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Homework Help: Rotational Kinetic Energy

  1. Oct 23, 2006 #1
    A 2.4 kg cylinder (radius = 0.09 m, length = 0.50 m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.74 m high and 5.0 m long. What is its rotational kinetic energy?

    I just wanted to make sure that I was on the right track...

    I=2/5 mr^2
    KE=1/2 mv^2 + 1/5 mv^2

    Using conservation of energy: mgh=1/2 mv^2 + 1/5 mv^2 and then solve for v and plug into equation for KE.

    Does this sound like I'm doing this correctly?
  2. jcsd
  3. Oct 23, 2006 #2

    Doc Al

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    Staff: Mentor

    You are on the right track, but you are using the wrong formula for rotational inertia. (2/5 mr^2 is for a rolling ball, not a cylinder.)
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