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Bead on spinning hoop

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A bead of mass m is constrained to move on a hoop of radius R which is spinning with angular velocity ω. There is no friction.

    Determine the angle θ at which the bead does not move for angular velocity ω. Do not consider the solution θ = 0.

    YuMvt.png


    2. Relevant equations

    τ = r F cos θ = I α
    α = a / r

    3. The attempt at a solution

    Tried using formulas for torque - ended up with F = m R tan θ * α and α=α/ cos θ. Can't figure out what to do next. Also not sure about my free body diagram.

    Free body diagram:
    Force upwards due to angular velocity
    Force downwards due to gravity.
     
  2. jcsd
  3. Nov 10, 2012 #2

    lewando

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    Angular velocity is constant, so no torque involved. There is a normal force from the hoop acting on the bead. There is also a centripetal force.
     
    Last edited: Nov 10, 2012
  4. Nov 10, 2012 #3
    So there would be a downward force for gravity, a force towards the center of the hoop for centripetal force, and another force in the same direction for the normal force?
     
    Last edited: Nov 10, 2012
  5. Nov 10, 2012 #4

    lewando

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    The normal force will be pointing to the center of the hoop (normal to a tangent line where the bead and hoop intersect). Downwards for mg--correct, centripetal force is an outward force (think of the bead in orbit about the vertical axis of the hoop).
     
  6. Nov 10, 2012 #5
    So the centripetal force would be perpendicular against the vertical axis, and point torwards it?
     
  7. Nov 10, 2012 #6

    lewando

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    I always struggle on how to word this properly so here goes: The force that holds the bead in orbit is the centripetal force. You are correct that it is an inward force, not outward as I had incorrectly stated*. This force is provided by the horizontal component of the normal force. You can say that the centripetal force is equal to the horizontal component of the normal force.

    *I tend to look at these types of problems from the perspective of rotating along with the hoop...
     
  8. Nov 10, 2012 #7
    So there would be -mg and Fn cos θ in the vertical direction, and Fn sin θ = m v2 / r in the horizontal direction?
     
  9. Nov 10, 2012 #8

    lewando

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  10. Nov 10, 2012 #9
    So Fn = m v2 / (r sin θ) = m ω2 r / sin θ

    And solving Fn cos θ - m g = 0 :
    ω2 r cos θ/ sin θ - g = 0
    ω2 r cos θ/ sin θ = g
    cot θ = g / (ω2 r)
    and θ = arccot(g / (ω2 r))

    Is that right?
     
  11. Nov 10, 2012 #10

    lewando

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    I think the little r is not the same as the big R.
     
  12. Nov 10, 2012 #11
    Little r is the distance from the bead to the axis, correct?

    So that would be r = R sin θ

    ω2 r cos θ/ sin θ = g
    ω2 (R sin θ) cos θ/ sin θ = g
    cos θ = g / (ω2 R)
    and θ = arccos(g / (ω2 R))
     
  13. Nov 10, 2012 #12

    lewando

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    That looks right--good job by you!
     
  14. Nov 10, 2012 #13
    Thanks for your help
     
  15. Nov 10, 2012 #14

    lewando

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    You're welcome. This problem is almost exactly similar to a tetherball-type problem. Other than checking your math for errors (to validate your answer), and if you have any extra non-existant free time, grab a string, length R, and tie it to a mass of your choosing and see how fast it needs to rotate to achieve an eyeball-45 degree angle. See if this reality matches your result.
     
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