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Mass Slides Inside a Hoop

  1. Nov 17, 2011 #1
    The problem statement, all variables and given/known data

    A mass M of 5.20E-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 5.27 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.

    Picture attached at the bottom.


    The attempt at a solution

    E at the top of the circle should equal the energy at that particular part of the circle. So,

    mg(2r) + (1/2)mvtop2 = mg(r-rcosθ) + (1/2)mv^2
    m's cancel, so
    g(2r) + (1/2)vtop2 = g(r-rcosθ) + (1/2)v^2
    2(g(2r) + (1/2)vtop2 - g(r-rcosθ) = v^2

    a = v^2/r

    F = ma


    When I plug everything in, I get
    2(39.831) = v^2
    v^2 = 79.66

    a = 79.66/1.4
    a = 56.90

    F = (.52)(56.90)
    F = 29.59 N

    This answer is incorrect.

    Please, can someone let me know where I'm going wrong?
     

    Attached Files:

    Last edited: Nov 17, 2011
  2. jcsd
  3. Nov 17, 2011 #2

    gneill

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    Staff: Mentor

    You've calculated the force due to the circular motion, but what other force is also working on the mass?
     
  4. Nov 17, 2011 #3
    Okay. I'm forgetting the force of gravity acting on the mass, aren't I?

    Should I add mgcosθ to my answer? Is that all I'm missing?
     
  5. Nov 17, 2011 #4

    gneill

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    Staff: Mentor

    That looks right. Since the hoop has "negligible friction", only the component of the force due to gravity that is normal to the hoop's surface should matter -- the other component acts to accelerate the mass tangentially.
     
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