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Find the translational kinetic energy of its center of gravity

  1. Nov 14, 2005 #1
    A 10.7kg cylinder rolls without slipping on a rough surface. At the instant when its center of gravity has a speed of 11.8 m/s,

    a) Find the translational kinetic energy of its center of gravity.

    b) Find the rotational kinetic energy about its center of mass at that time.

    c) What is its kinetic energy?

    I already found the answer to a by using the formula Ke = .5mv^2 and it's 744.934 J. For b I want to use the rotational kinetic energy formula Ke = .5Iω^2, but I need inertia for that. I can't find the inertia without the radius. Besides, I wouldn't know what formula to use since they don't explicitly say whether or not the cylinder is hollow.
  2. jcsd
  3. Nov 14, 2005 #2


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    You do know that the moment of inertia will contain a factor of [itex]R^2[/itex] and it will cancel when you finally write out the rotational KE. The geometric factor in the moment of inertial will still be there, however, so you'll have to make an assumption as to whether the cylinder is hollow or not. I believe that if the cylinder were hollow that would have been spelled out in the problem.
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