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Homework Help: Circular motion of hoop and mass

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A mass M of 6.00E-1 kg slides inside a hoop of radius R=1.40 m with negligible friction. When M is at the top, it has a speed of 5.23 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 31.0°.

    I have no idea from where to begin

    I dont want you to do my homework,Can someone give me a hint?
     
  2. jcsd
  3. Dec 12, 2009 #2

    rock.freak667

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    Start by drawing a free body diagram. The use conservation of energy.
     
  4. Dec 12, 2009 #3
    When I do the diagram
    the total forces work on the mass is :
    mg=mv^2/r

    the low of energy is:
    Mgh=1/2mv^2r

    But I cant see how the angle is related to the sloution?
     
  5. Dec 12, 2009 #4

    ideasrule

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    Can you use the conservation of energy to figure out the speed of the mass at 31 degrees? If so, do that first, and make sure you can get a numerical value.

    Also, mg is not the only force on the block. N, the normal force, provides part of the centripetal acceleration.
     
  6. Dec 12, 2009 #5
    But I dont know which force I have to do to components
    I need cos ans sin right?
    How can I do that?
     
  7. Dec 12, 2009 #6
    I tried to find the speed of the mass at 31 degrees by:
    N+mgcosa=mv^2/r
    but i dont know what to but instead of N?
     
  8. Dec 12, 2009 #7
    I tried to slove the question by :
    N+mgcosa=mv^2/r
    but i dont know what to put instead of N!?
     
  9. Dec 12, 2009 #8

    rock.freak667

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    N is what you want to find.
     
  10. Dec 13, 2009 #9
    I tried to fins the speed at angle 31 by:
    I/2mv^2+mg2R=1/2mv^2+mgRsina
    all m goes together
    v^2+4gR=v^2+2gRsina
    the answer i get from solving it i plugg into this:
    N+mgcosa=mv^2/R
    N=mv^2/Rsina-mgcosa
    I get N=18.27N
    but it is a wrong answer
    Can you tell me where i went wrong?
     
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