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Work and energy - easy question

  1. Oct 22, 2005 #1
    a mass "m" slides without friction along a looped apparatus. If the object is to remain on the track, even at the top of the loop (whose radius is "r") from what minimum height "h" must it be released?
     
  2. jcsd
  3. Oct 22, 2005 #2

    arildno

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    Remember that at the top of the loop, gravity must provide the centripetal acceleration of the object.
     
  4. Oct 22, 2005 #3
    right. so

    mg = (mv^2)r, masses cancel, solve for v^2

    v^2 = 9.8r

    and E = 1/2mv^2 + mgh
    mgh = 1/2m (9.8r) + m(9.8)(2r)
    9.8h = 1/2 (9.8r) + (9.8)(2r)
    h = 1.55 --> = height of loop

    am i tackling this right so far?
     
  5. Oct 22, 2005 #4

    arildno

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    Right, when "h" is measured from the level of the bottom of the loop.
    :confused:

    just divide with g, and get h=2.5r.
     
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