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A uniform rod of length L and mass M is pivoted at one end (O) and is released from rest in a horizontal position. The moment of inertia of the rod about one end is I = (mL^2)/3 . The initial tangential acceleration fo the far end A of the rod would be?

This is part of my review for my AP Physics C test. I thought first maybe I could use torque in some way

T= rF = LMg

And [tex] T[/tex] [tex]= I \alpha [/tex]

So, I equated the torque statements thus having

[tex] LMg [/tex][tex] = (1/3)(ML^2)(\alpha) [/tex]

This lead me to the result of 3g/L = [tex] \alpha [/tex]

and by the identity [tex] a = r \alpha [/tex] I got 3g = acceleration. This is not the right answer. The correct answer is 3g/2, as opposed to my 3g. Where did the 1/2 come from? Thanks!

Wait...I think I figured it out...is it hte fact that in T = rF, the r is referring to 1/2 the Length, not the full length?

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# Homework Help: Simple Tangential Acceleration Problem

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