A massless rod S having length `2l` has equal point masses attached to its two ends as shown in figure. The rod is rotating about an axis passing through its centre and making angle `alpha` with the axis. The magnitude of change of momentum of rod i.e., `|(dL)/(dt)|` equals

Options

A. `2 ml^(3) omega^(2) sin alpha. Cos alpha`
B. `ml^(2)omega^(2) sin 2 alpha`
C. `ml^(2) sin 2 alpha`
D. `m^(1//2)l^(1//2) omega sin alpha. cos alpha`

Answer

B

Explanation

Refer to Fig.

The radius `r` of the circle traced by the masses is `r = l sin alpha`
Angular momentum `overset rarr(L) = overset rarr(r ) xx m overset rarr(v)`
`|overset rarr(L)| = r xx m omega r = m omega r^(2)`
`= momega (l sin alpha)^(2) = m omegal^(2) sin^(2)alpha`
`(dL)/(dt) = m omegal^(2) sin alpha cos alpha(d alpha)/(dt)`
`= m omegal^(2) (sin 2 alpha) omega = m omega^(2)l^(2) sin 2 alpha`