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SHM question

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment for figure)
    A small object is mounted to the perimeter of a hoop of radius r. The mass of the object and the hoop is same. The hoop is placed into a fixed semi-cylinder shaped rough trough of radius R, such that small mass is at top. Find the least r/R ratio such that the object performs simple harmonic motion.

    2. Relevant equations



    3. The attempt at a solution
    I honestly have no idea on how should I begin to solve this problem, this is very different from the SHM problems I have done.

    Any help is appreciated. Thanks!
     

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    Last edited: Jan 9, 2013
  2. jcsd
  3. Jan 9, 2013 #2

    mfb

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    After a rotation by an angle of θ, can you calculate the position of the mass and the circle? Can you calculate the potential energy?
     
  4. Jan 9, 2013 #3
    But then, how will I find the angle rotated by the small mass?
     
  5. Jan 9, 2013 #4

    mfb

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    What do you mean with "the angle rotated by the small mass"?
    Rotate the hoop by an angle θ in your trough, find the new positions of hoop and attached mass with geometry.
     
  6. Jan 9, 2013 #5
    (see attachment)
    I am having trouble finding the new position of the attached mass. Is angle α=θ?
     

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  7. Jan 9, 2013 #6

    mfb

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    α=θ would imply that the mass always stays on top. In reality, the hoop rotates quicker.
    Hmm, you use a different θ.

    I'll switch to your angles:
    After rolling a distance of d on the track, the contact position (on the hoop) changed by α=d/r. As seen from the center of the track, the hoop moved by an angle θ=d/R. Putting this together, the point mass is an angle of α-θ = d/r - d/R away from the top of the hoop now.
     
  8. Jan 9, 2013 #7
    Nice explanation mfb, thanks! :smile:

    I still cannot find the potential energy of the system. (see attachment)
    The blue dot represents the CM of hoop and the black dot represents the attached mass. How can I find h?
     

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  9. Jan 9, 2013 #8

    mfb

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    The blue dot has a fixed distance from the center of the track (which?), and you can use θ...
     
  10. Jan 9, 2013 #9
    R-r?

    I worked on it to find h, is h=r+rcosθ?
     
  11. Jan 9, 2013 #10

    haruspex

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    No, not quite.
    It's easier to think in terms of how far the blue dot is below the centre of the hoop. It starts R-r below.
     
  12. Jan 10, 2013 #11
    I gave it a try again. This time, I got h=R(1-cosθ)+rcosθ. Is this right?
     
  13. Jan 10, 2013 #12

    haruspex

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    Yes.
     
  14. Jan 10, 2013 #13
    What should be my next step?
     
  15. Jan 10, 2013 #14
    Can I get some more help?
     
  16. Jan 10, 2013 #15

    mfb

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    Your next step should be to think about the problem. In addition, see post 2:
     
  17. Jan 10, 2013 #16
    Potential energy or change in potential energy?
     
  18. Jan 10, 2013 #17

    mfb

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    Potential energy (as function of some angle), as I wrote.

    An expression for the kinetical energy can be interesting, too, but I think it is not required for the answer.
     
  19. Jan 10, 2013 #18
    Potential energy assuming it is zero at the bottom of the trough is
    [tex]U=mgh+mg(h+r\sin(\alpha-\theta))[/tex]
    [tex]U=2mgR(1-\cos \theta)+2mgr\cos \theta+mgr\cos(\alpha-\theta)[/tex]
    m is the mass of the hoop and the attached mass.
     
    Last edited: Jan 10, 2013
  20. Jan 10, 2013 #19

    haruspex

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    Wouldn't r.sin(α-θ) represent a horizontal displacement?
     
  21. Jan 10, 2013 #20
    Oops, sorry about that.

    What should I do next?
     
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