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Rotational and translational energy problem

  1. Apr 30, 2009 #1
    A solid sphere of radius R is placed at a height of 30cm on a 15 deg slope. It is released and rolls, without slipping to the bottom.

    a) From what height should circular hoop of radius R be released in order to equal the sphere's speed at the bottom?

    First i started with the sphere
    mgh=.5mv2 +.5Iw2
    I=.5mR2
    w=v/r
    so
    mgh=.5mv2+.25mv2
    gh=.75v2
    h=30
    v2 =40g

    next the hoop
    the only difference is we dont know h and I is different
    I=mR2
    so gh=v2
    therefore setting them equal, h=40.

    However the answer is 43. I did neglect to include R in the height because both objects have this height R. However when i included R in to h. I found 40+(1/3)R=h which leave me know where.

    Any ideas? There is no mention of friction...
     
  2. jcsd
  3. Apr 30, 2009 #2
    Check your number for the sphere's moment of inertia. The value you are using is for a disc or solid cylinder.
     
  4. Apr 30, 2009 #3
    Thankyou! jeez and to think i spent an hour for a little mistake. For anyone else I for a sphere is 2/5MR2
     
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