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A smooth wire is bent into a vertical ci...

A smooth wire is bent into a vertical circle of radius a. `A` bead `P` can slide smoothly on the wire. The circle is rotated about vertical diameter AB as axis with a speed omega as shown in figure. The bead `P` is ar rest w.r.t. the circular ring in the position shown. then `omega^(2)` is equal to:

A

`(2g)/(a)`

B

`(2g)/(asqrt(3))`

C

`(gsqrt(3))/(a)`

D

`(2a)/(gsqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
B
As `, cos theta =(a)/(2a)`
`theta =60^(@)`
`:. N sin 60^(@) =mg`
`N cos 60^(@) = m ( omega^(2)a)/(2)`

`w.r.t.` wire
`:. tan 60^(@)=(2g)/(omega^(2)a), omega^(2)=(2g)/(asqrt(3))`
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