Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

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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 angular-momentum-conservation 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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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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