Version 058 – 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 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 di!erent ways are there toarrange the total energy E = E1 +E2 so thatthere is E1 in the first object and E2 in theother?

1. 25

2. 15

3. cannot be determined

4. 10

5. 150 correct

6. 225

7. 1! 1015

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."total = "1"2

002 10.0 pointsA length of light nylon cord is wound arounda uniform cylinder of radius 0.375 m and mass0.537 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 3.09 N.How fast will the spool be rotating after the

string has been pulled for #t = 3.67 s?The moment of inertia of the cylinder is

I =1

2mr2.

1. 108.1652. 15.0845

3. 96.3724. 47.89415. 61.32976. 28.17817. 112.6298. 14.50479. 16.696310. 44.2199

Correct answer: 112.629 rad/s.

Explanation:

Let : F = 3.09 N ,

m = 0.537 kg ,

r = 0.375 m , and

t = 3.67 s .

The angular momentum after 3.67 s is

L = !#t

= Fr ·#t

= (3.09 N)(0.375 m)(3.67 s)

= 4.25261 kg ·m2/s .

The moment of inertia of the cylinder is

" =

!

1

2mr2

"

=1

2(0.537 kg) (0.375 m)2

= 0.0377578 kg ·m2 .

Therefore, the rotational velocity will be

" = L/I

= (4.25261 kg ·m2/s)/(0.0377578 kg ·m2)

= 112.629 rad/s .

003 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.

Version 058 – Test 4 – florin – (57850) 2

What is conserved?

1. #P only

2. #L, #P , and E

3. #L and E

4. #L only

5. #P and E

6. #L and #P correct

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.

004 10.0 pointsA force

#F = ("4̂ı+ 4$̂) N

acts at a distance from the point of rotationgiven by

#r = (8̂ı+ 3$̂) m,

where ı̂ 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 k̂ is a unit vectorpointing in the +z direction.)

1. (8̂ı+ 40k̂) N ·m

2. (8̂ı" 40k̂) N ·m

3. ("44$̂+ 40k̂) N ·m

4. (40̂ı+ 44$̂) N ·m

5. (20̂ı) N ·m

6. (20k̂) N ·m

7. (40̂ı" 20$̂) N ·m

8. (44k̂) N ·m correct

9. (44$̂) N ·m

10. ("44k̂) 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 = 44N ·m.

So the resultant vector is (44k̂) N ·m .

005 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. 2/7mg

3. 5/4mg

4. 2/3mg

5. mg

6. 7/4mg

7. 7/2mg correct

8. 3mg

9. 3/2mg

10. 4/5mg

Explanation:

Version 058 – Test 4 – florin – (57850) 3

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:

#

!L = "3

4L(4mg)"

1

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

Therefore TR = 7/2mg.

006 10.0 pointsTwo rigid rods of length % and mass M arerigidly attached as shown

pivot

"

What is the magnitude of the angular mo-mentum L of this system when it is rotating atan angular velocity " 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 %2, while the

rotational inertia about an axis through the

center of mass is ICM =1

12M %2.

1. L =1

3M %2 ".

2. L =9

12M %2 ".

3. L =17

12M %2 ". correct

4. L =11

12M %2 ".

5. L =2

3M %2 ".

6. L = M %2 ".

7. L =5

12M %2 ".

8. L =5

4M %2 ".

9. L =13

12M %2 ".

10. L =11

3M %2 ".

Explanation:Using the parallel axis theorem,

I =1

3M %2 +

1

12M %2 +M %2

=

!

4 + 1 + 12

12

"

M %2 =17

12M %2 .

Therefore, the angular momentum is

L = I" =17

12M %2 "

007 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.

P

W1

W2

%

d

T

&h

b

F

The weight of the ladder is W1 . Jill, with

a weight W2, is one-fourth the way

!

d =%

4

"

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

1.h

2W2 + hW1 = F b

2. (W1 +W2)b

2= F h

Version 058 – Test 4 – florin – (57850) 4

3.b

2W2 + bW1 = F h

4.h

4W2 +

h

2W1 = F b

5. (W1 +W2)h

2= F b

6.b

4W2 +

b

2W1 = F h correct

Explanation:

Pivot

F

T

Nf

W2

W1 &

Applying rotational equilibrium,

#

!P= W2 d cos &+W1

%

2cos &"F % sin & = 0 ,

where d is the distance of the person from the

bottom of the ladder, sin & =h

%and cos & =

b

%.

2F % sin & = 2W2 d cos & +W1 % cos &

2F h = 2W2d b

%+W1 b

= 2W2b

4+W1 b

F h =b

4W2 +

b

2W1

008 10.0 pointsA flywheel with a very low friction bearingtakes 2.3 h to stop after the motor poweris turned o!. The flywheel was originallyrotating at " = 5.96903 rad/s.Assuming the angular deceleration is con-

stant, how many revolutions does the flywheel

make before it stops? Remember that thenumber of revolutions is given by the #&/2'.1. 1617.02. 2700.03. 2277.04. 1674.05. 2046.06. 3240.07. 3933.08. 3000.09. 3384.010. 3420.0

Correct answer: 3933 rev.

Explanation:

Let : t = 2.3 h .

The angular acceleration of the flywheel is

( =""0

t,

so the angle through which it rotates beforestopping is given by

#& = "0 t+1

2( t2 = "0 t+

1

2

!

""0

t

"

t2

=1

2"0 t

=1

2(5.96903 rad/s) (2.3 h)

3600 s

1 h

1 rev

2 '

= 3933 rev .

009 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 rotational

speed?

1. Harry’s tires

2. It depends on the angular acceleration ofthe wheels.

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

Version 058 – Test 4 – florin – (57850) 5

4. It depends on which tires have moremass.

5. It depends on the center of mass speed.

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

7. The rotational speeds are the same.

8. Beth’s tires correct

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

a larger rotational speed to obtain the samelinear speed.

010 10.0 pointsA 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 & to the rod creating a torque #!about the pivot.

L

m

F

&

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 #! about the pivot?

1. d = L/2

2. d = L sin & correct

3. d = L cos &

4. d =#5 L

5. d =#2 L

6. d = L/ cos &

7. d = L tan &

8. d = L

9. d = L/ tan &

10. d = L/ sin &

Explanation:The force generates a torque of

! = F L sin & ,

so the distance is L sin & .

011 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 =1

2v1

2. v2 = v1

3. v2 =1

2#2v1

4. v2 =1

4v1

5. v2 =#2 v1

6. v2 = 2#2 v1

7. v2 =1#2v1

8. v2 = 4 v1

Version 058 – Test 4 – florin – (57850) 6

9. v2 = 0

10. v2 = 2 v1 correct

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 .

012 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 =49 m/s and v2 = 87 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

m

m

v1

v2

What is the magnitude of the rotationalangular momentum Lrot of the system?

1. 80.06252. 85.31253. 105.04. 22.31255. 78.756. 49.8757. 53.81258. 101.0629. 81.37510. 51.1875

Correct answer: 49.875 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,d

2, 0% ! $mv1, 0, 0%

+ $0,"d

2, 0% ! $mv2, 0, 0%

= $0, 0, (0.625 m)[(2.1 kg)(87 m/s)" (2.1 kg)(49

= $0, 0, 49.875 kgm2/s% .

The magnitude is thus 49.875 kgm2/s .

013 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 "i about its verticalfixed symmetry axis, i.e. the turntable ishorizontal.What is the post-collision angular speed of

the turntable plus putty?

1. "f ="i

3m+mR3

I02. "f =

"i

1 +mR2

I0

correct

Version 058 – Test 4 – florin – (57850) 7

3. "f ="i

1 +mR3

I04. "f =

"i

1 +mR4

I05. "f =

"i

m+mR3

I06. "f =

"i

2 +mR4

I07. "f =

"i

1 +mR

I08. "f =

"i

2 +mR

I09. "f =

"i

2m+mR3

I010. "f =

"i

2 +mR2

I0Explanation: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 = I0"i.

If "f = I0 "0

"f =I0 "i

If

"f =I0 "i

I0 +mR2

="i

1 +mR2

I0

.

014 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.

1.3g

4L

2.2g

L

3.g

L

4.5g

4L

5.3g

2Lcorrect

6.7g

4L

7.g

3L

8.g

4L

9.g

2L

10.5g

6L

Explanation:The equation of motion ! = I( gives:

mgL

2= I( =

1

3mL2(,

So

( =3g

2L.

015 (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 & 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.

Version 058 – Test 4 – florin – (57850) 8

Find the magnitude of the angular momen-tum of the mass m relative to the center ofthe plant.

1. mv0R cos &

2. mv0R sin & +Mv0R sin &

3. Mv0R cos &

4. Mv0R sin &

5. mv0R sin & "Mv0R sin &

6. mv0R cos & "Mv0R cos &

7. mv0R sin & correct

8. 0

Explanation:The magnitude of the angular momentum

vector,

|#Lm| = |#r !#p| = mv0R sin &.

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

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

1. GmM sin &/R2

2. GmM sin &/R

3. GmM cos &/R

4. Need more information

5. 0 correct

6. GmM cos &/R2

7. GmM/R

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

gravitational attraction of the planet,

#! = #r ! #Fgrav.

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

017 10.0 pointsA sizable quantity of soil is washed downthe Mississippi River and deposited in theGulf of Mexico each year. Thus, there is anet movement of mass southward towards theequator.What e!ect does this tend to have on the

length of a day?

1. no change

2. lengthen the day correct

3. Impossible to determine

4. shorten the day

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 oppositee!ect; shortening our days!)

018 10.0 pointsTwo masses of 27 kg and 15 kg are suspendedby a pulley that has a radius of 15 cm anda mass of 9 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.

Version 058 – Test 4 – florin – (57850) 9

h

15 cm

9 kg

(

27 kg

15 kg

Determine the angular acceleration ( of thepulley after the masses are released and beforethey fall o! of the pulley.The acceleration of gravity is 9.8 m/s2 .1. 36.33672. 61.253. 46.89934. 13.19875. 33.72046. 22.96887. 15.32758. 26.07929. 16.860210. 29.7601

Correct answer: 16.8602 rad/s2.

Explanation:

Let : M = 9 kg ,

R = 15 cm ,

m1 = 27 kg , and

m2 = 15 kg

The tension in the cord attached to thefirst mass T1 is di!erent 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

I( = !net= 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 = (R sincethere the cord does not slip.

I( = (m1g "m1(R)R" (m2g +m2(R)R

= +(m1 "m2)gR" (m1R2 +m2R

2)(

( = [(m1 "m2)gR]/(I +m1R2 +m2R

2)

= 16.8602 rad/s2 .

019 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

!

20!

7! · 5!

"

2. #S = kB ln

!

21!

7! · 5!

"

3. #S = kB ln

!

17

6

"

4. #S = kB ln

!

10

7

"

5. #S = kB ln

!

21!

14! · 7!

"

6. #S = kB ln

!

16!

7! · 7

"

7. #S = kB ln (10) correct

8. #S = kB ln

!

20!

14! · 5!

"

Version 058 – Test 4 – florin – (57850) 10

9. #S = kB ln

!

21

7 · 5

"

10. #S = kB ln

!

10!

7!

"

Explanation:Entropy is defined as S = kB ln" where "

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

" =(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)

020 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.

1. the slopes do not change

2. m1 < m2

3. m1 = m2 = 0

4. m1 > m2

5. m1 = m2 = &

6. more information is needed

7. m1 =1

2m2

8. m1 = m2 correct

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.