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Work and Energy at an angle

  1. Oct 22, 2008 #1
    1. An ice cube is placed on top of an overturned spherical bowl of radius r. (a half circle with a radius r). If the cube slides down from rest at the top of the bowl, at what angle(o) does it separate from the bowl? (hint: what happens to normal force when it leaves the bowl?)

    2. Relevant equations



    3. The attempt at a solution

    When the normal force leaves the bowl, it doesn't exist anymore.

    At when the ice cube is about leave, the height is rcos(o), so PE = mgrcos(o)

    When the ice cue is about to hit the ground, all the PE is Ke, so:

    1/2mv^2 = mgrcos(o)
    1/2v^2 = grcos(o)

    the x distance the ice cube slides before separates is rsin(o)

    Um...so I'm not too sure how I'm suppose do, so far I have found random information, but I don't know what to do know.
  2. jcsd
  3. Oct 23, 2008 #2


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    Hi aimslin22! :smile:

    I'd have sad that your equations were right, if there wasn't some extra stuff that isn't relevant.

    Yes, the ice cube loses contact when the normal force is zero, so you have to use F = ma (with the gravitational force and the centripetal acceleration)
  4. Oct 23, 2008 #3
    so would it be
    mg = mv^2/r
    g = v^2/r

    1/2v^2 = v^2/r*rcos(o)
    1/2 = cos (o)
    o = 60 degrees?
  5. Oct 24, 2008 #4


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    No … what happened to theta? :confused:
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