1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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.)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook