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    Circular Motion and Gravitation: bead can slide without friction on a circular hoop

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    A bead can slide without friction on a circular hoop with a radius of .100 m in a vertical plane. The hoop rotates at a constant rate of 3.00 rev/s about a vertical diameter.

    a) find the angle θ at which the bead is a vertical equilibrium.

    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 1 rev/s?

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    Solution Preview

    Radius of hoop (R) = 1.00 m (I think .100m is typo, but any way! I have solved for both values.)
    angular rotation of hoop (w)= 2*pi*f

    pi = 22/7, f = rev/sec = 3.00, w == omega (angular frequency)
    w = 2*pi*3 = 18.85 rad/sec
    see attached file.
    because, radius of circular motion of bead = R*sin(Q)
    here Q == theta.
    centrifugal force:
    Fp(pseudo ...

    Solution Summary

    The solution explains the problem and shows the workings to arrive at the answers.