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Circular motion, find speed

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle is at rest at the apex A of a smooth fixed hemisphere whose base is horizontal. The hemisphere has centre O and radius a. The particle is then displaced very slightly from rest and moves on the surface of the hemisphere. At the point P on the surface where angle AOP = α the particle has speed v. Find an expression for v in terms of a, g and α.

    2. Relevant equations



    3. The attempt at a solution

    So I’ve worked like this:
    Total energy at A = PE + KE = amg + 0
    Total energy at P = PE + KE = (0.5m(v^2)) + xmg
    x = a – y
    (cos α)/y = (sin90)/a => y = a(cos α)
    => x = a – (a(cos α)) = a (1 - cos α)
    => Total energy at P = PE + KE = (0.5m(v^2)) + amg (1 - cos α)

    Therefore (Total energy at A) = (Total energy at P) gives
    amg = (0.5m(v^2)) + amg (1 - cos α)
    ag = 0.5(v^2) + ag (1 – cos α)
    ag – ag (1 – cos α) = 0.5(v^2)
    2ag (1 – 1 + cos α) = v^2
    v = sqrt (2ag (cos α))

    However, the correct answer is v = sqrt (2ag (1 - cos α))

    Where’s the problem?
     

    Attached Files:

  2. jcsd
  3. Oct 27, 2011 #2

    ehild

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    Homework Helper
    Gold Member

    You calculated the initial potential energy with respect to the base of the semiphere. PE=mg*(height above the base). The height is y in your picture.

    ehild
     
  4. Oct 27, 2011 #3
    thanks
     
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