(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A cylindrical rod 36.4 cm long has mass 0.655kg and radius 1.1cm. A 18.5kg ball of diameter 11.4cm is attached to one end. The arrangement is originally vertical with the ball at the top and is free to pivot about the other end. After the ball-rod system falls a quarter turn, what is its rotational kinetic energy?

2. Relevant equations

KEr = 1/2 Iw^2.

mgh = 1/2Iw^2 + 1/2 mv^2.

3. The attempt at a solution

I found the moment of inertia of the ball-rod system, but then I don't know what to do next.

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# Homework Help: Torque on a cylindrical rod

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