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March 27, 2017

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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 - ,

    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 - ,

    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 - ,

    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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