Posted by **kelsey** on Monday, October 20, 2008 at 4:32am.

A 2.0 kg cylinder (radius = 0.08 m, length = 0.50 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.77 m high and 5.0 m long.

(b) What is its rotational kinetic energy?

? J

(c) What is its translational kinetic energy?

? J

How do I find the velocity in this equation so I can use RotKE= 1/2 I w ^2 and trans =1/2mv^2

- PHYSICS 2 -
**drwls**, Monday, October 20, 2008 at 4:40am
Set the potential energy decrease M g H equal to the total kinetic energy. Then divide that up between translational and rotational parts. One is

(1/2) M V^2, and the other is (1/2)(1/2)M R^2(V^2/R^2) = (1/4) M V^2, so the translational part is 2/3 of the total, etc.

- PHYSICS 2 -
**kelsey**, Monday, October 20, 2008 at 4:54am
okay so i have mgh =15.1 and i set it equal to KErot + KE trans?

15.1= 1/2[(2/5 X 2 X .08^2) X (1/2 X 2 X v)^2]

that right?

- PHYSICS 2 -
**drwls**, Monday, October 20, 2008 at 6:56am
You have an X where you need a + between the translational and rotational terms. Both terms should include V. Then you could solve for V. But they don't ask for V.

AOnce yopu have MgH, all you have to do is compute KE(tr) = (2/3) M g H and (KE)rot) = (1/3) M g H

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