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Motion of ring/body down an incline

  1. Oct 25, 2006 #1
    A body of mass 'm' slides down an incline and reaches the bottom with a velocity 'v'. If the same mass were in the form of a ring which rolls down the incline, what would have been the velcity of the ring?

    (A)[tex]v[/tex]

    (B)[tex]\sqrt{2}v[/tex]

    (C)[tex]\frac{1}{\sqrt{2}}v[/tex]

    (D)[tex]\frac{\sqrt{2}}{\sqrt{5}}v[/tex]

    How do I do this? please help.
     
    Last edited: Oct 25, 2006
  2. jcsd
  3. Oct 26, 2006 #2

    OlderDan

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    Rolling implies rotational energy as well as translational energy. The knietic energy of the ring involves two terms, one for translation and one for rotation.
     
  4. Oct 26, 2006 #3
    ok

    [tex] mgh = \frac{mv^2}{2} + \frac{I\omega^2}{2}[/tex]

    solving, [tex] velocity = \frac{v}{\sqrt{2}}[/tex]

    correct? thanks for your help.
     
  5. Oct 26, 2006 #4

    OlderDan

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    Looks good.
     
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