A uniform rod $\mathrm{AB}$ of mass ' $\mathrm{m}^{\prime}$ and length ' $\ell^{\prime}$ is at rest on a…

A uniform rod $\mathrm{AB}$ of mass ' $\mathrm{m}^{\prime}$ and length ' $\ell^{\prime}$ is at rest on a smooth horizontal surface. An impulse ' $\mathrm{P}^{\prime}$ is applied to the end $\mathrm{B}$. The time taken by the rod to turn through a right angle is
  1. $\frac{\pi}{12} \frac{\mathrm{m} \ell}{\mathrm{P}}$
  2. $\frac{\pi \mathrm{P}}{\mathrm{m} \ell}$
  3. $2 \pi \frac{\mathrm{m} \ell}{\mathrm{P}}$
  4. $2 \frac{\pi \mathrm{P}}{\mathrm{m} \ell}$

Solution

$\begin{aligned} & \mathrm{I}_{\mathrm{cm}}=\frac{1}{12} \mathrm{ml}^2 \\ & \mathrm{~L}=\mathrm{P} \cdot \frac{\mathrm{l}}{2} \Rightarrow \frac{1}{12} \mathrm{ml}^2 \omega=\mathrm{P} \frac{\mathrm{l}}{2} \\ & \Rightarrow \omega=\frac{6 \mathrm{P}}{\mathrm{ml}} \\ & \theta=\frac{\pi}{2}=\omega \mathrm{t} \Rightarrow \mathrm{t}=\frac{\pi}{2 \omega} \\ & =\frac{\pi \mathrm{ml}}{2.6 \mathrm{P}}=\frac{\pi \mathrm{ml}}{12 \mathrm{P}}\end{aligned}$
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Asked in: MHT CET 2020 (15 Oct Shift 2)

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