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Velocity of Center of Mass for a Downwardly-Rotating Sphere

  1. Mar 4, 2015 #1

    CARNiVORE

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    1. The problem statement, all variables and given/known data
    mass = M
    radius = r
    rot. inertia = i
    height = h
    Sphere of mass M is released from rest at the top of an inclined plane.
    The speed of the center of mass at the bottom of the incline, without friction, is sqrt(2gh).

    I need to find the velocity of the center of mass assuming the plane has friction and the sphere rolls without slipping.

    2. Relevant equations

    I'm perplexed. My only guess would be vcm = (Σmivi)/(Σmi)

    3. The attempt at a solution

    The correct answer is √(2MgHr2)/(I+Mr2).
     
  2. jcsd
  3. Mar 4, 2015 #2

    haruspex

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    A more useful relevant equation would be work conservation. What does that have to say?
     
  4. Mar 4, 2015 #3

    CARNiVORE

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    Thanks for the response. That's a good idea. Let's see what I can do with that. It starts from rest, so Ki = 0. It ends at h=0, so Uf = 0.
    Wi = Ki
    Ui = Kf + Wƒ
    mgh = .5mv2 + μmgh
    2gh = v2 + 2μgh
    √2gh - 2μgh = v

    Definitely incorrect. Since μ isn't even in the answer, I'm guessing that I either need to solve for μ, or I approached this completely incorrectly. Maybe both. What do you think?
     
  5. Mar 4, 2015 #4

    haruspex

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    It is rolling now, not sliding, so the friction does no work. On the other hand, you have not counted all the KE of the ball.
     
  6. Mar 4, 2015 #5

    CARNiVORE

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    Oops, that was a silly mistake. I also forgot to add .5iω2 to the final kinetic energy. I've figured it out now, though:

    mgh = .5mv2 + .5iω2
    2mgh = mv2 + i(v2/r2)
    2mghr2 = mv2r2 + iv2
    2mghr2 = v2(mr2 + i)
    2mghr2/(mr2 + i) = v2

    Then just take the square root of both sides. Thanks for your help!
     
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