problem10_22

University Physics with Modern Physics with Mastering Physics (11th Edition)

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10.22: a) The acceleration down the slope is , sin M f θ g a - = the torque about the center of the shell is , 3 2 3 2 2 MRa R a MR R a I Rf τ = = = = = . so 3 2 a M f = Solving these relations a for f and simultaneously gives or , sin 3 5 θ g a = . N 83 . 4 ) s m kg)(3.62 00 . 2 ( 3 2 3 2 , s m 62 . 3 0 . 38 sin ) s m 80 . 9 ( 5 3 sin 5 3 2 2 2 = = = = ° = = Ma f θ g a The normal force is Mg cos θ , and since , s n μ f . 313 . 0 tan 5 2 cos sin 3 2 cos 3 2 cos 5 3 3 2 s = = = = = θ θ g
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