Angular Velocity from Potential and Mechanical Energy of rotating rod

AI Thread Summary
To find the angular velocity of a falling rod when it meets the horizontal, the potential energy (U) can be equated to the mechanical energy at that point. The relevant equation is U = (1/2)MgL, where L represents the length to the center of mass. The moment of inertia (I) is also necessary for calculating kinetic energy (K), given by K = (1/2)Iw². The initial potential energy should use the altitude of the center of mass without the (1/2) factor. This approach correctly leads to solving for angular velocity (ω).
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Homework Statement



I am asked to find the angular velocity of a falling rod in the instant that the rod meets the horizontal. The system is set up in this image:

1sdszo.jpg


The only force acting on the rod is gravity

Homework Equations



U = (1/2)MgL
K = (1/2)Iw2
Kf + Uf = Ki + Ui

The Attempt at a Solution



Can I solve this by using the potential energy of the rod with this equation:

U = (1/2)MgL

and then equating that to the Mechanical Energy at the point that it meets the horizontal?

I know the moment of inertia of the rod so if I do (1/2)MgL = (1/2)Iw2 and solve for omega, is that correct?

Also, when finding the initial potential energy, should i use the full length of the rod or only the length to the center of mass?

Thank you!
 
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yes, you are correct. to calculate the initial potential energy, just use the altitude of the center of mass and drop the (1/2), that should not be there.

good luck
 
Oh okay, so let me just make sure I understand:

so this is the correct way to find the initial potential energy?

U = mgLsin\theta

where L is the length to the center of mass
 
correct.
 
thanks so much!
 
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