A solid sphere pendulum, with mass M, is released from rest and allowed to swing

Question

A solid sphere pendulum, with mass M, is released from rest and allowed to swing down and rotate about a frictionless pivot located at the surface of the sphere. The axis of rotation extends into and put of the page in regards to the figure provided. The initial center of the sphere is located at the same height as the pivot. Calculate the moment of inertia for the sphere about the given pivot point given a final magnitude of angular velocity of 10 radius per second, calculate the radius of the sphere in meters.

Explanation / Answer

A) using parallel axis theorem,

I = Icm + md2

=> I = 2mr2/5 + mr2 = 7mr2/5

B) Loss in PE = Gain in rotational KE

=> mgr = I2/2

=> mgr = 7mr22/10

=> r = 10g/72 = (10 * 9.8) / (7 * 102) = 0.14 m

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