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 Km v I =+ 1 2 1 2 22 , 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 sR = 4 af θ and its speed is v ds dt R d dt == F H G I K J 4 . Also, since the ball rolls without slipping, v ds dt R so Ω= = F H G I K J v R d dt 4 The kinetic energy is then 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 θθ h 5 R R s FIG. P15.56 When the ball has an angular displacement , its center is distance hR =− 41c o s higher than when at the equilibrium position. Thus, the potential energy is Um g hm g R g 41 c o s . For small angles, 1 2 2 −≈ cos (see Appendix B). Hence, g R g 2 2 , and the total energy is EKU
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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