HW 8 Solutions

2m 5 1 5m 5m 2m b substitute for a in equation 2 4 and

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Unformatted text preview: hermore, the moment of inertia for a sphere is I = 2 M R2 . So, nd 5 Newton’s of law to the sphere and we find the following system2 equations and the hanging object: 2 T= Ma 5 T = m ( g − a) . ∑F x I sphereα = mg − T = ma ( Setting these two ubstitute for Isphere andother and solving for the acceleration2 S expressions equal to each α in 2 TR = 5 MR gives equation (1) to obtain: g a= . 1 + 2M 5m a )R 2 Eliminate T between equations (2) g (b) Now we just substitute this result back in to the expression T = 5 M a to find and (3) and solve for a to obtain: a= 1+ 2 Mg 2M mg T= = . 2M 5 1 + 5m 5m + 2M (b) Substitute for a in equation (2) 4 and solve for T to obtain: T= 2M 5m 2mMg 5m + 2 M 5. A basketball rolls without slipping down an incline of angle θ. The coefficient of static friction is µs . Model the ball as a thin spherical shell. Find Picture center of mass of the ball, (a) the acceleration of the the Problem The three forces acting on the basketball a the ball, the the ball, force, and the force of friction....
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This document was uploaded on 03/14/2014 for the course PHYS 18 at UC Merced.

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