InterviewSolution
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InterviewSolution
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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 angle `alpha` with the axis. The magnitude of change of momentum of rod i.e., `|(dL)/(dt)|` equals A. `2 m l^(3) omega^(2) sin theta. cos theta`B. `ml^(2) omega^(2) sin 2 theta`C. `ml^(2) sin 2 theta`D. `m^(1//2) l^(1//2) omega sin theta. Cos theta` |
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Answer» Correct Answer - B The radius of the circular followed by the masses is `r = l sin alpha` As, angular momentum, `L = r xx p = r xx mv` `rArr |L| = l sin theta (m omega l sin theta)` On differentiating, we get `(d|L|)/(dt)=momega l^(2)2 sin theta. cos theta (d theta)/(dt)` `rArr |(dL)/(dt)|=2ml^(2) omega^(2)sin theta. cos theta =ml^(2)omega^(2) sin 2 theta`. |
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