315_Physics ProblemsTechnical Physics

315_Physics ProblemsTechnical Physics - 317 Chapter 10...

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Chapter 10 317 P10.80 Consider the free-body diagram shown. The sum of torques about the chosen pivot is τα =⇒= F H G I K J F H G I K J = F H G I K J IF m l a ml a l A 1 3 2 3 2 2 CM CM (1) (a) A == l 124 . m : In this case, Equation (1) becomes a F m CM 2 N kg . ms = 3 2 3147 20630 35 0 . . af bg Fm a FHm a xx =⇒ + = CM CM or Hm a F x =− CM Thus, H x = + 0630 350 147 735 . kg . N 2 ej or Hi x = . ± N . (b) A 1 2 0620 . m: For this situation, Equation (1) yields a F m CM 2 N kg = 3 4 40630 17 5 . . . H y H x F = 14.7 N CM mg pivot A l FIG. P10.80 Again, a a F CM CM , so H x = 175 368 . 2 or x . ± N . (c) If H x = 0, then a a x = CM CM , or a F m CM = . Thus, Equation (1) becomes l F m A = 2 3 F H G I K J F H G I K J so A = = m m from the top 2 3 2 3 0827 l .. = . P10.81 Let the ball have mass m and radius r . Then Im r = 2 5 2 . If the ball takes four seconds to go down twenty-meter alley, then v = ms 5 . The translational speed of the ball will decrease somewhat as the ball loses energy to sliding friction and some translational kinetic energy is converted to rotational kinetic energy; but its speed will always be on the order of 500 . m s , including at the starting point. As the ball slides, the kinetic friction force exerts a torque on the ball to increase the angular speed.
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