Physics Solution Manual for 1100 and 2101

90 ms 162 n 216 n 220 kg chapter 9 problems 471 48

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Unformatted text preview: nertia for the three-ball system is I = m1r12 + m2 r2 + m3r32 (Equation 9.6), where m1, m2, and m3 are the masses of the balls and r1, r2, and r3 are the distances from the axis. In system A, the ball whose mass is m1 does not contribute to the moment of inertia, because the ball is located on the axis and r1 = 0 m. In system B, the ball whose mass is m3 does not contribute to the moment of inertia, because it is located on the axis and r3 = 0 m. b. The magnitude of the torque is equal to the magnitude F of the force times the lever arm l (see Equation 9.1). In system A the lever arm is l = 3.00 m. In B the lever arm is l = 0 m, since the line of action of the force passes through the axis of rotation. c. According to Newton’s second law for rotational motion, Equation 9.7, the angular acceleration α is given by α = ( Στ ) / I , where Στ is the net torque and I is the moment of inertia. The angular velocity ω is given by Equation 8.4 as ω = ω0 + α t, where ω0 is the initial angular velocity and t is the time. SOLUTION a. The moment of inertia for each system is System A I = m1r12 + m2 r22 + m3r32 = ( 9.00 kg )( 0 m ) + ( 6.00 kg )( 3.00 m ) + ( 7.00 kg )( 5.00 m ) = 229 kg ⋅ m 2 2 System B 2 2 I = m1r12 + m2 r22 + m3r32 = ( 9.00 kg )( 5.00 m ) + ( 6.00 kg )( 4.00 m ) + ( 7.00 kg )( 0 m ) = 321 kg ⋅ m 2 2 2 2 Chapter 9 Problems 467 b. A torque that tends to produce a counterclockwise rotation about the axis is a positive torque. The torque produced by the force has a magnitude that is equal to the product of the force magnitude and the lever arm: System A τ = − F l = − ( 424 N ) ( 3.00 m ) = −1270 N ⋅ m The torque is negative because it produces a clockwise rotation about the axis. System B τ = F l = ( 424 N ) ( 0 m ) = 0 N ⋅ m c. The final angular velocity ω is related to the initial angular velocity ω0, the angular acceleration α, and the time t by ω = ω0 + α t (Equation 8.4). The angular acceleration is given by Newton’s second law for rotational motion as α = ( Στ ) / I (Equation 9.7), where Στ is the net torque and I is the moment of inertia. Since there is only one torque acting on each system, it is the net torque, so Στ = τ . Substituting this expression for α into Equation 8.4 yields τ ω = ω0 + α t = ω0 + t I In both cases the initial angular velocity is ω0 = 0 rad/s, since the systems start from rest. The final angular velocities after 5.00 s are: System A −1270 N ⋅ m τ ω = ω0 + t = ( 0 rad/s ) + ( 5.00 s ) = −27.7 rad/s 2 I System B τ ω = ω0 + t = ( 0 rad/s ) + I 229 kg ⋅ m 0 N⋅m ( 5.00 s ) = 0 rad/s 321 kg ⋅ m 2 468 ROTATIONAL DYNAMICS 44. REASONING To determine the angular acceleration of the pulley and the tension in the cord attached to the block, we will apply Newton’s second law to the pulley and the block separately. Only one external forces acts on the pulley, as its free-body diagram at the right shows. This force T is due to the tension in the cord. The torque that results from this force is the net torque acting on the pulley and obeys Newton’s second law for rotational motion (Equation 9.7). Two external forces act on the block, as its free-body diagram at the right indicates. These are (1) the force T due to the tension in the cord and (2) the weight mg of the block. The net force that results from these forces obeys Newton’s second law for translational motion (Equation 4.2b). R +y +τ T +x Free-body diagram for pulley T′ mg Free-body diagram for block SOLUTION Applying Newton’s second law for rotational motion to the pulley gives Σ τ = +T R = I α (1) where we have written the torque as the magnitude of the tension force times the lever arm (the radius) as specified by Equation 9.1. In addition, we have assigned this torque a positive value, since it causes a counterclockwise rotation of the pulley. To obtain a value for T, we note that the tension has the same magnitude everywhere in the massless cord, so that T = T ′ . Thus, by applying Newton’s second law for translation (Equation 4.2b), we obtain ΣF = T ′ − mg = −ma or T = T ′ = mg − ma where we have used a to denote the magnitude of the vertical acceleration of the block and included the minus sign to account for the fact that the block is accelerating downward. Substituting this result for T into Equation (1) gives ( mg − ma ) R = Iα (2) To proceed further, we must deal with a. Note that the pulley rolls without slipping against the cord, so a and α are related according to a = Rα (Equation 8.13). With this substitution, Equation (2) becomes mg − m ( Rα ) R = Iα Solving for α, we find that or mgR = mR 2α + Iα Chapter 9 Problems ( 469 ) ( 2.0 kg ) 9.80 m/s2 ( 0.040 m ) mgR α= = = 180 rad/s 2 2 2 2 −3 mR + I ( 2.0 kg )( 0.040 m ) + 1.1×10 kg ⋅ m The value for the tension can be now obtained by substituting this value for α into Equation (1): TR = Iα or ( )( ) 1.1×10−3 kg ⋅ m 2 180 rad/s 2 Iα T= = = 5.0 N R 0.040 m 45. SSM REASONING The angular acceler...
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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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