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A long uniform rod length L and mass M is pivoted about a horizonal, frictionless pin passing through one end. The rod is released from rest in a vertical position. The instant the rod is horizontal, what is the magnitue of its angular acceleration.

It's angular speed I know is [tex]\sqrt{\frac{3g}{L}}[/tex]

using the formula [tex]\alpha=rw^2[/tex]

I get then angular acceleration to be 3g. However the book tells me that the answer is 3g/2L. What am I missing?

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# Homework Help: Rod - angular acceleration

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