I've also attached the picture of this problem at the end of the post.(adsbygoogle = window.adsbygoogle || []).push({});

A uniform rod of mass 4:5 kg is 10mlong. The

rod is pivoted about a horizontal, frictionless

pin at the end of a thin extension (of negligible

mass) a distance 10 m from the center of

mass of the rod. Initially the rod makes an

angle of 60± with the horizontal. The rod is

released from rest at an angle of 60± with the

horizontal, as shown in the ¯gure below

The acceleration of gravity is 9:8 m=s2 :

Hint: The moment of inertia of the rod

about its center-of-mass is Icm = (1/12)m*l^2

What is the angular speed of the rod at

the instant the rod is in a horizontal position?

Answer in units of rad=s.

I tried doing mgh = 1/2 I * omega^2, but it doesn't feel right at all. Especially since I was unsure of whether I could use the height of the center of mass of the rod for the h in mgh. Can I? There's gotta be a better way to solve this problem...

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# Rod in rotational motion. Not sure how to account for potential energy.

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