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457_Physics ProblemsTechnical Physics

# 457_Physics ProblemsTechnical Physics - Chapter 15 P15.56 1...

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Chapter 15 459 P15.56 The kinetic energy of the ball is K mv I = + 1 2 1 2 2 2 , where is the rotation rate of the ball about its center of mass. Since the center of the ball moves along a circle of radius 4 R , its displacement from equilibrium is s R = 4 a f θ and its speed is v ds dt R d dt = = F H G I K J 4 θ . Also, since the ball rolls without slipping, v ds dt R = = so Ω = = F H G I K J v R d dt 4 θ The kinetic energy is then K m R d dt mR d dt mR d dt = F H G I K J + F H G I K J F H G I K J = F H G I K J 1 2 4 1 2 2 5 4 112 10 2 2 2 2 2 θ θ θ h 5 R θ R s FIG. P15.56 When the ball has an angular displacement θ , its center is distance h R = 4 1 cos θ a f higher than when at the equilibrium position. Thus, the potential energy is U mgh mgR g = = 4 1 cos θ a f . For small angles, 1 2 2 cos θ θ a f (see Appendix B). Hence, U mgR g 2 2 θ , and the total energy is
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