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]?(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Frictionless, rotating circular hoop

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