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## Homework Statement

A cylindrical rod 26.0 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball of diameter of 7.80 cm and a mass of 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top. The apparatus is free to pivot about the bottom end of the rod.

(a) After it falls through 90°, what is its rotational kinetic energy?

## Homework Equations

KE=1/2I[tex]omega[/tex]^2

I=(2mr^2)/5

I=(2mL^2)/3

## The Attempt at a Solution

Moment of inertia=I of rod + I of ball?

I=(2*m*(r+L)^2)/5)+(2mL^2)/3

I'm lost as to finding [tex]\omega[/tex] so I can find KE.

Please point me in the right direction.

Thank you

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