Finding angular velocity using work energy theorem

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theBEAST
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Homework Statement


Multiple-choice question diagram: a horizontal rod of mass m and length L pivoted at its left end, with options a. (g/3L)^1/2, b. (g/L)^1/2, c. (12g/L)^1/2, d. (3g/L)^1/2, and d selected.


Homework Equations


Inertia about pin = ml^2/3
Work done by rotation = M*theta = 0.5(Inertia about pin) *ω2

The Attempt at a Solution


I tried two different methods and both came out to be the same answer. However they did not correspond to any of the multiple choice answers... What am I doing wrong?

Handwritten notes on lined paper showing two methods for finding the final angular velocity of a rod pivoted at one end, with a diagram of the rod and mg, and the resulting expressions for omega final.


Thanks!
 
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Method 1: The angular acceleration varies with [itex]\cos(\theta)[/itex]. So using constant acceleration formulas doesn't work. You simply can't use that method.

Method 2: The potential energy is still [itex]U=mgh[/itex], not [itex]U=M\theta[/itex].
 
frogjg2003 said:
Method 1: The angular acceleration varies with [itex]\cos(\theta)[/itex]. So using constant acceleration formulas doesn't work. You simply can't use that method.

Method 2: The potential energy is still [itex]U=mgh[/itex], not [itex]U=M\theta[/itex].

Thanks, but why does Mtheta not work?
 
frogjg2003 said:
I'm not even sure what M is.

Moment about the pin.
 
It's still wrong. It should be [itex]\frac{1}{2}M\omega[/itex]. But that by itself contains no information about the fact it's in gravity.