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Rotational Dynamics of a solid sphere

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A solid sphere starts from rest at the upper end of the track, as shown in figure below, and rolls without slipping until it rolls off the right-hand end. If H = 12.0 m and h = 2.0 m and the track is horizontal at the right-hand end, how far horizontally from point A does the sphere land on the floor?


    2. Relevant equations

    mgH = 1/2 I w^2 + mgh
    y = vot + 1/2gt^2

    3. The attempt at a solution

    I'm not exactly sure how i went wrong here.
    the moment of inertia for a solid sphere is 2/5 mr^2 and w^2 = v^2/r^2 so the radius cancels out....after cancelling out the mass as well i come out with
    gH = 1/5 v^2 + gh which becomes sqrt(5(gH-gh)) = 22.136 m/s
    Since it starts at rest... you use the equation 2 = -4.9t^2 to get a time of 0.639s

    Since its asking for horizontal distance i multiplied the velocity by time to get 14.145m, but that's wrong somehow...is there something i'm missing?
     
  2. jcsd
  3. Oct 25, 2009 #2

    Doc Al

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    Staff: Mentor

    Don't forget translational kinetic energy.
     
  4. Oct 25, 2009 #3
    i thought there is no translational kinetic energy since the sphere isn't slipping
     
  5. Oct 26, 2009 #4

    Doc Al

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    Staff: Mentor

    No, the sphere's center of mass is definitely moving as it rolls without slipping, so it has both translational and rotational kinetic energy. If it was just spinning in place without moving, then it would only have rotational KE. But here it's rolling along just fine. (Hint: Since it's rolling without slipping, the translational and rotational speeds are related.)
     
  6. Oct 26, 2009 #5
    ah right..i just got mixed up on a concept, that makes perfect sense now, thanks a lot
     
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