = (I1 + I2)ωf

ωf =I1 − I2I1 + I2

ω0

=2mr2 − 1

2mr2

2mr2 +1

2mr2

ω0

=3

5ω0 .

Now the initial kinetic energy is Ki =(1/2)(I1 + I2)ω

20, while the final kinetic en-

ergy is Kf = (1/2)(I1 + I2)(3/5)2ω2

0. Hence

the ratio is just (3/5)2 = 9/25.

012 10.0 points

Two blocks are connected by a massless, in-extensible string wrapped around a massive,frictionless pulley as shown below. The blockresting on the plane which is inclined at an-gle θ from the horizontal, has mass m. Thepulley is a uniform disk and has mass 2m andradius r. And the hanging block has mass3m. The coefficient of kinetic friction betweenblock and the incline is µk. The accelerationof gravity is g .

3m

2m, r

m

µ k

ℓ

θ

If the system starts from rest, how longdoes it take the block on the incline to slide adistance ℓ?

1. t =

√

(

5ℓ

g

)(

1

3 + µk cos θ

)

2. t =

√

(

5ℓ

g

)(

1

3− µk cos θ

)

3. t =

√

(

5ℓ

g

)(

1

3− µk sin θ

)

4. t =

√

(

10ℓ

g

)(

1

3 + µk cos θ

)

5. t =

√

(

10ℓ

g

)(

1

3− µk cos θ

)

correct

6. t =

√

(

10ℓ

g

)(

1

3 + µk sin θ

)

7. t =

√

(

3ℓ

g

)(

1

3 + µk sin θ

)

8. t =

√

(

3ℓ

g

)(

1

3− µk cos θ

)

9. t =

√

(

10ℓ

g

)(

1

3− µk sin θ

)

10. t =

√

(

5ℓ

g

)(

1

3 + µk sin θ

)

Explanation:


Let T1 be the tension in the string attachedto the block with mass m and T2 be the ten-sion in the string attached to the block withmass 3m. Now the equation of motion forblock m is

T1 − fk = ma

T1 − µk mg cos θ = ma.

The equation of motion for the disk withI = mr2 is

(T2 − T1)r = mr2α

T2 − T1 = ma,

where the last equality follows from rα = a.The equation of motion for 3m is simply

3mg − T2 = 3ma.

Summing the three equations of motion to-gether eliminates the tensions and leads to

mg (3− µk cos θ) = 5ma

a =(g

5

)

(3− µk cos θ) .

Finally, from kinematics

ℓ =1

2at2

t =

√

(

10ℓ

g

)(

1

3− µk cos θ

)

013 10.0 points

A 3 kg bicycle wheel rotating at a2608 rev/min angular velocity has its shaftsupported on one side as shown. When viewedfrom the left (from the positive x-axes), thewheel is rotating in a clockwise manner. Thedistance from the center of the wheel to thepivot point is 0.5 m. The wheel is a hoop ofradius 0.4 m with its shaft is horizontal.

y

z

x

0.5m

mg

ω

2608

rev/m

inω

0.4 m

radius

Find the change in the precession angleafter a 1.2 s time interval. Assume all of themass of the system is located at the rim of thebicycle wheel. The acceleration of gravity is9.8 m/s2 .1. 7.96282. 7.709813. 9.574864. 6.101275. 10.43376. 3.866197. 7.562928. 4.893739. 3.4789410. 4.39085

Correct answer: 7.70981.

Explanation:

Let : m = 3 kg ,

ω = 2608 rev/min ,

b = 0.5 m , and

R = 0.4 m .

+x −x

L+∆L∆L

∆φ ∆Lτ

wheel

Viewed from Above

+y

L

The magnitude of the angular momentumI of the wheel is L = I ω = mR2 ω and isdirected along the negative x-axis.

∆φ =∆L

Land


ω = (2608 rev/min)2 π

rev

1 min

60 s= 273.109 rad/s .

The torque is ~τ =d ~L

dtso ∆L = τ ∆t and

the precession angle is

∆φ =∆L

L=

τ ∆t

L=

mg b∆t

mR2 ω

=g b∆t

R2 ω=

(9.8 m/s2) (0.5 m) (1.2 s)

(0.4 m)2(273.109 rad/s)

= 7.70981 .

014 10.0 points

Initially a wheel rotating about a fixedaxis with a constant angular accelerationof −0.4 rad/s2, has an angular velocity of1.2 rad/s, and an angular position of 6.7 rad.What is the angular position of the wheel

after 2.9 s?1. 6.22. 8.8973. 10.8594. 9.155. 11.26. 11.57. 8.4988. 12.45959. 13.57210. 9.468

Correct answer: 8.498 rad.

Explanation:

Let : α = −0.4 rad/s2 ,

ω0 = 1.2 rad/s ,

θ0 = 6.7 rad , and

t = 2.9 s .

From kinematics,

θ = ω0 t+1

2α t2 + θ0

= (1.2 rad/s) (2.9 s)

+1

2(−0.4 rad/s2) (2.9 s)2 + 6.7 rad

= 8.498 rad .

015 10.0 points

A rotating uniform-density disk of radius Rand mass M is mounted in the vertical plane,as shown in the following figure.

b

R

×A

Clay

A lump of clay with mass m falls and sticksto the outer edge of the wheel at the positionA given by the vector ~A = 〈−x0, y0, 0〉 (choiceof axes: x to the right, y up, z out of thepage; origin at the center of the axle). Justbefore the impact, the clay has velocity ~v =〈0,−v0, 0〉, and the disk is rotating clockwisewith angular speed ω0.Just after the impact, what is the angular

speed ωf of the combined system of wheelplus clay?

1. ωf =

∣

∣mv0R− 1/2MR2ω0

∣

∣

(1/2M +m)R2

2. ωf =

∣

∣mv0R+ 1/2MR2ω0

∣

∣

(1/2M −m)R2

3. ωf =

∣

∣mv0x0 + 1/2MR2ω0

∣

∣

(1/2M +m)R2

4. ωf =

∣

∣mv0y0 + 1/2MR2ω0

∣

∣

(1/2M −m)R2

5. ωf =

∣

∣mv0x0 − 1/2MR2ω0

∣

∣

(1/2M +m)R2correct

6. ωf =

∣

∣mv0x0 − 1/2MR2ω0

∣

∣

(1/2M −m)R2

7. ωf =

∣

∣mv0R− 1/2MR2ω0

∣

∣

(1/2M −m)R2


8. ωf =

∣

∣mv0y0 − 1/2MR2ω0

∣

∣

(1/2M +m)R2

9. ωf =

∣

∣mv0y0 − 1/2MR2ω0

∣

∣

(1/2M −m)R2

10. ωf =

∣

∣mv0x0 + 1/2MR2ω0

∣

∣

(1/2M −m)R2

Explanation:

To calculate the angular speed of the com-bined system, one needs to compute the mo-ment of inertia of the system.

Isys =1

2M R2 +mR2.

Therefore,

ωsys =

∣

∣

∣

~Ltot,f

∣

∣

∣

Isys

=

∣

∣mv0x0 − 1/2MR2ω0

∣

∣

(1/2M +m)R2

016 10.0 points

A disk of radius 9.5 cm and mass 0.52 kg ispulled along a frictionless surface with a forceof 15 N by a string wrapped around the edge(Fig. 11.43, displayed below). 24 cm of stringhas unwound off the disk. At the instantshown the angular velocity is 21 rad/s in theclockwise sense.

9.5 cm

15 N24 cm

At time ∆t = 0.2 s later, what is the mag-nitude of the angular momentum about thecenter of the disk?1. 0.266242. 0.292853. 0.231374. 0.276403

5. 0.2073196. 0.218217. 0.1936178. 0.2045329. 0.25996610. 0.334277

Correct answer: 0.334277 kg ·m2/s.

Explanation:

Use ~τ = ~r × ~F . Therefore, |~τ | = |~r⊥||~F | =(0.095 m) (15 N). Thus, |~τ | = 1.425 N ·m.

~L = I~ω =< 0, 0,−Iω >, where I =(1/2)MR2 for the solid disc and ~ω is directedalong the negative z-axis. Using the valuesgiven from the beginning of the problem, theequation becomes

~L =< 0, 0,−(1/2)MR2ω >

=< 0, 0,−0.0492765 kg ·m2/s >

|~L| = 0.0492765 kg ·m2/s

Since the torque acting on the disc is con-stant over time, we conclude the change inangular momentum is given by

∆~L = ~τ∆t

∆~L =< 0, 0,−RF∆t >

The final angular momentum is then given by

~Lf = ~L+∆~L

= ~L+ < 0, 0,−RF∆t >

=< 0, 0,−(1/2)MR2ω −RF∆t >

=< 0, 0,−0.0492765− 0.285 > kg ·m2/s

=< 0, 0,−0.334277 kg ·m2/s >

|~Lf | = 0.334277 kg ·m2/s

017 10.0 points

A uniform beam of mass M supports twoweights that hang motionless.


672 N

W

L

4

L

4

L

2

L

A B

Find W such that the supporting force atA is zero. Assume the force acting at B hasno horizontal component.1. 445.52. 351.03. 336.04. 491.05. 363.06. 342.07. 478.08. 438.59. 382.510. 378.5

Correct answer: 336 N.

Explanation:

Let : W1 = 672 N and

FA = 0 .

∑

τ = 0, so for a fulcrum at B,

W1L

4= W L

2

W =1

2W1 =

1

2(672 N) = 336 N .

018 10.0 points

A sticky ball collides with a rod exactly at thepoint where the rod can rotate on a frictionlessaxle as shown. The axle is permanently fixedin place and the rod is initially at rest.

Consider the system of the putty and therod, and let E be the kinetic energy of the sys-tem, ~P the linear momentum of the system,and ~L the angular momentum of the system.List which of these quantities are conserved,first for a point about the axle and second fora point about the center of the rod.

1. axle: ~P ; center: none

2. axle: none; center: none

3. axle: ~L ,E; center: ~L ,E

4. axle: ~L ,E; center: ~L

5. axle: ~L , ~P , E; center: ~L , ~P

6. axle: ~L; center: none correct

7. axle: ~L , ~P ; center: ~L , ~P

8. axle: ~L , ~P , E; center: ~L , ~P , E

9. axle: ~L , ~P ; center: ~L

10. axle: ~L; center: ~L

Explanation:

First, notice the rod doesn’t rotate or trans-late. The mechanical energy of the system isnot conserved because initially there was ki-netic energy and finally there was not; me-chanical became internal energy or was dis-sipated as heat. Linear momentum is notconserved since there was motion before butwas none after; a force was exerted on the sys-tem by the axle. Finally, for the point aboutthe axle angular momentum is conserved sinceinitially it was zero and finally it was zero; theforce from the axle acted through the point ofrotation and applied no torque. And for the

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point about the center, angular momentum isnot conserved because it was first nonzero andthen later was zero; here the force due to theaxle did apply a torque. Hence ~L is conservedabout the axle and nothing is conserved aboutthe center.

019 10.0 points

There is a moon in a circular orbit around anEarth-like planet. The mass of the moon is8.26×1022 kg, the center-to-center separationof the planet and the moon is 9.29× 105 km,the orbital period of the moon is 26 days, andthe radius of the moon is 1500 km.What is the translational angular momen-

tum of the moon about the planet?1. 1.71642e+342. 5.64551e+343. 2.07747e+354. 4.57364e+335. 1.67004e+356. 2.60541e+347. 1.9939e+358. 1.26096e+359. 2.78853e+3410. 3.13592e+34

Correct answer: 1.9939× 1035 kgm2/s.

Explanation:

The angular speed ω in radians per unittime, for a complete circle is

ω =2 π

T.

Translational angular momentum Ltrans is

Ltrans = mv r = mr2 ω =2 πmr2

T= 2 π (8.26× 1022 kg)

×

[

(9.29× 105 km)1000 m

1 km

]2

(26 days)

(

24 hr

1 day

)

= 1.9939× 1035 kgm2/s .

020 10.0 points

A uniform rod of mass m and length ℓ is

pivoted about a horizontal, frictionless pinat the end of a thin extension (of negligiblemass) a distance ℓ from the center of mass ofthe rod. The rod is released from rest at anangle of θ with the horizontal, as shown in thefigure.

ℓℓ

m

O

θ

What is the magnitude of the vertical forceFy exerted on the rod by the pivot at theinstant the rod is in a horizontal position?The acceleration due to gravity is g and themoment of inertia of the rod about its center

of mass is1

12mℓ2.

1. Fy =1

5mg

2. Fy = mg

3. Fy =1

3mg

4. Fy =1

2mg

5. Fy =13

12mg

6. Fy = 3mg

7. Fy = 2mg

8. Fy =1

13mg correct

9. Fy =12

13mg

10. Fy =25

13mg

Explanation:


I = ICM +md2 =1

12mℓ2+mℓ2 =

13

12mℓ2 .

90

Fy

Fx

mg

O

Since the rod is uniform, its center of massis located at ℓ . The weight mg acts at thecenter of mass, so the magnitude of the torqueis

τ = r × F = I α(

13

12mℓ2

)

α = ℓmg sin 90 = ℓmg

α =12 g

13 ℓ.

The vertical component of the tangential ac-celeration is

ay = α r =

(

12 g

13 ℓ

)

ℓ =12

13g

and the sum of the forces acting on the rod inthe vertical direction is

Fy −mg = −may ,

so the vertical reaction force Fy on the pivotis

Fy = m (g − ay) = m

(

g − 12

13g

)

=mg

13.

Answer, Key Homework 8 Rubin H Landau 1

This print-out should have 14 questions.Check that it is complete before leaving theprinter. Also, multiple-choice questions maycontinue on the next column or page: nd allchoices before making your selection.We are back in order (let's hope). Some

solutions may be found on the class homepage.

Rolling of a Cylinder

11:01, trigonometry, numeric, > 1 min.004

An m = 3:34 kg hollow cylinder with Rin =0:3 m and Rout = 0:5 m is pulled by a horizon-tal string with a force F = 22:8 N, as shownin the diagram.

Fm

Rin

Rout

What must the magnitude of the force offriction be if the cylinder is to roll withoutslipping?Correct answer: 4:34286 N.Explanation:

Suppose the cylinder has a length of l.Then the density of the cylinder is =

m

(R2out R2

in)l. The moment of inertia about

the center is:

Icm =

Zr2 dr

=m

(R2out R2

in)l

Z Rout

Rin

r2 r dr

Z2

0

d

Z l

0

dz

=m

(R2out R2

in)l

R4out R4

in

4(2)(l)

=m

2(R2

out +R2in)

By the parallel axis theorem, the moment ofinertia about the ground is:

I = Icm +mR2out

=m

2(3R2

out +R2in)

From the force equation, we have:

F + f = ma:

From the torque equation, we have:

F (2Rout) = I = Ia

Rout:

Solving this pair of equations, we get:

f = F

2mR2

out

I 1

= 22:8 N

2(3:34 kg)(0:5 m)2

1:4028 kgm2 1

= 4:34286 N

005

In what direction is the frictional force?

1. To the left

2. To the right correct

3. Force is zeroExplanation:

The frictional force calculated has the samesign as the applied force, so it must be in thesame direction as F , i.e. to the right.

006


What is the acceleration of the cylinder's cen-ter of mass?Correct answer: 8:1266 m=s2.Explanation:

From the pair of equations above, we cansolve for a:

a =2FR2

out

I

=2(22:8 N)(0:5 m)2

1:4028 kgm2

= 8:1266 m=s2

:

Algorithm

r=0:3 m0:10:3

(1)

R=0:5 m0:40:6

(2)

m=3:34 kg2

5

(3)

F=22:8 N20

50

(4)

I=0:5m3:0 R2:0+r2:0

(5)

=0:5 h3:34i3:0 h0:5i2:0+h0:3i2:0

=1:4028 kgm2

hkgm2i=hi hkgihi hmi2:0+hmi2:0

units

a=2:0 F R2:0

I(6)

=2:0 h22:8i h0:5i2:0

h1:4028i

=8:1266 m=s2

hm=s2i=hi hNi hmi2:0

hkgm2iunits

f=F

2:0m R2:0

I1

(7)

=h22:8i

2:0 h3:34i h0:5i2:0

h1:4028i1

=4:34286 N

hNi=hNi

hi hkgi hmi2:0

hkgm2ihi

units

Contacting a Surface

11:01, advanced, numeric, > 1 min.008

Determine the distance the disk travels beforepure rolling occurs.Correct answer: 1:11296 m.Explanation:

From the perspective of the ground andnoting that the acceleration is constant, thedistance traveled by the disk before the purerolling occurs is,

s =1

2a t2

=1

2 g

R!0

3 g

2

=R2 !2

0

18 g

=(0:187 m)2 (19 rad=s)2

18 (0:0643) (9:8 m=s2)

= 1:11296 m :

Algorithm

hm

cmi = 0:01 m=cm (1)

g = 9:8 m=s2 (2)

r = 18:7 cm10

30

(3)

!0 = 19 rad=s10

20

(4)

= 0:06430:050:2

(5)

R = r hm

cmi (6)

= h18:7i h0:01i

= 0:187 m

hmi = hcmi hm=cmi units

t =R!0

3:0 g(7)

=h0:187i h19i

3:0 h0:0643i h9:8i= 1:87948 s

hsi =hmi hrad=si

hi hi hm=s2iunits

s =R2:0 !2:0

0

18:0 g(8)

=h0:187i2:0 h19i2:0

18:0 h0:0643i h9:8i= 1:11296 m

hmi =hmi2:0 hrad=si2:0

hi hi hm=s2iunits

009


M1

l1 l2

m

M2

l1 l2

m

A rod of negligible mass is pivoted at a pointthat is o-center, so that length `1 is diferentfrom length `2. the gures above show twocases in which masses are suspended from theends of the rod. In each case the unknownmass m is balanced by a known mass, M1

or M2, so that the rod remains horizontal.What is the value of m in terms of the knownmasses?

1.M1 +M2

2.M1 +M2

2

3.M1M2

4.pM1M2 correct

5.M1M2

2Explanation:

The balance in the rst case requiresm`1 =M1`2. And the balance in the second caserequires M2`1 = m`2. Cancel `1 and `2 fromthe above equations. So m2 = M1M2, i.e.

m =pM1M2.

010

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 angularmomentum of the object with respect to anaxis perpendicular to the circle and throughits center?

1. 24kg m2

scorrect

2. 12m2

s

3. 13:5kg m2

s2

4. 18N m

kg

5. 9N m

kgExplanation:

The angular momentum is

L = mvr

= (2 kg)(3m=s)(4m)

= 24kg m2

s:

Mass on Solid Cylinder

11:01, calculus, numeric, > 1 min.011

A 7:9 kg mass is attached to a light cord,which is wound around a pulley. The pulleyis a uniform solid cylinder of radius 11:6 cmand mass 1:89 kg. What is the resultant nettorque on the system about the center of thewheel?Correct answer: 8:98072 kg m2=s2.Explanation:

The net torque on the system is the torqueby the external force, which is the weight ofthe mass. So it is given by:

= r F sin = r mg sin 90 = rmg

= 0:116 m 7:9 kg 9:8 m=s2

= 8:98072 kg m2=s2:

012

When the falling mass has a speed of 5:47m=s,the pulley has an angular velocity of v=r.Determine the total angular momentum ofthe system about the center of the wheel.Correct answer: 5:61233 kgm2=s.Explanation:

The total angular momentum has twoparts, one of the pulley and one of the mass.


So it is

j~Lj = j~rm~v + I ~!j

= r mv +1

2M r2

vr

= r

m+

M

2

v

= 11:6 cm

7:9 kg +

1:89 kg

2

v = 1:02602 kgm v

= 1:02602 kgm 5:47 m=s = 5:61233 kgm2=s :

Mass on Solid Cylinder

11:02, calculus, multiple choice, < 1 min.013

Using the fact that = dL=dt and your resultfrom the previous part, calculate the acceler-ation of the falling mass.Correct answer: 8:75297 m=s2.Explanation:

Use the torque-angular momentum rela-tion, we have

=dL

dt=

d

dt(1:02602 kgm v) = 1:02602 kgma ;

Solving for acceleration:

a =

1:02602 kgm

=8:98072 kg m2=s2

1:02602 kgm

= 8:75297 m=s2:

Algorithm

hm

cmi = 0:01 m=cm (1)

m = 7:9 kg4

8

(2)

r = 11:6 cm

8

15

(3)

ru = rhm

cmi (4)

= h11:6ih0:01i

= 0:116 m

hmi = hcmihm=cmi units

M = 1:89 kg1

3

(5)

g = 9:8 m=s2 (6)

= rumg (7)

= h0:116ih7:9ih9:8i

= 8:98072 kg m2=s2

hkg m2=s2i = hmihkgihm=s2i units

v = 5:47 m=s

5

10

(8)

I =1:0Mr2:0u

2:0(9)

=1:0h1:89ih0:116i2:0

2:0= 0:0127159 kgm2

hkgm2i =hihkgihmi2:0

hiunits

! =v

ru(10)

=h5:47i

h0:116i

= 47:1552 s1

hs1i =hm=si

hmiunits

L1 = rumv (11)

= h0:116ih7:9ih5:47i

= 5:01271 kgm2=s

hkgm2=si = hmihkgihm=si units

L2 = I! (12)

= h0:0127159ih47:1552i

= 0:599621 kgm2=s

hkgm2=si = hkgm2ihs1i units

L = L1 + L2 (13)

= h5:01271i+ h0:599621i

= 5:61233 kgm2=s

hkgm2=si = hkgm2=si+ hkgm2=siunits

b =L

v(14)

=h5:61233i

h5:47i

= 1:02602 kgm

hkgmi =hkgm2=si

hm=siunits

a =v

L(15)

=h8:98072ih5:47i

h5:61233i

= 8:75297 m=s2

hm=s2i =hkg m2=s2ihm=si

hkgm2=siunits

Child on a MerryGoRound

11:03, trigonometry, multiple choice, < 1 min.


014

A playground merry-go-round of radius 2:5 mhas a moment of inertia 200 kgm2 and isrotating at 9:5 rev=min. A child with mass21 kg jumps on the edge of the merry-go-round.What is the new moment of inertia of the

merry-go-round and child, together?Correct answer: 331:25 kgm2.Explanation:

The moment of inertia will be the combi-nation of the individual moments of inertia ofeach component.

Imerrygoround + Ichild = Itotal

Child on a MerryGoRound


Assuming that the boy's initial speed is neg-ligible, what is the new angular speed of themerry-go-round?Correct answer: 5:73585 rev=min.Explanation:

Basic Concepts:

X~L = const

The net angular momentum of the system re-mains constant, therefore, from conservationof the angular momentum we have:

I1!1 = (I1 +mR2)!2

And

!2 = !1

I1I1 +mR2

=(9:5 rev=min) (200 kgm2)

(200 kgm2) + (21 kg) (2:5 m)2

= 5:73585 rev=min

Algorithm

R = 2:5 m1:52:5

(1)

I1 = 200 kgm2100

300

(2)

!1 = 9:5 rev=min

5

15

(3)

m = 21 kg20

35

(4)

I2 = I1 +m R2:0 (5)

= h200i+ h21i h2:5i2:0

= 331:25 kgm2

hkgm2i = hkgm2i+ hkgi hmi2:0 units

!2 =!1 I1I2

(6)

=h9:5i h200i

h331:25i

= 5:73585 rev=min

hrev=mini =hrev=mini hkgm2i

hkgm2iunits

018

A bicycle wheel of mass m rotating at anangular velocity ! has its shaft supportedon one side, as shown in the gure. Whenviewing from the left, one sees that the wheelis rotating in a counterclockwise manner. Thedistance from the center of the wheel to thepivot point is b. We assume the wheel is ahoop of radius R, and the shaft is horizontal.

b

τ

W=mg

ω

The magnitude of the angular momentumof the wheel is given by

1.1

4mR2 !

2. mR2 !2

3.1

2mR2 !2

4.1

4mR2 !2


5.1

2mR2 !

6. mR2 ! correct

Explanation:

Solution: Basic Concepts:

~ =d~L

dt

Top view

L

τL+∆L

∆L∆φ

The magnitude of the angular momentum ofthe wheel, L, is

L = I ! = mR2 !;

since the moment of inertia of the wheel, I, ismR2.

019

Given: the mass 3 kg, the angular velocity15 rad=s, the axil length b = 0:5 m, andthe radius of the wheel R = 0:49 m. Find theprecession angle in the time interval t = 1:1 s.Correct answer: 85:7488 .Explanation:

From the gure below, we get =L

L.

Using the relation, L = t, where is themagnitude of the torque, mg b, we get

=L

L

= t

L

=mg bt

mR2 !

=g bt

R2 !

=(9:8 m=s2)(0:5 m)(1:1 s)

(0:49 m)2(15 rad=s)

= 1:4966 rad

= 85:7488:

Precession


The direction of precession as viewed from thetop is:

1. along the direction of rotation of thewheel

2. counterclockwise correct

3. clockwise

4. opposite to the direction of rotation of thewheelExplanation:

From the gure, we can see the direction ofprecession is counterclockwise.Algorithm

hdegradi = 57:2958 deg=rad (1)

g = 9:8 m=s2 (2)

m = 3:0 kg (3)

! = 15 rad=s10

15

(4)

b = 0:5 m0:40:6

(5)

R = 0:49 m0:40:6

(6)

t = 1:1 s1

2

(7)

=gbt

R2:0!(8)

=h9:8ih0:5ih1:1i

h0:49i2:0h15i

= 1:4966 rad

hradi =hm=s2ihmihsi

hmi2:0hrad=siunits

deg = hdegradi (9)

= h1:4966ih57:2958i

= 85:7488

hi = hradihdeg=radi units

Version 001 – circular and gravitation – holland – (2383) 1


AP B 1993 MC 57001 10.0 points

Two objects of masses 14 kg and 35 kg arehung from the ends of a stick that is 70 cmlong and has marks every 10 cm, as shown.

14 kg 35 kg

ABCDEFG

10 20 30 40 50 60

If the mass of the stick is negligible, atwhich of the points indicated should a cord beattached if the stick is to remain horizontalwhen suspended from the cord?

1. F

2. A

3. B

4. C

5. E

6. G correct

7. D

Explanation:

Let : ℓ = 70 cm ,

m1 = 14 kg , and

m2 = 35 kg .

For static equilibrium, τnet = 0.Let x be the distance from the left end of

the stick to the point of attachment of thecord:

∑

T = m1 g x−m2 g (ℓ− x) = 0

(m1 +m2) x = m2 ℓ

x =m2 ℓ

m1 +m2

=(35 kg) (70 cm)

14 kg + 35 kg

= 50 cm .

Therefore the point should be point G .

AP B 1998 MC 7002 10.0 points

Three forces act on an object.If the object is in translational equilibrium,

which of the following must be true?I. The vector sum of the three forces must

equal zero;II. The magnitude of the three forces must

be equal;III. The three forces must be parallel.

1. I only correct

2. I, II and III

3. II only

4. II and III only

5. I and III only

Explanation:If an object is in translational equilibrium,

the vector sum of all forces acting on it mustequal zero.

AP M 1993 MC 35 A003 10.0 points

A rod of negligible mass is pivoted at a pointthat is off-center, so that length ℓ1 is differentfrom length ℓ2. The figures show two cases inwhich masses are suspended from the ends ofthe rod. In each case the unknown mass m isbalanced by a known mass M1 or M2 so thatthe rod remains horizontal.

m M1

ℓ1 ℓ2


mM2

ℓ1 ℓ2

What is the value of m in terms of theknown masses?

1. m =M1 +M2

2

2. m = M1M2

3. m =√

M1M2 correct

4. m = M1 +M2

5. m =M1M2

2

Explanation:

Applying∑

τ = 0 to balance the masses

in both cases,

mℓ1 = M1 ℓ2 and

M2 ℓ1 = mℓ2 .

Dividing,

m

M2

=M1

m

m2 = M1M2

m =√

M1M2 .

AP M 1998 MC 30004 10.0 points

Consider the wheel-and-axle system shownbelow.

a

b

m1 m2

Which of the following expresses the con-dition required for the system to be in staticequilibrium?

1. b2m1 = a2m2

2. am2 = bm1

3. a2m1 = b2m2

4. am1 = bm2 correct

5. m1 = m2

Explanation:In equilibrium, the total torque is zero,

which gives

am1 = bm2 .

AP B 1993 MC 8005 10.0 points

Two spheres have equal densities and are sub-ject only to their mutual gravitational attrac-tion.

Which quantity must have the same mag-nitude for both spheres?

1. displacement from the center of mass

2. gravitational force correct

3. acceleration

4. velocity

5. kinetic energy

Explanation:Two spheres with the same density have

different masses due to their relative sizes.Using Newton’s third law, ~F1 = − ~F2.All of the other quantities (acceleration, ve-

locity, kinetic energy, and displacement fromthe center of mass) have different magnitudesbecause the two spheres have different masses.


AP M 1998 MC 7 8006 (part 1 of 2) 10.0 points

A ball is tossed straight up from the surfaceof a small, spherical asteroid with no atmo-sphere. The ball rises to a height equal to theasteroid’s radius and then falls straight downtoward the surface of the asteroid.What forces, if any, act on the ball while it

is on the way up?

1. Only a decreasing gravitational force thatacts downward correct

2. Only a constant gravitational force thatacts downward

3. Both a constant gravitational force thatacts downward and a decreasing force thatacts upward

4. Only an increasing gravitational forcethat acts downward

5. No forces act on the ball.

Explanation:There is no friction in the system, and the

ball doesn’t have any contact with other ob-jects, so the only force acting on the ball isthe attractive gravitational force, which actsdownward.

From ~F = −GM m

r2r, the force will de-

crease as the ball rises.

007 (part 2 of 2) 10.0 pointsThe acceleration of the ball at the top of itspath is

1. zero.

2. equal to one-fourth the acceleration at thesurface of the asteroid. correct

3. equal to one-half the acceleration at thesurface of the asteroid.

4. equal to the acceleration at the surface ofthe asteroid.

5. at its maximum value for the ball’sflight.

Explanation:

F = ma ∝ 1

r2, so a ∝ 1

r2and

a′ ∝ 1

(2 r)2=

1

4

1

r2∝ 1

4a .

Weight of Spacecraft in Space008 (part 1 of 2) 10.0 points

The radius of Earth is about 6440 km. A7770 N spacecraft travels away from Earth.What is the weight of the spacecraft at a

height 6440 km above Earth’s surface?

Correct answer: 1942.5 N.

Explanation:

Let : rE = 6440 km ,

W = 7770 N , and

h = 6440 km .

By Newton’s Universal Law of Gravitation

W ∝ 1

r2, so

Wh

W =

1

r2h

1

r2

=r2

r2h

Wh = W r2

r2h

= W r2E(rE + h)2

= (7770 N)(6440 km)2

(6440 km + 6440 km)2

= 1942.5 N .

009 (part 2 of 2) 10.0 pointsWhat is the weight 53400 km above Earth’ssurface?



Explanation:

Let : h = 53400 km .

W2 = W r2

r2h

= W r2E(rE + h)2

= (7770 N)(6440 km)2

(6440 km + 53400 km)2

= 89.9932 N .

AP B 1993 MC 1010 10.0 points

Consider the following situations.A) An object moves in a straight line at con-

stant speed.B) An object moves with uniform circular

motion.C) An object travels as a projectile in a

gravitational field with negligible air re-sistance.

In which of the situations would the objectbe accelerated?

1. C only

2. A only

3. B and C only correct

4. B only

5. A and C only

6. None exhibits acceleration.

7. A and B only

8. All exhibit acceleration.

Explanation:A) The velocity of the object (its direction

and magnitude) is unchanged, so it is notaccelerated.B) The direction of the velocity constantly

changes; the centripetal acceleration is di-rected toward the center of the motion.

C) The projectile undergoes gratitationalacceleration.

AP B 1993 MC 6011 10.0 points

If Spacecraft X has twice the mass of Space-craft Y , then what is true about X and Y ?I) On Earth, X experiences twice the grav-

itational force that Y experiences;II) On the Moon, X has twice the weight of

Y ;III) When both are in the same circular orbit,

X has twice the centripetal accelerationof Y .

1. III only

2. I only

3. I and II only correct

4. II and III only

5. I, II, and III

Explanation:I) gravitational force ∝ mass.II) weight ∝ mass.III) The centripetal acceleration is deter-

mined by

ac =v2

r,

so X and Y should have the same centripetalacceleration when they are in the same circu-lar orbit.

AP M 1998 MC 14 15012 (part 1 of 2) 10.0 points

A spring has a force constant of 593 N/m andan unstretched length of 6 cm. One end isattached to a post that is free to rotate in thecenter of a smooth table, as shown in the topview below. The other end is attached to a4 kg disk moving in uniform circular motionon the table, which stretches the spring by4 cm.Note: Friction is negligible.


593 N/m4 kg

10 cm

What is the centripetal force Fc on the disk?


Explanation:

Let : r = 6 cm = 0.06 m ,

∆r = 4 cm = 0.04 m ,

m = 4 kg , and

k = 593 N/m .

The centripetal force is supplied only bythe spring. Given the force constant and theextension of the spring, we can calculate theforce as

Fc = k∆r

= (593 N/m) (0.04 m)

= 23.72 N .

013 (part 2 of 2) 10.0 pointsWhat is the work done on the disk by thespring during one full circle?

1. W = 0 J correct

2. W = 0.00512352 J

3. W = 4.47111 J

4. W = 1.4232 J

5. W = 8.94223 J

Explanation:Since the force is always perpendicular to

the movement of the disk, the work done by

the spring is zero .

AP B 1993 MC 48014 10.0 points

The planet Krypton has a mass of8.5× 1023 kg and radius of 4× 106 m.What is the acceleration of an object in free

fall near the surface of Krypton? The gravita-tional constant is 6.6726× 10−11 N ·m2/kg2.

Correct answer: 3.54482 m/s2.

Explanation:

Let : M = 8.5× 1023 kg ,

R = 4× 106 m , and

G = 6.6726× 10−11 N ·m2/kg2 .

Near the surface of Krypton, the gravita-tion force on an object of mass m is

F = GM m

R2,

so the acceleration a of a free-fall object is

a = gKrypton =F

m

= GM

R2

= (6.6726× 10−11 N ·m2/kg2)

× 8.5× 1023 kg

(4× 106 m)2

= 3.54482 m/s2 .

AP B 1998 MC 39015 10.0 points

An object has a weight W when it is on thesurface of a planet of radius R.What will be the gravitational force on the

object after it has been moved to a distanceof 4R from the center of the planet?

1. F = 4W

2. F =1

4W

3. F = W


4. F =1

16W correct

5. F = 16WExplanation:On the surface of the planet,

W =GM m

R2.

When the object is moved to a distance 4Rfrom the center of the planet, the gravitationalforce on it will be

F =GM m

(4R)2

=GM m

16R2

=1

16

GM m

R2

=1

16W .

AP M 1993 MC 22016 10.0 points

A newly discovered planet has twice the massof the Earth, but the acceleration due to grav-ity on the new planet’s surface is exactly thesame as the acceleration due to gravity on theEarth’s surface.What is the radius Rp of the new planet in

terms of the radius R of Earth?

1. Rp =1

2R

2. Rp = 4 R

3. Rp = 2 R

4. Rp =

√2

2R

5. Rp =√2 R correct

Explanation:From Newton’s second law and the law of

universal gravitation, the gravitational forcenear the surface is

Fg = mg = GM m

r2

g =GM

r2.

Mp = 2Me and gp = ge, so

GMe

R2=

GMp

R2p

=2GMe

R2p

1

R2=

2

R2p

Rp =√2 R .

Gravity on Planet X short017 10.0 points

Planet X has a mass 4.6 times that of theEarth and a radius 2.51 times the radius ofthe Earth.What is the ratio of the acceleration due

to gravity on the surface of Planet X to theacceleration due to gravity on the surface ofthe Earth?

Correct answer: 0.730147.

Explanation:

Let : MX = 4.6ME and

RX = 2.51RE .

The acceleration due to gravity is

a =GM

R2∝ M

R2, so

gXgE

=R2

E

RX2

MX

ME

gXgE

=

(

RE

2.51RE

)2(

4.6ME

ME

)

= 0.730147 .

Gravity on Ceres018 (part 1 of 2) 10.0 points

The asteroid Ceres has a mass 6.601× 1020 kgand a radius of 476.9 km.What is g on the surface? The value

of the universal gravitational constant is6.67259× 10−11 N ·m2/kg2.


Correct answer: 0.193664 m/s2.

Explanation:

Let : M = 6.601× 1020 kg ,

r = 476.9 km , and

G = 6.67259× 10−11 N ·m2/kg2 .

The weight on the surface for an object ofmass M is

W = mg = GM m

r2

g = GM

r2

= (6.67259× 10−11 N ·m2/kg2)

× 6.601× 1020 kg

(476.9 km)2

(

1 km

1000 m

)2

= 0.193664 m/s2 .

019 (part 2 of 2) 10.0 pointsHow much would an 81.9 kg astronaut weighon this asteroid?


Explanation:The weight of the astronaut will be

W = mg

= (81.9 kg) (0.193664 m/s2)

= 15.8611 N .

∆𝑆 = 𝑆𝑓 − 𝑆𝑖

= 𝑘𝑏 ln14+ 9 !

14! 9!− 𝑘𝑏 ln

14+ 6 !

14! 6!

= 𝑘𝑏 ln

14 + 9 !14! 9!14 + 6 !14! 6!

= 𝑘𝑏 ln23!

14!9!×

14! 6!

20!

= 𝑘𝑏 ln23 ∙ 22 ∙ 21 ∙ 20 ∙ 19 ∙ 18 ∙ 17 ∙ 26 ∙ 15 ∙ 14 ∙ 13 ∙ 𝑒𝑡𝑐

(14 ∙ 13 ∙ 𝑒𝑡𝑐) 9!

×14 ∙ 13 ∙ 𝑒𝑡𝑐 6!

20 ∙ 19 ∙ 18 ∙ 17 ∙ 16 ∙ 15 ∙ 14 ∙ 13 ∙ 𝑒𝑡𝑐

= 𝑘𝑏 ln23 ∙ 22 ∙ 21 ∙ 20 ∙ 19 ∙ 18 ∙ 17 ∙ 16 ∙ 15 (6!)

20 ∙ 19 ∙ 18 ∙ 17 ∙ 16 ∙ 15 (9 ∙ 8 ∙ 7 ∙ 6)

= 𝑘𝑏 ln23 ∙ 22 ∙ 21

9 ∙ 8 ∙ 7

= 𝑘𝑏 ln23 ∙ 11 ∙ 3

9 ∙ 4 ∙ 1

= 𝑘𝑏 ln23 ∙ 11 ∙ 1

1 ∙ 4 ∙ 3

= 𝑘𝑏 ln23 ∙ 11

4 ∙ 3

= 𝑘𝑏 ln23 ∙ 11

12

𝜔𝑓

𝜔0=

3

5

𝐾𝑓

𝐾𝑖=

𝜔𝑓2

𝜔02

𝐾𝑓

𝐾𝑖=

3

5

2

𝐾𝑓

𝐾𝑖=

9

25

l

00

𝑟 × 𝐹 = 𝑙 𝑗 × − 𝑖𝐹1 − 𝑗𝐹2 + 𝑘𝐹3

𝑟 × 𝐹 =

𝑖 𝑗 𝑘𝑟𝑥 𝑟𝑦 𝑟𝑧𝐹𝑥 𝐹𝑦 𝐹𝑧

𝑟 × 𝐹 = 𝑖 𝑗 𝑘

0 𝑙 0−𝐹1 −𝐹2 𝐹3

𝑟 × 𝐹 = 𝑙𝐹3 𝑖 + 𝑙𝐹1 𝑘

𝑟 × 𝐹 = 𝑙(𝐹3 𝑖 + 𝐹1 𝑘)

p

r

𝑟 × 𝐹 = 𝑟𝑠𝑖𝑛𝜃 𝐹

𝑟 × 𝐹 = 𝑟⊥𝐹

002 10.0 points A ball of putty with mass m falls vertically onto the outer rim of a hori zontal turntable of radius R and moment of inertia Io that is rotating freely with angular sp eed w i about its vertical axis.

What is the post -collision angular speed of the turntable plus putty?

Wi 1. Wf = 2

l +m R Io

E xplanation: The final rotational inertia of t he turntable

plus-putty is

ince tlwr<' is no <'xt<'rna l torque on the sy5t<'m of the putty plus the turntable, we know L1 = L; = Iow;.

f JWJ = lowo low,

WJ= -f J

lowi w I - -=--- -=.., - lo + m R2

w, m R2 ·

1+ - Io

003 10.0 points A uniform bar of m ass M and length f is

propped against a very slick vertical wall as shown. The angle b etween the wall and the upper end of the bar is B. The force of static friction between the upper end of the bar and the wall is negligible, but the bar remains at rest (in equilibrium).

If we take the pivot point to be at the middle of the rod, which expression below is z-component of the net torque, wh ere p ositive x is along the floor to the right and positive y is upward along the wall?

R 5. 2 (Fw cos B + f s cos B - n sinB)

E xpla nat ion : Taking the piYot at the center of rhe hnr.

the torque d ue to Frr is ( i/ 2) F rr. cos fJ. t he torque d ne to f., is ( I / 2) f, co,; fJ . and the torque due to 11 is (l/ 2)n sin 0.

004 10.0 points A spool floating in space has radius r mass m and moment of inertia ab out its center I = (3m r 2 The spool is unwound by a constant force F .

F

6 If initially the spool is motionless, at some

later time what is the ratio of translational kinetic energy t o r otational kinetic energy?

1. K trans = (3 K rot

Expla nation: From the moment um principle

F a=

m

and from the angular momentum principle

r F a: = [ ·

From kinematics for any timet

and

F Vtrans = - t

m

rF Wrot =It.

Now equation (l ) implies

K - l 2 - 1 (p)2 2 trans - 2 mvtrans - 2m m t ,

while equation (2) gives

_ 1 2 _ 1 (r F) 2 2

K rot - 2 I Wtrans - 2 I I t ,

(1)

(2)

(3)

( 4).

Finally, dividing equation (3) by equation ( 4),

I< t rans = I /r2 = 3

K rot m ·

005 10.0 points A sticky ball collides with a rod exactly at the point where the rod can rotat e on a frictionless axle as shown. The axle is permanently fixed in place and the rod is initially at rest.

0---~

Consider the system of the putty and the rod, ~d let E be the kinetic energy of the system, ..f the linear momentum of the system, and L the angular momentum of the syst em. List which of these quantiti es are conserved , first for a point ab out the axle and second for a point about the center of the rod.

6. axle: L; center: n one

Explanation: First , notice the rod doesn 't r otate or t rans

late. The mechanical energy of the system is not conserved because initially there was kinetic energy and finally t here was not; mechanical became internal energy or was dissipated as heat. Linear momentum is not conserved since there was motion before but was none aft er ; a force was exerted on the syst em by the axle. Finally, for the point ab out the axle angular momentum is conserved since init ially it was zero and finally it was zero; the force from the axle acted through the p oint of rotation and applied no torque. And for the point about the center, angular momentum is not conserved because it was first nonzero and then lat er was zero; her e the force due t o the axle did apply a t orque. Hence Lis conserved about the axle and nothing is conserved about the center .

006 10.0 points A small disk with radius a, is coaxial with a large disk of radius of b. Three for ces of magnitudes H, F2, and Fs act tangentially on the small and large disks as shown. The force of m agnitude F2 acts B b elow the horizontal.

F l

Find the m agnitude of the n et t orque on the syst em with r espect t o the center of the axle.

10. T = laF2 - bH- bFsl

Expla nation:

T he angle doesn't matter because t he radial dj p lacement h·om t he center of the circle is perpendicular to any tangent force. And the torque h·om F2 acting at a distan ce a is opposite the direction of the torques h·om F1 and F3 acting at a distance b.

The magnitude of the total torque is

T = laF2 - bFi -b F3I

E=qħω

PHY 303K - Florin - Exam IV - Fall 2013

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Transcript of PHY 303K - Florin - Exam IV - Fall 2013