Physics Solution Manual for 1100 and 2101

1 thus the reels angular displacement is given by lr

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Unformatted text preview: 9.2) indicates that the net external torque is equal to the moment of inertia times the angular acceleration. To determine the angular acceleration, we will use Equation 8.7 from the equations of rotational kinematics. This equation indicates that the angular displacement θ is given by θ = ω0t + 1 α t 2 , where ω0 is the initial angular velocity, t is the time, and α is the angular 2 acceleration. Since both wheels start from rest, ω0 = 0 rad/s for each. Furthermore, each wheel makes the same number of revolutions in the same time, so θ and t are also the same for each. Therefore, the angular acceleration α must be the same for each. SOLUTION Using Equation 9.2, the net torque Στ that acts on each wheel is given by Στ = I α , where I is the moment of inertia and α is the angular acceleration. Solving 2 (θ − ω0t ) Equation 8.7 for the angular acceleration α = and substituting the result into t2 Equation 9.2 gives 2 (θ − ω0t ) Στ = I α = I t2 Table 9.1 indicates that the moment of inertia of a hoop is I hoop = MR 2 , while the moment 1 of inertia of a disk is I disk = 2 MR 2 . The net external torques acting on the hoop and the disk are: 460 ROTATIONAL DYNAMICS Hoop 2 (θ − ω0t ) Στ = I hoopα = MR 2 t2 2 2 (13 rad ) − 2 ( 0 rad/s )( 8.0 s ) = 0.20 N ⋅ m = ( 4.0 kg )( 0.35 m ) (8.0 s )2 Disk 2 (θ − ω0t ) Στ = I diskα = 1 MR 2 2 t2 = 1 2 2 (13 rad ) − 2 ( 0 rad/s )(8.0 s ) = 0.10 N ⋅ m (8.0 s )2 ( 4.0 kg )( 0.35 m )2 36. REASONING The ladder is subject to three vertical forces: the P FN upward pull P of the painter on the top end of the ladder, the upward normal force FN that the ground exerts on the bottom end of the ladder, and the downward Axis of W force W of the ladder’s weight, rotation which acts at the ladder’s center of gravity, halfway between the Free-body diagram of the ladder ends (see the free-body diagram of the ladder). The bottom end serves as the axis of rotation. The normal force FN is applied at the axis, so it has no lever arm. The net torque acting on the ladder can be obtained with the aid of Equation 9.1: Στ = Pl P − W l W + FN ( 0) = PL − W { 1 3 1 24 2 43 Torque due to P Torque due to W ( 1 L) = PL − 1 mgL 2 2 (1) Torque due to FN where L is the length of the ladder and m is its mass. Once we know the net torque Σ τ acting on the ladder, we will use Newton’s second law for rotation, Στ = Iα (Equation 9.7) to determine the moment of inertia I of the ladder. SOLUTION a. From Equation (1), the net torque acting on the ladder is ( ) Στ = PL − 1 mgL = ( 245 N )( 9.75 m ) − 1 ( 23.2 kg ) 9.80 m/s 2 ( 9.75 m ) = 1280 N ⋅ m 2 2 Chapter 9 Problems 461 b. The ladder’s moment of inertia is found from α = Στ I (Equation 9.7). Using the result found in part a, we obtain I= Στ = α 1280 N ⋅ m = 711 kg ⋅ m 2 2 1.80 rad/s 37. SSM REASONING The rotational analog of Newton's second law is given by Equation 9.7, ∑τ = Iα . Since the person pushes on the outer edge of one section of the door with a force F that is directed perpendicular to the section, the torque exerted on the door has a magnitude of FL, where the lever arm L is equal to the width of one section. Once the moment of inertia is known, Equation 9.7 can be solved for the angular acceleration α. The moment of inertia of the door relative to the rotation axis is I = 4IP, where IP is the moment of inertia for one section. According to Table 9.1, we find I P = 1 ML2 , so that the 3 rotational inertia of the door is I = 4 ML2 . 3 SOLUTION Solving Equation 9.7 for α, and using the expression for I determined above, we have α= FL = 4 ML2 3 4 3 68 N F =4 = 0.50 rad/s 2 ML 3 (85 kg)(1.2 m) 462 ROTATIONAL DYNAMICS 38. REASONING The angular acceleration α that results from the application of a net external torque Στ to a rigid object with a moment of inertia I is given by Newton’s second law for Στ (Equation 9.7). By applying this equation to each of the two rotational motion: α = I situations described in the problem statement, we will be able to obtain the unknown angular acceleration. SOLUTION The drawings at the right illustrate the two types of rotational motion of the object. In applying Newton’s second law for rotational motion, we need to keep in mind that the moment of inertia depends on where the axis is. For axis 1 (see top drawing), piece B is rotating about one of its ends, so according to Table 9.1, the moment of inertia is B Axis 1 1 3 I for axis 1 = M B L2 , where MB and LB are the mass B Axis 2 and length of piece B, respectively. For axis 2 (see bottom drawing), piece A is rotating about an axis through its midpoint. According to Table 9.1 the moment of inertia is I for axis 2 = 1 M A L2 . A 12 A Applying the second law for each of the axes, we obtain α for axis 1 = Στ I for axis 1 and α for axis 2 = Στ I for axis 2 As given, the net torque Στ is the same in both expressions. Dividing the expression for axis 2 by the expression for axis 1 gives Στ α for axis 2 I for axis 2 I for axis 1 = = Στ α for a...
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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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