A uniform rod of length l is free to rotate in a vertical plane about a fixed horizontal axis through O. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle `theta`, its angular velocity `omega` is given as

Options

A. `sqrt((6g)/(l))sintheta`
B. `sqrt((6g)/(l))sin.(theta)/(2)`
C. `sqrt((6g)/(l))cos.(theta)/(2)`
D. `sqrt((6g)/(l))costheta`

Answer

B

Explanation

Here applying the law of conservation of energy, we have
`(1)/(2)Iomega^(2)` = gravitation of P.E.
`(1)/(2).(ml^(2))/(3)xxomega^(2)=(mg)/(2)xxh`
But `h=l-lcostheta`
`=l(1-costheta)`
`therefore (1)/(6)ml^(2)omega^(2)=(mgl)/(2)(1-costheta)`
`(lomega^(2))/(3)=g[1-(1-2sin^(2).(theta)/(2))]`
`=2gsin^(2)theta//2`
or `omega^(2)=(6g)/(l)sin^(2)theta//2`
or `omega=sqrt((6g)/(l)).sintheta//2`