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Beads on a rotating hoop

  1. Jun 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A vertical hoop of mass M can rotate round the z axis without friction. 2 beads of mass m each are at the top and start sliding, one on each side. the hoop starts to rotate with angular velocity ω0. the coefficient of friction is μks=μ and R is the hoop's radius.
    What's ω when the beads are at angle θ. Hint: think which parameters are conserved.

    2. Relevant equations
    Centripetal force: ##F=m\frac{v^2}{R}=m\omega^2 R##
    Angular momentum: ##L=m\omega R^2##
    Moment of inertia of a hoop about it's diameter: ##I=\frac{1}{2}MR^2##

    3. The attempt at a solution
    The angular momentum is conserved since there is no torque round ##\hat{z}##, but i guess the initial angular momentum is 0 since the beads are at the top.
    The angular momentum at angle θ: ##L=2m\omega R^2\sin^2\theta##
     

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    Last edited: Jun 27, 2015
  2. jcsd
  3. Jun 27, 2015 #2

    Nathanael

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    I'm curious about the problem statement.

    Is the hoop is being forced to spin at ω0? (Otherwise won't we need the hoop's mass?)

    Are we supposed to find the angular speed of the beads about the z-axis? Or are we trying to find ##\dot \theta##?
     
  4. Jun 27, 2015 #3
    I think the hoop is forced to spin at the start at ω0, and i suppose it decreases as the beads fall. we search for ω at angle θ, so we seek the angular speed of the beads about the z-axis.
     
  5. Jun 27, 2015 #4

    Nathanael

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    But how should we be able to determine how quickly the hoop slows down unless we know the mass?
    (We would use conservation of angular momentum like you tried, but the initial angular momentum would not be zero, it would be that of the hoop, which depends on it's mass.)
     
  6. Jun 27, 2015 #5
    You are right, i forgot the hoop's mass M!
    Moment of inertia of a hoop about it's diameter: ##I=\frac{1}{2}MR^2##
    I will try to solve:
    $$\frac{1}{2}MR^2\omega_0=\frac{1}{2}MR^2\omega+2m\omega R^2\sin^2\theta$$
    $$\omega=\frac{MR^2\omega_0}{MR^2+4mR^2\sin^2\theta}$$
    It's wrong
     
    Last edited: Jun 27, 2015
  7. Jun 27, 2015 #6

    Nathanael

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    I agree with your answer, I don't know why it would be wrong. (You can cancel the R2 though.)
     
  8. Jun 27, 2015 #7
    You are right Nathanael this is one of the possible answers (with R2 canceled)
     
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