Cap10

Chapter 10 Angular Momentum Conceptual Problems True or false:

1

•

(a)

If two vectors are exactly opposite in in direction, their vector product must be zero. The magnitude of the vector product of two vectors is at a minimum when when the two vectors are perpendicular. Knowing the magnitude of the vector product of two nonzero vectors and the vectors′ individual magnitudes uniquely determines the angle between them.

(b) (c)

r

r

Determine the Concept The vector product of and  B is defined to be r r r  A × B =  AB sinφ  sin φ nˆ where nˆ  is a unit vector normal to the plane defined by and r

r

 B  and φ is φ  is the angle between r

(a) True. If r

(b) False. If  product of

r

r

and  B .

r

and  B are in opposite direction, then sinφ  sinφ  =  = sin 180° = 0. 0. r

and  B are perpendicular, then sinφ  sinφ  =  = sin 90° = 1 and the vector r

and  B is a maximum.

(c) False. Knowing the magnitude of the vector product and the vectors′ 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 us if the sine of the a ngle is positive or negative. r

r

Consider two nonzero vectors and  B . Their vector product has the 2 • r r greatest magnitude if and  B are (a (a) parallel, (b (b) perpendicular, (c (c) antiparallel, (d ) at an angle of 45° to each other. r

r

Determine the Concept The vector product of the vectors and  B is defined to r r  be  A × B =  AB sinφ  sin φ nˆ   where nˆ   is a unit vector normal to the plane defined by r r r

r

and  B and φ  is the angle between r

r

r

and  B . Hence, the vector product of r

and  B is a maximum when sinφ  sinφ  =  = 1. This condition is satisfied provided

 B are perpendicular  are perpendicular . (b) is correct.

969

and

970  Chapter 10 r

3

•

What is the angle between a force vector F and a torque vector

r

r

 produced by F ? Determine the Concept Because

r τ 

r

r

= r × F  = rF sinφ  sin φ nˆ , where r

r

normal to the plane defined by r  and

nˆ  is a unit vector

r

, the angle between

 and

r

 is 90°.

A point particle of mass m is moving with a constant speed v along a 4 • straight line that passes through point P  point P . What can you say about the angular momentum of the particle relative to point po int P   P ? (a (a) Its magnitude is mv. mv. (b (b) Its magnitude is zero. (c (c) Its magnitude changes sign as the particle passes through  point P   point P . (d  (d ) It varies in magnitude as the particle pa rticle approaches point point P   P . r

r

r

r

r

Determine the Concept  L and  p   are related according to  L = r × p   and the r

r

r

magnitude of  L is  L = rp sin φ   where φ  is   is the angle between r   and  p . Because the motion is along a line that passes through point P  point  P , r  =   = 0 and so is L is  L..

(b) is

correct. point P is at the 5 • [SSM] A particle travels in a circular path and point P r center of the circle. (a (a) If the particle’s linear momentum  p  is doubled without changing the radius of the circle, how is the magnitude of its angular momentum about P about P affected? (b (b) If the radius of the circle is doubled but the speed of the  particle is unchanged, how is the magnitude of its angular momentum about P about P affected? r

r

r

r

r

Determine the Concept  L and  p  are related according to  L = r × p  and the r

r

r

magnitude of  L  is  L = rp sin φ   where φ  is  is the angle between r   and  p . r

r

(a) Because  L  is directly proportional to  p , L is  L is doubled. r

r

(b) Because  L  is directly proportional to r  , L is  L is doubled. A particle moves along a straight line at constant speed. How does its 6 • angular momentum about any fixed point vary with time? Determine the Concept We can determine how the angular momentum of the  particle about any fixed point varies with time by examining the derivative of the r r cross product of r  and  p .

The angular momentum of the  particle is given by:

r

r

r

 L = r × p

Angular Momentum r

r

Differentiate L with respect to time to obtain: r

r

Because  p = mv , r

d r  dt 

r

d   p dt 

r

r

r

 L d 

= F net , and

r

r d   p ⎞ ⎛ d r  r ⎞ = ⎛  ⎜ r × ⎟ + ⎜ ×  p ⎟   dt  ⎝  dt  ⎠ ⎝  dt   ⎠

 L d 

dt 

r

971

r

r

(1)

r

= (r × F net ) + (v × p )

r

=v:

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

r

r

r

r

r

= 0 ⇒ r × F net = 0

 F net

r

Because v and  p(= mv )  are parallel:

v × p = 0

Substitute in equation (1) to obtain:

 L d 

r

dt 

r

= 0 ⇒ L =

constant

An alternate solution The following diagram shows a particle whose mass is m moving along a straight line at constant speed. The point identified as  P   is any r fixed point and the vector r   is the position vector, relative to this fixed point, of the particle moving with constant velocity.

m

v

φ 

r 

d  φ   P  From the definition of angular momentum, the magnitude of the angular momentum of the particle, relative to the fixed point at P, at P, is  is given by:

 = r sin φ  : Because d  =

r

 L

r

 L

= rmv sin φ  = mv(r sin φ )

= mvd 

or, because m, v and v and d  are  are constants, r

 L

=

constant

True or false: If the net torque on a rotating system is zero, the angular 7 •• velocity of the system cannot change. cha nge. If your answer is false, give an example of such a situation.

972  Chapter 10

False. The net torque acting on a rotating system equals the change in the system’s angular momentum; that is, τ net = dL dt   where  L =  I ω  ω.  Hence, if τ net is zero, all we can say for sure is that the angular momentum (the product of I of I and ω ) is constant. If I  If  I  changes,   changes, so must ω . An example is a high diver going from a tucked to a layout position. You are standing on the edge of a turntable with frictionless bearings 8 •• 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. Assume you do not move relative to the turntable. (a (a) Does the angular speed of the turntable increase, decrease, or remain the same during the catch? (b (b) Does the magnitude of your angular momentum (about the rotation axis of the table) increase, decrease, or remain the same after the catch? (c (c) How does the ball’s angular momentum (about the rotation axis of the table) change after the catch? (d ) How does the total angular momentum of the system, you-table-ball (about the rotation axis of the table) change after the catch? Determine the Concept You can apply conservation of angular momentum to the you-table-ball system to answer each of these questions.

(a) Because the ball is moving in the same direction that you are moving, your rd

angular speed will increase when you catch it. Your Newton’s 3   law interaction with the ball causes a torque that acts on the you-table-ball system to increase ω . (b) The ball has angular momentum relative to the rotation axis of the table before you catch catch it it and so so catchin catching g it increa increases ses your your angular angular mome momentum ntum relati relative ve to the the rotation axis of the table. (c) The ball will slow down as a result of your catch and so its angular momentum relative to the center of the table will decrease. (d ) Because there is zero net torque on the you-table-ball system, its angular momentum remains the same. If the angular momentum of a system about a fixed point P point P is constant, 9 •• which one of the following statements must be true? (a) (b) (c) (d ) (e)

No torque about P about P acts on any part of the system. A constant torque about P about P acts on each part of the system. Zero net torque about P about P acts on each part of the system. A constant external torque about P about P acts on the system. Zero net external torque about P about P acts on the system.

Angular Momentum

973

If L is  is constant, we know that the net  torque  torque acting on Determine the Concept If L the system is zero. There may be multiple constant or time-dependent torques acting on the system as long as the net torque is zero. (e ) is correct. A block sliding on a frictionless table is attached to a string that passes 10 •• through a narrow hole through the tabletop. Initially, the block is sliding with speed v0 in a circle of radius r 0. A student under the table pulls slowly on the string. What happens as the block spirals inward? Give supporting arguments for your choice. (The term angular momentum refers to the angular momentum about a vertical axis through the hole.) (a (a) Its energy and angular momentum are conserved. (b (b) Its angular momentum is conserved and its energy increases. (c (c) Its angular momentum is conserved and its energy decreases. (d  (d ) Its energy is conserved and its angular momentum increases. (e (e) Its energy is conserved and its angular momentum decreases. Determine the Concept The pull that the student exerts on the block is at right

angles to its motion and exerts no torque (recall that

r

r

r

= r × F and τ  = rF sin φ  ).

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 radial force and so the block’s energy increases. (b ) is correct. One way to tell if an egg is hardboiled or uncooked without 11 •• [SSM]  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 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 two types of eggs. Determine the Concept The hardboiled egg is solid inside, so everything rotates with a uniform uniform angular speed. By contrast, when you start start an uncooked egg 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.

Explain why a helicopter h elicopter with just one main rotor has a second smaller 12 •• 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. Determine the Concept The purpose of the second smaller rotor is to prevent the  body of the helicopter from rotating. If the rear rotor fails, the body of the helicopter will tend to rotate on the main axis due to angular momentum being conserved.

974  Chapter 10 The spin angular-momentum vector for a spinning wheel is parallel 13 •• 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? (a (a) up, (b) down, (c (c) north, (d  (d ) south, (e (e) east. r

Determine the Concept  The vector

Δ L

r

r

= Lf  − Li  

(and the torque that is

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 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. 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 the force that must be exerted on the east end of the axle, point upward. (a ) is correct. You are walking toward the north and in your left hand you are 14 •• 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  points north. You now begin to turn to walk toward the east. As a result, the front end of the suitcase will (a ( a) resist your attempt to turn it and will try to maintain its original orientation, (b (b) resist your attempt to turn and will pull to the west, (c (c) rise upward, (d  (d ) dip downward, (e (e) show no effect whatsoever. Determine the Concept  In turning toward the east, you redirect the angular 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 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 and the force the suitcase exerts on you is downward. Consequently, the front end

of the suitcase will dip downward. (d ) is correct. small single15 •• [SSM] The angular momentum of the propeller of a small engine airplane points forward. The propeller rotates clockwise if viewed from  behind. (a (a) Just after liftoff, as the nose of the plane tilts upward, the airplane tends to veer to one side. To which side does it tend to veer and why? (b (b) If the  plane is flying horizontally and suddenly turns to the right, does the nose of the  plane tend to veer upward or downward? Why? r

(a) The plane tends to veer to the right. The change in angular momentum Δ L prop r

for the propeller is upward, so the net torque τ   on the propeller is upward as well. 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

Angular Momentum

975

change in the angular momentum of the plane. This means the plane tends to rotate clockwise as viewed from above. (b) The nose of the plane tends to veer downward. The change in angular r

momentum Δ L prop   for the propeller is to the right, so the net torque

r

τ 

  on the

 propeller is toward the right as well. The propeller must exert an equal but opposite torque on the plane. This leftward directed torque exerted by the  propeller on the plane tends to cause a leftward-directed change in angular momentum for the plane. This means the plane tends to rotate clockwise as viewed from the right. 16

••

You have designed a car that is powered by the energy stored in a r

single flywheel with a spin angular momentum L momentum L . In the morning, you plug the 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 translational kinetic energy of the car during the day. Having taken a physics course involving angular momentum and torques, you realize that problems would arise during various maneuvers of the car. Discuss some of these problems. r

For example, suppose the flywheel is mounted so L  points vertically upward when the car is on a horizontal road. What would happen as the car travels over a r

hilltop? Through a valley? Suppose the flywheel is is mounted so L  points forward, 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, consider the direction of the torque exerted on the car by the road. r

Determine the Concept If  L   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) r

will increase and car will tend to tip. If  L  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  by having two identical flywheels that rotate on the same shaft in opposite directions. stool with your arms 17 •• [SSM] You are sitting on a spinning piano stool folded. (a (a) When you extend your arms out to the side, what happens to your kinetic energy? What is the cause of this change? (b (b) Explain what happens to your moment of inertia, angular speed and angular momentum as you extend your arms. Determine the Concept The rotational kinetic energy of the you-stool system is  L2 2 1 given by  K rot = 2 I ω  = .  Because the net torque acting on the you-stool system 2 I  r

is zero, its angular momentum  L  is conserved.

976  Chapter 10 (a) Your kinetic energy decreases. Increasing your moment of inertia  I while  L2 conserving your angular momentum L momentum L decreases you kinetic energy  K  = . 2 I  (b) Extending your arms out to the side increases your moment of inertia, decreases your angular speed. The angular momentum of the system is unchanged. A uniform rod of mass M mass M and length L length L rests  rests on a horizontal 18 •• frictionless table. A blob of putty of mass m = M   = M /4 /4 moves along a line  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. Determine the Concept The center of mass of the rod-and-putty system moves in a straight line, and the system rotates about abo ut its center of mass.

Estimation and Approximation 19 •• [SSM] An ice-skater starts her pirouette with arms outstretched, rotating at 1.5 rev/s. Estimate her rotational speed (in revolutions per second) when she brings her arms flat against her body. Picture the Problem Because we 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 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 acting on her is zero, her angular momentum will remain constant during her  pirouette.

Because the net external torque acting on her is zero:

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

Δ L

= Lf  − Li = 0

or  I arms in

arms in

 I arms out

=  I  body + I arms

 I  body

− I arms out

arms out

= 0   (1)

= 12 mr 2 = 12 (50 kg )(0.20 m )2 = 1.00 kg ⋅ m 2

Angular Momentum

= 2 13 (4 kg)( ) (1.0 m )2 = 2.67 kg ⋅ m 2

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

 I arms

Substitute to determine her total moment of inertia with her arms outstretched:

 I arm s out

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

 I arms in

Solve equation (1) for

arms in

 to

obtain: Substitute numerical values and evaluate arms in :

977

= 1.00 kg ⋅ m 2 + 2.67 kg ⋅ m 2 = 3.67 kg ⋅ m 2 = I  body + I arms = 1.00 kg ⋅ m 2 + 2[(4.0 kg )(0.20 m )2 ] = 1.32 kg ⋅ m 2

ω arms in

ω arms in

=

= ≈

 I arms out  I arms in

ω arms out

3.67 kg ⋅ m 2 1.32 kg ⋅ m 2

(1.5 rev/s)

4 rev/s

Estimate the ratio of angular velocities for the rotation of a diver 20 ••  between the full tuck position and the full-layout position. Picture the Problem 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 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 radius 0.50 m.

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

Solving for the ratio of the angular velocities gives:

Δ L

= Llayout − Ltuck  = 0

or  I layout ω tuck  ω layout

layout

=

− I tuck 

 I layout  I tuck 

tuck 

=0

978  Chapter 10

Substituting for the moment of inertia of a thin rod relative to an axis through its center of mass and the moment of inertia of a sphere relative to its center of mass and simplifying yields: Substitute numerical values and evaluate tuck  layout :

ω tuck  ω layout

ω tuck  ω layout

=

ml 2

1 12 2 5

2

mr 

=

5l 2 2

24r 

5(2.5 m )

2

=

24(0.50 m )

2

≈ 5

The days on Mars and Earth are of nearly identical length. Earth’s 21 •• 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 times farther away from the Sun than Earth is. The Martian year is 1.88 times longer than Earth’s year. Assume that they are both uniform spheres and that their orbits about the Sun are circles. Estimate the ratio (Earth to Mars) of (a (a) their spin angular momenta, (b (b) their spin kinetic energies, (c) their orbital angular momenta, and (d  (d ) their orbital kinetic energies. Picture the Problem We can use the definitions of spin angular momentum, spin kinetic energy, orbital angular momentum, and orbital kinetic energy to evaluate these ratios.

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

⎛  LE  ⎞  I  ω  ⎜⎜ ⎟⎟ = E E ⎝  LM  ⎠spin  I Mω M

Because Mars and Earth have nearly identical lengths of days, ω E ≈ ω M:

⎛  LE  ⎞  I  ⎜⎜ ⎟⎟ ≈ E ⎝  LM  ⎠spin  I M

Substituting for the moments of inertia and simplifying yields:

⎛  LE  ⎞ ⎜⎜ ⎟⎟ ≈ ⎝  LM  ⎠ spin

Substitute numerical values for the

⎛  LE  ⎞ ⎜⎜ ⎟⎟ ≈ 9.35(1.88)2 ≈ ⎝  LM  ⎠spin

⎛  L  ⎞ ratios and evaluate ⎜⎜ E ⎟⎟ : ⎝  LM  ⎠ spin (b) The ratio of the spin kinetic energies of Earth and Mars is:

 M E RE2

2 5

 M M RM2

2 5

=

 M E ⎛  RE  ⎞

 L2E

⎛  K E  ⎞ ⎜⎜ ⎟⎟ =  K  ⎝  M  ⎠ spin

2 I E 2

 LM 2 I M

=

⎜

⎟

⎜ R ⎟  M M ⎝  M  ⎠

 L2E  I M 2  LM  I E

33

2

Angular Momentum Because

⎛  LE  ⎞  I  ⎜⎜ ⎟⎟ ≈ E ⎝  LM  ⎠spin  I M

From (a (a)

⎛  LE  ⎞ ⎜⎜ ⎟⎟ ≈ 33 . Hence: ⎝  LM  ⎠ spin

979

⎛  K E  ⎞  L2E  LM ⎛  LE  ⎞ ⎜⎜ ⎟⎟ = 2 = ⎜⎜ ⎟⎟  K   L  L E M ⎝  M  ⎠ spin ⎝  LM  ⎠ spin

:

⎛  K E  ⎞ ⎜⎜ ⎟⎟ ≈  K  M ⎝   ⎠ spin

33

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

 I  ω  ⎛  LE  ⎞ ⎜⎜ ⎟⎟ = E orb, E ⎝  LM  ⎠ orb  I Mω orb, M

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

⎛ 2π  ⎞ ⎟⎟ T  ⎛  LE  ⎞ ⎝  E  ⎠ ⎜⎜ ⎟⎟ = 2π  ⎞ ⎝  LM  ⎠ orb  M  r 2 ⎛  ⎜ ⎟⎟ M M⎜ ⎝ T M  ⎠  M E r E2 ⎜⎜

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

⎛  LE  ⎞ ⎛  M   ⎞ ⎛  r   ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ E ⎟⎟ ⎜⎜ E ⎟⎟ ⎝  LM  ⎠ orb ⎝  M M  ⎠ ⎝ r M  ⎠

Substitute numerical values for the

2 ⎛  LE  ⎞ 1  ⎞ ⎛  ⎜⎜ ⎟⎟ = (9.35)⎜ ⎟ (1.88) ≈  L 1 . 52 ⎝   ⎠ ⎝  M  ⎠ orb

⎛  L  ⎞ three ratios and evaluate ⎜⎜ E ⎟⎟ : ⎝  LM  ⎠ orb

2

⎛ T M  ⎞ ⎜⎜ ⎟⎟ ⎝ T E  ⎠ 8

1  I  ω 2 ⎛  K E  ⎞ 2 E orb, E ⎜⎜ ⎟⎟ = 1 2  K  ⎝  M  ⎠ orb 2 I Mω orb, M

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

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

⎛ 2π  ⎞ ⎟⎟  M  r  ⎜⎜ 2 2 T   ⎞ ⎛   ⎞ ⎛   ⎞ ⎛  K E  ⎞ ⎛   M  r  T  ⎝  E  ⎠ = ⎜ E ⎟ ⎜ E ⎟ ⎜ M ⎟ ⎟⎟ = ⎜⎜ 2 ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎛   ⎞ ⎝  M M  ⎠ ⎝ r M  ⎠ ⎝ T E  ⎠ ⎝  K M  ⎠ orb 2 π  2 ⎟⎟  M M r M ⎜⎜ T  ⎝  M  ⎠ 2 E E

Substitute numerical values for the

⎛  K   ⎞ ratios and evaluate ⎜⎜ E ⎟⎟ : ⎝  K M  ⎠ orb

2 ⎛  K E  ⎞ 1  ⎞ ⎛  2 ⎜⎜ ⎟⎟ = (9.35)⎜ ⎟ (1.88) ≈ ⎝ 1.52 ⎠ ⎝  K M  ⎠ orb

14

980  Chapter 10 The polar ice caps contain about 2.3 × 10  kg of ice. This mass 22 •• 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 that would be expected if the polar ice caps were to melt and the water were distributed uniformly over the surface of Earth. 19

Picture the Problem  The change in the length of the day is the difference  between its length when the ice caps have melted and the water has been 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 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 of inertia of the Earth-plus-spherical shell the ice caps melt.

= T after  − T  before  

Express the change in the length of a day as:

ΔT 

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

Δ L

Substituting for L for Lafter  and L  and L before yields:

 I sphere + I shell

Because

= 2π  T  :

(1)

= Lafter  − L before = 0

( I 

sphere

after 

+ I shell )

2π  T after 

− I sphere

− I sphere

or, simplifying,  I sphere + I shell  I sphere

−

T after  Solve for T after  to obtain:

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

⎛  ⎜ ⎝ 

T after  = ⎜1 +

ΔT 

T  before

 before

2π  T  before

=0

 I shell  ⎞ ⎟T  before  I sphere  ⎠⎟

⎛   I   ⎞ = ⎜⎜1 + shell ⎟⎟T  before − T  before ⎝   I sphere ⎠ =

 I shell  I sphere

T  before

=0

=0

Angular Momentum Substitute for I  for I shell and I sphere shell  and I  sphere and simplify to obtain:

ΔT 

=

2 3 2 5

mr 2 2 E

 M E R

T  before

=

5m 3 M E

981

T  before

Substitute numerical values and evaluate ΔT :

ΔT 

=

5(2.3 × 1019 kg )

3(5.98 × 10

24

⎛ 1d × 24 h × 3600 s ⎞ = ⎜ ⎟ d h  ⎠ kg ) ⎝ 

0.55 s

speed of 3.0 mm/s 23 •• [SSM] A 2.0-g particle moves at a constant speed around a circle of radius 4.0 mm. (a (a) Find the magnitude of the angular momentum of the particle. (b (b) If  L = l(l + 1)h , where l  is an integer, find the

( )

value of l l + 1  and the approximate value of l . (c (c) By how much does l  change 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 is not noticed in macroscopic physics. use  L = mvr  to   to find the angular momentum of the Picture the Problem We can use L  particle. In (b) we can solve L = l (l + 1)h for l(l + 1) and the approximate value of l . (a) Use the definition of angular momentum to obtain:  L = mvr = 2.0 × 10 −3 kg ) 3.0 ×10 −3 m/s ) 4.0 × 10 −3 m ) = 2.40 × 10 −8 kg ⋅ m 2 /s

=

2.4 × 10 −8 kg ⋅ m 2 /s

(b) Solve the equation  L = l(l + 1)h for l(l + 1) :

l(l + 1) =

 L2 h

2

 

(1)

Substitute numerical values and evaluate l(l + 1) :

⎛ 2.40 ×10 −8 kg ⋅ m 2 /s ⎞ l(l + 1) = ⎜ ⎜ 1.05 ×10 −34 J ⋅ s ⎟⎟ ⎝   ⎠ = 5.2 ×10 52

Because l >>1, approximate its value with the square root of l(l + 1) :

l

(c) The change in l is:

Δl

2

≈ 2.3 ×1026

= l new − l  

(2)

982  Chapter 10

If the particle’s speed increases by one-millionth of a percent while nothing else changes: Equation (1) becomes:

v → v + 10 −8 v = (1 + 10 −8 )v and −8 −8  L → L + 10  L = (1 + 10 ) L  L

[ 1 + 10 − ) L ] ( + 1) = 8

l new (l new

and

h

2

2

( 1 + 10 − ) L ≈ 8

l new

Substituting in equation (2) yields:

Substitute numerical values and evaluate Δl :

Δl

h

= l new

1 + 10 − ) L  L  L ( −l ≈ − = 10 − 8

8

h

h

h

⎛ 2.40 × 10 −8 kg ⋅ m 2 /s ⎞ Δl = 10 ⎜ ⎜ 1.05 × 01−34 J ⋅ s ⎟⎟ ⎝   ⎠ = 2.3 × 1018 −8

and Δl

l

=

2.3 × 1018 2.3 × 10

26

≈ 10 −6%

The quantization of angular momentum is not noticed in macroscopic physics −  because no experiment can detect a fractional change in l of 10 6 % . the existence existence and structure 24 ••• Astrophysicists in the 1960s tried to explain the of pulsars of pulsars —extremely regular astronomical sources of radio pulses whose periods 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 signals of extraterrestrial civilizations. The explanation given today is no less interesting. Consider the following. following. Our Sun, which is a fairly fairly typical star, has a 30 8 mass of 1.99 × 10  kg and a radius of 6.96 × 10  m. Although it does not rotate rotate uniformly, because it is not a solid body, bod y, its average rate of rotation is about 1 rev/25 d. Stars larger than the Sun can end their their life in spectacular explosions called supernovae, leaving behind a collapsed remnant of the star called a neutron  star . These neutron-stars have masses comparable to the original masses of the stars but radii of only a few kilometers! The high rotation rates are due to the conservation of angular momentum during the collapses. These stars emit beams 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  pulses, the star has to rotate at rates that range from about 1 rev/s to 1000 rev/s. (a) Using data from the textbook, estimate the rotation rate of the Sun if it were to collapse into a neutron star of radius 10 km. The Sun is not a uniform sphere of

Angular Momentum

983

2

gas, and its moment of inertia is given by I  by  I  =  = 0.059 MR . Assume that the neutron star is spherical and has a uniform un iform mass distribution. (b (b) Is the rotational kinetic 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? (a) Picture the Problem We can use conservation of angular momentum in Part (a to relate the before-and-after collapse rotation rates of the Sun. In Part (b (b), we can 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 greater initially or after the collapse. (a) Use conservation of angular momentum to relate the angular momenta of the Sun before and after its collapse:

 I  b

 b

=  I a

a

⇒ ω a =

 I  b  I a

ω  b  

(1)

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

2 = 0.059 MRsun = 0.059(1.99 ×10 30 kg )(6.96 ×10 5 km ) = 5.69 ×10 46 kg ⋅ m 2 2

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

 I a

= 52 MR 2 = 52 (1.99 ×1030 kg )(10 km )2 = 7.96 ×1037 kg ⋅ m 2

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

ω a

=

5.69 × 1046 kg ⋅ m 2 7.96 × 1037 kg ⋅ m 2

ω  b

= 7.15 × 108 ω  b Given that ω  b = 1 rev/25 d, evaluate ω a:

ω a

⎛ 1rev ⎞ ⎟⎟ = 2.86 rev/d = 7.15 × 108 ⎜⎜ 25 d ⎝   ⎠ = 2.9 × 10 7 rev/d

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

=

2π  ω a

= 2.86 × 10 7

rev d

×

2π  2π  rad rev

×

1d 24 h

×

1h 3600 s

= 3.0 ×10 −3 s

984  Chapter 10 (b) Express the fractional change in the Sun’s rotational kinetic energy as a consequence of its collapse: Substituting for the kinetic energies and simplifying yields:

Δ K 

 K  b Δ K 

 K  b

=

 K a − K  b  K  b  I a ω a2

1 2

=

 I  ω  b2

1 2  b

=

 K a  K  b

−1 =

−1

 I a ω a2  I  bω  b2

−1

Substitute numerical values and evaluate Δ K / K   K  b: Δ K 

 K  b

2

⎛ 2.86 ×10 7 rev/d ⎞ 1 ⎛   ⎞ ⎟ −1 = =⎜ ⎟⎜⎜ 1rev/25 d  ⎠⎟ ⎝ 7.15 ×10 8  ⎠⎝ 

7.1× 108

That is, the rotational kinetic energy increases by increases by a factor of approximately 8 7 × 10 . The additional rotational kinetic energy comes at the expense of gravitational potential energy, which decreases as the Sun gets smaller. The moment of inertia of Earth about its spin axis is approximately 25 •• 37 2 8.03 × 10  kg⋅m . (a (a) Because Earth is nearly spherical, assume a ssume that the moment 2 of inertia can be written as I  as I  =  = CMR , where C is a dimensionless constant, 24  M = 5.98 × 10  kg is the mass of Earth, and R and  R =  = 6370 km is its radius. Determine C . (b (b) If Earth’s mass were distributed uniformly, C would equal 2/5. From the value of C calculated in Part (a (a), is Earth’s density greater near its center or near its surface? Explain your reasoning.

 = CMR for C   and substitute numerical Picture the Problem We can solve  I  = values in order to determine an experimental value of C  for   for Earth. We can then compare this value to those for a spherical shell and a sphere in which the mass is uniformly distributed to decide whether Earth’s mass density is greatest near its core or near its crust. 2

(a) Express the moment of inertia of Earth in terms of the constant C : Substitute numerical values and evaluate C :

 = CMR  I  = C  =

=

2

⇒ C  =  =

 I   R 2

8.03 ×1037 kg ⋅ m 2

(5.98 ×10

24

kg )(6370 km )

2

0.331

(b) Because experimentally C < 0.4, the mass density must be greater near the center of Earth. 26 ••• Estimate Timothy Goebel’s initial takeoff speed, rotational velocity, and angular momentum when he performs a quadruple Lutz (Figure 10-41). Make any assumptions you think reasonable, but justify them. Goebel’s mass is about 60 kg and the height of the jump is about 0.60 m. Note that his angular speed will

Angular Momentum

985

change quite a bit during the jump, as he begins with arms outstretched and pulls them in. Your answer should be accurate to within a factor of 2, if you’re careful. Picture the Problem  We’ll assume that he launches himself at an angle of 45° with the horizontal with his arms spread wide, and then pulls them in to increase his rotational speed during the jump. We’ll also assume that we can model him as a 2.0-m long cylinder with an average radius of 0.15 m and a mass of 60 kg. We can then find his take-off speed and ″air time″  using constant-acceleration equations, and use the latter, together with the definition of rotational velocity, to find his initial rotational velocity. Finally, we can apply conservation of angular momentum to find his initial angular momentum.

Using a constant-acceleration equation, relate his takeoff speed v0 to his maximum elevation Δ y:  y:

Solving for v0 and simplifying yields: Substitute numerical values and evaluate v0:

v2

= v02 y + 2a y Δ y

or, because v0y = v0sin(45°), v = 0, and a y = − g , 0 = v02 sin 2 45° − 2 g Δ y

v0

v0

2 g Δ y

=

sin 2 45°

2 g Δ y sin 45°

2(9.81m/s 2 )(0.60 m )

= =

=

sin45° 4.9 m/s

Δθ  Δt 

Use its definition to express Goebel’s angular velocity:

ω  =

Use a constant-acceleration equation to express Goebel’s ″air time″ Δt :

Δt  = 2Δt rise 0.6 m = 2

Substitute numerical values and evaluate Δt :

Δt 

Substitute numerical values and evaluate ω :

ω  =

Use conservation of angular momentum to relate his take-off angular velocity ω 0 to his average angular velocity ω  as  as he performs a quadruple Lutz:

 I 0

=2

0

2(0.60 m ) 9.81m/s 2

4 rev 0.699 s

= I 

×

2Δ y  g 

= 0.699 s

2π rad rev

=

36 rad/s

986  Chapter 10 Assuming that he can change his moment of inertia by a factor of 2 by  pulling his arms in, solve for and evaluate ω 0:

=

ω 0

 I   I 0

ω  =

1 2

(36 rad/s) =

18 rad/s

Express his take-off angular momentum:

 L0 = I 0

Assuming that we can model him as a solid cylinder of length l  with an average radius r  and  and mass m, express his moment of inertia with arms outstretched (his take-off configuration):

 I 0

Substituting for I  for I 0 gives:

 L0

= mr 2ω 0

Substitute numerical values and evaluate L evaluate L0:

 L0

= (60 kg )(0.15 m )2 (18 rad/s )

0

= 2(12 mr 2 ) = mr 2

where the factor of 2 represents our assumption that he can double his moment of inertia by extending his arms.

= 24 kg ⋅ m 2 /s

The Vector Product and the Vector Nature of Torque and Rotation 27

•

[SSM]

A force of magnitude F magnitude F is applied horizontally in the negative r

 x direction to the rim of a disk of radius R radius R as  as shown in Figure 10-42. Write F  and r r   in terms of the unit vectors iˆ , ˆ , and kˆ , and compute the torque produced by this force about the origin at the center of the disk. r

r ˆ  and Picture the Problem We can express F and r   in terms of the unit vectors i  ˆ and then use the definition of the vector product to find r .  j 

r

Express F in terms of F  of F  and  and the unit ˆ: vector i  r

Express r in terms of R of R and  and the unit ˆ: vector  j  Calculate the vector product of r

r

r  and F  :

r

ˆ F  = − F i 

r ˆ r  =  R j 

r

r

r

= r × F  = FR ( j ˆ × −i ˆ ) = FR (i ˆ ×  j ˆ ) =

ˆ  FR k 

Angular Momentum

987

Compute the torque about the origin of the gravitational force r F = − mg   jˆ acting on a particle of mass m located at r = x iˆ + y jˆ and show that this torque is independent of the y the y coordinate.  coordinate.

28 r

•

r

r

Picture the Problem We can find the torque from the vector product of r  and F .

Compute the vector product of r

r

r

r  and F  :

r

r

= r × F  = ( xi ˆ + y j ˆ )(− mg  j ˆ ) = −mgx(i ˆ ×  j ˆ )− mgy ( j ˆ ×  j ˆ ) ˆ = − mgxk 

r

r

r

r

Find × B for the following choices: (a (a) = 4iˆ and B = 6iˆ + 6 jˆ , 29 •  r r r r ˆ ˆ ˆ ˆ ˆ (b) = 4i and  B = 6i + 6 k , and (c (c)  A = 2i + 3 and  B = 3iˆ + 2 ˆ . Picture the Problem We can use the definitions of the vector products of the unit r r ˆ , and k  ˆ to evaluate × B in each case. vectors i ˆ ,  j  r

r

r

(a) Evaluate  B  for r ˆ + 6 j  ˆ:  B = 6 i 

ˆ and  = 4 i 

r

r

ˆ × 6i  ˆ + 6 j  ˆ  A × B = 4i 

= 24(i ˆ × i ˆ ) + 24(i ˆ ×  j ˆ ) = 24(0) + 24k ˆ = 24k ˆ

r

(b) Evaluate r ˆ + 6 k  ˆ:  B = 6 i 

r

r

ˆ and  B  for   = 4 i 

r

r

 A × B

= 4i ˆ × (6i ˆ + 6k ˆ ) = 24(i ˆ × i ˆ )+ 24(i ˆ × k ˆ ) = 24(0) + 24(−  j ˆ ) = − 24 j ˆ

r

r

 B  for (c) Evaluate r r ˆ  + 3 j  ˆ and  B =3 i  ˆ + 2 j  ˆ: = 2 i 

r

r

 A × B

= (2i ˆ + 3 j ˆ )× (3i ˆ + 2 j ˆ ) = 6(i ˆ × i ˆ)+ 4(i ˆ ×  j ˆ )+ 9( j ˆ × i ˆ ) + 6( j ˆ ×  j ˆ ) = 6(0) + 4(k ˆ )+ 9(− k ˆ )+ 6(0 ) = − 5k ˆ r

30

••

r

r r

For each case in Problem 31, compute  A × B . Compare it to  A  B  to

estimate which of the pairs of vectors vec tors are closest to being perpendicular. Verify your answers by calculating the angle using the dot product.

988  Chapter 10 r

r r

r

Picture the Problem Because  A × B

=  A  B sin φ  , if vectors r

r

r

 perpendicular, then  A × B

r r

=  A B

 or

r

r

and B and B are

r

 A × B r r

= 1 . The scalar product of vectors

 A  B r

r

r

r

r r

and B is A ⋅ B =  A  B cos φ  . We can verify our estimations using this definition

to calculate φ for φ  for each pair of vectors. r

(a) For

r

ˆ and  B = 6 i ˆ + 6 j  ˆ:  = 4 i 

r

r

 A × B r r

=

 A  B

( ) 24k ˆ = = 24 2 (4)(6 2 )

ˆ 4i ˆ × 6i ˆ + 6 j 

≈ 0.707

r

1 2

r

and the vectors  A and B and B are not  perpendicular. r

The angle between

r

and B is:

⎛  Ar ⋅ Br  ⎞ ⎛ 4i ˆ ⋅ (6i ˆ + 6 j ˆ ) ⎞ ⎜ ⎟ ⎟ φ  = cos −1 r r = cos −1 ⎜ ⎜⎜  A  B ⎟⎟ ⎜ 24 2 ⎟ ⎝   ⎠ ⎝   ⎠ = cos −1

24

= 45°, 24 2 a result confirming that obtained above. r

r

ˆ and  B = 6 i  ˆ + 6 k  ˆ: (b) For   = 4 i 

r

r

 A × B r r

=

 A  B

( ) − 24 j ˆ = = 24 2 (4)(6 2 )

ˆ × 6i  ˆ + 6k  ˆ 4i 

≈ 0.707 and the vectors  perpendicular. r

The angle between

r

and B is:

φ  = cos

−1

= cos −1

r

1 2

r

and B are not

⎛  Ar ⋅ Br  ⎞ ˆ ˆ ˆ  ⎞ ⎜ r r ⎟ = cos −1 ⎛  ⎜ 4i ⋅ (6i  + 6k )⎟ ⎜⎜  A  B ⎟⎟ ⎜ 24 2 ⎟  ⎠ ⎝  ⎝   ⎠ 24

= 45°, 24 2 a result confirming that obtained above.

Angular Momentum r

r

ˆ  + 3 j  ˆ and (c) For = 2 i  r ˆ: ˆ + 2 j   B =3 i 

r

 A × B r r

=

(2i ˆ + 3 j ˆ )× (3i ˆ + 2 j ˆ ) 13 13

 A  B

=

5 13

r

r

− 5k ˆ 13

≈ 0.385

and the vectors  perpendicular. The angle between

=

989

r

r

and B are not

⎛  Ar ⋅ Br  ⎞ ⎜ ⎟ φ  = cos −1 r r ⎜⎜ ⎟⎟ ⎝  A  B  ⎠ ˆ + 3 j  ˆ )⋅ (3i ˆ + 2 j  ˆ ) ⎞ ( 2i  −1 ⎛  ⎜ ⎟ = cos ⎜ ⎟ 13 13 ⎝   ⎠ 12 ⎞ = cos −1 ⎛  ⎜ ⎟ = 23°, ⎝ 13 ⎠

and B is:

a result confirming that obtained above. While none of these sets of vectors ve ctors are perpendicular, those in (a (a) and (b (b) are the closest, with φ = φ  = 45°, to being perpendicular. 31 •• A particle moves in a circle that is centered at the origin. The particle has r r r r r  position r and angular velocity . (a (a) Show that its velocity is given by v = ω × r . r r r r r r (b) Show that its centripetal acceleration is given by a c = ω × v = ω × (ω × r ) . r

Picture the Problem Let r  be in the  xy   xy  plane and point in the + x r direction. Then ω   points in the + z  direction. We can establish the results called for in this problem by forming the appropriate vector r  products and by differentiating v . r

r

(a) Express ω using unit vector notation:

ω

r

Express r  using unit vector notation: Form the vector product of

r ω

r

and r  :

ˆ = ω k 

r ˆ r  = r i  r

r

× r  = ω k ˆ × r i ˆ = r ω (k ˆ × i ˆ ) = r ω  j ˆ = v j ˆ r r r and v = ω × r  ω

990  Chapter 10 r

r

(b) Differentiate v with respect to t  to  to express a : r

a

r

=

d v

d  r

=

r

r

( × r ) = ω

dt  dt  r = a t + ac r

r

where a c

=

r

r

ω

d ω dt 

r

× (ω × r ) and

r

r

r

d r 

× r  + ω ×

r

dt 

r

=

d ω dt 

r

r

r

r

r

r

r

× r  + ω × v = a t + ω × (ω × r )

r

a t and a c are the tangential and centripetal

accelerations, respectively. 32 r

 A =

You are given three vectors and their components in the form: •• r r a x iˆ + a y jˆ + a z kˆ ,  B = b x iˆ + b y jˆ + b z kˆ , and C = c x iˆ + c y jˆ + c z kˆ . Show that the r

(

r

r

r

r

r

r

r

r

) = C ⋅ ( A × B) = B ⋅ (C × A)

following equalities hold:  A ⋅  B × C 

Picture the Problem We can establish these equalities by carrying out the details of the vector- and scalar-products and comparing the results of these operations. r

r

Evaluate the vector product of  B and C  to obtain: r

r

ˆ + (b c  B × C  = (b y c z  − b z c y )i   z   x r

Form the scalar product of r

(

r

r

)= a b c

 A ⋅  B × C 

 x  y  z 

r

r

with  B × C  to obtain:

− a x b z c y + a y b z c x − a y b x c z  + a z b x c y − a z b y c x   r

Evaluate the vector product of r

r

r

r

r

Form the scalar product of C  with

(

r

r

C ⋅  A × B

)= c a b

 x  y  z 

r

− c x a z b y + c y a z b x − c y a x b z  + c z a x b y − c z a y b x   to obtain:

r

ˆ + (c a C × A = (c y a z  − a z a y )i   z   x r

r

Form the scalar product of  B with C  ×

(

r

r

 B ⋅ C × A

)= b c a

 x  y  z 

(2)

r

r

r

− a x b z  ) j ˆ + (a x b y − a y b x )k ˆ

× B to obtain:

Evaluate the vector product of C  and r

(1)

and  B to obtain:

ˆ + (a b  A × B = (a y b z  − a z b y )i   z   x

r

− b x c z  ) j ˆ + (b x c y − b y c x )k ˆ

− a x a z  ) j ˆ + (c x a y − a y a x )k ˆ

r

to obtain:

− b x c z a y + b y c z a x − b y c x a z  + b z c x a y − b z c y a x  

(3)

Angular Momentum

991

The equality of equations (1), (2), and (3) establishes the equalities. r

33

If  A =

••

r

3ˆ ,

r

× B = 9iˆ , and

r

r

r

⋅ B = 12, find  B .

r

r

ˆ + B  j  ˆ ˆ Picture the Problem  We can write B in the form  B =  B x i   y + B z k and use r

the scalar product of

r

and B to find B find B y and their vector product to find B find B x and B  and B z .

r

r

ˆ + B  j  ˆ + B k  ˆ  B =  B x i   y  z   

Express  B  in terms of its components: r

Evaluate

r

r

⋅ B :

r

r

r

Evaluate

r

 A ⋅ B = 3 B y r

(1)

= 12 ⇒ B y = 4

(

)

ˆ ×  B i  ˆ ˆ ˆ  A × B = 3 j   x + 4 j + B z k 

 B :

ˆ + 3 B i  ˆ = −3 B x k   z  r

r

ˆ:  B  = 9 i 

Because

 B x = 0 and B and B z  = 3.

Substitute for B for B y and B and B z  in equation (1) to obtain: r

34

••

If

r

= 4iˆ , B z  = 0,  B = 5 , and

r

ˆ + 3k  ˆ  B = 4 j  r

r

r

× B = 12kˆ , determine  B . r

r

ˆ + B  j  ˆ Picture the Problem Because  B z   = 0, we can express  B as B =  B x i   y   and r

form its vector product with

 to determine B determine B x and B and B y.

r

r

Express  B in terms of its components: r

ˆ + B  j  ˆ  B =  B x i   y   r

r

Solving for B for B y yields:

 B y

=3

Relate B Relate B to  to B  and B y:  B x and B

 B 2

=  B x2 + B y2

Solve for and evaluate B evaluate B x:

 B x

=

Substitute for B for B x and B and B y in equation (1) to obtain:

 B

Express

r

 B :

(1)

(

)

ˆ ×  B i  ˆ ˆ ˆ ˆ  A × B = 4i   x + B y j  = 4 B y k  = 12k 

r

=

 B 2 − B y2 ˆ 4i ˆ + 3 j 

=

52

− 32 = 4

992  Chapter 10 r

35

•••

Given three noncoplanar vectors

r

r

r

r

r

( )

,  B , and C , show that  A ⋅  B × C

is the volume of the parallelepiped formed by the three vectors. r

the x Picture the Problem Let, without loss of generality, the vector C  lie along the x r

axis and the vector  B the xy plane  plane as shown below to the left. The diagram to  B lie in the xy the right shows the parallelepiped spanned by the three vectors. We can apply the r

(

r

r

)

definitions of the vector- and scalar-products to show that A ⋅  B × C    is the volume of the parallelepiped.

 B × C 

 A

 A cos θ 

 Bsin φ 

C 

θ  φ   B

The magnitude of the vector product r

r

of  B and C  is:

r

r

 B × C  = BC sin φ 

and r

r

 B × C  = ( B sin φ )C 

= area of  the base parallelogram Forming the scalar-product of r r

r

r

with the vector-product of  B and

C  gives:

(

r

r

) =  A( B sin φ )C cosθ 

 A ⋅  B × C 

= ( BC sin φ )( A cosθ ) = (area of  base)(height ) = V  parallelepiped

36 ••• Using the cross product, prove the law of sines for the triangle shown in Figure 10-43. That is, if A if A,, B,  B, and C  are  are the lengths of each side of the triangle, show that A that A/sin /sin a = B  = B/sin /sin b = C /sin /sin c. Picture the Problem Draw the triangle using the three vectors as shown below. r

r

 Note that  A + B r

 B  and of

r

r

= C  . We can find the magnitude of the vector product of r r r

 and C  and then use the vector product of

and C ,  using

r

 and

Angular Momentum

993

r r r  B C   , to show that  AC sin b =  AB sin c or  . Proceeding similarly, =  A + B = C  sin b sin c we can extend the law of sines to the third side of the triangle and the angle opposite it. c

 A

 B

a

b C 

Express the magnitude of the vector r

 product of

r

r

r

r

r

r

=  AB sin (180° − c ) =  AB sin c

and  B :

Express the magnitude of the vector r

 product of

r

 AC sin b  A × C  =  AC sin

and C : r

Form the vector product of r

with

C  to obtain:

r

r

 A × B

r

(

r

r

 A × C  =  A ×  A + B r

r

r

)

r

=  A × A + A × B r r =  Ar × Br  because  A × A = 0 . r

r

r

Because  A × C  =  A × B :

r

r

r

r

 A × C  =  A × B

and  AC sin b =  AB sin c Simplify and rewrite this expression to obtain: Proceed similarly to extend this result to the law of sines:

 B sin b  A sin a

=

=

C  sin c  B sin b

=

C  sin c

Torque and Angular Momentum 37 • [SSM] A 2.0-kg particle moves directly eastward at a constant speed of 4.5 m/s along an east-west e ast-west line. (a (a) What is its angular momentum (including direction) about a point that lies 6.0 m north of the line? (b ( b) What is its angular momentum (including direction) about a point that lies 6.0 m south of the line?

994  Chapter 10 (c) What is its angular momentum (including direction) about a point that lies 6.0 m directly east of the particle? r

r

r

Picture the Problem The angular momentum of the particle is  L = r ×  p  where r r r   is the vector locating the particle relative to the reference point and  p  is the

 particle’s linear momentum. (a) The magnitude of the particle’s angular momentum is given by:

 L = rp sin φ  = rmv sin φ  = mv(r sin φ )

Substitute numerical values and evaluate L evaluate L::

 L = (2.0 kg ) (4.5 m/s )(6.0 m )

Use a right-hand rule to establish

 L = 54 kg ⋅ m /s, upward

r

= 54 kg ⋅ m 2 /s 2

the direction of  L : (b) Because the distance to the line along which the particle is moving is the same, only the direction of

 L = 54 kg ⋅ m 2 /s, downward

r

 L differs: r

r

(c) Because r × p = 0 for a point on

r

 L = 0

the line along which the particle is moving: You observe a 2.0-kg particle moving at a constant speed of 38 • 3.5 m/s in a clockwise direction around a circle of radius 4.0 m. (a) What is its angular momentum (including direction) about the center of the circle? (b (b) What is its moment of inertia about an axis a xis through the center of the circle c ircle and  perpendicular to the plane of the motion? (c (c) What is the angular velocity of the  particle? r

r

r

Picture the Problem The angular momentum of the particle is  L = r ×  p  where r r r  is the vector locating the particle relative to the reference point and  p  is the

 particle’s linear momentum. (a) The magnitude of the particle’s angular momentum is given by:

 L = rp sin φ  = rmv sin φ  = mv(r sin φ )

Substitute numerical values and evaluate the magnitude of L of L::

 L = (2.0 kg ) (3.5 m/s )(4.0 m )

= 28 kg ⋅ m 2 /s

Angular Momentum Use a right-hand rule to establish the r

995

 L = 28 kg ⋅ m 2 /s, away from you

direction of  L : (b) Treat the 2.0-kg particle as a  point particle to obtain:

 I  = mr 2

Substitute numerical values and evaluate I  evaluate I :

 I  = (2.0 kg ) (4.0 m )

(c) Because L Because L = I ω  ω ,  the angular speed of the particle is the ratio of its angular momentum and its moment of inertia:

ω  =

Substitute numerical values and evaluate ω :

2

= 32 kg ⋅ m 2

 L  I 

ω  =

28 kg ⋅ m 2 /s 32 kg ⋅ m

2

=

0.88 rad/s 2

(a) A particle moving at constant velocity has zero angular momentum 39 •• about a particular point. Use the definition of angular momentum to show that under this condition the particle is moving either directly toward or directly away from the point. (b (b) You are a right-handed batter b atter and let a waist-high fastball go  past you without swinging. What is the direction of its angular momentum relative to your navel? (Assume the ball travels in a straight horizontal line as it  passes you.) r

r

r

r

r

r

Picture the Problem  L and  p  are related according to L = r ×  p. If  L = 0, then r r examination of the magnitude of r ×  p   will allow us to conclude that sin φ  = 0

and that the particle is moving either directly toward the point, directly away from the point, or through the point. r

r

r

r

or r r r × v

=0

r

r ×  p = r × mv

(a) Because  L = 0:

r

r

r

r

Express the magnitude of r × v :

r × v

Because neither r  nor  nor v is zero:

sin φ  = 0

r

r

= mr × v = 0

= rv sin φ  = 0 r

r

where φ  is  is the angle between r  and v . Solving for φ  yields:  yields:

φ  = sin −1 (0 ) = 0° or 180°

996  Chapter 10 (b) Use the right-hand rule to establish that the ball’s angular momentum is downward. r

A particle that has a mass m is traveling with a constant velocity v 40 •• along a straight line that is a distance b from the origin O (Figure 10-44). Let dA  be the area swept out by the position vector from O to the particle during a time interval dt . Show that dA/ dA/dt  is  is constant and is equal to L to L 2m , where L where L is the magnitude of the angular momentum of the particle about the origin. Picture the Problem The area swept out by the position vector (the shaded area in Figure 10-44) is the difference between the area of a trapezoid and the area of a r r triangle. Let x Let  x1 be the x the  x component of r 1  and x  and  x1 + Δ x be  x be the x the x component of r 2 .

Use the formulas for the areas of a trapezoid and a triangle to express Δ A and  A and then r take the limit as Δt → 0 to express dA/ dA/dt. Letting dt. Letting θ  be  be the angle between r 1  and r

the horizontal axis, we can express b as a function of r 1 and θ . The area swept out by the position vector (the shaded area in Figure 1044) is given by:

Δ A =  Atrap − Atriangle = 12 b[Δ x + ( x1 + Δ x )] − 12 b( x1 + Δ x ) = 12 bΔ x

In the limit as Δt → 0:

dA dt 

r

Because b = r 1 sin θ  :

dA dt 

= 12 b

dx dt 

= 12 bv = constant r

= 12 bv = 12 ( r 1 sin θ )v = =

1 2m

( r r  p sin θ ) = 1

( r r sin θ )mv 1

2m

 L 2m

A 15-g coin that that has a diameter diameter of 1.5 cm is spinning at 10 rev/s about 41 •• a fixed vertical axis. The coin is spinning on edge with its center directly d irectly above the point of contact with the tabletop. As you look down on the tabletop, the coin spins clockwise. (a (a) What is the angular momentum (including direction) of the coin about its center of mass? (To find the moment of inertia about the axis, see Table 9-1.) Model the coin as a cylinder of length L length L and take the limit as L as L approaches zero. (b (b) What is the coin’s angular momentum (including direction) about a point on the tabletop 10 cm from the axis? (c) Now the coin’s center of mass travels at 5.0 cm/s in a straight line east e ast across the tabletop, while spinning the same way as in Part (a (a). What is the angular momentum (including direction) of the coin about a point on the line of motion of the center of mass? (d  (d ) When it is both spinning and sliding, what is the angular momentum of the coin (including direction) about a point 10 cm north of the line of motion of the center of mass?

Angular Momentum

997

Picture the Problem We can find the total angular momentum of the coin from the sum of its spin and orbital o rbital angular momenta.

(a) The spin angular momentum of the coin is:

 Lspin = I 

From Table 9-1, for L for L negligible compared to R to R::

 I  = 14 MR 2

Substitute for I  for I  to  to obtain:

 Lspin

spin

= 14 MR 2ω spin

Substitute numerical values and evaluate L evaluate Lspin:  Lspin

= 14 (0.015 kg )(0.0075 m )2 ⎛  ⎜10 ⎝ 

rev s

×

2π  rad ⎞ rev

⎟ = 1.33 ×10 −5 kg ⋅ m 2 /s  ⎠

r

Use a right-hand rule to establish the direction of  Lspin :  Lspin

= 1.3 × 10 −5 kg ⋅ m 2 /s, away from you.

(b)The total angular momentum of the coin is the sum of its orbital and spin angular momenta:

 Ltotal = Lorbital + Lspin

Substitute numerical values and evaluate L evaluate Ltotal:

 Ltotal

= 0 + Lspin = 1.3 × 10 −5 kg ⋅ m 2 /s r

Use a right-hand rule to establish the direction of  Ltotal :  Ltotal (c) Because  Lorbital

= 1.3 ×10 −5 kg ⋅ m 2 /s, away from you

= 0:  Ltotal

= 1.3 ×10 −5 kg ⋅ m 2 /s, away from you

(d ) When it is both spinning and sliding, the total angular momentum of the coin is:

 Ltotal = Lorbital + Lspin

998  Chapter 10

The orbital angular momentum of the coin is:

 Lorbital = MvR

The spin angular momentum of the coin is:

 Lspin = I spinω spin

Substituting for  Lorbital and Lspin yields:

 Ltotal = MvR + 14 MR 2ω spin

= 14 MR 2ω spin

Substitute numerical values and evaluate  Ltotal :  Ltotal

= (0.015 kg )(0.050 m/s )(0.10 m ) rev 2π  rad ⎞ + 14 (0.015 kg )(0.0075 m )2 ⎛  × ⎜10 ⎟ rev  ⎠ ⎝  s = 8.8 × 10 −5 kg ⋅ m 2 /s, pointing toward you

42

r

r

(a) Two stars of masses m1 and m2 are located at r 1  and r 2  relative to

••

some origin O, as shown in Figure 10-45. They exert equal and opposite attractive gravitational forces on each other. For this two-star system, calculate the net torque exerted by these internal forces about the origin O and show that it is zero only if both forces lie along al ong the line joining the particles. (b (b)The fact that the Newton’s third-law pair of forces are not only equal and oppositely directed  but also lie along the line connecting the two objects is sometimes called the strong form of Newton’s third law. Why is it important to add that last phrase?  Hint: Consider what would happen to these two objects if the forces were offset  from each other. Picture the Problem Both the forces acting on the particles exert torques with respect to an axis perpendicular to the page and through point O and the net torque about this axis is their vector sum.

(a) The net torque about an axis  perpendicular to the page and through point O is given by:

r

i

r

r

r

− F 1 :

r

or, because F 2 r

net

Because r 1 − r 2 points along

r

r

r

r

r

net = ∑ i = r  1 × F  1 + r  2 × F  2

r net

r

r

r

r

r

= − F 1 , r

= (r 1 − r 2 ) × F 1 r

= (r 1 − r 2 )× F 1 = 0

Angular Momentum

999

(b) If the forces are not along the same line, there will be a net ne t torque (but still no net force) acting on the system. This net torque would cause the system to accelerate angularly, contrary to observation, and hence makes no sense  physically. A 1.8-kg particle moves in a circle of radius 3.4 m. As you look down 43 •• on the plane of its orbit, it is initially moving clockwise. If we call the clockwise direction positive, its angular momentum relative to the center of the circle varies (a) Find the magnitude and with time according to L(t ) = 10 N ⋅ m ⋅ s − (4.0 N ⋅ m )t . (a direction of the torque acting on the particle. (b (b) Find the angular velocity of the  particle as a function of time. Picture the Problem The angular momentum of the particle changes because a net  torque   torque acts on it. Because we know how the angular momentum depends on time, we can find the net torque acting on the particle by differentiating its angular momentum. We can use a constant-acceleration equation and Newton’s second law to relate the angular speed of the particle to its angular acceleration.

(a) The magnitude of the torque acting on the particle is the rate at which its angular momentum changes: Evaluate dL/ dL/dt  to  to obtain:

τ net

=

dL

τ net

=

d 

dt 

[10 N ⋅ m ⋅ s − (4.0 N ⋅ m )t ]

dt 

= − 4.0 N ⋅ m  Note that, because L because L decreases  decreases as the particle rotates clockwise, the angular acceleration and the net torque are both upward. (b) The angular speed of the particle is given by: Treating the 1.8-kg particle as a  point particle, express its moment of inertia relative to an axis through the center of the circle and normal to it: Substitute for  I orbital  and Lorbital to obtain:

ω orbital

=

 Lorbital  I orbital

 I orbital = MR 2

ω orbital

=

10 N ⋅ m ⋅ s − (4.0 N ⋅ m )t   R 2

1000  Chapter 10 Substitute numerical values and evaluate ω orbital orbital: ω orbital

=

10 N ⋅ m ⋅ s − (4.0 N ⋅ m )t 

(1.8 kg )(3.4 m )

2

= 0.48 rad/s − (0.19 rad/s 2 )t 

 Note that the direction of the angular velocity is downward. You are designing a lathe motor and part of it consists of a uniform 44 •• cylinder whose mass is 90 kg and radius is 0.40 m that is mounted so that it turns without friction on its axis, which is fixed. The cylinder is driven by a belt that wraps around its perimeter and exerts a constant torque. At t = 0, the cylinder’s angular velocity is zero. At t = 25 s, its angular speed is 500 rev/min. (a (a) What is the magnitude of its angular momentum at t = 25 s? (b (b) At what rate is the angular momentum increasing? (c (c) What is the magnitude of the torque acting on the cylinder? (d  (d ) What is the magnitude of the frictional force acting on the rim of the cylinder? Picture the Problem The angular momentum of the cylinder changes because a net   torque acts on it. We can find the angular momentum at t   = 25 s from its definition and the magnitude of the net  torque  torque acting on the cylinder from the rate at which the angular momentum is changing. The magnitude of the frictional force acting on the rim can be found using the definition of torque.

 L =  I ω  = 12 mr 2ω 

(a) The angular momentum of the cylinder is given by:

Substitute numerical values and evaluate L evaluate L::  L =

1 2

⎛  ⎝ 

(90 kg )(0.40 m )2 ⎜⎜ 500

rev min

×

2π  rad 1min ⎞ rev

×

⎟

60 s  ⎠⎟

= 377 kg ⋅ m 2 /s = 3.8 × 10 2 kg ⋅ m 2 /s (b) The rate at which the angular momentum of the cylinder is increasing is given by:

dL dt 

(377 kg ⋅ m /s ) = 15 kg ⋅ m = 2

25 s

=

15 kg ⋅ m 2 /s 2

2

/s 2

Angular Momentum 1001

(c) Because the torque acting on the uniform cylinder is constant, the rate of change of the angular momentum is constant and hence the instantaneous rate of change of the angular momentum at any instant is equal to the average rate of change over the time during which the torque acts: (d ) The magnitude of the frictional force f force f acting on the rim is:

τ  =

dL

 f  =

τ 

dt 

l

=

=

15 kg ⋅ m 2 /s 2

15.1kg ⋅ m 2 /s 2 0.40 m

=

38 N

45 •• [SSM] In Figure 10-46, the incline is frictionless and the string  passes through the center of mass of each block. The pulley has a moment of inertia I inertia I and radius R radius R.. (a) Find the net torque acting ac ting on the system (the two masses, string, and pulley) about the center of the pulley. (b (b)Write an expression for the total angular momentum of the system about the center of the pulley. Assume the masses are moving with a speed v. (c (c) Find the acceleration of the masses by using your results for Parts (a (a) and (b (b) and by setting the net torque equal to the rate of change ch ange of the system’s angular momentum. Picture the Problem Let the system include the pulley, string, and the blocks and assume that the mass of the string is negligible. The angular momentum of this system changes because a net   torque acts on it. We’ll take clockwise to be  positive to be consistent with a positive upward velocity of the block whose mass is m1 as indicated in the figure.

(a) Express the net torque about the center of mass of the pulley: τ net

= −m1 gR + T 1 R − T 1 R + T 2 R + (m2 g sin θ ) R − Rm1 g − T 2 R =  Rg (m2 sin θ  − m1 )

(b) Express the total angular momentum of the system about an axis through the center of the pulley:

 L =  I 

+ m1vR + m2 vR

 I   ⎞ = vR⎛  ⎜ 2 + m1 + m2 ⎟ ⎝  R  ⎠

1002  Chapter 10 (c) Express τ  as  as the time derivative of the angular momentum:

τ  =

dL dt 

d  ⎡ ⎛   I  vR⎜ dt  ⎢⎣ ⎝  R 2

=

 I  = aR⎛  ⎜ 2 ⎝  R Equate this result to that of Part (a ( a) and solve for a to obtain:

a

⎤ + m1 + m2 ⎞⎟⎥  ⎠⎦ + m1 + m2 ⎞⎟  ⎠

 g (m2 sin θ  − m1 )  I  + m1 + m2  R 2

=

Figure 10-47 shows the rear view of a space capsule that was left 46 •• rotating rapidly about its longitudinal axis at 30 rev/min after a collision with another capsule. You are the flight controller and have just moments to tell the crew how to stop this rotation before they become ill from the rotation and the situation becomes dangerous. You know kno w that they have access acce ss to two small jets mounted tangentially at a distance of 3.0 m from the axis, as indicated in the figure. These jets can each eject 10 g/s of gas with a nozzle speed of 800 m/s. Determine the length of time these jets must run to stop the rotation. In flight, the moment of inertia of the ship about abou t its axis (assumed constant) is known to be 2 4000 kg⋅m . Picture the Problem The forces resulting from the release of gas from the jets will exert a torque on the spaceship that will slow and eventually stop its rotation. We can relate this net torque to the angular momentum of the spaceship and to the time the jets must fire.

Δ L

Relate the firing time of the jets to the desired change in angular momentum:

Δt  =

Express the magnitude of the net torque exerted by the jets:

τ net

= 2 FR

 F  =

Δm v Δt '

τ net

= 2vR

Letting Δm/Δt ′ represent the mass of gas per unit time exhausted from the  jets, relate the force exerted by the gas on the spaceship to the rate at which the gas escapes: Substituting for F  for F  yields:  yields:

τ net

=

 I Δ τ net

Δm Δt '

 

(1)

Angular Momentum 1003 Substitute for τ net in equation (1) to

 I Δ

Δt  =

obtain:

2vR

Δm Δt '

Substitute numerical values and evaluate Δt :

(4000 kg ⋅ m )⎛ ⎜⎜ 30 rev × 2

π 

2

Δt 

=

2(10

min

⎝ 

−2

rad 1 min ⎞

rev

×

⎟

60 s  ⎠⎟

kg/s )(800 m/s )(3.0 m )

=

2.6 × 10 2 s

A projectile (mass M  (mass M ) is launched at an angle θ  with  with an initial speed v0. 47 •• Considering the torque and angular momentum about the launch point, explicitly show that dL/ dL/dt  =  = τ . Ignore the effects of air resistance. (The equations for  projectile motion are found in Chapter 3.) Picture the Problem We can use constant-acceleration equations to express the  projectile’s position and velocity coordinates as functions of time. We can use r r these coordinates to express the particle’s position and velocity vectors r  and v . r

Using its definition, we can express the projectile’s angular momentum  L as a r

func functi tion on of time time and and the then n dif diffe fere rent ntia iate te this this expr expres essi sion on to obta obtain ind   L dt .  Finally, we can use the definition of the torque, relative to an origin located at the launch r  position, the gravitational force exerts on the projectile to express  and complete r

r

the demonstration that d   L dt  = . Using its definition, express the r

r

r

r

 L = r × mv  

(1)

angular momentum vector  L  of the  projectile: Using constant-acceleration equations, express the position coordinates of the projectile as a function of time:

Express the projectile’s position r vector r  : Using constant-acceleration equations, express the velocity of the  projectile as a function of time: Express the projectile’s velocity r vector v :

 x = v0 x t  = (v0 cosθ )t  and  y =  y 0

+ v0 y t + 12 a y t 2

= (v0 sin θ )t − 12 gt 2 r ˆ + (v sin θ )t − 1 gt 2  j  r  = [(v0 cosθ )t ]i   ˆj  0 2

[

v x

= v0 x = v0 cosθ 

and v y = v0 y r

v

+ a y t = v0 sin θ  − gt 

= [v0 cosθ ]i ˆ + [v0 sin θ  − gt ] j ˆ

]

1004  Chapter 10

Substituting in equation (1) and simplifying yields: r

{

[

]} {

}

ˆ + (V sin θ )t − 1 gt 2  j  ˆ × m [V cosθ ]i  ˆ + [V sin θ  − gt ] j  ˆ  L = [(V cosθ )t ]i  2

= (− 12 mgt 2V cosθ )k ˆ r

Differentiate L with respect to t  to  to obtain:

r

 L d 

d 

2 ˆ mgt  V cosθ )k    dt  dt  ˆ = (− mgtV cosθ ) k 

=

(−

1 2

(2)

Using its definition, express the torque acting on the projectile: r

r

= r × (− mg ) j ˆ = [(v0 cosθ )t ]i ˆ + [(v0 sin θ )t − 12 gt 2 ] j ˆ × (− mg ) j ˆ = (− mgtV cosθ )k ˆ

(3)

r

Comparing equations (2) and (3) we see that:

 L d 

dt 

=

r

Conservation of Angular Momentum A planet moves in an elliptical orbit about the Sun, with the Sun at one 48 • focus of the ellipse, as in Figure 10-48. (a ( a) What is the torque about the center of the Sun due to the gravitational force of attraction of the Sun on the planet? (b (b) At  position A, the planet has an orbital radius r 1 and is moving with a speed v1  perpendicular to the line from the sun to the planet. At position B, the planet has an orbital radius r 2 and is moving with speed v2, again perpendicular to the line from the sun to the planet. What is the ratio of v1 to v2 in terms of r 1 and r 2? Picture the Problem  Let m  represent the mass of the planet and apply the definition of torque to find the torque produced by the gravitational force of

attraction. We can use Newton’s second law of motion in the form r

r

r

= d  L

dt to

show that  L is constant and apply conservation of angular momentum to the motion of the planet at points A points A and  and B.  B. (a) Express the torque produced by the gravitational force of attraction of the sun for the planet: (b) Because

r

= 0:

r

r

r

r

= r × F  = 0  because F  acts along r

the direction of  r . r

d   L dt 

r

r

r

= 0 ⇒ L = r × mv = constant

Angular Momentum 1005  Noting that at points A points A and  and B  B r r r × v = rv , express the relationship

r 1v1

= r 2 v2 ⇒

v1 v2

=

r 2 r 1

 between the distances from the sun and the speeds of the planets: stand on a frictionless platform that is rotating at an 49 •• [SSM] You stand angular speed of 1.5 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the 2  platform is 6.0 kg⋅m . When you pull the weights in toward your body, the 2 moment of inertia decreases to 1.8 kg⋅m . (a (a) What is the resulting angular speed of the platform? (b (b) What is the change in kinetic energy of the system? (c) Where did this increase in energy come from? Picture the Problem Let the system consist of you, the extended weights, and the  platform. Because the net external torque acting on this system is zero, its angular momentum remains constant during the pulling in of the weights.

(a) Using conservation of angular momentum, relate the initial and final angular speeds of the system to its initial and final moments of inertia: Substitute numerical values and evaluate f  :

 I i

ω f 

(b) Express the change in the kinetic energy of the system:

i

− I f 

=

f 

= 0 ⇒ ω f  =

6.0 kg ⋅ m 2 1.8 kg ⋅ m 2

 I i  I f 

ω i

(1.5 rev/s) =

5.0 rev/s

Δ K  = K f  − K i = 12 I f ω f 2 − 12 I iω i2

Substitute numerical values and evaluate Δ K : Δ K 

= 12 (1.8 kg ⋅ m =

2

2

)⎛ ⎜ 5.0 rev × 2 rad  ⎞⎟ − 12 (6.0 kg ⋅ m 2 )⎛ ⎜1.5 rev × 2 rad  ⎞⎟ rev  ⎠ rev  ⎠ ⎝  s ⎝  s π 

2

π 

0.62 kJ

(c) Because no external agent does work on the system, the energy comes from your internal energy. A small blob of putty of mass m falls from the ceiling and lands on the 50 •• outer rim of a turntable of radius R radius R and  and moment of inertia I  inertia I 0 that is rotating freely (a) What is the postwith angular speed ω 0 about its vertical fixed-symmetry axis. (a collision angular speed of the turntable-putty system? (b (b) After several turns, the

1006  Chapter 10  blob flies off the edge of the turntable. What is the angular speed of the turntable after the blob’s departure? Picture the Problem Let the system consist of the blob of putty and the turntable. Because the net external torque acting on this system is zero, its angular momentum remains constant when the blob b lob of putty falls onto the turntable.

(a) Using conservation of angular momentum, relate the initial and final angular speeds of the turntable to its initial and final moments of inertia and solve for ω f f : Express the final rotational inertia of the turntable-plus-blob: Substitute for I  for I f f in equation (1) and   simplify to obtain:

 I 0

0

− I f 

f 

= 0 ⇒ ω f  =

 I f  = I 0 + I  blob

ω f 

=

+ mR

 I f 

ω 0  (1)

= I 0 + mR 2

 I 0  I 0

 I 0

2

ω 0

1

= 1+

mR 2

ω 0

 I 0

(b) If the blob flies off tangentially to the turntable, its angular momentum doesn’t change (with respect to an axis through the center of turntable). Because there is no external torque acting on the blob-turntable system, the total angular momentum of the system will remain constant and the angular momentum of the turntable will not change. The turntable will continue to spin at ω '  = ω f  . 51 •• [SSM] A lazy Susan consists of a heavy plastic disk mounted on a frictionless bearing resting on a vertical shaft through its center. The cylinder has a radius R radius R =  = 15 cm and mass M mass M = 0.25 kg. A cockroach (mass m = 0.015 kg) is on the lazy Susan, at a distance of 8.0 cm from the center. Both the cockroach and the lazy Susan are initially at rest. The cockroach then walks along a circular path concentric with the center of the Lazy Susan at a constant distance of 8.0 cm from the axis of the shaft. If the speed of the cockroach with respect to the lazy Susan is 0.010 m/s, what is the speed of the cockroach with respect to the room? Picture the Problem Because the net external torque acting on the lazy Susancockroach system is zero, the net angular momentum of the system is constant (equal to zero because the lazy Susan is initially at rest) and we can use conservation of angular momentum to find the angular velocity ω  of the lazy Susan. The speed of the cockroach relative to the floor vf   is the difference  between its speed with respect to the lazy Susan and the speed of the lazy Susan at the location of the cockroach with respect to the floor.

Angular Momentum 1007

= v − r  

Relate the speed of the cockroach with respect to the floor vf  to  to the speed of the lazy Susan at the location of the cockroach:

vf 

Use conservation of angular momentum to obtain:

 LLS − LC

Express the angular momentum of the lazy Susan:

 LLS =  I LSω  = 12 MR 2ω 

Express the angular momentum of the cockroach:

 LC

Substitute for L for LLS and L  and LC in equation (2) to obtain: Solving for ω  yields:  yields:

= 0 

(2)

v  ⎞ = I Cω C = mr 2 ⎛  ⎜ − ω ⎟ ⎝ r   ⎠

⎛ v − ω  ⎞ = 0 ⎟ ⎝ r   ⎠

 MR 2ω  − mr 2 ⎜

1 2

ω  =

Substitute for ω in ω  in equation (1) to obtain:

(1)

vf 

2mrv  MR 2 + 2mr 2

=v−

2mr 2 v  MR 2

+ 2mr 2

Substitute numerical values and evaluate vf : 2(0.015 kg )(0.080 m ) (0.010 m/s ) 2

vf 

= 0.010 m/s −

(0.25 m )(0.15 m )2 + 2(0.015 kg )(0.080 m )2

= 10 mm/s

Remarks: Because the moment of inertia of the lazy Susan is so much larger larg er than the moment of inertia of the cockroach, after the cockroach begins moving, the angular speed of the lazy Susan is very small. Therefore, the speed of the cockroach relative to the floor is almost the same as the speed relative to the lazy Susan.

Two disks of identical mass but different radii (r  and  and 2r  2r ) are spinning 52 •• on frictionless bearings at the same angular speed ω 0 but in opposite directions (Figure 10-49). The two disks are brought b rought slowly together. The resulting frictional force between the surfaces eventually brings them to a common angular velocity. (a) What is the magnitude of that final angular an gular velocity in terms of ω 0? (b (b) What is the change in rotational kinetic energy of the system? Explain.

1008  Chapter 10 Picture the Problem The net external torque acting on this system is zero and so we know that angular momentum is conserved as these disks are brought together. Let the numeral 1 refer to the disk to the left and the numeral 2 to the disk to the right. Let the angular momentum of the disk with the larger radius be p ositive.

(a) Using conservation of angular momentum, relate the initial angular speeds of the disks to their common final speed and to their moments of inertia: Solving for ω f f yields:  

Express I  Express I 1 and I   and I 2:

 I i

i

=  I f 

f 

or  I 1

0

− I 2

0

ω f 

=

 I 1

 I 1 − I 2  I 1 + I 2

= ( I 1 + I 2 )

f 

ω 0  

(1)

= 12 m(2r )2 = 2mr 2

and  I 2 = 12 mr 2 Substitute for I  for I 1 and I   and I 2 in equation (1) and simplify to obtain:

ω f  =

2mr 2 − 12 mr 2 2

2mr 

+

1 2

2

mr 

ω 0

=

3 5

ω 0

= K f  − K i  

(b) The change in kinetic energy of the system is given by:

Δ K 

The initial kinetic energy of the system is the sum of the kinetic energies of the two disks:

 K i = K 1 + K 2

Substituting for K  for K f f  and K   and K i in equation (2) yields:

Δ K 

= 12 ( I 1 + I 2 )ω f 2 − 12 ( I 1 + I 2 )ω 02

Substitute for ω f f from part (a (a) and   simplify to obtain:

Δ K 

= 12 ( I 1 + I 2 )( 53 ω 0 )2 − 12 ( I 1 + I 2 )ω 02 = − 1625 [12 ( I 1 + I 2 )ω 02 ]

Δ K 

= − 1625 K i

 Noting that the quantity in brackets is  K i, substitute to obtain:

(2)

= 12 I 1ω 02 + 12 I 2ω 02 = 12 ( I 1 + I 2 )ω 02

The frictional force between the surfaces is responsible for some of the initial kinetic energy being converted to thermal energy as the two disks come together.

Angular Momentum 1009 A block of mass m sliding on a frictionless table is attached to a string 53 •• that passes through a narrow hole through the center of the table. The block is sliding with speed v0 in a circle of radius r 0. Find (a (a) the angular momentum of the block, (b (b) the kinetic energy of the block, and (c (c) the tension in the string. (d ) A student under the table now slowly pulls the string downward. How much work is required to reduce the radius of the circle from r 0 to r 0/2? (b) We can express the angular momentum and Picture the Problem (a) and (b kinetic energy of the block directly from their definitions. (c (c) The tension in the string provides the centripetal force required for the uniform circular motion and can be expressed using Newton’s second law. (d  (d ) Finally, we can use the workkinetic energy theorem to express the work required to reduce the radius of the circle by a factor of two. (a) Express the initial angular momentum of the block:

 L0

= r 0 mv0

(b) Express the initial kinetic energy of the block:

 K 0

=

(c) Using Newton’s second law, relate the tension in the string to the centripetal force required for the circular motion: (d ) Use the work-kinetic energy theorem to relate the required work to the change in the kinetic energy of the block:

Substituting for L for L0 from Part (a (a) and simplifying gives:

T  = F c

1 2

mv02

=

m

v02 r 0

2

W  = Δ K  = K f  − K 0

=

 Lf  2 I f 

−

 L20 2 I 0

⎛  1 1  ⎞ ⎜⎜ − ⎟⎟ 2 I f  2 I 0 2 ⎝   I f   I 0  ⎠  L20 ⎛  1 1  ⎞ = ⎜⎜ 1 2 − 2 ⎟⎟ 2 ⎝ m( 2 r 0 ) mr 0  ⎠ 2  L20 ⎛  4 1  ⎞  L0 ⎛  3  ⎞ = ⎜⎜ 2 − 2 ⎟⎟ = ⎜⎜ 2 ⎟⎟ 2 ⎝ mr 0 mr 0  ⎠ 2 ⎝ mr 0  ⎠ =

W  =

 L20

−

 L20

=

(r 0 mv0 )2 ⎛  ⎜ 2

 L20

 ⎞ ⎜ mr 2 ⎟⎟ = ⎝  0  ⎠ 3

3 2

mv02

o n a frictionless horizontal surface is 54 ••• A 0.20-kg point mass moving on attached to a rubber band whose other end is fixed at point P  point P . The rubber band exerts a force whose magnitude is F is F = bx, bx, where x where x is the length of the rubber band

1010  Chapter 10 and b is an unknown constant. The rubber band force points inward towards P  towards P . The mass moves along the dotted line in Figure 10-50. When it passes point A point A,, its velocity is 4.0 m/s, directed as shown. The distance AP distance AP is 0.60 m and BP  and BP  is  is 1.0 m. (a) Find the speed of the mass at points B points  B and C . (b (b) Find b. Picture the Problem Because the force exerted by the rubber band is parallel to the position vector of the point mass, the net external torque acting on it is zero and we can use the conservation of angular momentum to determine the speeds of the ball at points B points B and  and C . We’ll use mechanical energy conservation to find b by relating the kinetic and elastic potential energies at A at A and  and B.  B.

(a) Use conservation of momentum to relate the angular momenta at  points A  points A,, B and  B and C : Solve for v B in terms of v A :

Substitute numerical values and evaluate v B : Solve equation (1) for vC  in terms of v A :

 L A

= L B =  LC 

or mv A r  A

= mv B r  B = mvC r C    r  A

v B

= v A

v B

= (4.0 m/s)

vC  = v A

(1)

r  B 0.60 m 1.0 m

=

2.4 m/s

=

4.0 m/s

r  A r C 

Substitute numerical values and evaluate vC  :

vC  = (4.0 m/s )

(b) Use conservation of mechanical energy between points A points A and  and B  B to  to relate the kinetic energy of the point mass and the energy stored in the stretched rubber band:

Δ E 

0.60 m 0.60 m

=  E  A − E  B = 0

or 1 mv A2 2

+ 12 br  A2 − 12 mv B2 − 12 br  B2 = 0

Solving for b yields:

m(v B

− v A2 ) b= 2 r  A2 − r   B

Substitute numerical values and evaluate b:

(0.20 kg ) (2.4 m/s )2 − (4.0 m/s )2 b= (0.60 m )2 − (1.0 m) 2

2

= 3 N/m

Angular Momentum 1011

*Quantization of Angular Momentum 55

••

The z component of the spin of an electron is − 12 h , but the [SSM]  The z

magnitude of the spin vector is 0.75h . What is the angle between the electron’s spin angular momentum vector and the positive zpositive z-axis? axis?  z 

Picture the Problem The electron’s spin angular momentum vector is related to its  z   component as shown in the diagram. The angle between r s and the + z-axis  z-axis is φ .

φ  θ 

0     . 7      5     

s

− 12 Express φ  in  in terms of θ  to  to obtain: Using trigonometry, relate the r magnitude of s to its − z  component:  component: Substitute for θ in θ  in the expression for φ  to  to obtain:

φ  = 180° − θ 

⎛  ⎜⎜ ⎝ 

 ⎞ ⎟⎟ 0.75h ⎠ 1 2

h

θ  = 180° − cos

−1

θ  = cos

−1

⎛  ⎜⎜ ⎝ 

 ⎞ ⎟⎟ = 0.75h ⎠ 1 2

h

125°

Show that the energy difference between one rotational state of a 56 •• molecule and the next higher state is proportional to l + 1. 10-29a  describes the quantization of rotational Picture the Problem Equation 10-29a energy. We can show that the energy difference between a given state and the next higher state is proportional to l + 1 by using Equation 10-27a 10-27a to express the energy difference.

= l(l + 1) E 0 r 

From Equation 10-29a 10-29a we have:

 K l

Using this equation, express the difference between one rotational state and the next higher state:

Δ E  = (l + 1)(l + 2) E 0 r  − l(l + 1) E 0 r  =

2(l + 1) E 0 r 

57 •• [SSM] You work in a bio-chemical research lab, where you are investigating the rotational energy levels of the HBr molecule. After consulting the periodic chart, you know that the mass of the bromine atom a tom is 80 times that of the hydrogen atom. Consequently, in calculating the rotational motion of the molecule, you assume, to a good approximation, that the Br nucleus remains  –27 stationary as the H atom (mass 1.67 × 10  kg) revolves around it. You also a lso know

1012  Chapter 10 that the separation between the H atom and bromine nucleus is 0.144 nm. Calculate (a (a) the moment of inertia of the HBr molecule about the bromine nucleus, and (b (b) the rotational energies for the bromine nucleus’s ground nucleus’s ground state (lowest energy) l = 0, and the next two states of higher energy (called the first and second excited states) states) described by l = 1, and l = 2. Picture the Problem The rotational energies of HBr molecule are related to l and  E 0 r  according to K l = l(l + 1) E 0 r   wh  where E 0 r  = h 2 2 I .

(a) Neglecting the motion of the  bromine molecule:

 I HBr  ≈ m p r 2

= mH r 2

Substitute numerical values and evaluate I  evaluate I HBr  HBr :

 I HBr  ≈ (1.67 × 10 −27 kg )(0.144 × 10 −9 m )

2

= 3.463 × 10 −47 kg ⋅ m 2 = 3.46 × 10 −47 kg ⋅ m 2

(b) Relate the rotational energies to l and E 0 r  : Substitute numerical values and evaluate  E 0 r  :

 K l

= l(l + 1) E 0 r  where  E 0 r  =

 E 0 r 

1.055 × 10 − ( = = 2 I  2(3.463 × 10 − h

2

34

47

= 1.607 ×10 −22 J ×

h

2

2 I HBr 

J ⋅ s)

2

kg ⋅ m 2 ) 1eV

1.602 × 10 −19 J

= 1.003 meV Evaluate E  Evaluate E 0 to obtain:

 E 0 = K 0

= 1.00 meV

Evaluate E  Evaluate E 1 to obtain:

 E 1 = K 1

= (1 + 1)(1.003 meV )

= Evaluate E  Evaluate E 2 to obtain:

2.01meV

 E 2 = K 2

=

= 2(2 + 1)(1.003 meV )

6.02 meV

58 ••• The equilibrium separation between the nuclei of the nitrogen molecule (N2) is 0.110 nm and the mass of each nitrogen nucleus is 14.0 u, where  –27 u = 1.66 × 10  kg. For rotational energies, the total energy is due to rotational kinetic energy. (a (a) Approximate the nitrogen molecule as a rigid dumbbell of two equal point masses and calculate the moment of inertia about its center of mass.

Angular Momentum 1013 (b) Find the energy E  energy E l  of the lowest three energy levels using  E l

= K  = l(l + 1)h 2 (2 I ) . (c (c) Molecules emit a particle (or quantum) quantum) of light l

called a photon a photon when  when they make a transition from transition from a higher energy state to a lower one. Determine the energy of a photon emitted when a nitrogen molecule drops from the l = 2 to the l = 1 state. Visible light photons each have between 2 and 3 eV of energy. Are these photons in the visible region? Picture the Problem We can use the definition of the moment of inertia of point  particles to calculate the rotational inertia of the nitrogen molecule. The rotational energies of nitrogen molecule are related to l   and  E 0 r  according to

 E l = K l

 where E 0 r  = h 2 = l(l + 1) E 0r   wh

(a) Using a rigid dumbbell model, express and evaluate the moment of inertia of the nitrogen molecule about its center of mass: Substitute numerical values and evaluate  I  N 2 :

2 I .

= ∑ mi r i 2 = m N r 2 + m N r 2

 I  N 2

i

= 2m N r 2

= 2(14)(1.66 ×10

 I  N 2

− 27

⎛ 0.110 nm ⎞ kg )⎜ ⎟ ⎝  2  ⎠

= 1.406 ×10 −46 kg ⋅ m 2 = (b) Relate the rotational energies to l and E 0 r  :

 E l = K l

= l(l + 1) E 0 r 

where  E 0 r  =

Substitute numerical values and evaluate  E 0 r  :

1.41× 10 − 46 kg ⋅ m 2

h

2

2 I  N 2

( 1.055 × 10 − = 2(1.406 ×10 −

34

 E 0 r 

46

J ⋅ s)

kg ⋅ m 2 )

= 3.958 × 10 −23 J ×

1eV 1.60 × 10 − J

= 0.2474 meV Evaluate E  Evaluate E 0 to obtain:

 E 0  = 0.247 meV

Evaluate E  Evaluate E 1 to obtain:

 E 1

= (1 + 1)(0.2474 meV ) =

0.495 meV

2

19

2

1014  Chapter 10 Evaluate E  Evaluate E 2 to obtain:

 E 2

= 2(2 + 1)(0.2474 meV ) =

(c) The energy of a photon emitted when a nitrogen molecule drops from the l = 2 to the l = 1 state is:

1.48 meV

Δ E l = 2→l =1

= E 2 − E 1 = 1.48 meV − 0.495 meV = 0.99 meV

 No. This energy is too low to produce radiation in the visible portion of the spectrum. a transition from  from a lower energy state to a higher one. That 59 ••• Consider a transition is, the absorption of a quantum qua ntum of energy resulting in an increase in the rotational energy of an N2 molecule (see Problem 64). Suppose such a molecule, initially in its ground rotational state, was exposed to photons each with energy equal to the three times the energy of its first excited state. (a) Would the molecule be able to absorb this photon energy? Explain why or why not and if it can, determine the energy level to which it goes. (b (b) To make a transition from its ground state to its second excited state requires how many times the energy of the first excited state? Picture the Problem The rotational energies of a nitrogen molecule depend on the quantum number l according to  E l =  L2 / 2 I  = l(l + 1)h 2 / 2 I .

(a) No. None of the allowed values of  E l are equal to E  to E 0r  0r . (b) The upward transition from the ground state to the second excited state requires energy given by:

Δ E l =0→l =2

Set this energy difference equal to a st constant n times the energy of the 1 excited state:

 E 2 − E 0

Substitute numerical values and evaluate n:

n=

= E 2 − E 0

= nE 1 ⇒ n =

2(2 + 1) E 0 r  − E 0 r 

(1 + 1) E 0r 

 E 2 − E 0  E 1

=

2.5

Collisions with Rotations A 16.0-kg, 2.40-m-long rod is supported at its midpoint on a knife 60 •• edge. A 3.20-kg ball of clay drops from rest from a height of 1.20 m and makes a  perfectly inelastic collision with the rod 0.90 m from the point of support (Figure 10-51). Find the angular momentum of the rod and clay system about the point of support immediately after the inelastic collision.

Angular Momentum 1015 Picture the Problem Let the zero of gravitational potential energy be at the elevation of the rod. Because the net external torque acting on this system is zero, we know that angular momentum is conserved in the collision. We’ll use the definition of angular momentum to express the angular momentum just after the collision and conservation of mechanical energy to determine the speed of the ball  just before it makes its perfectly inelastic collision collision with the rod.

Use conservation of angular momentum to relate the angular momentum before the collision to the angular momentum just after the  perfectly inelastic collision:

 Lf  = Li

Use conservation of mechanical energy to relate the kinetic energy of the ball just before impact to its initial potential energy:

 K f  − K i + U f  − U i

Letting h represent the distance the  ball falls, substitute for  K f  and U i

= mvr  

(1)

=0

or, because K  because K i = U f f  = 0,  K f  − U i = 0

1 2

mv 2

− mgh = 0 ⇒ v =

2 gh

to obtain: Substituting for v in equation (1) yields:

 Lf 

= mr  2 gh

Substitute numerical values and evaluate L evaluate Lf :  Lf 

= (3.20 kg )(0.90 m) 2(9.81m/s 2 )(1.20 m ) = 14 J ⋅ s

length L and mass 61 •• [SSM] Figure 10-52 shows a thin uniform bar of length L  M and a small blob of putty of mass m. The system is supported by a frictionless r horizontal surface. The putty moves to the right with velocity v , strikes the bar at a distance d from the center of the bar, and sticks to the bar at the point of contact. Obtain expressions for the velocity of the system’s center of mass and for the angular speed following the collision. Picture the Problem The velocity of the center of mass of the bar-blob system does not change during the collision and so we can calculate it before the collision using its definition. Because there are no external forces or torques acting on the  bar-blob system, both linear and angular angu lar momentum are conserved in the collision. Let the direction the blob of putty is moving initially be the + x   x  direction. Let lower-case letters refer to the blob of putty and upper-case letters refer to the bar.

1016  Chapter 10 The diagram to the left shows the blob of putty approaching the bar and the diagram to the right shows the bar-blob system rotating about its center of mass and translating after the perfectly inelastic collision.  M 

 M 

cm

 ycm

d 

vcm

d 

m

ω 

v

r

r

r

( M  + m )v cm = mv + M V 

The velocity of the center of mass  before the collision is given by:

r

or, because V  = 0 , r

v cm

Using its definition, express the location of the center of mass relative to the center of the bar:

=

m  M  + m

r

v

( M  + m ) ycm = md  ⇒ ycm =

md 

 M  + m  below the center of the bar.

Express the angular momentum, relative to the center of mass, of the bar-blob system:

 Lcm

 Lcm

= I cm ⇒ ω  =

Express the angular momentum about the center of mass:

 Lcm

= mv(d  − ycm ) md   ⎞ mMvd  = mv⎛  ⎜ d  − ⎟= ⎝   M  + m ⎠  M  + m

Using the parallel axis theorem, express the moment of inertia of the system relative to its center of mass:

 I cm

2 = 121  ML2 + Mycm + m(d  −  ycm )2

 I cm

 

Substitute for y for ycm and simplify to obtain:

 I cm

=

2 1 12 ML

2

2

2 md   ⎞ md   ⎞ ⎛  ⎛  2 mMd  1 + M ⎜ ⎟ = 12 ML + ⎟ + m⎜ d  − + +  M  m  M  m  M  + m  ⎠ ⎝   ⎠ ⎝ 

(1)

Angular Momentum 1017 Substitute for I  for I cm and Lcm in equation cm and L (1) and simplify to obtain:

ω  =

mMvd   ML ( M  + m ) + Mmd 2

1 12

2

Remarks: You can verify the expression for I  for I cm cm by letting m 2 1  I cm = 12 ML and letting M  letting M  0 to obtain I  obtain I cm cm = 0.

0 to obtain

Figure 10-52 shows a thin uniform bar whose length is L is L and  and mass is M is M and a compact hard sphere whose mass is m. The system is supported by a frictionless horizontal surface. The sphere moves to the right r with velocity v , strikes the bar at a distance 14 L  from the center of the bar. 62

••

The collision is elastic, and following the collision the sphere is at rest. Find the value of the ratio m/M . Picture the Problem Because there are no external forces or torques acting on the bar-sphere system, both linear and angular momentum are conserved in the collision. Kinetic energy is also conserved in the elastic collision of the hard sphere with the bar. Let the direction the sphere is moving initially be the + x direction Let lower-case letters refer to the compact hard sphere and upper-case characters refer to the bar. Let unprimed characters refer to before the collision and primed characters to after the collision. The diagram to the left shows the path of the sphere before its collision with the bar and the diagram to the right shows the sphere at rest after the collision and the bar rotating about its center of mass and translating to the right.  M 

 M 

V '  cm

cm

d= 14 L m

m

v

Apply conservation of linear momentum to the collision to obtain:

mv = 0 + MV ' ⇒ V ' =

Apply conservation of angular momentum to the collision to obtain:

mvd  = 0 + I cm

 

ω 

m  M 

v

(1)

(2)

1018  Chapter 10 1 2

Apply conservation of mechanical energy to the elastic collision to obtain:

mv 2

= 0 + 12 MV '2 + 12 I cmω 2  

(3)

= 121  ML2

Use Table 9-1 to find the moment of inertia of a thin bar about an axis through its center:

 I cm

Substitute for  I cm in equation (2) and

mvd  =  = 121  ML2ω  ⇒ ω  = ⎜

⎛ 12vd  ⎞ m ⎟ ⎝   L2  ⎠ M 

simplify to obtain: Substitute for  I cm  and V ' in equation

mv

(3) and simplify to obtain:

2

m  ⎞ = M ⎛  ⎜ ⎟ ⎝  M  ⎠

2

+ 121  ML2ω 2

v2

Substituting for ω  yields:  yields:

mv

Solve this equation for

Because d = L/4:  L/4:

2

m  ⎞ = M ⎛  ⎜ ⎟ ⎝  M  ⎠

m  M 

2

v

 to obtain:

2

+ 121  ML

2

m  M 

m  M 

=

=

⎡⎛ 12vd  ⎞ m ⎤ ⎜ 2 ⎟ ⎥ ⎢⎝  ⎣  L  ⎠ M ⎦

2

1

⎛ d  ⎞ 1 + 12⎜ ⎟ ⎝  L ⎠

2

1

⎛ 1 ⎞ 1 + 12⎜ ⎟ ⎝ 4 ⎠

2

=

4 7

Figure 10-53 shows a uniform rod whose length is L is L and whose mass 63 •• is M is M pivoted  pivoted at the top. The rod, which is initially at rest, is struck by a particle whose mass is m at a point x point x = 0.8 L below  L below the pivot. Assume that the particle sticks to the rod. What must be the speed v of the particle so that following the collision the maximum angle between the rod and the vertical is 90°? distance x Picture the Problem Let the zero of gravitational potential energy be a distance x  below the pivot and ignore friction between the rod and the pivot. Because Beca use the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its  potential energy at the top of its swing.

Angular Momentum 1019

Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ : θ  :

Δ K  + ΔU  = 0 or, because K  because K f f  =  = U i = 0, − K i + U f  = 0

Substitute for K  for K i and U f f to obtain:    L  ⎞ − 12 I ω 2 + ⎛  ⎜ Mg  + mgx ⎟(1 − cosθ ) = 0   ⎝  2  ⎠ Applying conservation of angular momentum to the collision gives:

Δ L

The moment of inertia of the system about the pivot is given by:

 I f 

Substituting for I  for I f f  and I   and I i in equation (2) yields:

1 3

Solving for ω  yields:  yields:

(1)

= Lf  − Li = 0  

(2)

= m(0.8 L )2 + 13 ML2  

 ML2

ω  =

+ (0.8 L )2 m ω − 0.8 Lmv = 0 0.8 Lmv

 ML2

1 3

+ 0.64mL2

 

Substitute equations (3) and (4) in equation (1) and simplify to obtain: 0.32( Lmv )

2

1 3

2

 ML

+ 0.64mL

2

(3)

 L  ⎞ − ⎛  ⎜ Mg  + mg (0.8 L) ⎟(1 − cosθ ) = 0 ⎝  2  ⎠

(4)

1020  Chapter 10 Solving for v gives:

v=

(0.5 M  + 0.8m ) (13 ML2 + 0.64mL2 ) g (1 − cosθ ) 0.32 Lm 2

Evaluate v for θ  =  = 90° to obtain:

v=

(0.5 M  + 0.8m )(13 ML2 + 0.64mL2 ) g  0.32 Lm 2

If, for the system of Problem 63, L 63, L = 1.2 m, M m, M = 0.80 kg, m = 0.30 kg, 64 •• and the maximum angle between the rod and the vertical following the collision is 60°, find the speed of the particle before impact. Picture the Problem Let the zero of gravitational potential energy be a distance  x  below the pivot and ignore friction between the rod and pivot. Because the net external torque acting on the system is zero, angular momentum is conserved in this  perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing.

Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ  :  :

 K f  − K i + U f  − U i

=0

or, because K  because K f f  =  = U i = 0, − K i + U f  = 0

Substitute for K  for K i and U f f to obtain:    L  ⎞ − 12 I ω 2 + ⎛  ⎜ Mg  + mgx ⎟(1 − cosθ ) = 0   ⎝  2  ⎠

(1)

Angular Momentum 1021

= Lf  − Li = 0  

Apply conservation of momentum to the collision:

Δ L

The moment of inertia of the system about the pivot is:

 I  = m(0.80 L )

Substituting for L for Lf  and L  and Li in equation (2) gives:

(13 M  + 0.64m) L2ω  − 0.80 Lmv = 0

Solving for ω  yields:  yields:

ω  =

2

(2)

+ 13 ML2

= (0.64m + 13 M ) L2

0.80 Lmv  ML2

1 2

+ 0.64mL2

Substitute for ω  in equation (1) and simplify to obtain: 0.32( Lmv )

2

−

 I 

 L  ⎞ + ⎛  ⎜ Mg  + 0.80 Lmg ⎟ (1 − cosθ ) = 0 ⎝  2  ⎠

Solving for v yields:

 g (0.50 M  + 0.80m )(1 − cosθ ) I 

v=

Substitute numerical values and evaluate I  evaluate I :

0.32 Lm

2

 I  = [0.64(0.30 kg ) + 13 (0.80 kg )](1.2 m )

2

= 0.660 kg ⋅ m 2

Substitute numerical values and evaluate v for θ  =  = 60° to obtain: v=

(9.81m/s ) [0.50 (0.80 kg ) + (0.80)(0.30 kg )](0.50)(0.660 kg ⋅ m ) = 2

2

)(0.30 kg ) 0.32 (1.2 m )(

2

7.7 m/s

A uniform rod is resting on a frictionless table when it is suddenly 65 •• struck at one end by a sharp horizontal blow in a direction perpendicular to the rod. The mass of the rod is M is  M and the magnitude of the impulse applied by the  blow is J  is J . Immediately after the rod is struck, struck, (a (a) what is the velocity of the center of mass of the rod, (b ( b) what is the velocity of the end that is struck, (c ( c) and what is the velocity of the other end of the rod? (d  (d ) Is there a point on the rod that remains motionless?

1022  Chapter 10 Picture the Problem Let the length of the uniform stick be l. We can use the impulse-change in momentum theorem to express the velocity of the center of mass of the stick. By expressing the velocity V  of   of the end of the stick in terms of the velocity of the center of mass and applying the angular impulse-change in angular momentum theorem we can find the angular velocity of the stick and, hence, the velocity of the end of the stick.

(a) Apply the impulse-change in momentum theorem to obtain:

 J  = Δ p =  p −  p0

(b) Relate the velocity V  of  of the end of the stick to the velocity of the center of mass vcm :

V  = vcm

Relate the angular impulse to the change in the angular momentum of the stick:

 J ( 12 l ) = Δ L = L − L0

Refer to Table 9-1 to find the moment of inertia of the stick with respect to its center of mass:

 I cm

Substitute for I  for I cm cm in equation (2) to obtain:

 J ( 12 l ) = 121  M l 2ω  ⇒ ω  =

Substituting for ω in ω  in equation (1) yields:

V  =

(c) Relate the velocity V ′ of the other end of the stick to the velocity of the center of mass vcm :

=  p

or, because p because p0 = 0 and p and p =  = Mv  Mvcm,  J   J  = Mvcm ⇒ vcm =  M 

+ vrel to cm = vcm + ( 12 l )  

= I cm

or, because L because L0 = 0,  J ( 12 l ) = I cm  

(2)

= 121  M l 2

6 J   M l

6 J  ⎞ l 4 J  + ⎛  ⎜ ⎟ =  M  ⎝  M l ⎠ 2  M   J 

V  = vcm

=

(1)

− vrel to cm = vcm − ( 12 l )

6 J  ⎞ l 2 J  − ⎛  ⎜ ⎟ = −  M  ⎝  M l ⎠ 2  M   J 

(d ) Yes, one point remains motionless, but only for a very brief time. r

A projectile of mass m p is traveling at a constant velocity v 0  toward a 66 •• stationary disk of mass M mass M and radius R radius R that  that is free to rotate about its axis O (Figure 10-54). Before impact, the projectile is traveling along a line displaced a distance b below the axis. The projectile strikes the disk and sticks to point B point B.. Model the projectile as a point mass. (a (a) Before impact, what is the total angular momentum L momentum L0 of the disk-projectile system about the axis? Answer the following questions in terms of the symbols given at the start of this problem. (b (b) What is

Angular Momentum 1023 the angular speed ω  of  of the disk-projectile system just after the impact? (c ( c) What is the kinetic energy of the disk-projectile system after impact? (d  (d ) How much mechanical energy is lost in this collision? Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision.

(a) Use its definition to express the total angular momentum of the disk and projectile just before impact: (b) Use conservation of angular momentum to relate the angular momenta just before and just after the collision: The moment of inertia of the disk projectile after the impact is: Substitute for I for I in the expression for ω to ω  to obtain:

(c) Express the kinetic energy of the system after impact in terms of its angular momentum:

 L0

=

 L0

= L = I  ⇒ ω  =

m p v0 b

 I  = 12 MR 2 + m p R 2

 L0  I 

= 12 ( M  + 2m p ) R 2

2m p v0b

ω  =

( M  + 2m ) R

2

 p

 K f 

= =

 L2

=

2 I 

(m v b )

2

 p 0

[

2 12 ( M  + 2m p ) R

(m v b )

2

]

2

 p 0

( M  + 2m ) R

2

 p

(d ) Express the difference between the initial and final kinetic energies, substitute, and simplify to obtain:

Δ E 

= K i − K f 

(m v b ) − ( M  + 2m ) R 2

= 12 m

2  p 0

v

 p 0

2

 p

=

1 2

2  p 0

mv

⎡ ⎤ 2m p b 2 ⎢1 − 2 ⎥ ⎢⎣ ( M  + 2m p ) R ⎦⎥

length L1 and mass M mass M equal to 0.75 kg is 67 •• [SSM] A uniform rod of length L attached to a hinge of negligible mass at one end and is free to rotate in the vertical plane (Figure 10-55). The rod is released from rest in the position shown. A particle of mass m = 0.50 kg is supported by a thin string of length L length L2 from the hinge. The particle sticks to the rod on contact. What should the ratio L ratio L2/ L  L1 be so that θ max max = 60° after the collision?

1024  Chapter 10 Picture the Problem Assume that there is no friction between the rod and the hinge. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational kinetic of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational  potential energy be at a distance  L1  below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L distances L1 and L  and  L2 and the masses M  masses M  and  and m.

Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the  particle: Substitute for K  for K f f,  U f f , and U i to obtain: Solving for ω  yields:  yields:

Letting ω′  represent  represent the angular speed of the rod-and-particle system  just after impact, use conservation of angular momentum to relate the angular momenta before and after the collision: Solve for ω′  to  to obtain:

Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θ max max:

 K f  − K i

+ U f  − U i = 0

or, because K  because K i = 0,  K f  + U f  − U i = 0

(  ML ) ω  2 1

1 1 2 3

 L

+ Mg  1 − MgL1 = 0 2

3 g 

ω  =

Δ L

2

 L1

= Lf  − Li = 0

or (13 ML12 + mL22 )ω '−(13 ML12 )ω  = 0

ω ' =

 ML12

1 3

 ML12

1 3

 K f  − K i

+ mL22

ω 

+ U f  − U i = 0

Angular Momentum 1025

Because K  Because K f f  =  = 0:

− 12 I ω '2 + Mg ( 12 L1 )(1 − cosθ max ) + mgL2 (1 − cosθ max ) = 0   Express the moment of inertia of the system with respect to the pivot: Substitute for θ max  and ω′  in  in max, I  and equation (1):

 I  = 13 ML12

+ mL

2 1

2 2

 ML

Simplify to obtain:

+ mL22

2  g  3 (13 ML12 )  L1 1 3

 L31

=2

(1)

= Mg ( 12 L1 ) + mgL2

m

 L12 L2  M 

m  L22 L1 + 6  L32  M   M 

+3

m

Dividing both sides of the equation by  L13  yields: 2

2 m  ⎞ ⎛  L2  ⎞ ⎛  m  ⎞ ⎛  L2  ⎞ m  ⎞ ⎛  L2  ⎞ ⎛  ⎛  1 = 2⎜ ⎟ ⎜⎜ ⎟⎟ + 3⎜ ⎟ ⎜⎜ ⎟⎟ + 6⎜ ⎟ ⎜⎜ ⎟⎟  M  ⎝   ⎠ ⎝  L1  ⎠ ⎝  M  ⎠ ⎝  L1  ⎠ ⎝  M  ⎠ ⎝  L1  ⎠

Let α = α  = m/ M   M  and β   and  β  = L  = L2/ L  L1 to obtain:

6α 2 β 3 + 3αβ 2

Substitute for α  and  and simplify to obtain the cubic equation in β  in  β :

8 β 3 + 6 β 2

Use the solver function of your calculator to find the only real value of β  of  β :

 β  = 0.39

3

+ 2αβ  − 1 = 0

+ 4β  − 3 = 0

A uniform rod that has a length L length L1 equal to 1.2 m and a mass M mass M equal 68 •• to 2.0 kg is supported by a hinge at one end and is free to rotate in the vertical  plane (Figure 10-55). The rod is released from rest in the position shown. A  particle whose mass is m is supported by a thin string that has a length L length L2 equal to 0.80 m from the hinge. The particle sticks to the rod on contact, and after the collision the rod continues to rotate until θ max (a) Find m. (b (b) How much max = 37°. (a energy is dissipated during the collision? Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational

1026  Chapter 10 kinetic energy of the after-collision system is then transformed into gravitational  potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L distance  L1  below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L distances L1 and L and L2 and the mass M  mass M  to  to m. (a) Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the particle: Substitute for K  for K f f,  U f f , and U i to obtain: Solving for ω  yields:  yields:

Letting ω′  represent  represent the angular speed of the system after impact, use conservation of angular momentum to relate the angular momenta before and after the collision: Solving for ω ′ and simplifying yields:

 K f  − K i

+ U f  − U i = 0

or, because K  because K i = 0,  K f  + U f  − U i = 0

(  ML ) ω  2 1

1 1 2 3

 L

+ Mg  1 − MgL1 = 0 2

3 g 

ω  =

Δ L

2

 L1

= Lf  − Li = 0

or (13 ML12

ω ' =

=

+ mL22 )ω '−(13 ML12 )ω  = 0   (1)

 ML12

1 3

 ML12

1 3

+ mL22

 ML12

1 3 1 3

2 1

 ML

ω  3 g 

+ mL

2 2

 L1

Substitute numerical values and simplify to obtain:

(2.0 kg )(1.2 m )2 ω ' = 1 (2.0 kg )(1.2 m )2 + m(0.80 m )2 3 1 3

3(9.81m/s 2 ) 1.2 m

=

4.75 kg / s 0.960 kg + 0.64m

Angular Momentum 1027

Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θ max max:

 K f  − K i + U f  − U i

=0

Substitute for K  for K i, U f f , and U i to obtain:

− 12 I ω '2 + Mg ( 12 L1 )(1 − cosθ max ) + mgL2 (1 − cosθ max ) = 0

Express the moment of inertia of the system with respect to the pivot:

 I  = 13 ML12

or, because K  because K f f  = 0, − K i + U f  − U i = 0

+ mL22

Substitute for θ max  and ω′  in  in max, I  and equation (1) and simplify to obtain:

(4.75 kg/s )2 = 0.2 g ( ML1 + mL2 ) 0.960 kg + 0.64m

Substitute for M  for M , L1 and L  and L2 and simplify to obtain:

(4.75 kg/s )2 0.960 kg + 0.64m = 0.2 g (2.4 kg ⋅ m + (0.80 m )m )

Solve for m to obtain:

m = 1.18 kg = 1.2 kg

(b) The energy dissipated in the inelastic collision is:

Δ E  = U i − U f   

Express U i:

 L U i = Mg  1 2

Express U f f :

1 2

1 2

U f 

 L  ⎞ = (1 − cosθ max ) g ⎛  ⎜ M  1 + mL2 ⎟ ⎝  2  ⎠

Substitute for U i and U f f  in equation (2) to obtain: Δ E 

(2)

 L  L  ⎞ = Mg  1 − (1 − cosθ max ) g ⎛  ⎜ M  1 + mL2 ⎟ 2 ⎝  2  ⎠

1028  Chapter 10

Substitute numerical values and evaluate Δ E :

Δ E 

=

(2.0 kg )(9.81m/s 2 )(1.2 m ) 2

(2.0 kg )(1.2 m ) − (1 − cos37° )(9.81m/s 2 )⎛  + (1.18 kg )(0.80 m ) ⎞⎟ ⎜ 2 ⎝   ⎠ = 7.5 J Precession 69 •• [SSM] A bicycle wheel that has a radius equal to 28 cm is mounted at the middle of an axle 50 cm long. The tire and rim weigh 30 N. The wheel is spun at 12 rev/s, and the axle is then placed in a horizontal position with one end resting on a pivot. (a (a) What is the angular momentum due to the spinning of the wheel? (Treat the wheel as a hoop.) (b (b) What is the angular velocity of  precession? (c (c) How long does it take for the axle to swing through 360° around the pivot? (d  (d ) What is the angular momentum associated with the motion of the center of mass, that is, due to the precession? In what direction is this angular momentum? Picture the Problem We can determine the angular momentum of the wheel and the angular velocity of its precession from their definitions. The period of the  precessional motion can be found from its angular velocity and the angular momentum associated with the motion of the center of mass from its definition.

(a) Using the definition of angular momentum, express the angular momentum of the spinning wheel: Substitute numerical values and evaluate L evaluate L::

 L =  I ω  = MR 2ω  =

w 2  R ω   g 

⎛  30 N  ⎞ ⎟(0.28 m )2 2 ⎟ ⎝ 9.81m/s  ⎠ rev 2 rad ⎞ × ⎛  ⎜12 × ⎟ rev  ⎠ ⎝  s = 18.1J ⋅ s = 18 J ⋅ s

 L = ⎜⎜

π 

(b) Using its definition, express the angular velocity of precession:

ω  p

=

d φ   MgD dt 

=

 L

Angular Momentum 1029 Substitute numerical values and evaluate ω  p:

ω  p

=

(30 N )( )(0.25 m ) = 0.414 rad/s 18.1 J ⋅ s

= (c) Express the period of the  precessional motion as a function of the angular velocity of  precession:

T  =

(d ) Express the angular momentum of the center of mass due to the precession:

 L p

Substitute numerical values and evaluate L p :

 L p

0.41rad/s

2π  ω  p

=

2π  0.414 rad/s

= 15 s

=  I cmω  p = MD 2ω  p

⎛  30 N  ⎞ ⎟(0.25 m )2 (0.414 rad/s ) = ⎜⎜ 2 ⎟ ⎝ 9.81m/s  ⎠ = 0.079 J ⋅ s

The direction of  L p is either up or down, depending on the direction of L. of L. A uniform disk, whose mass is 2.50 kg and radius is 6.40 cm is 70 •• mounted at the center of a 10.0-cm-long axle and spun at 700 rev/min. The axle is then placed in a horizontal position with one end resting on a pivot. The other end is given an initial horizontal velocity such that the precession is smooth with no nutation. (a (a) What is the angular velocity of precession? (b (b) What is the speed of the center of mass during the precession? (c (c) What is the acceleration (magnitude and direction) of the center of mass? (d  ( d ) What are the vertical and horizontal components of the force exerted by the pivot on the axle? Picture the Problem The angular velocity of precession can be found from its definition. Both the speed and the magnitude of the acceleration of the center of mass during precession are related to the angular velocity of precession. We can use Newton’s second law to find the vertical and horizontal components of the force exerted by the pivot on the axle.

(a) The angular velocity of precession is given by:

ω  p

=

Substituting for I  for I s and simplifying yields:

ω  p

=

d φ   MgD dt 

=

 I sω s

 MgD  MR 2ω s

1 2

=

2 gD  R 2ω s

1030  Chapter 10

Substitute numerical values and evaluate ω  p:

ω  p

=

2(9.81 m/s 2 ) (0.050 m )

⎛  rev 2π rad 1min ⎞ ⎟⎟ (0.064 m ) ⎜⎜ 700 × × min rev 60 s ⎝   ⎠

= 3.27 rad/s =

3.3 rad/s

2

(b) Express the speed of the center of mass in terms of its angular speed of precession:

vcm = D

(c) Relate the acceleration of the center of mass to its angular speed of  precession:

acm

(d ) Use Newton’s second law to relate the vertical component of the force exerted by the pivot to the weight of the disk:

 F v = Mg  = (2.5 kg )(9.81m/s 2 )

Relate the horizontal component of the force exerted by the pivot on the axle to the acceleration of the center of mass:

 F H = Macm

=

 p

= (0.050 m )(3.27 rad/s )

16 cm/s

= Dω  p2 = (0.050 m )(3.27 rad/s )2 = 0.535 m/s 2 =

=

54 cm/s 2

25 N

=

= (2.5 kg )(0.535 m/s 2 )

1.3 N

General Problems particle whose mass is 3.0 kg moves moves in in the the xy  xy plane  plane with 71 • [SSM] A particle r ˆ velocity v = (3.0 m / s)i along the line y line y =  = 5.3 m. (a (a) Find the angular momentum r

 L about the origin when the particle is at (12 m, 5.3 m). (b (b) A force r F = (–3.9 N) iˆ is applied to the particle. Find the torque about the origin due to

this force as the particle passes through the point (12 m, 5.3 m). Picture the Problem While the 3-kg particle is moving in a straight line, it has r

r

r

r

r

angular momentum given by  L = r ×  p where r  is its position vector and  p is its r

linear momentum. The torque due to the applied force is given by τ  (a) The angular momentum of the  particle is given by:

r

r

r

 L = r ×  p

r

r

= r × F .

Angular Momentum 1031 r

r

r ˆ + (5.3 m ) j  ˆ r  = (12 m) i 

Express the vectors r  and  p :

and

r ˆ  p = mvˆi = (3.0 kg ) (3.0 m/s )i 

= (9.0 kg ⋅ m/s )i ˆ r

r

r

ˆ + (5.3 m ) j  ˆ × (9.0 kg ⋅ m/s )i  ˆ  L = (12 m ) i 

Substitute for r  and  p :and simplify r

= (47.7 kg ⋅ m 2 /s )( j ˆ × i ˆ )

to find  L :

= − (48 kg ⋅ m 2 /s )k ˆ r

(b) Using its definition, express the torque due to the force:

τ 

r

r

r

Substitute for r  and F and simplify to r find τ  :

τ 

r

r

= r × F 

= (12 m ) i ˆ + (5.3 m ) j ˆ × (− 3.9 N )i ˆ = −(15.9 N ⋅ m )( j ˆ × i ˆ ) = (21 N ⋅ m ) k ˆ

The position vector of a particle whose mass is 3.0 kg is given by r 2 ˆ  + 3.0t  3.0t   jˆ , where r is in meters and t is in seconds. Determine the r = 4.0 i  angular momentum and net torque, about the origin, acting on the particle.

72 r

•

Picture the Problem The angular momentum of the particle is given by r

r

r

r

r

 L = r ×  p where r  is its position vector and  p  is its linear momentum. The torque r

r

 L dt . acting on the particle is given by τ  = d  r

Evaluating

dt 

= mr × r

[

dt  r

Substitute for mr  and r

d r 

yields:

r

{

d r  r  dt 

r

r

r

r  d r 

r

r

r

 L = r ×  p = r × mv

The angular momentum of the  particle is given by:

=

d 

r

r

r

= mr × v

d r  dt 

ˆ ] = (6.0t ) j  ˆ [ 4.0iˆ + 3.0t   j  dt  2

r

and simplify to find  L :

(

) }]× (6.0t m/s) j ˆ = (72t J ⋅ s )k ˆ

ˆ + 3.0t 2 m/s 2  j  ˆ  L = (3.0 kg ) (4.0 m ) i 

1032  Chapter 10 r

Find the net torque due to the force:

r τ 

=

net

 L d 

dt 

=

d 

[ (72t J ⋅ s )k ˆ ] dt 

= (72 N ⋅ m )k ˆ Two ice skaters, whose masses are 55 kg and 85 kg, hold hands and 73 •• rotate about a vertical axis that passes between them, making one revolution in 2.5 s. Their centers of mass are separated by 1.7 m and their center of mass is stationary. Model each skater as a point particle and find (a (a) the angular momentum of the system about their center ce nter of mass and (b (b) the total kinetic energy of the system. Picture the Problem The ice skaters rotate about their center of mass; a point we can locate using its definition. Knowing the location of the center of mass we can determine their moment of inertia with respect to an axis through this point. The angular momentum of the system is then given by  L =  I cm  and its kinetic energy

can be found from  K  =  =  L2 (2 I cm ). (a) Express the angular momentum of the system about the center of mass of the skaters: Using its definition, locate the center of mass, relative to the 85-kg skater, of the system:

 L =  I cm

 xcm

=

(55 kg )(1.7 m ) + (85 kg )(0 ) 55 kg + 85 kg

= 0.668 m = (55 kg )(1.7 m − 0.668 m )2 + (85 kg )(0.668 m )2 = 96.5 kg ⋅ m 2

Calculate  I cm :

 I cm

Substitute to determine L determine L::

 L = (96.5 kg ⋅ m 2 )⎜⎜

⎛ 1rev 2π rad ⎞ ⎟⎟ × 2.5 s rev ⎝   ⎠ = 243 J ⋅ s = 0.24 kJ ⋅ s  L2

(b) Relate the total kinetic energy of the system to its angular momentum and evaluate K  evaluate K :

 K  =

Substitute numerical values and evaluate K  evaluate K :

(243 J ⋅ s )2  K  = = 2(96.5 kg ⋅ m 2 )

2 I cm

0.31kJ

Angular Momentum 1033 A 2.0-kg ball attached to a string whose length is 1.5 m moves 74 •• counterclockwise (as viewed from above) in a horizontal circle (Figure 10-56). The string makes an angle θ  =  = 30° with the vertical. (a ( a) Determine both the r

horizontal and vertical components of the angular momentum  L of the ball about r

 L dt and verify that it equals the point of support P  support P . (b (b) Find the magnitude of d  the magnitude of the torque exerted by gravity about the point of support.  y

Picture the Problem Let the origin of the coordinate system be at the pivot. The diagram shows the forces acting on the ball. We’ll apply Newton’s second law to the ball to determine its speed. We’ll then use the derivative of its  position vector to express its velocity and the definition of angular

 P 

 x

θ   z  r 

r

momentum to show that  L   has both horizontal and vertical components. We r

can use the derivative of  L   with respect to time to show that the rate at which the angular momentum of the  ball changes is equal to the torque, relative to the pivot point, acting on it.

θ 

T 

m

m g 

r

r

r

r

r

(a) Express the angular momentum of the ball about the point of support:

 L = r ×  p = mr × v  

Apply Newton’s second law to the  ball:

∑

= T sin θ  = m

and  F  z 

= T cosθ  − mg  = 0

 F  x

∑

Eliminate T  between  between these equations and solve for v to obtain:

v =

Substitute numerical values and evaluate v:

v

Express the position vector of the  ball:

(1)

v2 r sin θ 

rg sin θ tan θ  tan θ 

= (1.5 m )(9.81m/s2 ) sin30°tan30° = 2.06 m/s

r ˆ + sin ω t  ˆ r  = (1.5 m )sin 30° cos ω t i   j 

(

− (1.5 m )cos 30°k ˆ

)

1034  Chapter 10 r

The velocity of the ball is:

v

r

=

d r  dt 

= (0.75ω m/s )(− sin ω t i ˆ + cos ω t  j ˆ ) Evaluating ω  yields:  yields:

ω  = r

Substitute for ω  to  to obtain:

v

2.06 m/s

(1.5 m )sin 30°

= 2.75 rad/s

= (2.06 m/s ) − sin ω t i ˆ + cos ω t  j ˆ

r

Substitute in equation (1) and evaluate  L : r

(

)

ˆ + sin ω t  j  ˆ − (1.5 m )cos 30°k  ˆ  L = (2.0 kg ) (1.5 m )sin 30° cos ω t i 

× [(2.06 m/s )(− sin ω t i ˆ + cos ω t  j ˆ )] = (5.35 J ⋅ s )cos ω t i ˆ + (5.35 J ⋅ s )sin ω t  j ˆ + (3.09 J ⋅ s )k ˆ r

The horizontal component of  L is the component in the xy the xy plane:  plane: r

 Lhor  =

(5.4 J ⋅ s )cos ω t i ˆ + (5.4 J ⋅ s )sin ω t  j ˆ r

The vertical component of  L is its z its z component:

(b) Evaluate

r

 Lvertical

r

r

d   L

d   L

dt 

:

dt  r

Evaluate the magnitude of

 L d 

dt 

ˆ = (3.1J ⋅ s )k 

= [5.36ω (− sin ω t i ˆ + cos ω t  j ˆ )] J

r

:

 L d 

dt 

= (5.36 N ⋅ m ⋅ s )(2.75 rad/s ) = 15 N ⋅ m

Express the magnitude of the torque exerted by gravity about the point of support:

τ  = mgr sin θ 

Substitute numerical values and evaluate τ : τ  :

τ  = (2.0 kg )(9.81 m/s 2 )(1.5 m )sin 30°

=

15 N ⋅ m

Angular Momentum 1035 A compact object whose mass is m resting on a horizontal, frictionless 75 •• surface is attached to a string that wraps around a vertical cylindrical post attached to the surface. Thus, when the object is set into motion, it follows a path that spirals inward. (a (a) Is the angular momentum of the object o bject about the axis of the post conserved? Explain your answer. (b (b) Is the energy of the object conserved? Explain your answer. (c (c) If the speed of the ob object ject is v0 when the unwrapped length of the string is r , what is its speed when the unwrapped length has shortened to r /2? /2? Picture the Problem The pictorial representation depicts the object rotating counterclockwise around the cylindrical post. Let the system be the object. In Part (a (a) we need to decide whether a net torque acts on the object and in Part (b (b) the issue is whether any external forces act on the object. In Part (c (c) we can apply the definition of kinetic energy to find the speed of the object when the unwrapped length has shortened to r /2. /2.

(a) The net torque acting on the object is given by:

 R

r  T  m

τ net

=

dL dt 

= RT 

Because τ net net ≠ 0, angular momentum is not conserved. (b) Because, in this frictionless environment, the net external force acting on the object is the tension force and it acts at right angles to the object’s velocity, the energy of the object is conserved. (c) Apply conservation of mechanical energy to the object to obtain:

Substituting for the kinetic energies yields:

= Δ K + ΔU  = 0 or, because ΔU  =  = 0, Δ K rot = 0 Δ E 

 I' ω ' 2 − 12 I ω 02

1 2

or  I' ω ' 2 − I ω 02

=0

=0

1036  Chapter 10 Substitute for I, for I, I ′, ω ′, and ω 0 to obtain:

⎛   ⎞ 2 2⎜ ⎟ v r  v'  ⎛   ⎞ ⎛   ⎞ 2 0 1 m⎜ ⎟ ⎜ ⎟ − 12 mr  ⎜ ⎟ = 0 2 ⎝ 2 ⎠ ⎜⎜ r  ⎟⎟ ⎝  r  ⎠ ⎝  2 ⎠

Solving for v′ yields:

 = v0 v'  =

2

Figure 10-57 shows a hollow cylindrical tube that has a mass M  mass M , a 76 •• 2 length L length L,, and a moment of inertia ML inertia ML /10. Inside the cylinder are two disks each of mass m, separated by a distance l, and tied to a central post by a thin string. The system can rotate about a vertical axis through the center of the cylinder. You are designing this cylinder-disk apparatus to shut down the rotations when the strings break by triggering an electronic ″shutoff ″ signal (sent to the rotating motor) when the disks hit the ends of the cylinder. During development, you notice that with the system rotating at some critical angular speed ω , the string suddenly breaks. When the disks reach the ends of the cylinder, they stick. Obtain expressions for the final angular speed and the initial and final kinetic energies of the system. Assume that the inside walls of the cylinder are frictionless. Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular velocities of the system. See Table 9-1 for the moment of inertia of a disk.

Using conservation of angular momentum, relate the initial and final angular speeds to the initial and final moments of inertia: Solving for 

f 

yields:

Δ L

= Lf  − Li = 0

or  I f 

f 

ω f 

=

− I i  I i  I f 

i

ω i

=0

=

 I i  I f 

Use the parallel-axis theorem to express the moment of inertia of each of the disks with respect to the axis of rotation:

 I i, each disk  = I cm

Express the initial moment of inertia  I i of the cylindrical tube plus disks system:

 I i = I cylindrical

ω   

+ m( 12 l )2

= 14 mr 2 + 14 ml 2 = 14 m(r 2 + l 2 ) + 2 I i, each disk 

tube

= 101  ML2 + 2[14 m(r 2 + l 2 )] = 101  ML2 + 12 m(r 2 + l 2 )

(1)

Angular Momentum 1037 When the disks have moved out to the end of the cylindrical tube:

 I f  = 101  ML2 + 12 m(r 2 + L2 )

Substitute for I  for I i and I   and I f f in   in equation (1) and simplify to obtain:

+ 12 m(r 2 + l 2 ) ω f  = ω  2 2 2 1 1  ML m r   L + + ( ) 10 2  ML2

1 10

+ 5m(r 2 + l 2 ) ω  =  ML2 + 5m(r 2 + L2 )  ML2

The initial kinetic energy of the system is:

 K i

= 12 I iω 2

Substituting for I  for I i and simplifying yields:

 K i

= 12 101  ML2 + 12 m(r 2 + l 2 ) ω 2

The final kinetic energy of the system is:

 K f 

=

[

 ML2 + 14 m(r 2

1 20

+ l 2 )]ω 2

= 12 I f ω f 2

Substitute for I  for I f f  and  and ω f f  and simplify to obtain: 2 2 2 2 2 2 ⎛  ML + 5m(r  + l )  ⎞ 1 1 1 ω ⎟⎟  K f  = 2 [10 ML + 2 m(r  + L )]⎜⎜ 2 2 2  ML + 5 m r  +  L ( ) ⎝   ⎠ ⎡ [ ML2 + 5m(r 2 + l 2 )] 2 ⎤ 2 1 = 20 ⎢ ⎥ ω  2 2 2 ⎢⎣  ML + 5m(r  + L ) ⎥⎦

2

the disks and 77 •• [SSM] Repeat Problem 76, this time friction between the the walls of the cylinder is not negligible. However, the coefficient of friction is not great enough to prevent the disks from reaching the ends of the cylinder. Can the final kinetic energy of the system be determined without knowing the coefficient of kinetic friction? Determine the Concept Yes. The solution depends only upon conservation of angular momentum of the system, so it depends only upon the initial and final moments of inertia.

Suppose that in Figure 10-57 l = 0.60 m, L m, L =  = 2.0 m, M m, M = 0.80 kg, and 78 •• m = 0.40 kg. The string breaks when the system’s angular speed approaches the critical angular speed ω i, at which time the tension in the string is 108 N. The masses then move radially outward until they undergo perfectly inelastic collisions with the ends of the cylinder. De termine the critical angular speed and the angular speed of the system after the inelastic collisions. Find the total kinetic

1038  Chapter 10 energy of the system at the critical angular speed, and again after the inelastic collisions. Assume that the inside walls of the cylinder are frictionless. Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular speeds of the system.

Using conservation of angular momentum, relate the initial and final angular speeds to the initial and final moments of inertia: Express the tension in the string as a function of the critical angular speed of the system: Substitute numerical values and evaluate i :

Δ L

= Lf  − Li = 0

or  I f 

f 

− I i

T  = mr ω i2

=

ω i

i

= 0 ⇒ ω f  =

=m

l

2

 I i  I f 

ω i2 ⇒ ω i

2(108 N )

(0.40 kg )(0.60 m )

ω i   (1)

=

2T  ml

= 30.0 rad/s

= 30 rad/s Express I  Express I i:

 I i

= 101  ML2 + 2(14 ml 2 )

Substitute numerical values and evaluate I  evaluate I i:

 I i

= 101 (0.80 kg )(2.0 m )2

Express I  Express I f f :

 I f  = 101  ML2

Substitute numerical values and evaluate I  evaluate I f f: 

 I f 

Substitute numerical values in equation (1) and evaluate ω f f :

+ 12 (0.40 kg )(0.60 m )2 = 0.392 kg ⋅ m 2 + 2(14 mL2 )

= 101 (0.80 kg )(2.0 m )2 + 12 (0.40 kg )(2.0 m )2 = 1.12 kg ⋅ m 2

ω f 

=

0.392 kg ⋅ m 2 1.12 kg ⋅ m 2

= 10.5 rad/s =

(30.0 rad/s )

11 rad/s

Angular Momentum 1039

The total kinetic energy of the system at the critical angular speed is:

 K i

= 12 I iω i2

Substitute numerical values and evaluate K i :

 K i

= 12 (0.392 kg ⋅ m 2 )(30.0 rad/s )2

The total kinetic energy of the system after the inelastic collisions is:

 K f 

= 12 I f ω f 2

Substitute numerical values and evaluate K f  :

 K f 

= 12 (1.12 kg ⋅ m 2 )(10.5 rad/s )2

= 176 J = 0.18 kJ

= 62 J

79 •• [SSM] Kepler’s second law states: The line from the center of the Sun to the center of a planet sweeps out equal areas in equal times. Show that this law follows directly from the law of conservation of angular momentum and the fact that the force of gravitational attraction between a planet and the Sun acts along the line joining the centers of the two celestial objects. Picture the Problem The pictorial representation shows an elliptical orbit. The triangular element of the area is dA = 12 r (rd θ ) = 12 r 2 d θ .

r  d θ 

Differentiate dA with dA with respect to t  to  to obtain: Because the gravitational force acts along the line joining the two objects, τ  = 0. Hence:

dA θ 

dA

d θ 

= 12 r 2ω   

(1)

 L = mr 2ω  = constant  

(2)

dt 

= 12 r 2

dt 

1040  Chapter 10 2

Eliminate r  ω  between  between equations (1) and (2) to obtain:

dA dt 

=

 L 2m

= constant

Consider a cylindrical turntable whose mass is M  is M  and  and radius is R is R,, 80 •• turning with an initial angular speed 1 . (a (a) A parakeet of mass m, hovering in flight above the outer edge of the turntable, gently lands on it and stays in one  place on it as shown in Figure 10-58. What is the angular speed of the turntable after the parakeet lands? (b (b) Becoming dizzy, the parakeet jumps off (not flies off) r r with a velocity v relative to the turntable. The direction of v is tangent to the edge of the turntable, and in the direction of its rotation. What will be the angular speed of the turntable afterwards? Express your answer in in terms of the two two masses m and M   and M , the radius R radius R,, the parakeet speed v and the initial angular speed 1. Picture the Problem The angular momentum of the turntable-parakeet is conserved in both parts of this problem.

= Lf  − Li = 0  

(a) Apply conservation of angular momentum to the turntable-parakeet system as the parakeet lands to obtain:

Δ L

The final angular momentum of the system is given by:

 Lf  = Lturntable + L parakeet

Because  I turntable r

r

r ×  p parakeet

= 12 MR 2 and

r

 and Li in equation Substituting for L for Lf  and L (1) yields: Solve for ω f f to obtain:  

r

= I turntableω f  + r × p parakeet  Lf 

= Rmv parakeet :

The initial angular momentum of the system is given by:

(1)

= 12 MR 2ω f  + Rmv parakeet = 12 MR 2ω f  + Rm( Rω f  ) = 12 MR 2ω f  + mR 2ω f 

 Li = I turntableω i

= 12 MR 2ω i

 MR 2ω f  + mR 2ω f  − 12 MR 2ω i

1 2

ω f 

=

 M   M  + 2m

ω i

=0

Angular Momentum 1041

= Lf  − Li = 0  

(b) Apply conservation of angular momentum to the turntable-parakeet system as the parakeet jumps off to obtain:

Δ L

The final angular momentum of the system is given by:

 Lf  = Lturntable + L parakeet

Because  I turntable r

r

r ×  p parakeet

= 12 MR 2 and

r

(2)

r

= I turntableω f  + r × p parakeet  Lf 

= 12 MR 2ω f  + Rmv parakeet  

(3)

= Rmv parakeet : v parakeet

= vturntable + v = R

Using the expression derived in (a (a), substitute for ω f f to obtain:  

v parakeet

=

Substituting for v parakeet in equation (3) and simplifying yields:

 Lf 

 M   ⎞ = 12 MR 2ω f  + mR⎛   Rω i + v ⎟ ⎜ ⎝  M  + 2m  ⎠

 Li

= 12 MR 2ω i

Express the speed of the parakeet relative to the ground:

The initial angular momentum of the system is the same as the final angular momentum in (a (a):

 M 

 Rω i  M  + 2m

f 

+v

+v

Substituting for L for Lf  and L  and Li in equation (2) yields:

⎡⎛   M   ⎞  MR 2ω f  + mR ⎢⎜ ⎟ Rω i  M  2 m +  ⎠ ⎣⎝ 

1 2

Solving for ω f f gives:   gives:

ω f 

=

⎤ + v ⎥ − 12 MR 2ω i = 0 ⎦  M   M  + 2m

ω i

m  ⎞ v − 2⎛  ⎜ ⎟ ⎝  M  ⎠ R

You are given a heavy but thin metal disk (like a coin, but larger; 81 •• Figure 10-59). (Objects like this are called Euler called Euler disks.) disks.) Placing the disk on a turntable, you spin the disk, on edge, about a vertical axis through a diameter of the disk and the center of the turntable. As you do this, you hold the turntable still with your other hand, letting it go immediately after you spin the disk. The turntable is a uniform solid cylinder with a radius eq ual to 0.250 m and a mass equal to 0.735 kg and rotates on a frictionless bearing. The disk has an initial angular speed of 30.0 rev/min. (a (a) The disk spins down and falls fa lls over, finally

1042  Chapter 10 coming to rest on the turntable with its symmetry axis coinciding with the turntable’s. What is the final angular speed of the turntable? (b (b) What will be the final angular speed if the disk’s symmetry axis ends up 0.100 m from the axis of the turntable?   denote the disk and the letters t , Picture the Problem Let the letters d , m, and r  denote  M , and R and  R the  the turntable. We can use conservation of angular momentum to relate the final angular speed of the turntable to the initial angular speed of the Euler disk and the moments of inertia of the turntable and the disk. In part (b (b) we’ll need to use the parallel-axis theorem to express the moment of inertia of the disk with respect to the rotational axis of the turntable. You can find the moments of inertia of the disk in its two orientations and that of the turntable in Table 9-1. (a) Use conservation of angular momentum to relate the initial and final angular momenta of the system:

 I df 

df 

+ I tf 

tf 

− I di

di

=0

Because ω tf   = ω df  tf  = df :

 I df 

tf 

+ I tf 

tf 

− I di

di

=0

Solving for ω tf   yields: tf  yields:

ω tf  =

 I di  I df  + I tf 

 I di

Express the final moment of inertia of the disk:

 I df  = 12 mr 2

Express the final moment of inertia of the turntable:

 I tf  = 12 MR2

ω tf 

= =

Express ω di di in rad/s:

ω di

(1)

= 14 mr 2

Ignoring the negligible thickness of the disk, express its initial moment of inertia:

Substitute in equation (1) and simplify to obtain:

ω di  

1 2

1 4 2

mr 

mr 2

+ 12 MR 2

1  MR 2 2+2 mr 2

= 30.0

rev min

= π  rad/s

×

ω di

ω di

 

(2)

2π  rad 1 min rev

×

60 s

Angular Momentum 1043 Substitute numerical values in equation (2) and evaluate ω tf  tf :

=

ω tf 

= (b) Use the parallel-axis theorem to express the final moment of inertia of the disk when it is a distance L distance  L from the center of the turntable: Substitute in equation (1) to obtain:

π  rad/s

(0.735 kg )(0.250 m )2 2+2 (0.500 kg )(0.125 m )2 0.228 rad/s

 I df  = 12 mr 2 + mL2

ω tf  =

=

1 4

= m(12 r 2 + L2 )

mr 2

m(12 r 2 + L2 ) + 12 MR 2 1  L2 2+4 2 r 

 MR 2 +2 2 mr 

ω di

ω di

Substitute numerical values and evaluate ω tf  tf : ω tf 

=

π  rad/s 2

(0.100 m ) (0.735 kg )(0.250 m )2 2+ 4 +2 2 (0.125 m ) (0.500 kg )(0.125 m )2

=

0.192 rad/s

(a) Assuming Earth to be a homogeneous sphere that has a radius r 82 •• and a mass m, show that the period T (time for one daily rotation) of Earth’s 2 rotation about its axis is related to its radius by T  =  = br  , where b = (4/5)π  (4/5)π  m/  m/ L.  L. Here L Here L is the magnitude of the spin angular ang ular momentum of Earth. (b (b) Suppose that the radius r changes by a very small amount Δr  due  due to some internal cause such as thermal expansion. Show that the fractional change in the period ΔT  is  is given approximately by ΔT /T  =  = 2Δr /r . (c (c) By how many kilometers would r need to increase for the period to change by 0.25 d/y (so that leap years would no longer  be necessary)? Picture the Problem We can express the period of Earth’s rotation in terms of its angular velocity of rotation and relate its angular velocity to its angular momentum and moment of inertia with respect to an axis through its center. We can differentiate this expression with respect to T   and then use differentials to approximate the changes in r  and  and T .

(a) Express the period of Earth’s rotation in terms of its angular velocity of rotation:

T  =

2π  ω 

1044  Chapter 10 Relate Earth’s angular speed to its angular momentum and moment of inertia:

ω  =

 L  I 

T  =

(b) Find dT /dr :

dT 

=

Solving for dT /T  yields:  yields:

dT 

=2

Substitute numerical values and evaluate Δr :

mr 2

 L

=

4π  m 5 L

r 2

 ⎞ ⎛ 4π  m ⎞r  r 2 ⎟ = 2⎜ ⎜ ⎟ dr  dr ⎝  5 L  ⎠ ⎝  5 L  ⎠ T  ⎞ 2T  = 2⎛  ⎜ 2 ⎟r = r  ⎝ r   ⎠

T 

Solving for Δr  yields:  yields:

2 5

2π (25 mr 2 )

Substitute for ω and ω  and simplify to obtain:

(c) Using the equation we just derived, substitute for the change in the period of Earth:

 L

=

ΔT  T 

=

 =  =

Δr 

Δr 

d  ⎛ 4π  m

=

1 4

dr  r  d

y

⇒

×

ΔT 

T 

≈ 2

1y 365.24 d

=

Δr 

r  1 1460

=2

Δr  r 

r  2(1460) 6.37 × 10 3 km 2(1460)

= 2.18 km

term precession of the equinoxes refers equinoxes refers to the fact that 83 •• [SSM] The term precession Earth’s spin axis does not stay fixed but sweeps out a cone once every 26,000 y. (This explains why our pole star, Polaris, will not remain the pole star forever.) The reason for this instability is that Earth is a giant gyroscope. The spin axis of Earth precesses because of the torques exerted on it by the gravitational forces of the Sun and moon. The angle between the direction of Earth’s spin axis and the normal to the ecliptic plane (the plane of Earth’s orbit) is 22.5 degrees. Calculate an approximate value for this torque, given that the period of rotation of o f Earth is 37 2 1.00 d and its moment of inertia is 8.03 × 10  kg⋅m . Picture the Problem Let ω P  be the angular velocity of precession of Earth-asgyroscope, ω s its angular velocity about its spin axis, and  I  its   its moment of inertia with respect to an axis through its poles, and relate ω P to ω s  and  I   using its definition.

Angular Momentum 1045 Use its definition to express the  precession rate of Earth as a giant gyroscope:

ω P

=

τ   L

Substitute for I for I and solve for τ  to obtain:

τ  =  L

The angular velocity ω s of Earth about its spin axis is given by:

ω  =

2π 

τ  =

2π  I 

Substitute for ω to ω  to obtain:

P

=  I ω 

P

where T  is  is the period of T  rotation of Earth. P

T 

Substitute numerical values and evaluateτ  evaluateτ : τ  =

2π  (8.03 × 1037 kg ⋅ m 2 ) (7.66 ×10−12 s −1 ) = 4.47 ×1022 N ⋅ m 24 h 3600 s 1d × × d h

As indicated in the text, according to the Standard Model of Particle 84 •• Physics, electrons are point-like particles particles having no spatial extent. (This assumption has been confirmed experimentally, and the radius of the electron has 18  been shown to be less than 10−  m.) The intrinsic spin of an electron could in  principle be  principle be due to its rotation. Let’s check to see if this conclusion is feasible. 18 (a) Assuming that the electron is a uniform sphere whose radius is 1.00 × 10−  m, what angular speed would be necessary to produce the observed intrinsic angular momentum of h/2? (b (b) Using this value of angular speed, show that the speed of a point on the ″equator ″ of a ″spinning″ electron would be moving faster than the speed of light. What is your conclusion about the spin angular momentum being analogous to a spinning sphere with spatial extent? Picture the Problem We can use the definition of the angular momentum of a spinning sphere, together with the expression for its moment of inertia, to find the angular speed of a point on the surface of a spinning electron. The speed of such a  point is directly proportional to the angular speed of the sphere.

(a) Express the angular momentum of the spinning electron:

 L = I 

= 12 h

Assuming a spherical electron of radius R, radius R, its moment of inertia, relative to its spin axis, is:

 I  =  = 52 MR 2

1046  Chapter 10 Substituting for I  for I  yields:  yields:

Substitute numerical values and evaluate ω :

 MR 2ω  = 12 h ⇒ ω  =

2 5

5h 4 MR 2

5(1.05 × 10 −34 J ⋅ s )

ω  =

4(9.11× 10 −31 kg )(10 −18 m )

2

= 1.44 × 10 32 rad/s (b) The speed of a point on the ″equator ″ of a spinning electron of radius R radius R is  is given by:

v = R

Substitute numerical values and evaluate v:

v = (10 −18 m )(1.44 ×1032 rad/s )

=

1.44 ×1014 m/s

>c

Given that our model predicts a value for the speed of a point on the ″equator ″ of a spinning electron that is greater than the speed of light, the idea that the spin angular momentum of an electron is analogous to that of a spinning sphere with spatial extent lacks credibility. An interesting phenomenon occurring in certain pulsars certain pulsars (see  (see Problem 85 •• 24) is an event known as a ″spin glitch,″ that is, a quick change in the spin rate of the pulsar due to a shift in mass location and a resulting rotational inertia chang e. Imagine a pulsar whose radius is 10.0 km and whose period of rotation is 25.032 ms. The rotation period is observed to suddenly decrease from 25.032 ms to 25.028 ms. If that decrease was related related to a contraction of the star, star, by what amount would the pulsar radius have had to change? Picture the Problem We can apply the conservation of angular momentum to the shrinking pulsar to relate its radii to the observed periods.

The change in the radius of the  pulsar is: Apply conservation of angular momentum to the shrinking pulsar to obtain: Substituting for I  for I f and I  and I i yields:

Δ R

= Rf  − Ri  

Δ L

= Lf  − Li = 0

or  I f 

f 

− I i

i

=0

 MRf 2ω f  − 52 MR12ω i

2 5

(1)

=0

Angular Momentum 1047 Solve for ω f to obtain:

Because

= 2π  T  , where T  is  is the

ω f 

2π 

rotation period:

T f 

Substitute for R for Rf in equation (1) and simplify to obtain:

Δ R

Substitute numerical values and evaluate Δ R:  R:

Δ R

=

 Ri2  Rf 2

=

=

ω i

 Ri2 2π  2 f 

 R

T i

⇒ Rf  =

T f   Ri − Ri T i

⎛  = ⎜⎜ ⎝ 

T f   Ri T i T f  T i

 ⎞ − 1⎟⎟ Ri  ⎠

⎛  25.028 ms  ⎞ = ⎜⎜ − 1⎟⎟ (10.0 km ) ⎝  25.032 ms  ⎠ = − 79.9 cm

a 0.008% decrease in radius. the form of a uniform disk with a rope 86 ••• Figure 10.60 shows a pulley in the hanging over it. The circumference of the pulley is 1.2 m and a nd its mass is 2.2 kg. The rope is 8.0 m long and its mass is 4.8 kg. At the instant shown in the figure, the system is at rest and the difference in height of the two ends of the rope is 0.60 m. (a) What is the angular speed of the pulley when the difference in height  between the two ends of the rope is 7.2 m? (b (b) Obtain an expression for the angular momentum of the system as a function of time while neither end of the rope is above the center of the pulley. There is no slippage between rope and  pulley wheel. Picture the Problem Let the origin of the coordinate system be at the center of the pulley with the upward direction positive. Let λ   be the linear density (mass  per unit length) of the rope and L and L1 and L  and  L2 the lengths of the hanging parts of the rope. We can use conservation of mechanical energy to find the angular speed of the pulley when the difference in height between the two ends of the rope is 7.2 m.

(a) Apply conservation of energy to relate the final kinetic energy of the system to the change in potential energy:

Δ K  + ΔU  = 0 or, because K  because K i = 0,  K  + ΔU  = 0  

(1)

Express the change in potential energy of the system: ΔU 

= U f  − U i = − 12 L1f  ( L1f λ ) g − 12 L2f  ( L2f λ ) g − [− 12 L1i ( L1i λ ) g − 12 L2i ( L2i λ ) g ] 2 2 = − 12 ( L1f  + L22f  )λ  g + 12 ( L1i2 + L22i )λ  g  = − 12 λ  g [( L1f  + L22f  ) − ( L1i2 + L22i )]

1048  Chapter 10

Because L Because L1 + L  + L2 = 7.4 m,  – L1i = 0.6 m, and  L2i – L  L2f  – L  – L1f  = 7.2 m, we obtain:

 L1i = 3.4 m, L m, L2i = 4.0 m, and L2f  = 7.3 m.  L1f  = 0.1 m, and L

Substitute numerical values and evaluate ΔU : ΔU 

= − 12 (0.60 kg/m )(9.81m/s 2 ) (0.10 m )2 + (7.3 m )2 − (3.4 m )2 − (4.0 m )2 = −75.75 J

Express the kinetic energy of the system when the difference in height  between the two ends of the rope is 7.2 m:

 K  = 12 I  pω 2 + 12 Mv 2

Substitute numerical values and simplify:

⎛ 1.2 m ⎞ ω 2  K  = [ (2.2 kg ) + 4.8 kg ] ⎜ ⎟ ⎝  2π   ⎠ = (0.1076 kg ⋅ m 2 )ω 2

Substitute in equation (1) and solve for ω :

(0.1076 kg ⋅ m )ω 

=

1 2

=

1 2

(  M  R )ω  +  MR ω  (  M  + M ) R ω  2

1 2

 p

1 2

 p

2

2

1 2

2

2

2

2

1 1 2 2

2

2

− 75.75 J = 0

and ω  =

75.75 J

= 27 rad/s

0.1076 kg ⋅ m 2

(b) Noting that the moment arm of each portion of the rope is the same, express the total angular momentum of the system:

 L = L p + Lr  = I  pω  + M r  R 2ω 

Letting θ  be  be the angle through which the pulley has turned, express U (θ ): ):

U (θ ) = − 12 ( L1i − Rθ )

Express ΔU  and  and simplify to obtain:

ΔU  = U f  − U i = U (θ ) − U (0 ) = − 12 [( L1i − Rθ )2 + ( L2i + Rθ )2 ]λ  g  + 12 ( L1i2 + L22i )λ  g  = − R 2θ 2λ  g + ( L1i − L2i ) Rθλ  g 

Assuming that, at t = 0, L 0, L1i ≈ L2i:

ΔU  ≈ − R 2θ 2λ  g 

= (12 M  p R 2 + M r  R 2 )ω 

  (2)

= (12 M  p + M r ) R 2ω  2

+ ( L2i + Rθ )2 λ  g 

Angular Momentum 1049

Substitute for K for K and ΔU in equation (1) to obtain: Solving for ω  yields:  yields:

(0.1076 kg ⋅ m )ω  2

2

− R 2θ 2λ  g  = 0

 R 2θ 2 λ  g 

ω  =

0.1076 kg ⋅ m 2

Substitute numerical values to obtain:

⎛ 1.2 m ⎞ (0.6 kg/m )(9.81m/s 2 ) ⎜ ⎟ 2π   ⎠ ⎝  ω  = θ  0.1076 kg ⋅ m 2 = (1.41s -1 )θ 

Express ω  as  as the rate of change of θ : θ  :

d θ 

Integrate θ  from 0 to θ to θ  to obtain:

ln θ  = (1.41s −1 )t 

2

= (1.41s −1 )θ  ⇒

dt 

d θ  θ 

= (1.41s −1 )dt 

−1 t  θ (t ) = e (1.41s )

Transform from logarithmic to exponential form to obtain: Differentiate to express ω  as  as a function of time:

ω (t ) =

d θ  dt 

− = (1.41s −1 )e (1.41s )t  1

−1  L = ( 12 M  p + M r ) R 2 (1.41s −1 )e(1.41s )t 

Substitute for ω  in  in equation (2) to obtain:

Substitute numerical values and evaluate L evaluate L::

⎛ 1.2 m ⎞  L = [ (2.2 kg ) + (4.8 kg )]⎜ ⎟ ⎝  2π   ⎠ 1 2

2

[(1.41s )e ( −1

)

1.41s −1 t 

] = (0.30 kg ⋅ m / s)e( 2

)

1.41s −1 t 

1050  Chapter 10