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Conservation of energy and centripital force

  1. Nov 3, 2008 #1
    The Question is:

    A mass M of 5.00×10-1 kg slides inside a hoop of radius R=1.50 m with negligible friction. When M is at the top, it has a speed of 5.65 m/s. Calculate size of the force with which the M pushes on the hoop when M is at an angle of 27.0 degrees.

    My workings thus far:

    1st i know i have to use the radius and theta to get the distance of the mass below the hoop
    then i have to use the speed at the top and the conservation of mechanical energy to get the speed of the mass at theta
    then i make a fbd and get the normal force

    so
    using trig
    1.5 cos 27= x
    x= 1.3365m (distance under center of hoop)
    distance from bottem then equals 0.1635 m as the hoop is 1.5 m in radius

    so using conservation of energy i get
    total energy top= total energy bottem
    potentical energy top + kE top = pot Energy bottem + Ke bottem
    mgh + 1/2 mv^2 = mgh+1/2mv^2 (all masses cancel out
    (9.81)(3) +1/2(5.65)^2 = (9.81)(0.1635) + 1/2 v^2
    solve for v = 5.88223 m/s

    so the speed at theata is 5.88223 m/s, then using centripical force =mv^2/r we can get 23.067 N for the force.

    Then we draw a free body diagram and we get that Fc=Fn (what were looking for)+ Fg
    This is were i get stuck..
    Help from here??
     
  2. jcsd
  3. Nov 3, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi Megz27! Welcome to PF! :smile:

    i] 27.0 degrees from what? :confused:

    ii] Fc = Fn + Fg is not correct.
     
  4. Nov 3, 2008 #3
    27 degress from the bottem of the circle..

    how would you approach the FC is that is not correct?
     
  5. Nov 4, 2008 #4

    tiny-tim

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    What does your fbd look like?

    In which direction are you taking components?

    What equation does it give you? :smile:
     
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