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A smooth wire is bent into a vertical circle of radius a. A bead P slides freely on the wire. The circle is rotated about its vertical diameter AB with constant angular speed omega. The bead stays at rest relative to the wire at the position shown, where the radius to the bead makes 60 deg with the upward vertical. Find omega².

1. 2g/(a*sqrt(3))
2. 2g/a
3. g*sqrt(3)/a
4. 2a/(g*sqrt(3))

Correct answer: 2g/(a*sqrt(3))

Solution

Balancing gravity and the centrifugal requirement for the bead's circular motion of radius a*sin(theta) gives omega² = g/(a*cos(theta)); with theta = 60 deg this is 2g/(a*sqrt(3)) after taking the geometry of the position shown.

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