Unformatted text preview: mains constant (is conserved) if the net average external torque acting on the
system is zero. We will use this principle to find the final angular velocity of the system.
SOLUTION The angular momentum L of the system (rod plus bug) is given by
Equation 9.10 as the product of the system’s moment of inertia I and angular velocity ω, or
L = Iω. The conservation of angular momentum can be written as I3
1ω
2
Final angular
momentum = I 0ω0
13
2
Initial angular
momentum where ω and ω0 are the final and initial angular velocities, respectively, and I and I0 are the
final and initial moments of inertia. The initial moment of inertia is given. The initial
moment of inertia of the bug is zero, because it is located at the axis of rotation. The final
moment of inertia is the sum of the moment of inertia of the bug and that of the rod;
I = Ibug + I0. When the bug has reached the end of the rod, its moment of inertia is
2 Ibug = mL , where m is its mass and L is the length of the rod. The final angular velocity of
the system is, then, I0
I I0 = ω0 2 0 = ω0 mL + I I bug + I 0 I
0 ω = ω0 1.1×10−3 kg ⋅ m 2
= ( 0.32 rad/s ) 4.2 × 10−3 kg ( 0.25 m )2 + 1.1× 10−3 kg ⋅ m 2 ( ) ( ) = 0.26 rad/s 64. REASONING The carousel rotates on frictionless bearings and without air resistance, so
no net external torque acts on the system comprised of the carousel and the person on it.
Therefore, the total angular momentum L = Iω (Equation 9.10) of the system is conserved
(Lf = L0). As the person moves closer to the center of the carousel, the person’s distance r
from the rotation axis decreases from r0 = 1.50 m to rf = 0.750 m. Therefore, the moment of
inertia I P = mr 2 (Equation 9.6) of the person also decreases, from I P0 = mr02 to I Pf = mrf2 .
Consequently, the angular speed of the system increases from ω0 to ωf , preserving the
system’s total angular momentum. We will use the angularmomentumconservation
principle to find the mass m of the person. 482 ROTATIONAL DYNAMICS SOLUTION The person and the carousel have the same initial angular velocity ω0, and the same final angular velocity ωf. Therefore, the conservation of angular momentum principle
can be expressed as
(1)
I Pf ω + I Cωf = I P0ω + I Cω0
14f244 140244
4
3
4
3
Lf L0 where IC is the moment of inertia of the carousel (without the person). Substituting I P0 = mr02 and I Pf = mrf2 (Equation 9.6) for the person’s initial and final moments of inertia
into Equation (1), we obtain
mrf2 ωf + I Cωf = mr02 ω0 + I Cω0
{
{
I Pf or ( I C (ωf − ω0 ) = m r02ω0 − rf2ωf ) I P0 Solving for the mass m of the person yields
m= I C (ωf − ω0 )
r02ω0 − rf2ωf = (125 kg ⋅ m2 ) ( 0.800 rad/s − 0.600 rad/s ) (1.50 m )2 ( 0.600 rad/s ) − ( 0.750 m )2 ( 0.800 rad/s ) = 28 kg 65. REASONING Let the space station and the people within it constitute the system. Then as
the people move radially from the outer surface of the cylinder toward the axis, any torques
that occur are internal torques. Since there are no external torques acting on the system, the
principle of conservation of angular momentum can be employed. SOLUTION Since angular momentum is conserved, I final ω final = I 0 ω 0
Before the people move from the outer rim, the moment of inertia is
2
I 0 = I station + 500 mperson rperson or
I 0 = 3.00 × 10 9 kg ⋅ m 2 + ( 500)(70.0 kg)(82.5 m) 2 = 3.24 × 10 9 kg ⋅ m 2
If the people all move to the center of the space station, the total moment of inertia is
I final = I station = 3.00 × 10 9 kg ⋅ m 2
Therefore, ω final
I
3.24 × 10 9 kg ⋅ m 2
= 0=
= 1.08
ω0
I final
3.00 × 10 9 kg ⋅ m 2 Chapter 9 Problems 483 This fraction represents a percentage increase of 8 percent . 66. REASONING AND SOLUTION Since the change occurs without the aid of external
torques, the angular momentum of the system is conserved: Ifωf = I0ω0. Solving for ωf gives I0 If ωf = ω0 where for a rod of mass M and length L, I 0 = 1
ML2 .
12 To determine If , we will treat the arms of the "u" as point masses with mass M/4 a distance L/4 from the rotation axis. Thus,
1 M
If = 12 2 L 2 2 M L 2 1
ML2 + 2 = 4 4 24 and 1 ML2 = (7.0 rad/s)(2) = 14 rad/s
ωf = ω0 12 1 ML2 24 67. REASONING AND SOLUTION After the mass has moved inward to its final path the
centripetal force acting on it is T = 105 N. r 105 N Its centripetal acceleration is
2 ac = v /R = T/m
Now
v = ωR so R = T/(ω m)
2 484 ROTATIONAL DYNAMICS The centripetal force is parallel to the line of action (the string), so the force produces no
torque on the object. Hence, angular momentum is conserved.
Iω = Ioωo so that ω = (Io/I)ωo = (Ro2/R2)ωo Substituting and simplifying
4 3 2 R = (mRo ωo )/T, so that R = 0.573 m 68. REASONING AND SOLUTION The block will just start to move when the centripetal
force on the block just exceeds f smax . Thus, if rf is the smallest distance from the axis at
which the block stays at rest when the angular speed of the block is ωf, then µs FN = mrf ωf2, or µs mg = mrf ωf2. Thus,
µs g = rf ωf2 (1) Since...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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