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A smooth wire of length 2pir is bent in...

A smooth wire of length `2pir` is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed about the vertical diameter AB, as shown in the figure, the bead is at rest with respect to the circular ring at potion P as shown. Then the value of `omega^(2)` is equal to:

A

`(2g)/(r )`

B

`(sqrt(3)g)/(2r)`

C

`((gsqrt(3)))/(r)`

D

`(2g)/(rsqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
D
`theta=30^(@)`
`Nsin30^(@)=m omega^(2)(r )/(2)`
`N cos 30^(@)=mg`
`tan30^(@)=(omega^(2)r)/(2g)`
implies `omega^(2)=(2g)/(rsqrt(3))`
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