A massless rod of length 2l has equal point m | vedclass

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(B) The rate of change of angular momentum is equal to the net torque acting on the system,i.e.,$\left|\frac{dL}{dt}ight| = 	au_{	ext{net}}$.For e

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$ml^2\omega^2\sin 2\alpha$

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(B) The rate of change of angular momentum is equal to the net torque acting on the system,i.e.,$\left|\frac{dL}{dt}ight| = 	au_{	ext{net}}$.For each mass $m$ at a distance $l$ from the centre,the radius of the circular path is $r = l\sin\alpha$.The centripetal force acting on each mass is $F = mr\omega^2 = m(l\sin\alpha)\omega^2$.The torque $	au$ due to one mass about the centre is $	au = F 	imes r_{\perp}$,where $r_{\perp} = l\cos\alpha$ is the perpendicular distance from the axis of rotation to the force vector.Thus,$	au = (ml\sin\alpha\omega^2) 	imes (l\cos\alpha) = ml^2\omega^2\sin\alpha\cos\alpha$.Since there are two such masses on opposite sides of the centre,the torques produced by them are in the same direction.Therefore,the net torque is $	au_{	ext{net}} = 2	au = 2ml^2\omega^2\sin\alpha\cos\alpha$.Using the trigonometric identity $\sin 2\alpha = 2\sin\alpha\cos\alpha$,we get:$	au_{	ext{net}} = ml^2\omega^2\sin 2\alpha$.Hence,$\left|\frac{dL}{dt}ight| = ml^2\omega^2\sin 2\alpha$.

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