457_Physics ProblemsTechnical Physics

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 .

Ask a homework question - tutors are online