Chapter 15459P15.56The kinetic energy of the ball is KmvI=+121222Ω,where Ωis the rotation rate of the ball about itscenter of mass. Since the center of the ball movesalong a circle of radius 4R, its displacement fromequilibrium is sR=4a fθand its speed isvdsdtRddt==FHGIKJ4θ. Also, since the ball rolls withoutslipping,vdsdtR==ΩsoΩ ==FHGIKJvRddt4θThe kinetic energy is thenKmRddtmRddtmRddt=FHGIKJ+FHGIKJFHGIKJ=FHGIKJ124122541121022222θθθh5R θR sFIG. P15.56When the ball has an angular displacement θ, its center is distance hR=−41cosθafhigher thanwhen at the equilibrium position. Thus, the potential energy is UmghmgRg==−41cosθaf. For smallangles, 122−≈cosθθaf(see Appendix B). Hence, UmgRg≈22θ, and the total energy is
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