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Frictionless, rotating circular hoop

  1. Nov 22, 2005 #1
    A small bead can slide without friction on a circular hoop that is in a vertical plane and has a radius of [i tex]0.100 m[/itex]. The hoop rotates at a constant rate of [itex] 4.00 \frac{rev}{s} [/itex] about a vertical diameter. (a) Find the angle [itex] \theta [/itex] at which the bead is in vertical equilibrium. (It has a radial acceleration toward the axis) (b) Is it possible for the bead to "ride" at the same elevation as the center of the hoop? (c) What will happen if the hoop rotates at [itex]1.00 \frac{rev}{s} [/itex]?

    All I really know is that you have to find the velocity [itex] \frac{2\pi r}{T} [/itex]. You have to use the equation [itex] F = m\frac{v^{2}}{r} [/itex]. Is it possible to use the arc length formula [itex] s = r\theta [/itex]?

    Any help would be appreciated!

    Thanks :smile:
  2. jcsd
  3. Nov 22, 2005 #2


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    HINT: The hoop can only provide the radial component of the centripetal force - i.e. gravity must supply the tangential component. :)
  4. Nov 22, 2005 #3
    So the hoop is rotating about a vertical circle, while the bead is rotating in its own circle within the vertical circle? The direction of the acceleration is toward the center.

    Any help would be appreciated

    Thanks Tide for your hint
  5. Nov 22, 2005 #4
    can someone please help me out
  6. Nov 22, 2005 #5


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    If you measure the angle from the vertical line connecting the center of the hoop to the lowest point on the hoop, then the radius of the "orbit" is [itex]R \sin \theta[/itex]. This makes the component of centripetal force on the bead tangential to the hoop [itex]m \omega^2 R \sin \theta \cos \theta[/itex]. The component of the gravitational force tangential to the hoop is [itex]m g\sin \theta[/itex].

    Draw some pictures to verify the above and then proceed.
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