A stick with a mass of 0.170Kg and a length of 1.00m is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released.(adsbygoogle = window.adsbygoogle || []).push({});

1) As it swings through the vertical, calculatethe change in gravitational potential energy that has occurred. Gravity = 9.81m/s^2

Alright So for this one I have no idea really where to begin except to find the Moment of inertia. 1/3 ML^2. Substituting I get 1/3 (.170kg) (1)^2

The problem with that is is that I dont know how to find the potential energy because there is no height given to use U=mgh. I tried to use Mgy(y is in cm) and that didnt work well because it gave me (.170kg)(9.81m/s^2)(100cm). Where would I go from here considering there is no w (angular velocity) or anything given.

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# Rotational Energy Problem

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