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

A rod of length L and mass M is pivoted about one end. The rod is released from rest in a horizontal position, and allowed to swing downward without friction or air resistance. When the rod is vertical, what is its angular speed w?

2. Relevant equations

Moment of Inertia, I = 1/3 ML^2

3. The attempt at a solution

I would like to apologize in advance if my notations are difficult to read.

Ki + Ui = Kf + Uf

0 + MgL = (1/2) Iw^2 + 0

MgL = (1/2)(1/3 ML^2)w^2

√6g/L = w

I think I'm supposed to use the rod's center of mass because my current answer is wrong and since the diagram in my book points out that the center of mass is at L/2 from the pivot point. I don't think they would point that out if it was unnecessary, but I don't understand why. Putting Ui as Mg(L/2) gives me the right answer though, according to the back of the book.

Any help clarifying this would be much appreciated.

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# Homework Help: Angular speed of falling rod

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