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Angular Momentum
CHAPTER OUTLINE 11.1 The Vector Product and Torque 11.2 Angular Momentum 11.3 Angular Momentum of a Rotating Rigid Object 11.4 Conservation of Angular Momentum 11.5 The Motion of Gyroscopes and Tops
ANSWERS TO QUESTIONS
Q11.1 No to both questions. An axis of rotation must be defined to calculate the torque acting on an object. The moment arm of each force is measured from the axis, so the value of the torque depends on the location of the axis.
*Q11.2 (i) Down–cross–left is away from you: − ˆ j^ × −( ˆ i^ (^) )= − k ˆ answer (f), as in the first picture.
(ii) Left–cross–down is toward you: − ˆ i^ × (^) ( −ˆ j^ ) = k ˆ answer (e), as in the second picture.
*Q11.3 (3 m down) × (2 N toward you) = 6 N ⋅ m left. The answers are (i) a (ii) a (iii) f
*Q11.4 The unit vectors have magnitude 1, so the magnitude of each cross product is |1 ⋅ 1 ⋅ sin θ | where θ is the angle between the factors. Thus for (a) the magnitude of the cross product is sin 0° = 0. For (b), |sin 135°| = 0.707 (c) sin 90° = 1 (d) sin 45° = 0.707 (e) sin 90° = 1. The assem- bled answer is c = e > b = d > a = 0.
Q11.5 Its angular momentum about that axis is constant in time. You cannot conclude anything about the magnitude of the angular momentum.
Q11.6 No. The angular momentum about any axis that does not lie along the instantaneous line of motion of the ball is nonzero.
*Q11.7 (a) Yes. Rotational kinetic energy is one contribution to a system’s total energy.
(b) No. Pulling down on one side of a steering wheel and pushing up equally hard on the other side causes a total torque on the wheel with zero total force.
(c) No. A top spinning with its center of mass on a fixed axis has angular momentum with no momentum. A car driving straight toward a light pole has momentum but no angular momentum about the axis of the pole.
283
FIG. Q11.
284 Chapter 11
Q11.8 The long pole has a large moment of inertia about an axis along the rope. An unbalanced torque will then produce only a small angular acceleration of the performer-pole system, to extend the time available for getting back in balance. To keep the center of mass above the rope, the performer can shift the pole left or right, instead of having to bend his body around. The pole sags down at the ends to lower the system center of gravity.
*Q11.9 Her angular momentum stays constant as I is cut in half and ω doubles. Then (12) I ω^2 doubles. Answer (b).
Q11.10 Since the source reel stops almost instantly when the tape stops playing, the friction on the source reel axle must be fairly large. Since the source reel appears to us to rotate at almost constant angular velocity, the angular acceleration must be very small. Therefore, the torque on the source reel due to the tension in the tape must almost exactly balance the frictional torque. In turn, the frictional torque is nearly constant because kinetic friction forces don’t depend on velocity, and the radius of the axle where the friction is applied is constant. Thus we conclude that the torque exerted by the tape on the source reel is essentially constant in time as the tape plays.
As the source reel radius R shrinks, the reel’s angular speed ω = v R
must increase to keep the tape speed v constant. But the biggest change is to the reel’s moment of inertia. We model the reel as a roll of tape, ignoring any spool or platter carrying the tape. If we think of the roll of tape as a uniform disk, then its moment of inertia is I = MR
(^2). But the roll’s mass is proportional to its base area π R^2. Thus, on the whole the moment of inertia is proportional to R^4. The moment of inertia decreases very rapidly as the reel shrinks!
The tension in the tape coming into the read-and-write heads is normally dominated by balancing frictional torque on the source reel, according to TR ≈τfriction. Therefore, as the tape plays the tension is largest when the reel is smallest. However, in the case of a sudden jerk on the tape, the rotational dynamics of the source reel becomes important. If the source reel is full, then the moment of inertia, proportional to R^4 , will be so large that higher tension in the tape will be required to give the source reel its angular acceleration. If the reel is nearly empty, then the same tape acceleration will require a smaller tension. Thus, the tape will be more likely to break when the source reel is nearly full. One sees the same effect in the case of paper towels; it is easier to snap a towel free when the roll is new than when it is nearly empty.
*Q11.11 The angular momentum of the mouse-turntable system is initially zero, with both at rest. The fric- tionless axle isolates the mouse-turntable system from outside torques, so its angular momentum must stay constant with the value zero.
(i) The mouse makes some progress north, or counterclockwise. Answer (a).
(ii) The turntable will rotate clockwise. The turntable rotates in the direction opposite to the motion of the mouse, for the angular momentum of the system to remain zero. Answer (b).
(iii) No, mechanical energy changes as the mouse converts some chemical into mechanical energy, positive for the motions of both the mouse and the turntable.
(iv) No, momentum is not conserved. The turntable has zero momentum while the mouse has a bit of northward momentum. The sheave around the turntable axis exerts a force northward to feed in this momentum.
(v) Yes, angular momentum is constant with the value zero.
284 Chapter 11
286 Chapter 11
P11.
A B ⋅ = − 3 00 6 00. (. ) + 7 00. (− 10 0. ) + −( 4 00. )( 9 00. )== − 124 AB = ( −3 00. ) + ( 2 7 00. ) + − 2 ( 4 00. ) 2 ⋅ ( 6 00. ) + −^2 ( 10 0. )) + (^2 9 00. ) 2 = 127
(a) cos −^ ⋅ cos −.
(^1) = 1 (− 0 979 ) = 168 °
A B
AB
(b)
A B
i j k × = − − −
23 0 0 ˆ i^ + 3 00. ˆ j^ −12 0. k ˆ
A × B = ( 23 0. ) + ( 2 3 00. ) + − 2 ( 12 0. ) 2 =26 1.
sin −^ sin −..
⎛ ×
A B
AB
° or 168°
(c) Only the first method gives the angle between the vectors unambiguously.
P11.
ττ =
= ( ) ( − ) ×
r × F 0 450. m 0 785. N sin 90 ° 14 ° up eastt = 0 343. N m⋅ horizontally north
P11.6 The cross-product vector must be perpendicular to both of the factors, so its dot product with either factor must be zero: Does (^) ( 2 ˆ i^ − 3 ˆ j^ + 4 k ˆ^ (^) )⋅ (^) ( 4 ˆ i^ + 3 ˆ j^ − k ˆ) = 0? We have 8 − 9 − 4 = − 5 ≠ 0 so the answer is No. The cross product could not work out thhat way.
P11.
A × B = A B ⋅ ⇒ AB sin θ =ABcosθ ⇒ tan θ = 1 or
θ = 45 0. °
FIG. P11.
Angular Momentum 287
*P11.8 (a)
ττ = r × F = = ( − ) − ( − ) + −
i j k i j k
4 6 0 ˆ^ ˆ^ ˆ
0 0 0 0 ( 8 188 ) = ( − 10 0. N m⋅ ) k ˆ
(b) Yes. The line of action of the force is the dashed line in the diagram. The point or axis must be on the other side of the line of action, and half as far from this line along which the force acts. Then the lever arm of the force about this new axis will be half as large and the force will produce counterclockwise instead of clockwise torque. There are infinitely many such points, along the dotted line in the diagram. But the locus of these points intersects the y axis in only one point, which we now determine. Let (0, y ) represent the coordinates of the special axis of rotation located on the y axis of coordinates. Then the displacement from this point to the particle feeling the force is r new (^) = 4 i ˆ^ + ( 6 − y ) j ˆ^ meters. The torque of the force about this new axis is
ττ new = r new (^) × F = − y = ( − ) − −
i j k i j
4 6 0 ˆ^ ˆ
0 0 (( 0 0 ) + k ˆ^ ( 8 − 18 + 3 y ) = + ( 5 N m⋅ ) k ˆ
Then we need only − 10 + 3 y = 5 y = 5 m. The position vector of the new axis is 5 00. ˆ j m.
P11.
F 3 = F 1 + F 2
The torque produced by
F 3 depends on the perpendicular distance OD , therefore translating the point of application of
F 3 to any other point along BC will not change the net torque.
P11.10 ˆ i^ × ˆ i^ = 1 1⋅ ⋅ sin 0 ° = 0
ˆ j (^) × ˆ j and k ˆ (^) × k ˆare zero similarly since the vectors being multiplied are parallel. ˆ i (^) × ˆ j (^) = 1 1⋅ ⋅ sin 90 ° = 1
Lever arm of force about origin
x
y Line of action
Particle
Locus of points about which F produces torque +5 k^ ^^ N··m
F
FIG. P11.8(b)
A
B
C
D O
F 1
F 2
F 3
FIG. P11.
j i
k k ˆ^ × ˆ i^ =ˆ j
ˆ i (^) × ˆ j (^) = k ˆ
ˆ j (^) × k ˆ (^) =ˆ i
ˆ i (^) × k ˆ (^) = −ˆ j
ˆ j (^) × ˆ i (^) = − k ˆ
k^ ˆ^ × ˆ j^ = −ˆ i
FIG. P11.
Angular Momentum 289
P11.15 The angular displacement of the particle around the circle is θ = ω t = t R
v (^).
The vector from the center of the circle to the mass is then R cos θ ˆ i^ + R sin θˆ j.
The vector from point P to the mass is
r i i j
r
R R R
R t R
ˆ (^) cos ˆ (^) sin ˆ
cos
θ θ
1 v^ ⎞⎞ ⎠⎟^
ˆ i (^) sin ˆ j ⎥ vt R The velocity is
v =^ d r^ = − ⎛⎝⎜ ⎞⎠⎟ i + ⎛⎝⎜ ⎞⎠⎟ j dt
t R
t R
v sin v^ ˆ^ v cos v ˆ
So (^) L = r × m v
L = m Rv ⎡⎣( 1 +cos ω t ) ˆ i^ +sin ω t ˆ j ⎤⎦^ × − sin ω t i ˆ^ +cosω tt
m R t R
ˆ (^) cos
j
L k
⎣⎢^
v v 1
P11.16 (a) The net torque on the counterweight-cord-spool system is:
(^) τ = r × F = 8 00. × 10 − 2 m 4.00 kg( )( 9 80. m s (^2) ) = 3 1. 4 4 N m⋅
(b)
L = r × m v + I ω
L = Rm + MR ⎛⎝⎜ ⎞⎠⎟ = ⎛⎝⎜ + ⎞⎠⎟ = ⋅ R R m
M
v v v
(^2) (0 400. kg mm) v
(c) τ =^ dL = ( ⋅ ) dt
0 400. kg m a a =
N m 0.400 kg m m s 2
P11.17 (a) zero
(b) At the highest point of the trajectory,
x R g = 1 = i 2
v^2 sin θ and
y h g
= = (^ i ) max
v sin θ 2 2 (^) L r v
i j
1 1 1 2 2 2 2 2
= ×
⎡ ( ) ⎣
m
g g
vi sin θ (^) ˆ vi sin θ (^) ˆ⎤⎤ ⎦
×
− ( )
m
m g
xi
i i
v
v v
sin cos (^) ˆ
i
k θ θ 2 2
continued on next page
x
y
θ
R^ m P Q
v
FIG. P11.
O R
v i v 2
v i = vxi i
FIG. P11.
290 Chapter 11
(c)
L (^) 2 = R i ˆ^ × m v 2
R =
g
, where vi^ sin^ θ
= mR i^ ˆ^ × vi cos θ ˆ ii j
k k
( − )
= − =
v
v v
i
mR (^) i i m g
sin ˆ
sin ˆ^ sin sin (^) ˆ
θ
θ
(^3 2) θ θ
(d) The downward force of gravity exerts a torque in the^ − z^ direction.
P11.18 Whether we think of the Earth’s surface as curved or flat, we interpret the problem to mean that the plane’s line of flight extended is precisely tangent to the mountain at its peak, and nearly parallel to the wheat field. Let the positive x direction be eastward, positive y be northward, and positive z be vertically upward.
(a) r = ( 4 30. km) k ˆ^ = (^) ( 4 30. × 10 3 m) k ˆ p = m v = 12 000 kg (^) ( − 175 i ˆ^ m s (^) ) = − 2 10. × 10 6 i ˆ^ kg m⋅ ss
m kg m s
L = r × p = (^) ( 4 30. × 10 3 k ˆ^ (^) ) × −(( 2 10. × 106 i ˆ ⋅ ) = (^) ( − 9 03. × 10 9 kg m⋅ 2 s)ˆ j
(b) No. L = = m ( r )
r p sin θ v sinθ, and r sin θ is the altitude of the plane. Therefore, L = constant as the plane moves in level flight with constant velocity.
(c) Zero. The position vector from Pike’s Peak to the plane is anti-parallel to the velocity of the plane. That is, it is directed along the same line and opposite in direction. Thus, L = m rv sin180 ° = 0.
*P11.19 (a) The vector from P to the falling ball is
r r v a
r i j
= (^) ( + ) + −
i i t^ t
2
cos θ ˆ^ sin θ ˆ ⎛⎝ ggt^2 ⎞⎠ ˆ j
The velocity of the ball is v = v (^) i + a t = 0 − gt j ˆ
So (^) L = r × m v L = ⎡( i + j (^) ) + − ⎛⎝ ⎞⎠ j ⎣⎢^
m cos θ ˆ^ sin θ ˆ^0 1 gt ˆ × − 2
(^2) ggt
m gt
cos ˆ
j
L k
( )
= −
θ
(b) The Earth exerts a gravitational torque on the ball.
(c) Differentiating with respect to time, we have − mg cos θ k ˆ for the rate of change of angular momentum, which is also the torque due to the gravitational force on the ball.
m
l
P
θ
FIG. P11.
292 Chapter 11
*P11.24 (a) I = (25) MR^2 = (25)(5.98 × 1024 kg)(6.37 × 106 m) 2 = 9.71 × 1037 kg ⋅ m^2 ω = 1 rev24 h = 2 π rad86400 s = 7.27 × 10 −^5 s L = I ω = (9.71 × 1037 kg ⋅ m 2 )(7.27 × 10 −^5 s) = 7.06 × 10 33 kg m⋅ 2 s The earth turns toward the east, counterclockwise as seen from above north, so the vector angular momentum points north along the earth’s axis, toward the north celestial pole or nearly toward the star Polaris.
(b) I = MR^2 = (5.98 × 1024 kg)(1.496 × 1011 m)^2 = 1.34 × 1047 kg ⋅ m 2 ω = 1 rev365.25 d = 2 π rad(365.25 × 86400 s) = 1.99 × 10 −^7 s L = I ω = (1.34 × 1047 kg ⋅ m 2 )(1.99 × 10 −^7 s) = 2.66 × 10 40 kg m⋅^2 s The earth plods around the Sun, counterclockwise as seen from above north, so the vector angular momentum points north perpendicular to the plane of the ecliptic, toward the north ecliptic pole or 23.5° away from Polaris, toward the center of the circle that the north celestial pole moves in as the equinoxes precess. The north ecliptic pole is in the constellation Draco.
(c) The earth is so far from the Sun that the orbital angular momentum is much larger, by 3.78 × 106 times.
P11.25 (a) I^ =^ m L^ +^ m (^ )^ =^ (^ )(^ ) +
1 2 2.^2 0 100.^ 1 00.^2 0 40.^0 0 0 500^ 0 108 3
kg m 2
L I ω 4 433 kg m ⋅ 2 s
(b) I = m L + m R = ( )( ) + ( )
12 2 2 0 100.^ 1 00.^2 0 400 1 00.^.^2 ==
L I ω... kg m 2 s
P11.26 The total angular momentum about the center point is given by L = I (^) h ω h (^) + Im ω m
with I m L h^ =^ h^ h =^
. kg 2.70 m kg m 2 and I m L m^ =^ m^ m =^
kg 4.50 m kg m 2 In addition, ω π h =^
⎝⎜^
= × −
rad 12 h
h 3 600 s
. rad s while
ω π m =^
⎝⎜^
= × −
rad 1 h
h 3 600 s
. rad s
Thus, L = 146 kg m⋅ (^2) ( 1 45. × 10 −^4 rad s) + 675 kg m⋅ (^2) (1 75. × 100 −^3 rad s)
or L = 1 20. kg m⋅ 2 s The hands turn clockwise, so their vector angular momentum is perpendicularly into the clock face.
Angular Momentum 293
P11.27 We require a g r c =^ =^ = r
v^22 ω
ω = = ( ) =
= = ×
g r I Mr
m s m
rad s
2
kkg 100 m ( ) 2 = 5 × 108 kg m⋅ 2
(a) L = I ω = 5 × 10 8 kg m⋅ 2 0 313. s = 1 57. × 108 kg m⋅ 2 s
(b) τ α
ω ω ∑ =^ =^
I
I
t
f i ∆ ∑ τ^ ∆^ t^ =^ I^ ω^ f −^ I ω^ i =^ L^ f − Li
This is the angular impulse-angular momentum theorem.
(c) ∆ t
L (^) f
× ⋅
( )( )^
∑
8 τ
kg m s N m
2 226 × 10 3 s =1 74. h
P11.28 (^) ∑ Fx^ =^ max : +^ f^ s = max
We must use the center of mass as the axis in
∑ τ^ =^ I α^ :^ Fg^ (^0 ) −^ n^ (^ 77 5.^ cm^ ) +^ fs (^88 cm) =^0
∑^ Fy^ =^ may :^ +^ n^ −^ Fg =^0
We combine the equations by substitution:
( )
mg ma
a
x
x
cm cm m s 2 ccm 88 cm
= 8 63. m s 2
Section 11.4 Conservation of Angular Momentum
P11.29 (a) From conservation of angular momentum for the system of two cylinders:
( I (^) 1 + I (^) 2 ) ω (^) f = I 1 ω i or ω f ω i
I
I I
1 1 2
(b) K (^) f = ( I + I ) f
ω 2 and K I i =^ i
ω^2
so K K
I I
I
I
I I
I
I
f i i
= (^) i ( + )
⎠⎟^
(^12 1 ) 12 1 2
1 1 2
2 1 ω 1
ω II 2
which is less than 1
88 cm
Fg
n f^ s 155 cm 2 FIG. P11.
Angular Momentum 295
(b) K (^) i = 1 Ii i = (^) ( ⋅ )( ) = 2
ω 2 9 00. kg m 2 0 750. rad s^2 2 53. J
K (^) f = 1 If f = (^) ( ⋅ )( ) = 2
ω 2 3 54. kg m 2 1 91. rad s^2 6 44. JJ
P11.33 I (^) i ω i (^) = If ω f : (^) ( 250 kg m ⋅ (^2) )(10 0. rev min ) = ⎡⎣ 250 kg m⋅ 2 +25 0. kg 2( ..00 m)^2 ⎤⎦ ω 2
ω 2 = 7 14. rev min
*P11.34 (a) Let M = mass of rod and m = mass of each bead. From I (^) i ω i (^) = If ω f between the moment of release and the moment the beads slide off, we have 1 12
⎡ M 2 + mr 1 (^2) i M (^22) mr 22 f ⎣⎢^
⎦⎥^
⎣⎢^
ω ω
When M = 0.3 kg, = 0 500. m, r 1 = 0 100. m, r 2 = 0 250. m, ω i = 36 s, we f ind
[0.00625 + 0.02 m ]36 = [0.00625 + 0.125 m ] ω f
ω (^) f = (36 s)(1 + 3.2 m ) (1 + 20 m )
(b) The denominator of this fraction always exceeds the numerator, so ω f decreases smoothly from a maximum value of 36.0 rads for m = 0 toward a minimum value of (36 × 3.220) = 5.76 rad s as m → ∞.
As a bonus, we find the work that the bar does on the beads as a function of m. Consider the beads alone. Their kinetic energy increases because of work done on them by the bar.
initial kinetic energy + work = f inal kinetic energy
(12)(2 mr 12 )( ω i )^2 + Wb = (12)(2 mr 22 )(ω f )^2
m (0.1)^2 (36)^2 + Wb = m (0.25)^2 [(36s)(1 + 3.2 m )(1 + 20 m )]^2
Wb = m [81(1 + 3.2 m )^2 − 12.96(1 + 20 m )^2 ] (1 + 20 m ) 2 = (68.04 m )(1 − 64 m^2 )(1 + 20 m ) 2 joules
W (^) b increases from 0 for m = 0 toward a maximum value of about 0.8 J at about m = 0.035 kg, and then decreases and goes negative, diverging to −∞ as m →∞.
*P11.35 (a) Mechanical energy is not conserved; some chemical energy is converted into mechan- ical energy. Momentum is not conserved. The turntable bearing exerts an external northward force on the axle. Angular momentum is conserved. The bearing isolates the system from outside torques. The table turns opposite to the way the woman walks, so its angular momentum cancels that of the woman.
continued on next page
296 Chapter 11
(b) From conservation of angular momentum for the system of the woman and the turntable, we have L (^) f = Li = 0 so,
L (^) f = I (^) woman ω woman + I table ω (^) table= 0
and
ωtable woman ω table
= − woman woman
⎠⎟^
I
I
m r I
2 ttable
woman woman woman ta
⎝⎜^
v v r
m r I (^) bble
ωtable 2 kg 2.00 m m s kg m
( )( ) ⋅
0 0 360. rad s
or
ωtable = 0 360. rad s (counterclockwise )
(c) chemical energy converted into mechanical = ∆ KK = K (^) f − 0 = m + I
woman woman 2 2 table v ω^2
∆ K = 1 ( )( ) + (^) ( ⋅ ) 2
kg. m s 2 500 kg m 2 (0 360. raad s ) 2 = 99 9. J
P11.36 When they touch, the center of mass is distant from the center of the larger puck by
y CM g 4.00 cm^ cm g g
= +^ (^ + )
..00 cm
(a) L = r m 1 1 (^) v 1 (^) + r m 2 2 v 2 = 0 + (^) ( 6 00. × 10 − 2 m (^) ) ((80 0. × 10 −^3 kg)( ) = × − ⋅
m s kg m 2 s
(b) The moment of inertia about the CM is
I m r m d m r m d
I
2 1 1 2 2 2 2 2 2 2
11 2
2 (. kg ) (^) (. × −^ m (^) ) + (. kg). × 110 1 2
2 2
3 2 2
−
− −
( )
- (^) (. × kg) (^) (. × m (^) ) +. 00 10 6 00 10
7 60 10
( ×^3 ) ( ×^2 )^2 = × ⋅
− −
−
kg m (^4) kg m 2
I.
Angular momentum of the two-puck system is conserved: L = I ω
ω = =
× ⋅
× ⋅
− −
L
I
3 4
kg m s kg m r
2 2 aad s
298 Chapter 11
(b) No; some mechanical energy changes into internal energy.
(c) Momentum is not conserved. The axle exerts a backward force on the cylinder.
P11.40 (a) Let ω be the angular speed of the signboard when it is vertical.
1 2 1 2
2
2 2
I Mgh
ML Mg L
g
ω
ω θ
ω
cos
( ) (^ − )
cos
. cos. . .
θ L 3 9 80 1 25 0 0 50 2 3
m s m
5 5 rad s
(b) I (^) i ω i − m Lv = If ω f represents angular momentum conservation for the sign-snowball system. In more detail, 1 3
ML^2^ mL^2^ ML^2
⎝⎜^
⎠⎟^ ω^ = ω ii^ −^ m Lv Solving,
f i
ML m M m L
( + )
v ω
(^13) ω (^13)
(^13) ( 2.. 40 kg )(0 5. m )( 2 347. rad s ) − (0 4. kg )((1 6. m s) ⎡⎣ ( ) + ⎤⎦( )^
3 2 40^ 0 4^ 0 5^
kg kg m
rad s
(c) Let h CM = distance of center of mass from the axis of rotation.
h CM kg m kg m = (^ )( 2 40 0 25 ) + ( 0 4 (^) )( 0 50 ) 2 40
. kkg kg
m
Apply conservation of mechanical energy:
( M + m gh ) ( − ) = ⎛⎝⎜ ML + mL ⎞⎠⎟
CM 1
cos^2 2
cos
θ ω
θ −−
−
( + )
1 13 2 2
1 13
M m L M m gh
ω CM
cos
⎡(^ kg^ ) + kg^ m^2 rad s^2 ⎣ ⎤⎦(^ ) (^ ) (^2 2 40.^ kg^ +0 4.^ kg^ )(9 80.^ m s^2 )( 0 285 7. m)
θ
Mg m v
L
FIG. P11.
Angular Momentum 299
P11.41 The meteor will slow the rotation of the Earth by the largest amount if its line of motion passes farthest from the Earth’s axis. The meteor should be headed west and strike a point on the equator tangentially. Let the z axis coincide with the axis of the Earth with + z pointing northward. Then, conserving angular momentum about this axis, (^) ∑^ L^ f =^ ∑ L^ i ⇒^ I^ ω^ f =^ I^ ω i +^ m v^ × r
or 2 5
MR^2 MR^2 m R ω (^) f k ˆ^ = ω i k ˆ^ − v k ˆ
Thus, ω (^) i ω f m R MR
m MR
− = 2^ v^ = v 5 2
or
ω (^) i − ω f = ( × ) ( × ) ×
13 3 2
kg m s 44 6
14 6 37 10
kg m rad s ( ) ( × )
= × −
∆ωmax ~ −−^13 rad s
Section 11.5 The Motion of Gyroscopes and Tops
P11.42 Angular momentum of the system of the spacecraft and the gyroscope is conserved. The gyro- scope and spacecraft turn in opposite directions.
0 = I 1 ω 1 + I 2 ω 2 : − I = I (^1 1 2) t ω θ
− ⋅ ( − ) = × ⋅
kg m rad s kg m 2 2 r t
π aad 180 s 2 000
s
t = 2 62^ ×^10 = 131
.^5
P11.43 I = 2 MR = (^) ( × ) ( × ) = × 5
(^2) 5 98. 10 24 kg 6 37. 10 6 m 2 9 71. 110 37 kg m ⋅ 2
L = I = × ⋅ ⎛
⎝⎜^
ω 9 71. 10 37 kg m 2 π^ rad =7 0. 86 400 s
33
33
× ⋅
= = ( × ⋅ )
kg m s
kg m s r
2 2
τ L ω^2 π p.^
aad 2.58 10 yr
yr 365.25 d
d ×^4 86 40
⎝⎜^
0 0 s N m
⎝⎜^
= 5 45. × 10 22 ⋅
Angular Momentum 301
*P11.45 (a) Momentum is conserved for the system of two men: (162 kg)(+ 8 m/s) + (81 kg)(−11 m/s) = (243 kg)
v (^) f v^ (^) f = 1.67 i ˆm s
(b) original mechanical energy = (12)(162 kg)(+8 ms) 2 + (12)(81 kg)(−11 ms) 2 = 10 084 J fi nal mechanical energy = (12)(243 kg)(1.67 ms) 2 = 338 J Thus the fraction remaining is 33810 084 = 0.033 5 = 3.35%
(c) The calculation in part (a) still applies: v (^) f = 1.67 i ˆm s
(d) With half the mass of Perry, Flutie is distant from the center of mass by (23)(1.2 m) = 0.8 m. His angular speed relative to the center of mass just before they link arms is ω = v r = (11 + 1.67)(ms)0.8 m = 15.8 rads. That of Perry is necessarily the same (8 − 1.67)0.4 m = 15.8 rad s. In their linking of arms, angular momentum is conserved. Their total moment of inertia stays constant, so their angular speed also stays constant at 15.8 rad/s.
(e) Only the men's direction of motion is changed by their linking arms. Each keeps constant speed relative to the center of mass and the center of mass keeps constant speed, so all of the kinetic energy is still present. The fraction remaining mechanical is 1.00 = 100%. We can compute this explicitly: the fi nal total kinetic energy is (12)(243 kg)(1.67 ms) 2 + (12)[(81 kg)(0.8 m)^2 + (162 kg)(0.4 m) 2 ]((15.8 s) 2 = 338 J + 6498 J + 3249 J = 10 084 J, the same as the original kinetic energy.
P11.46 (a) ( K + U (^) s ) A (^) = ( K + Us ) B
0
= = (^) ( ) =
mgy m
gy
A B
B A
v
v. m s 2. m. 1 1 m s
(b) L = m rv = 76 kg 11.1 m s 6.3 m = 5 32. × 10 3 kg m⋅ 2 s toward you along the axis of the channel.
(c) The wheels on his skateboard prevent any tangential force from acting on him. Then no torque about the axis of the channel acts on him and his angular momentum is constant. His legs convert chemical into mechanical energy. They do work to increase his kinetic energy. The normal force acts forward on his body on its rising trajectory, to increase his linear momentum.
(d) L = m rv v =
× ⋅
kg m s kg 5.85 m m s
2
(e) ( K + U g ) B + U chemical B , = ( K + Ug ) C
kg 11.1 m s( ) +^2 + U (^) chem = 76 kg 12.0 m s( )^2 ++
= −
kg 9.8 m s 0.45 m
kJ kJ
2
U (^) chem.. ++ 335 J = 1 08. kJ
(f) ( K + U g ) C = ( K + Ug ) D
kg 12.0 m s( (^) ) +^2 = 76 kg v^2 D + 76 kg 9.8 m s^22 5.85 m
vD = 5 34. m s
continued on next page
302 Chapter 11
(g) Let point E be the apex of his flight:
(^ K^^ + U^ g ) D =^ ( K^ + Ug ) E
76 kg 5.34 m s ( ) +^2 0 = 0 + 76 kg 9.8 m s( 2 ) ( y (^) E − yD )
( yy^ E − yD ) =^ 1 46.^ m
(h) For the motion between takeoff and touchdown
y y t a t
t
f =^ i +^ yi + y − = + −
v^1 2 2 34 0 5 34 4 9
2
. m. m s. m s 2 tt
t
2
5 34 5 34 2 4 4 9 2 34 9 8
. s
(i) This solution is more accurate. In Chapter 8 we modeled the normal force as con- stant while the skateboarder stands up. Really it increases as the process goes on.
P11.47 (a) I m r
m d m d m d
m d
= i i
= ⎛⎝ ⎞⎠ + ⎛⎝ ⎞⎠ + ⎛⎝ ⎞⎠
∑ 2 2 2 2
2
(b) Think of the whole weight, 3 mg , acting at the center of gravity.
ττ =^ ^ ^ × = ⎛ ⎝
r F (^) ⎠ ( − i (^) ) × (^) (− j (^) ) = ( (^) ) k d (^) mg mgd 3
(c) α = τ= = I
mgd md
g d
counterclockwise
(d) a r g d
= = ⎛ d^ g ⎝
α (^) ⎠ =
up
The angular acceleration is not constant, but energy is. K U E K U
m g d I
i f
f
ω^20
(e) maximum kinetic energy (^) = mgd
(f) ω (^) f g d
(g) (^) L I md^ g d
g (^) md f =^ f =^ =^
ω (^) ⎠
2 1 2 3 2
(h) v (^) f f r g d
d gd = ω = =
m m (^) m 1 2 3 d d
P
d
FIG. P11.
