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Moment of inertia

  1. Dec 7, 2008 #1
    This is a general question:
    Does a higher coefficient for moment of inertia imply that it is harder for the object to rotate than for an object with a smaller coefficient?

    We did this experiment in class where we rolled an empty can and then a can full of beans. The can of beans rolled faster than the empty can. We said that the can of beans was more massive, the mass was distributed closer to the axis, and I=1/2 mr^2 but ....meanwhile the empty can is less massive, the mass was distributed further from the axis and I=mr^2.

    Does the coefficient have to do with it or is it just because of the mass distribution?
  2. jcsd
  3. Dec 7, 2008 #2


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    The "coefficient" you are talking about is a consequence of its distribution.

    Actually it is the sum of all the little weights times their respective distances from the rotation that make up an object summed across the volume of the object. With most of the mass close to the radius then an empty can will be more like mr² and a full can more like 1/2mr².

    Looking at the conservation of energy then an object will carry at the bottom of a ramp a velocity that satisfies the potential to kinetic relationship. If it's a frictionless ramp then v2 = 2gh.

    Note that this is independent of total mass and assumes no rotational kinetic energy. Just like dropping objects with no air friction results in the same times regardless of mass.

    Now to the empty and full cans, the full can will have greater mass, that you can ignore, but its distribution results in just half as much (1/2mr²)of the potential energy going to rotational kinetic energy and hence more will be in the translational kinetic energy and will be faster at the bottom.

    The empty can having closer to a full mr² will take more kinetic rotational energy and less will be left for translational velocity.
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