Cap10

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969 969

Angular Momentum

Conceptual Problems

1

1 •• True or false:True or false:

((aa) ) If If two vectors two vectors are exactly are exactly opposite iopposite in direction, n direction, their vector their vector product must product must bebe zero.

zero.

((bb) ) The magnitude The magnitude of the of the vector product vector product of two of two vectors is vectors is at a at a minimum wminimum whenhen the two vectors are perpendicular.

the two vectors are perpendicular.

((cc) ) Knowing the Knowing the magnitude of magnitude of the vector the vector product of product of two two nonzero vectors nonzero vectors andand the vectors

the vectors

′′

individual magnitudes uniquely determines the angle between individual magnitudes uniquely determines the angle between them.

them.

Determine the Concept

Determine the Concept The vector product of The vector product of

r r and and B B r r is defined to be is defined to be n n B B A

A

××

==

AB ABsinsinφ φ ˆˆ

r r r r

where

where nnˆˆ is a unit vector normal to the plane defined by is a unit vector normal to the plane defined by

r r and and B B r r and

and φ φ is the angle betweenis the angle between

r r and and B B r r .. ((aa) True. If) True. If

r r and and B B r r

are in opposite direction, then sin

are in opposite direction, then sinφ φ = = sin 180sin 180

°°

= 0 = 0.. ((bb) False. If) False. If r r and and B B r r

are perpendicular, then sin

are perpendicular, then sinφ φ = = sin 90sin 90

°°

= 1 and the vector = 1 and the vector product of product of r r and and B B r r is a maximum. is a maximum.

((cc) False. Knowing the magnitude of the vector product and the vectors) False. Knowing the magnitude of the vector product and the vectors

′′

individual magnitudes only gives the magnitude of the sine of the angle between individual magnitudes only gives the magnitude of the sine of the angle between the vectors. It does not determine the angle uniquely, nor does this knowledge tell the vectors. It does not determine the angle uniquely, nor does this knowledge tell us if the sine of the a

us if the sine of the angle is positive or negative.ngle is positive or negative.

2

2 •• Consider Consider two two nonzero nonzero vectorsvectors

r r and and r r B

B. Their vector product has the. Their vector product has the greatest magnitude if greatest magnitude if r r and and r r B

B are (are (aa) parallel, () parallel, (bb) perpendicular, () perpendicular, (cc) antiparallel,) antiparallel, ((d d ) at an angle of 45° to each other.) at an angle of 45° to each other.

Determine the Concept The vector product of the vectorsThe vector product of the vectors

r r and and B B r r is defined to is defined to be

be A A

××

B B

==

AB ABsinφ sinφ nnˆˆ

r r r r

where

where nnˆˆ is a unit vector normal to the plane defined by is a unit vector normal to the plane defined by

r r and and B B r r and

and φ φ is the angle betweenis the angle between

r r and and B B r r

. Hence, the vector product of . Hence, the vector product of

r r and and B B r r

is a maximum when sin

is a maximum when sinφ φ = = 1. This condition is satisfied provided1. This condition is satisfied provided

r r and and B B r r are

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3

3 •• What What is is the the angle angle between between a a force force vectorvector F F and a torque vectorand a torque vector produced by

produced byF F

r r

??

Determine the Concept BecauseBecause τ τ

==

r r

××

F F

==

rF rF sinsinφ φ nnˆˆ

r r r r r r , where

, where nnˆˆ is a unit vector is a unit vector normal to the plane defined by

normal to the plane defined by r r rrandand

r r

, the angle between , the angle between

r r

and

and rr is is 9090

°°

..

4

4 •• A A point point particle particle of of massmass mm is moving with a constant speed is moving with a constant speed vv along aalong a straight line that passes through point

straight line that passes through point P P . What can you say about the angular. What can you say about the angular momentum of the particle relative to po

momentum of the particle relative to pointint P P ? (? (aa) Its magnitude is) Its magnitude is mvmv. (. (bb) Its) Its magnitude is zero. (

magnitude is zero. (cc) Its magnitude changes sign as the particle passes through) Its magnitude changes sign as the particle passes through point

point P P . (. (d d ) It varies in magnitude as the pa) It varies in magnitude as the particle approaches pointrticle approaches point P P ..

Determine the Concept Determine the Concept L L

r r

and

and p prr are related according to are related according to L L r r rr p prr

r r

××

==

and the and the magnitude of

magnitude of L L

r r

is

is L L

=

rprpsinsinφ φ where where φ φ is the angle between is the angle between r r rr and and p prr . Because. Because the motion is along a line that passes through point

the motion is along a line that passes through point P P ,, r r = 0 and so is = 0 and so is L L.. ((bb)) isis correct.

correct.

5

5 •• [SSM][SSM] A A particle particle travels travels in in a a circular circular path path and and pointpoint P P is at theis at the center of the circle. (

center of the circle. (aa) If the particle’s linear momentum) If the particle’s linear momentum p prr is doubled without is doubled without changing the radius of the circle, how is the magnitude of its angular momentum changing the radius of the circle, how is the magnitude of its angular momentum about

about P P affected? (affected? (bb) If the radius of the circle is doubled ) If the radius of the circle is doubled but the speed of thebut the speed of the particle is unchanged, how is the magnitude of its angular momentum about particle is unchanged, how is the magnitude of its angular momentum about P P

affected? affected?

Determine the Concept Determine the Concept L L

r r

and

and p prr are related according to are related according to L L r r rr p prr

r r

××

==

and the and the magnitude of

magnitude of L L

r r

is

is L L

=

rprpsinsinφ φ where where φ φ is the angle between is the angle between r r rr and and p prr .. ((aa) Because) Because L L

r r

is directly proportional to

is directly proportional to p prr ,, L L is doubled. is doubled. ((bb) Because) Because L L

r r

is directly proportional to

is directly proportional tor r rr,, L L is doubled. is doubled.

6

6 •• A particle moves along a A particle moves along a straight line at constant speed. How does itsstraight line at constant speed. How does its angular momentum about any fixed point vary with time?

angular momentum about any fixed point vary with time?

Determine the Concept We can determine how the angular momentum of the We can determine how the angular momentum of the particle about any fixed point varies with time by examining the derivative of the particle about any fixed point varies with time by examining the derivative of the

cross product of

cross product of r r rrandand p prr .. The angular momentum of the The angular momentum of the particle is given by:

particle is given by:

p p r r L L rr rr r r

××

==

(3)

Differentiate Differentiate L L

r r

with respect to time with respect to time to obtain: to obtain:

_⎠

⎟⎟

⎞

⎜⎜

⎝

⎛

_××

++

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎛ ××

==

r r p p r r p p L L rr rr rr rr r r dt dt d d dt dt d d dt dt d d (1) (1) Because

Because p prr

==

mmvvrr,, p p F F _net_net

r r r r

==

dt dt d d , and , and vv r r rr rr

==

dt dt d d ::

( (

r r F F

))

( (

vv p p

))

L L rr rr rr rr r r

××

++

××

==

_net_net dt dt d d

Because the particle moves along a Because the particle moves along a straight line at constant speed: straight line at constant speed:

0 0 net net

==

F F

⇒

r r

××

F F _net_net

==

00 r r r r Because

Because vvrrandand p prr

( (

==

mmvvrr

))

are parallel: are parallel: vvrr

××

p prr

==

00 Substitute in equation (1) to obtain:

Substitute in equation (1) to obtain:

0 0

==

dt dt d d L L r r

⇒

L L

==

constantconstant r r An alternate solution

An alternate solution The following diagram shows a particle whose mass isThe following diagram shows a particle whose mass is mm moving along a straight line at constant speed. The point identified as

moving along a straight line at constant speed. The point identified as P P is any is any fixed point and the vector

fixed point and the vector r r rr is the position vector, relative to this fixed point, of is the position vector, relative to this fixed point, of the particle moving with constant velocity.

the particle moving with constant velocity.

φ φ φ φ r r d d v v m m P P

From the definition of angular From the definition of angular momentum, the magnitude of the momentum, the magnitude of the angular momentum of the particle, angular momentum of the particle, relative to the fixed point at

relative to the fixed point at P, P, is is given by: given by:

( (

φ φ

))

φ φ sinsin sin sin mvmv r r rmv rmv

==

L L r r Because

Because d d

=

r r sinsinφ φ :: _L_Lrr

₌₌

_mvd_mvd or, because

or, because m, vm, v and and d d are constants, are constants, constant constant

==

L L r r 7 ••

7 •• True or false: If the net torque True or false: If the net torque on a rotating system is zero, the on a rotating system is zero, the angularangular velocity of the system cannot cha

velocity of the system cannot change. If your answer is false, give an nge. If your answer is false, give an example ofexample of such a situation.

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False. The net torque acting on a rotating system equals the change in the False. The net torque acting on a rotating system equals the change in the system’s angular momentum; that is,

system’s angular momentum; that is,τ τ _net_net

==

dLdL dt dt where L where L == I I ω ω . Hence, if. Hence, if τ τ is_net_netis zero, all we can say for sure is that the angular momentum (the product of

zero, all we can say for sure is that the angular momentum (the product of I I andand ω

ω ) ) is is constant. constant. IfIf I I changes, so must changes, so must ω ω .. An example is a high diver going from a An example is a high diver going from a tucked to a layout position.

tucked to a layout position.

8

8 •••• You You are are standing standing on on the the edge edge of of a a turntable turntable with with frictionless frictionless bearingsbearings that is initially rotating when you catch a ball that is moving in the same direction that is initially rotating when you catch a ball that is moving in the same direction but faster than you are moving and on a line tangent to the edge of the turntable. but faster than you are moving and on a line tangent to the edge of the turntable.

Assume you do not move relative to the turntable. (

Assume you do not move relative to the turntable. (aa) Does the angular speed of) Does the angular speed of the turntable increase, decrease, or remain the same during the catch? (

the turntable increase, decrease, or remain the same during the catch? (bb) Does the) Does the magnitude of your angular

magnitude of your angular momentum (about the rotation axis of the table)momentum (about the rotation axis of the table) increase, decrease, or remain the same after the catch? (

increase, decrease, or remain the same after the catch? (cc) How does the ball’s) How does the ball’s angular momentum (about the rotation axis of the table) change after the catch? angular momentum (about the rotation axis of the table) change after the catch? ((d d ) How does the total angular momentum of the system, you-table-ball (about the) How does the total angular momentum of the system, you-table-ball (about the rotation axis of the table) change after the catch?

rotation axis of the table) change after the catch?

Determine the ConceptYou can apply conservation of angular momentum to theYou can apply conservation of angular momentum to the you-table-ball system to answer each of these

you-table-ball system to answer each of these questions.questions.

((aa) Because the ball is moving in the same direction that you are moving, your) Because the ball is moving in the same direction that you are moving, your angular speed will

angular speed will increaseincrease when you catch it. Your Newton’s 3when you catch it. Your Newton’s 3rdrd law law interaction with the ball causes a torque that acts on the you-table-ball system to interaction with the ball causes a torque that acts on the you-table-ball system to increase

increase ω ω ..

((bb) The ball has angular momentum relative to the rotation axis of the table before) The ball has angular momentum relative to the rotation axis of the table before you ca

you catch itch it and st and so catco catchinhing itg it incincreareasesses youyour angur angular mlar momeomentum ntum relrelatiative to tve to thehe rotation axis of the table.

rotation axis of the table.

((cc) The ball will slow down as a result of your catch and so its angular momentum) The ball will slow down as a result of your catch and so its angular momentum relative to the center of the table will

relative to the center of the table will decrease.decrease.

((d d ) Because there is zero net torque on the you-table-ball system, its angular) Because there is zero net torque on the you-table-ball system, its angular m

moommeennttuumm rreemmaaiinnssthethessaammee..

9 ••

9 •• If the angular momentum of a system about a fixed pointIf the angular momentum of a system about a fixed point P P is constant,is constant, which one of the following statements must be true?

which one of the following statements must be true? ((aa) ) No No torque torque aboutabout P P acts on any part of the system.acts on any part of the system.

((bb) ) A A constant constant torque torque aboutabout P P acts on each part of the system.acts on each part of the system. ((cc) ) Zero Zero net net torque torque aboutabout P P acts on each part of acts on each part of the system.the system. ((d d ) ) A A constant constant external external torque torque aboutabout P P acts on the system.acts on the system. ((ee) ) Zero Zero net net external external torque torque aboutabout P P acts on the system.acts on the system.

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Determine the ConceptIfIf L L is constant, we know that the is constant, we know that the net net torque acting on torque acting on the system is zero. There may be multiple constant or time-dependent torques the system is zero. There may be multiple constant or time-dependent torques acting

acting on on the the system system as as long long as as the the net net torque torque is is zero. zero. ((ee)) is correct.is correct.

10 ••

10 •• A A block block sliding sliding on on a a frictionless frictionless table table is is attached attached to to a a string string that that passespasses through a narrow hole through the tabletop. Initially, the block is sliding with through a narrow hole through the tabletop. Initially, the block is sliding with speed

speed vv00 in a circle of radius in a circle of radius r r 00. A student under the table pulls slowly on the. A student under the table pulls slowly on the string. What happens as the block spirals inward? Give supporting arguments for string. What happens as the block spirals inward? Give supporting arguments for your choice. (The term angular momentum refers to the angular momentum about your choice. (The term angular momentum refers to the angular momentum about a vertical axis through the hole.) (

a vertical axis through the hole.) (aa) Its energy and angular momentum are) Its energy and angular momentum are conserved. (

conserved. (bb) Its angular momentum is conserved and its energy increases. () Its angular momentum is conserved and its energy increases. (cc) Its) Its angular momentum is conserved and its energy decreases. (

angular momentum is conserved and its energy decreases. (d d ) Its energy is) Its energy is conserved and its angular momentum increases. (

conserved and its angular momentum increases. (ee) Its energy is conserved and its) Its energy is conserved and its angular momentum decreases.

angular momentum decreases.

Determine the ConceptThe pull that the student exerts on the block is at rightThe pull that the student exerts on the block is at right angles to its motion and exerts no torque (recall that

angles to its motion and exerts no torque (recall that r r F F

r r r r r r

××

==

andand τ τ

==

rF rF sinsinφ φ ).). Therefore, we can conclude that the angular momentum of the block is conserved. Therefore, we can conclude that the angular momentum of the block is conserved. The student does, however, do work in displacing the block in the direction of the The student does, however, do work in displacing the block in the direction of the radial

radial force force and and so so the the block’s block’s energy energy increases. increases. ((bb)) is correct.is correct.

11

11 •••• [SSM][SSM] One One way way to to tell tell if if an an egg egg is is hardboiled hardboiled or or uncooked uncooked withoutwithout breaking the egg is to lay the egg flat on a hard surface and try to spin it. A

breaking the egg is to lay the egg flat on a hard surface and try to spin it. A hardboiled egg will spin easily, while an uncooked egg will not. However, once hardboiled egg will spin easily, while an uncooked egg will not. However, once spinning, the uncooked egg will do something unusual; if you stop it with your spinning, the uncooked egg will do something unusual; if you stop it with your finger, it may start spinning again. Explain the difference in the behavior of the finger, it may start spinning again. Explain the difference in the behavior of the two types of eggs.

two types of eggs.

Determine the Concept The hardboiled egg is solid inside, so everything rotatesThe hardboiled egg is solid inside, so everything rotates with a unif

with a uniform angular orm angular speed. speed. By contrast, By contrast, when you stwhen you start an uncooked art an uncooked eggegg spinning, the yolk will not immediately spin with the shell, and when you stop it spinning, the yolk will not immediately spin with the shell, and when you stop it from spinning the yolk will continue to spin for a while.

from spinning the yolk will continue to spin for a while.

12 ••

12 •• Explain why a hExplain why a helicopter with just one main rotor has a elicopter with just one main rotor has a second smallersecond smaller rotor mounted on a horizontal axis at the rear as in Figure 10-40. Describe the rotor mounted on a horizontal axis at the rear as in Figure 10-40. Describe the resultant motion of the helicopter if this rear rotor fails during flight.

resultant motion of the helicopter if this rear rotor fails during flight.

Determine the ConceptThe purpose of the second smaller rotor is to prevent theThe purpose of the second smaller rotor is to prevent the body

body of of the the helicopter helicopter from from rotating. rotating. If If the the rear rear rotor rotor fails, fails, the the body body of of thethe helicopter will tend to rotate on the main axis due to angular momentum being helicopter will tend to rotate on the main axis due to angular momentum being conserved.

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13 ••

13 •• The The spin spin angular-momentum angular-momentum vector vector for for a a spinning spinning wheel wheel is is parallelparallel with its axle and is pointed east. To cause this vector to rotate toward the south, in with its axle and is pointed east. To cause this vector to rotate toward the south, in which direction must a force be exerted on the east end of the axle? (

which direction must a force be exerted on the east end of the axle? (aa) up,) up, ((bb) down, () down, (cc) north, () north, (d d ) south, () south, (ee) east.) east.

Determine the Concept The vector The vector ΔΔ_L_L _L_L_f_f _L_L_ii

r r r r r r

−−

==

(and the torque that is (and the torque that is responsible for this change in the direction of the angular momentum vector) responsible for this change in the direction of the angular momentum vector) initially points to the south. One can use a right-hand rule to determine the initially points to the south. One can use a right-hand rule to determine the direction of this torque, and hence the force exerted on the east end of the axle, direction of this torque, and hence the force exerted on the east end of the axle, required to cause the angular-momentum vector to rotate toward the south. required to cause the angular-momentum vector to rotate toward the south. Letting the fingers of your right hand point east, rotate your wrist until your Letting the fingers of your right hand point east, rotate your wrist until your thumb points south. Note that your curled fingers, which point in the direction of thumb points south. Note that your curled fingers, which point in the direction of the

the force force that that must must be be exerted exerted on on the the east east end end of of the the axle, axle, point point upward. upward. ((aa)) isis correct.

correct.

14 ••

14 •• You You are are walking walking toward toward the the north north and and in in your your left left hand hand you you areare carrying a suitcase that contains a massive spinning wheel mounted on an axle carrying a suitcase that contains a massive spinning wheel mounted on an axle attached to the front and back of the case. The angular velocity of the gyroscope attached to the front and back of the case. The angular velocity of the gyroscope points north. You now begin to turn to walk toward the east. As a result, the front points north. You now begin to turn to walk toward the east. As a result, the front

end of the suitcase will (

end of the suitcase will (aa) resist your attempt to turn it and will try to maintain its) resist your attempt to turn it and will try to maintain its original orientation, (

original orientation, (bb) resist your attempt to turn and will pull to ) resist your attempt to turn and will pull to the west, (the west, (cc)) rise upward, (

rise upward, (d d ) dip downward, () dip downward, (ee) show no effect whatsoever.) show no effect whatsoever.

Determine the Concept In turning toward the east, you redirect the angular In turning toward the east, you redirect the angular momentum vector from north to east by exerting a torque on the spinning wheel. momentum vector from north to east by exerting a torque on the spinning wheel. The force that you must exert to produce this torque (use a right-hand rule with The force that you must exert to produce this torque (use a right-hand rule with your thumb pointing either east of north and note that your fingers point upward) your thumb pointing either east of north and note that your fingers point upward) is upward. That is, the force you exert on the front end of the suitcase is upward is upward. That is, the force you exert on the front end of the suitcase is upward and the force the suitcase exerts on you is downward. Consequently, the front end and the force the suitcase exerts on you is downward. Consequently, the front end of

of the the suitcase suitcase will will dip dip downward. downward. ((d d is correct.)) is correct.

15 ••

15 •• [SSM][SSM] The The angular momentum angular momentum of of the the propeller propeller of of a sa small mall single- single-engine airplane points forward. The propeller rotates clockwise if viewed from engine airplane points forward. The propeller rotates clockwise if viewed from behind. (

behind. (aa) Just after liftoff, as the nose of the plane ) Just after liftoff, as the nose of the plane tilts upward, the airplanetilts upward, the airplane tends to veer to one side. To which side does it tend to veer and why? (

tends to veer to one side. To which side does it tend to veer and why? (bb) If the) If the plane is flying horizontally and suddenly turns to the right, does the nose of the plane is flying horizontally and suddenly turns to the right, does the nose of the plane tend to veer upward or downward? Why?

plane tend to veer upward or downward? Why?

((aa) The plane tends to veer to the right. The change in angular momentum) The plane tends to veer to the right. The change in angular momentum L L_prop_prop

r r

Δ

for the propeller is upward, so the net torque for the propeller is upward, so the net torque τ τ

r r

on the propeller is upward as well. on the propeller is upward as well. The propeller must exert an equal but opposite torque on the plane. This The propeller must exert an equal but opposite torque on the plane. This downward torque exerted on the plane by the propeller tends to cause a downward downward torque exerted on the plane by the propeller tends to cause a downward

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change in the angular momentum of the plane. This means the plane tends to change in the angular momentum of the plane. This means the plane tends to rotate clockwise as viewed from above.

rotate clockwise as viewed from above.

((bb) The nose of the plane tends to veer downward. The change in angular) The nose of the plane tends to veer downward. The change in angular momentum

momentum L L_prop_prop

r r

Δ

for the propeller is to the right, so the net torque for the propeller is to the right, so the net torque τ τ

r r

on the on the propeller

propeller is is toward toward the the right right as as well. well. The The propeller propeller must must exert exert an an equal equal butbut opposite torque on the plane. This leftward directed torque exerted by the opposite torque on the plane. This leftward directed torque exerted by the propeller

propeller on on the the plane plane tends tends to to cause cause a a leftward-directed leftward-directed change change in in angularangular momentum for the plane. This means the plane tends to rotate clockwise as momentum for the plane. This means the plane tends to rotate clockwise as viewed from the right.

viewed from the right.

16 ••

16 •• You You have have designed designed a a car car that that is is powered powered by by the the energy energy stored stored in in aa single flywheel with a spin angular

single flywheel with a spin angular momentummomentum L L

r r

. In the morning, you plug the . In the morning, you plug the car into an electrical outlet and a motor spins the flywheel up to speed, adding a car into an electrical outlet and a motor spins the flywheel up to speed, adding a huge amount of rotational kinetic energy to it—energy that will be changed into huge amount of rotational kinetic energy to it—energy that will be changed into translational kinetic energy of the car during the day. Having taken a physics translational kinetic energy of the car during the day. Having taken a physics course involving angular momentum and torques, you realize that problems course involving angular momentum and torques, you realize that problems would arise during various maneuvers of the

would arise during various maneuvers of the car. Discuss some of these problems.car. Discuss some of these problems. For example, suppose the flywheel is mounted so

For example, suppose the flywheel is mounted so L L

r r

points vertically upward points vertically upward when the car is on a horizontal road. What would happen as the car travels over a when the car is on a horizontal road. What would happen as the car travels over a hilltop? Through a valley?

hilltop? Through a valley? Suppose the flywheel iSuppose the flywheel is mounted sos mounted so L L

r r

points forward, points forward, or to one side, when the car is on a horizontal road. Then what would happen as or to one side, when the car is on a horizontal road. Then what would happen as the car attempts to turn to the left or right? In each case that you examine,

the car attempts to turn to the left or right? In each case that you examine, consider the direction of the torque exerted on the car by the road.

consider the direction of the torque exerted on the car by the road.

Determine the Concept Determine the Concept IfIf L L

r r

points upward and the car travels over a hill or points upward and the car travels over a hill or through a valley, the force the road exerts on the wheels on one side (or the other) through a valley, the force the road exerts on the wheels on one side (or the other) will increase and car will tend to tip. If

will increase and car will tend to tip. If L L

r r

points forward and the car turns left or points forward and the car turns left or right, the front (or rear) of the car will tend to lift. These problems can be averted right, the front (or rear) of the car will tend to lift. These problems can be averted by

by having having two two identical identical flywheels flywheels that that rotate rotate on on the the same same shaft shaft in in oppositeopposite directions.

directions.

17 ••

17 •• [SSM][SSM] You are You are sitting sitting on a on a spinning spinning piano spiano stool tool with with your your armsarms folded. (

folded. (aa) When you extend your arms out to the side, what happens to your) When you extend your arms out to the side, what happens to your kinetic energy? What is the cause of this change? (

kinetic energy? What is the cause of this change? (bb) Explain what happens to) Explain what happens to your moment of inertia, angular speed and angular momentum as you extend your your moment of inertia, angular speed and angular momentum as you extend your arms.

arms.

Determine the Concept The rotational kinetic energy of the you-stool system isThe rotational kinetic energy of the you-stool system is given given by by .. 2 2 2 2 2 2 2 2 1 1 rot rot I I L L I I K

K

==

ω ω

==

Because the net torque acting on the you-stool system Because the net torque acting on the you-stool system is zero, its angular momentum

is zero, its angular momentum L L

r r

is conserved. is conserved.

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((aa) Your kinetic energy decreases. Increasing your moment of inertia) Your kinetic energy decreases. Increasing your moment of inertia I I whilewhile conserving your angular momentum

conserving your angular momentum L L decreases you kinetic energydecreases you kinetic energy

I I L L K K 2 2 2 2

==

..

((bb) Extending your arms out to the side increases your moment of inertia,) Extending your arms out to the side increases your moment of inertia, decreases your angular speed. The angular momentum of the system is decreases your angular speed. The angular momentum of the system is unchanged.

unchanged.

18 ••

18 •• A A uniform uniform rod rod of of massmass M M and lengthand length L L rests on a horizontal rests on a horizontal frictionless table. A blob of putty of mass

frictionless table. A blob of putty of mass mm = = M M /4 moves along a line/4 moves along a line perpendicular to the rod, strikes the rod near its end, and sticks to the rod. perpendicular to the rod, strikes the rod near its end, and sticks to the rod.

Describe qualitatively the subsequent motion of the rod and putty. Describe qualitatively the subsequent motion of the rod and putty.

Determine the ConceptThe center of mass of the rod-and-putty system moves inThe center of mass of the rod-and-putty system moves in a straight line, and the system rotates abo

a straight line, and the system rotates about its center of mass.ut its center of mass.

Estimation and Approximation

19 ••

19 •• [SSM][SSM] An An ice-skater ice-skater starts starts her her pirouette pirouette with with arms arms outstretched,outstretched, rotating at 1.5 rev/s. Estimate her rotational speed (in

rotating at 1.5 rev/s. Estimate her rotational speed (in revolutions per second)revolutions per second) when she brings her arms flat against her body.

when she brings her arms flat against her body.

Picture the Problem

Picture the Problem BecauseBecause we have no information regarding the mass of thewe have no information regarding the mass of the skater, we’ll assume that her body mass (not including her arms) is 50 kg and that skater, we’ll assume that her body mass (not including her arms) is 50 kg and that each arm has a mass of 4.0 kg. Let’s also assume that her arms are 1.0 m long and each arm has a mass of 4.0 kg. Let’s also assume that her arms are 1.0 m long and that her body is cylindrical with a radius of 20 cm. Because the net external torque that her body is cylindrical with a radius of 20 cm. Because the net external torque acting on her is zero, her angular momentum will remain constant during her acting on her is zero, her angular momentum will remain constant during her pirouette.

pirouette.

Because the net external torque acting Because the net external torque acting on her is zero: on her is zero: 0 0 Δ Δ ii f f

−−

==

L L L L L L or or 0 0 out out arms arms out out arms arms in in arms arms in in arms

arms

−−

I I

==

I

I (1) (1)

Express her total moment of inertia Express her total moment of inertia with her arms outstretched:

with her arms outstretched:

arms arms body body out out arms

arms I I I I

I

==

++

Treating her body as though it is Treating her body as though it is cylindrical, calculate the moment of cylindrical, calculate the moment of inertia of her body, minus her arms: inertia of her body, minus her arms:

(

( ))(

(

))

2 2 2 2 2 2 1 1 2 2 2 2 1 1 body body m m kg kg 00 00 .. 1 1 m m 0.20 0.20 kg kg 50 50

⋅⋅

==

mr mr I I

(9)

Modeling her arms as though they Modeling her arms as though they are rods, calculate their moment of are rods, calculate their moment of inertia when they are outstretched: inertia when they are outstretched:

( )

( )(

(

))

2 2 2 2 3 3 1 1 arms arms m m kg kg 67 67 .. 2 2 m m 1.0 1.0 kg kg 4 4 2 2

⋅⋅

==

I I

Substitute to determine her total Substitute to determine her total moment of inertia with her arms moment of inertia with her arms outstretched: outstretched: 2 2 2 2 2 2 out out ss arm arm m m kg kg 67 67 .. 3 3 m m kg kg 67 67 .. 2 2 m m kg kg 00 00 .. 1 1

⋅⋅

==

⋅⋅

++

⋅⋅

==

I I

Express her total moment of inertia Express her total moment of inertia with her arms flat against her body: with her arms flat against her body:

(

))(

(

))

[ [

]]

2 2 2 2 2 2 arms arms body body in in arms arms m m kg kg 32 32 .. 1 1 m m 0.20 0.20 kg kg 4.0 4.0 2 2 m m kg kg 00 00 .. 1 1

⋅⋅

==

++

⋅⋅

==

++

==

I I I I I

I

Solve equation (1) for

Solve equation (1) for _arms_arms_in_in to to obtain:

obtain: armsarmsinin armsarmsoutout

out out arms arms in in arms arms ω ω ω ω I I I I

==

Substitute numerical values and Substitute numerical values and evaluate

evaluate _arms_arms_in_in::

( (

))

rev/s rev/s 4 4 rev/s rev/s 5 5 .. 1 1 m m kg kg 32 32 .. 1 1 m m kg kg 3.67 3.67 2 2 2 2 in in arms arms

≈≈

⋅⋅

==

ω ω 20 ••

20 •• Estimate the ratio of angular velocities for the rotation Estimate the ratio of angular velocities for the rotation of a diverof a diver between the full tuck position and the full-layout position.

between the full tuck position and the full-layout position.

Picture the Problem

Picture the Problem Because the net external torque acting on the diver is zero, Because the net external torque acting on the diver is zero, the diver’s angular momentum will remain constant as she rotates from the full the diver’s angular momentum will remain constant as she rotates from the full tuck to the full layout position. Assume that, in layout position, the diver is a thin tuck to the full layout position. Assume that, in layout position, the diver is a thin rod of length 2.5 m and that, in the full tuck position, the diver is a sphere of rod of length 2.5 m and that, in the full tuck position, the diver is a sphere of radius 0.50 m.

radius 0.50 m.

Because the net external torque acting Because the net external torque acting on the diver is zero:

on the diver is zero:

0 0 Δ Δ tuck tuck layout layout

−−

==

L L L L L L or or 0 0 tuck tuck tuck tuck layout layout layout

layout

−−

I I

==

I I Solving for the ratio of the angular

Solving for the ratio of the angular velocities gives: velocities gives: tuck tuck layout layout layout layout tuck tuck I I I I

==

ω ω ω ω

(10)

Substituting for the moment of Substituting for the moment of inertia of a thin rod relative to an inertia of a thin rod relative to an axis through its center of mass and axis through its center of mass and the moment of inertia of a sphere the moment of inertia of a sphere relative to its center of mass and relative to its center of mass and simplifying yields: simplifying yields: 2 2 2 2 2 2 5 5 2 2 2 2 12 12 1 1 layout layout tuck tuck 24 24 5 5 r r mr mr m mll ll

==

ω ω ω ω

Substitute numerical values and Substitute numerical values and evaluate

evaluate _tuck_tuck _layout_layout::

( (

_{( (}

))

₎₎

55 m m 50 50 .. 0 0 24 24 m m 5 5 .. 2 2 5 5 2 2 2 2 layout layout tuck tuck

==

≈≈

ω ω ω ω 21 ••

21 •• The days on Mars and Earth are of nearly identical length. Earth’sThe days on Mars and Earth are of nearly identical length. Earth’s mass is 9.35 times Mars’s mass, Earth’s radius is 1.88 times Mars’s radius, and mass is 9.35 times Mars’s mass, Earth’s radius is 1.88 times Mars’s radius, and Mars is on average 1.52

Mars is on average 1.52 times farther away from the Sun than Earth times farther away from the Sun than Earth is. Theis. The Martian year is 1.88 times longer than

Martian year is 1.88 times longer than Earth’s year. Assume that they are bothEarth’s year. Assume that they are both uniform spheres and that their orbits about the Sun are circles. Estimate the ratio uniform spheres and that their orbits about the Sun are circles. Estimate the ratio (Earth to Mars) of (

(Earth to Mars) of (aa) their spin angular momenta, () their spin angular momenta, (bb) their spin kinetic energies,) their spin kinetic energies, ((cc) their orbital angular momenta, and () their orbital angular momenta, and (d d ) their orbital kinetic energies.) their orbital kinetic energies.

Picture the Problem

Picture the Problem We can use the definitions of spin angular momentum, spin We can use the definitions of spin angular momentum, spin kinetic energy, orbital angular momentum, and orbital kinetic energy to evaluate kinetic energy, orbital angular momentum, and orbital kinetic energy to evaluate these ratios.

these ratios.

((aa) The ratio of the spin angular) The ratio of the spin angular momenta of Earth and Mars is: momenta of Earth and Mars is:

M M M M E E E E spin spin M M E E ω ω ω ω I I I I L L L L

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

Because Mars and Earth have nearly Because Mars and Earth have nearly identical lengths of days,

identical lengths of days, ω ω EE

≈≈

ω ω MM::

M M E E spin spin M M E E I I I I L L L L

_≈≈

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

Substituting for the moments of Substituting for the moments of inertia and simplifying yields: inertia and simplifying yields:

2 2 M M E E M M E E 2 2 M M M M 5 5 2 2 2 2 E E E E 5 5 2 2 spin spin M M E E

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

==

≈≈

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

R R R R M M M M R R M M R R M M L L L L

Substitute numerical values for the Substitute numerical values for the ratios and evaluate

ratios and evaluate

spin spin M M E E

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

L L L L ::

( (

11..8888

))

3333 35 35 .. 9 9 22 spin spin M M E E

≈≈

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

L L L L

((bb) The ratio of the spin kinetic) The ratio of the spin kinetic energies of Earth and Mars is: energies of Earth and Mars is:

E E M M 2 2 M M 2 2 E E M M 2 2 M M E E 2 2 E E spin spin M M E E 2 2 2 2 I I I I L L L L I I L L I I L L K K K K

₌₌

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

(11)

Because Because M M E E spin spin M M E E I I I I L L L L

≈≈

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

:: spin spin M M E E E E M M 2 2 M M 2 2 E E spin spin M M E E

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

L L L L L L L L L L L L K K K K From ( From (aa)) 3333 spin spin M M E E

≈≈

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

L L L L . . Hence: Hence: 3333 spin spin M M E E

≈≈

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

K K K K

((cc) Treating Earth and Mars as point) Treating Earth and Mars as point objects, the ratio of their orbital objects, the ratio of their orbital angular momenta is:

angular momenta is: MM orb,orb,MM

E E orb, orb, E E orb orb M M E E ω ω ω ω I I I I L L L L

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

Substituting for the moments of Substituting for the moments of inertia and orbital angular speeds inertia and orbital angular speeds yields: yields:

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

M M 2 2 M M M M E E 2 2 E E E E orb orb M M E E 2 2 2 2 T T r r M M T T r r M M L L L L π π π π where

where r r EE and and r r MM are the radii of the are the radii of the orbits of Earth and Mars, respectively. orbits of Earth and Mars, respectively. Simplify to obtain: Simplify to obtain:

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

E E M M 2 2 M M E E M M E E orb orb M M E E T T T T r r r r M M M M L L L L

Substitute numerical values for the Substitute numerical values for the three ratios and evaluate

three ratios and evaluate

orb orb M M E E

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

L L L L ::

(

( )

)

(

( ))

11..8888 88 52 52 .. 1 1 1 1 35 35 .. 9 9 2 2 orb orb M M E E

_⎟⎟

≈≈

⎠

⎞

⎜⎜

⎝

⎛

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

L L L L

((d d ) The ratio of the orbital k) The ratio of the orbital kineticinetic energies of Earth and Mars is:

energies of Earth and Mars is: 22

M M orb, orb, M M 2 2 1 1 2 2 E E orb, orb, E E 2 2 1 1 orb orb M M E E ω ω ω ω I I I I K K K K

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

Substituting for the moments of inertia and angular speeds and simplifying Substituting for the moments of inertia and angular speeds and simplifying gives: gives: 2 2 E E M M 2 2 M M E E M M E E 2 2 M M 2 2 M M M M 2 2 E E 2 2 E E E E orb orb M M E E 2 2 2 2

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

T T T T r r r r M M M M T T r r M M T T r r M M K K K K π π π π

Substitute numerical values for the Substitute numerical values for the ratios and evaluate

ratios and evaluate

orb orb M M E E

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

K K K K ::

(

( )

)

(

( ))

11..8888 1414 52 52 .. 1 1 1 1 35 35 .. 9 9 22 2 2 orb orb M M E E

_⎟⎟

≈≈

⎠

⎞

⎜⎜

⎝

⎛

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

K K K K

(12)

22 ••

22 •• The The polar polar ice ice caps caps contain contain about about 2.32.3

××

10 10 kg of ice. This mass kg of ice. This mass

contributes negligibly to the moment of inertia of Earth because it is located at the contributes negligibly to the moment of inertia of Earth because it is located at the poles, close to the axis of rotation. Estimate the change in the length of the day poles, close to the axis of rotation. Estimate the change in the length of the day

that would be expected if the polar ice caps were to melt and the water were that would be expected if the polar ice caps were to melt and the water were distributed uniformly over the surface of Earth.

distributed uniformly over the surface of Earth.

Picture the Problem

Picture the Problem The change in the length of the day The change in the length of the day is the differenceis the difference between

between its its length length when when the the ice ice caps caps have have melted melted and and the the water water has has beenbeen distributed over the surface of Earth and the length of the day before the ice caps distributed over the surface of Earth and the length of the day before the ice caps melt. Because the net torque acting on Earth during this process is zero, angular melt. Because the net torque acting on Earth during this process is zero, angular momentum is conserved and we can relate the angular speed (which are related to momentum is conserved and we can relate the angular speed (which are related to the length of the day) of Earth before and after the ice caps melt to the moments the length of the day) of Earth before and after the ice caps melt to the moments of inertia of the Earth-plus-spherical shell the ice caps melt.

of inertia of the Earth-plus-spherical shell the ice caps melt. Express the change in the length of a

Express the change in the length of a day as: day as: before before after after Δ Δ_T_T

==

_T_T

−−

_T_T ₍₁₎₍₁₎

Because the net torque acting on Because the net torque acting on Earth during this process is zero, Earth during this process is zero, angular momentum is conserved: angular momentum is conserved:

0 0 Δ Δ before before after after

−−

==

L L L L L L Substituting for

Substituting for L Lafter after and and L L before before yields: yields: 0 0 before before sphere sphere after after shell shell sphere

sphere

++

I I

−−

I I

==

I I Because Because

==

22π π T T ::

_{( (}

₎₎

0 0 2 2 2 2 before before sphere sphere after after shell shell sphere sphere

++

−−

==

T T I I T T I I I I π π π π or, simplifying, or, simplifying, 0 0 before before sphere sphere after after shell shell sphere sphere

++

₋₋

₌₌

T T I I T T I I I I Solve for

Solve for T T to obtain:_after_after to obtain:

before before sphere sphere shell shell after after 11 T T I I I I T T

_⎟⎟

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

++

==

Substituting for

Substituting for T T in equation (1)_after_after in equation (1) and simplifying yields:

and simplifying yields:

before before sphere sphere shell shell before before before before sphere sphere shell shell 1 1 Δ Δ T T I I I I T T T T I I I I T T

==

−−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

++

==

(13)

Substitute for

Substitute for I I shellshell and and I I spheresphere and and simplify to obtain:

simplify to obtain: before before

E E before before 2 2 E E E E 5 5 2 2 2 2 3 3 2 2 3 3 5 5 Δ Δ _T_T M M m m T T R R M M mr mr T T

==

Substitute numerical values and evaluate Substitute numerical values and evaluate

Δ

T T ::

( (

))

( (

))

hh 00..5555ss ss 3600 3600 d d h h 24 24 d d 1 1 kg kg 10 10 98 98 .. 5 5 3 3 kg kg 10 10 2.3 2.3 5 5 Δ Δ 24 24 19 19

==

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

_××

××

==

T T 23 ••

23 •• [SSM][SSM] A A 2.0-g particle 2.0-g particle moves moves at a at a constant sconstant speed of peed of 3.0 mm/s3.0 mm/s around a circle of radius 4.0 mm. (

around a circle of radius 4.0 mm. (aa) Find the magnitude of the angular) Find the magnitude of the angular momentum of the particle. (

momentum of the particle. (bb) If) If_L_L

==

ll

( (

ll

++

₁₁

))

hh_{, where}_{, where} ll_{is an integer, find the}_{is an integer, find the}

value of

value of ll l

( ( ))

l

+

₁₁ _{and the approximate value of}_{and the approximate value of} ll_{. (}_{. (cc) By how much does}_{) By how much does}ll_change_change

if the particle’s speed increases by one-millionth of a percent, and nothing else if the particle’s speed increases by one-millionth of a percent, and nothing else changing? Use your result to explain why the quantization of angular momentum changing? Use your result to explain why the quantization of angular momentum is not noticed in macroscopic physics.

is not noticed in macroscopic physics.

Picture the Problem

Picture the Problem We can use We can use L L == mvr mvr to find the angular momentum of the to find the angular momentum of the particle.

particle. In In ((bb) we can solve) we can solve_L_L

==

ll

( (

ll

++

₁₁

))

hh_for_for ll

( (

ll

++

₁₁

))

_{and the approximate value}_{and the approximate value}

of of ll_..

((aa) Use the definition of angular momentum to ) Use the definition of angular momentum to obtain:obtain:

) )

))

/s /s m m kg kg 10 10 2.4 2.4 /s /s m m kg kg 10 10 2.40 2.40 m m 10 10 4.0 4.0 m/s m/s 10 10 3.0 3.0 kg kg 10 10 2.0 2.0 2 2 8 8 2 2 8 8 3 3 3 3 3 3

⋅⋅

××

==

⋅⋅

××

==

××

==

−− −− −− −− −− mvr mvr L L

((bb) Solve the equation) Solve the equation

( (

ll

))

hh l l

++

₁₁

==

L L forfor ll

( (

ll

++

11

))

_::

( (

))

22 2 2 1 1 h h l l l l

++

==

L L ₍₁₎₍₁₎

Substitute numerical values and Substitute numerical values and evaluate evaluate ll

( (

ll

++

11

))

_::

( (

))

52 52 2 2 34 34 2 2 8 8 10 10 2 2 .. 5 5 ss JJ 0 0 1 1 1.05 1.05 /s /s m m kg kg 10 10 2.40 2.40 1 1

××

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⋅⋅

××

⋅⋅

××

==

++

−− ₋₋ l l l l Because

Because ll_{>>1, approximate its}_{>>1, approximate its}

value with the square root of value with the square root of

( (

ll

++

11

))

l l _:: 26 26 10 10 3 3 .. 2 2

××

≈≈

l l

((cc) The change in) The change in ll_is:_is: ll

==

ll

−−

ll

new new

Δ

(14)

If the particle’s speed increases by If the particle’s speed increases by one-millionth of a percent while one-millionth of a percent while nothing else changes:

nothing else changes:

( (

))

vv vv vv vv

→

++

1010−−88

==

11

++

1010−−88 and and

( (

))

L L L L L L L L

→

++

1010−−88

==

11

++

1010−−88 Equation (1) becomes: Equation (1) becomes:

( (

))

[ [

( (

₂₂

))

]]

2 2 8 8 new new new new 10 10 1 1 1 1 h h l l l l L L −−

++

==

++

and and

( (

))

h h l l L L 8 8 new new 10 10 1 1

++

−−

≈≈

Substituting in equation (2) yields:

( (

))

h h h h h h l l l l l l L L L L 88 L L 8 8 new new 1010 10 10 1 1 Δ Δ −− −−

==

−−

++

≈≈

−−

==

Substitute numerical values and Substitute numerical values and evaluate evaluate ΔΔll_:: 18 18 34 34 2 2 8 8 8 8 10 10 3 3 .. 2 2 ss JJ 01 01 1.05 1.05 /s /s m m kg kg 10 10 2.40 2.40 10 10 Δ Δ

××

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⋅⋅

××

⋅⋅

××

==

−− −− ₋₋ l l and and % % 10 10 10 10 3 3 .. 2 2 10 10 3 3 .. 2 2 Δ Δ ₆₆ 26 26 18 18 −−

≈≈

××

==

l l l l

The quantization of angular momentum is not noticed in macroscopic physics The quantization of angular momentum is not noticed in macroscopic physics because no experiment can detect a fractional change in

because no experiment can detect a fractional change inll_of_of₁₀₁₀−−66_%_%_..

24 •••

24 ••• Astrophysicists Astrophysicists in in the 1960s the 1960s tried tried to to explain texplain the exishe existence and tence and structurestructure of

of pulsars pulsars —extremely regular astronomical sources of radio pulses whose periods —extremely regular astronomical sources of radio pulses whose periods ranged from seconds to milliseconds. At one point, these radio sources were given ranged from seconds to milliseconds. At one point, these radio sources were given the acronym LGM (Little Green Men), a reference to the idea that they might be the acronym LGM (Little Green Men), a reference to the idea that they might be signals of extraterrestrial civilizations. The explanation given today is no

signals of extraterrestrial civilizations. The explanation given today is no lessless interesting. Consider the foll

interesting. Consider the following. owing. Our Sun, which is a fairOur Sun, which is a fairly typical star, has aly typical star, has a mass of 1.99

mass of 1.99

××

10 103030 kg and a radius of 6.96 kg and a radius of 6.96

××

10 1088 m. m. Although it Although it does not rotatdoes not rotatee uniformly, because it is not a solid bod

uniformly, because it is not a solid body, its average rate of rotation y, its average rate of rotation is aboutis about 1 rev/25 d.

1 rev/25 d. Stars larger than the Sun can end tStars larger than the Sun can end their life in spectacular heir life in spectacular explosionsexplosions called supernovae, leaving behind a collapsed remnant of the star called a

called supernovae, leaving behind a collapsed remnant of the star called a neutronneutron star

star . These neutron-stars have masses comparable to . These neutron-stars have masses comparable to the original masses of thethe original masses of the stars but radii of only a few

stars but radii of only a few kilometers! The high rotation rates are due to kilometers! The high rotation rates are due to thethe conservation of angular momentum during the

conservation of angular momentum during the collapses. These stars emit beamscollapses. These stars emit beams of radio waves. Because of the rapid angular speed of the stars, the beam sweeps of radio waves. Because of the rapid angular speed of the stars, the beam sweeps past Earth at regular, very short, intervals. To produce the observed radio-wave past Earth at regular, very short, intervals. To produce the observed radio-wave pulses, the star has to rotate at rates that range from about 1 rev/s to 1000 rev/s. pulses, the star has to rotate at rates that range from about 1 rev/s to 1000 rev/s.

((aa) Using data from the textbook, ) Using data from the textbook, estimate the rotation rate of the Sun if it were estimate the rotation rate of the Sun if it were toto collapse into a neutron star of

(15)

gas, and its moment of inertia is given by

gas, and its moment of inertia is given by I I = 0.059 = 0.059 MR MR22. Assume that the neutron. Assume that the neutron star is spherical and has a un

star is spherical and has a uniform mass distribution. (iform mass distribution. (bb) Is the rotational kinetic) Is the rotational kinetic energy of the Sun greater or smaller after the collapse? By what factor does it energy of the Sun greater or smaller after the collapse? By what factor does it change, and where does the energy go to or come from?

change, and where does the energy go to or come from?

Picture the Problem

Picture the Problem We can use conservation of angular momentum in Part ( We can use conservation of angular momentum in Part (aa)) to relate the before-and-after collapse rotation rates of the Sun. In Part (

to relate the before-and-after collapse rotation rates of the Sun. In Part (bb), we can), we can express the fractional change in the rotational kinetic energy of the Sun as it express the fractional change in the rotational kinetic energy of the Sun as it collapses into a neutron star to decide whether its rotational kinetic energy is collapses into a neutron star to decide whether its rotational kinetic energy is greater initially or after the collapse.

greater initially or after the collapse. ((aa) Use conservation of angular) Use conservation of angular momentum to relate the angular momentum to relate the angular momenta of the Sun before and after momenta of the Sun before and after its collapse: its collapse: aa aa b b b b I I I I

==

⇒

_b_b aa b b aa ω ω ω ω I I I I

==

(1)(1)

Using the given formula, approximate the moment of inertia

Using the given formula, approximate the moment of inertia I I b b of the Sun before of the Sun before collapse: collapse:

(

3030

))(

(

55

))

22 4646 22 2 2 sun sun b b

==

00..059059 MR MR

==

00..05905911..9999

××

1010 kgkg 6.966.96

××

1010 kmkm

==

55..6969

××

1010 kgkg

⋅⋅

mm I I

Find the moment of inertia

Find the moment of inertia I I aaof theof the Sun when it has collapsed into a Sun when it has collapsed into a spherical neutron star of radius spherical neutron star of radius 10 km and uniform mass

10 km and uniform mass distribution: distribution:

( (

))

( (

))

2 2 37 37 2 2 30 30 5 5 2 2 2 2 5 5 2 2 aa m m kg kg 10 10 96 96 .. 7 7 km km 10 10 kg kg 10 10 99 99 .. 1 1

⋅⋅

××

==

××

==

MR MR I I

Substitute numerical values in Substitute numerical values in equation (1) and simplify to obtain: equation (1) and simplify to obtain:

b b 8 8 b b 2 2 37 37 2 2 46 46 aa 10 10 15 15 .. 7 7 m m kg kg 10 10 96 96 .. 7 7 m m kg kg 10 10 69 69 .. 5 5 ω ω ω ω ω ω

××

==

⋅⋅

××

⋅⋅

××

==

Given that

Given that ω ω b b = 1 rev/25 d, evaluate = 1 rev/25 d, evaluate ω ω aa:: rev/d rev/d 10 10 9 9 .. 2 2 rev/d rev/d 86 86 .. 2 2 d d 25 25 rev rev 1 1 10 10 15 15 .. 7 7 7 7 8 8 aa

××

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

××

==

ω ω

Note that the rotational period decreases by the same factor of

Note that the rotational period decreases by the same factor of I I b b// I I aa and becomes: and becomes:

ss 10 10 0 0 .. 3 3 ss 3600 3600 h h 1 1 h h 24 24 d d 1 1 rev rev rad rad 2 2 d d rev rev 10 10 86 86 .. 2 2 2 2 2 2 33 7 7 aa aa −−