## Cylinder about side, finding angular velocity....

1. The problem statement, all variables and given/known data

A 5.0 kg, 60-cm-diameter disk rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released.

http://session.masteringphysics.com/...gure_13_72.jpg

What is the cylinder's angular velocity when it is directly below the axle?

2. Relevant equations

Law of conservation of energy (kinetic and potential of both sides)

3. The attempt at a solution

I tried to do this problem using mgy=1/2mv^2 but it didnt work...

I know it has something to do with moment of inertia, but do I just add the moment of inertia to the 1/2mv^2 and solve for v?
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 I tried to do this problem using mgy=1/2mv^2 but it didnt work...
Use rotational energy with the potential energy. The initial rotation energy about the axle is zero, so all the potential energy must go into rotational energy at the bottom point.
 I know it has something to do with moment of inertia, but do I just add the moment of inertia to the 1/2mv^2 and solve for v?
You do not just "add" in moment of inertia! You will need to find the moment of inertia of the disk about that pivot point.
 ah ic, i need to use rotational energy...i was treating this as a pendulum but i guess i cant do that haha lol yea, thats what i meant, i have to use Icm + Md^2 to find it at the pivot point thanks, ill see how it goes

## Cylinder about side, finding angular velocity....

yo thanks, i got it!

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