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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 an angle α with the axis. The magnitude of change of angular momentum of rod i.e. dLdt equals [AIIMS 2015]

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a

2ml3ω2sin⁡α⋅cos⁡α

b

ml2ω2sin⁡2α

c

ml2sin⁡2α

d

m1/2l1/2ωsin⁡α⋅cos⁡α

answer is B.

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Detailed Solution

The radius of the circle followed by the masses, r=lsin⁡αAs, angular momentum, L=r×p=r×mv⇒ |L|=lsin⁡α(mωlsin⁡α)On differentiating both sides, we get d|L|dt=mωl22sin⁡α⋅cos⁡αdαdt⇒ dLdt=2ml2ω2sin⁡α⋅cos⁡α=ml2ω2sin⁡2α

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