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Dynamics of rotational Motion

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A 392N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600m and its moment of inertia about its axis of rotation is 0.800MR^2. Friction does work on the wheel as it rolls up the hill to a stop, a height h above the bottom of the hill; this work has absolute value 3500J. Calculate h.

    2. Relevant equations

    KE = (1/2)MV^2 + (1/2)(Icm)(W^2)

    Principle of Conservation of energy

    3. The attempt at a solution

    What I did is use the fact that the sum of the translational and rotational kinetic energies at the bottom of the hill should be equal to the gain in the potential energy PLUS 3500J which is the work done against friction.

    In other words,

    (1/2)MV^2 + (1/2)(Icm)(W^2) = mgh + 3500.

    However, the correct answer is obtained if in the above equation, 3500 is subtracted from mgh rather than added.

    Can someone please explain why this is so. Subtracting makes no sense. It would simply mean that more the friction, h is greater since (1/2)MV^2 + (1/2)(Icm)(W^2) +3500 is greater then (1/2)MV^2 + (1/2)(Icm)(W^2) -3500.

    Thank-you for your assistance.
  2. jcsd
  3. Oct 20, 2007 #2


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    Homework Helper

    you're right.

    "(1/2)MV^2 + (1/2)(Icm)(W^2) = mgh + 3500. "

    is the correct equation...
  4. Oct 20, 2007 #3

    Thanks for the re-assurance!!:biggrin:
  5. Oct 20, 2007 #4


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    Homework Helper

    no prob. :)
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