Physics Solution Manual for 1100 and 2101

1 where f is the magnitude of the force exerted on

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Unformatted text preview: is greater than τ2, because the lever arm for the force F3 is greater than that for F2. The lines of action for the forces F1 and F4 pass through the axis of rotation. Therefore, the lever arms for these forces are zero, and the forces produce no torque. 4. (b) Since the counterclockwise direction is the positive direction for torque, the torque produced by the force F1 is τ1 = −(20.0 N)(0.500 m) and that produced by F2 is τ2 = +(35.0 N)[(1.10 m)(cos 30.0°)]. The sum of these torques is the net torque. 5. (e) The clockwise torque produced by F2 is balanced by the counterclockwise torque produced by F. The torque produced by F2 is (remembering that the counterclockwise direction is positive) τ2 = +F2[(80.0 cm − 20.0 cm)(sin 55.0°)], and the torque produced by F is τ = −(175 N)(20.0 cm). Setting the sum of these torques equal to zero and solving for F2 gives the answer. 6. (d) The sum of the forces (F − 2F + F) equals zero. Select an axis that passes through the center of the puck and is perpendicular to the screen. The sum of the torques [−FR + 2F(0) + FR ] equals zero, where R is the radius of the puck. Thus, the puck is in equilibrium. 7. Magnitude of F1 = 12.0 N, Magnitude of F2 = 24.0 N 8. (c) The horizontal component of F3 is balanced by F1, and the vertical component of F3 is balanced by F2. Thus, the net force and, hence, the translational acceleration of the box, is zero. For an axis of rotation at the center of the box and perpendicular to the screen, the forces F2 and F3 produce no torque, because their lines of action pass through the axis. The force F1 does produce a torque about the axis, so the net torque is not zero and the box will have an angular acceleration. 9. Distance of center of gravity from support = 0.60 m. Chapter 9 Answers to Focus on Concepts Questions 435 10. (b) The moment of inertia of each particle is given by Equation 9.6 as I = mr 2 , where m is its mass and r is the perpendicular distance of the particle from the axis. Using this equation, the moment of inertia of each particle is: A: 10m0 r02 , B: 8m0 r02 , C: 9m0 r02 . 2 11. I = 1.7 kg⋅m 12. Magnitude α of the angular acceleration = 1.3 rad/s . 2 13. (a) According to Newton’s second law for rotational motion, Equation 9.7, the angular acceleration is equal to the torque exerted on the wheel (the product of the force magnitude and the lever arm) divided by the moment of inertia. Thus, the angular acceleration of the 2 smaller wheel is α = FR/(MR ) = F/(MR), while that of the larger wheel is F ( 2R ) α= = 1 ( F / MR ) , so the smaller wheel has twice the angular acceleration. 2 2 M ( 2R ) 14. Magnitude α of the angular acceleration = 12.0 rad/s2 15. (c) The translational kinetic energy is 1 2 Mv 2 , where v is the speed of the center of mass of the wheel. The rotational kinetic energy is 1 2 I ω 2 , where I is the moment of inertia and ω is the angular speed about the axis of rotation. Since I = MR and ω = v/R for rolling motion 2 2 v ( MR 2 ) R = 12 Mv2 , which is the same as the translational kinetic energy. Thus, the ratio of the two energies is 1. (See Equation 8.12), the rotational kinetic energy is 1 2 Iω 2 = 1 2 16. (c) As each hoop rolls down the incline, the total mechanical energy is conserved. Thus, the loss in potential energy is equal to the gain in the total kinetic energy (translational plus rotational). Because the hoops have the same mass and fall through the same vertical distance, they lose the same amount of potential energy. Moreover, both start from rest. Therefore, their total kinetic energies at the bottom are the same. 17. (d) As discussed in Section 9.6, the angular momentum a system is conserved (remains constant) if the net external torque acting on the system is zero. 18. (b) The rotational kinetic energy of a rotating body is KE R = 1 I ω 2 (see Equation 9.9), 2 where I is the moment of inertia and ω is the angular speed. We also know that her angular momentum, L = Iω (Equation 9.10), is conserved. Solving the last equation for I and L substituting the result into the first equation gives KE R = 1 I ω 2 = 1 ω 2 = 1 Lω . Since L 2 2 2 ω is constant, the final rotational kinetic energy increases as ω increases. 19. I = 0.60 kg⋅m2 436 ROTATIONAL DYNAMICS CHAPTER 9 ROTATIONAL DYNAMICS PROBLEMS 1. REASONING AND SOLUTION According to Equation 9.1, we have Magnitude of torque = F l where F is the magnitude of the applied force and l is the lever arm. For an axis passing through the cable car at its center and perpendicular to the ground, the lever arm is one-half the length of the car or 4.60 m. Both people rotate the car in the same direction, so the torques that they create reinforce one another. Therefore, using Equation 9.1 to express the magnitude of each torque, we find that the magnitude of the net torque is Magnitude of net torque = (185 N )( 4.60 m ) + (185 N )( 4.60 m ) = 1.70 × 103 N ⋅ m 2. REASONING The drawing shows the wheel as it rolls to the right, so the torque applied by the engine is assumed to be clockwis...
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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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