Cylinder about side, finding angular velocity

AI Thread Summary
The discussion revolves around calculating the angular velocity of a rotating cylinder using conservation of energy principles. A 5.0 kg disk with a 60-cm diameter is analyzed as it rotates about an axle at one edge. Participants emphasize the importance of considering the moment of inertia about the pivot point rather than simply adding it to kinetic energy equations. The correct approach involves using rotational energy and potential energy, recognizing that all potential energy converts to rotational energy when the cylinder is at the lowest point. Ultimately, the problem is resolved by applying the appropriate formulas for moment of inertia and energy conservation.
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Cylinder about side, finding angular velocity...

Homework Statement



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/problemAsset/1001021/11/knight_Figure_13_72.jpg

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

Homework Equations



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

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 can't do that haha

lol yea, that's what i meant, i have to use Icm + Md^2 to find it at the pivot point

thanks, ill see how it goes
 
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yo thanks, i got it!
 
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