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Circular motion questions 2

  1. Dec 29, 2011 #1
    2) a particle P of mass m moves in a vertical circle along the smooth inner surface of a fixed hollow sphere of internal radius a and centre O,the plane of the circle passing through O. The particle is projected from the lowest point of the sphere with a horizontal velocity u. when OP makes an angle T with the upward vertical,the velocity of the particle is v and the normal reaction between the particle and the sphere is R.Find expressions for v and R...done
    v=[u^2-2ga(1+cos T)]^1/2 R=mu^2/a - mg(2+3cos T)
    show if u^2,5ga the particle leaves the sphere where

    cos T=(u^2-2ga)/3ga
    if the particle leaves the sphere at a point A and its trajectory meets the sphere again at a point B such that AM is a diameter of the sphere show OA makes an angle of 45

    from first part i have

    cos T=(u^2-2ga)/3ga


    v^2=[u^2-2ga(1+cos T)]

    so when particle leaves sphere i get v^2=agcosT

    now im saying the angle of projection is equal to angle OP makes with upwards vertical,so from P to O we have

    horizontal distance=asinT
    vertical distance=acosT

    for the projection

    x=-tvcosT
    y=tvsinT-gt^2/2

    when x=-asinT => t=a(tanT)/v
    so

    -acosT=atanTsinT-g/2[(atanT)^2/(v^2)]
    -acosT=atanTsinT-g/2[(atanT)^2/(agcosT)]
    -2=2tan^2T-tan^2Tsec^2T
    -2=tan^2T(2-sec^2T)
    -2=tan^2T(2-1-tan^2T)
    tan^4T-tan^2T-2=0
    (tan^2T-2)(tan^2+1)=0

    which is close as im after tan T=1. is there an obvious error in my working?i hoping im on right tracks but have dropped a sign somewhere?
     
  2. jcsd
  3. Dec 31, 2011 #2
    anyone able to spot my error here?
     
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