Two rotating bodies $\mathrm{P}$ and $\mathrm{Q}$ of masses ' $\mathrm{m}$ ' with moment of inertia $I_P$…

Two rotating bodies $\mathrm{P}$ and $\mathrm{Q}$ of masses ' $\mathrm{m}$ ' with moment of inertia $I_P$ and $I_Q\left(I_Q>I_P\right)$ have equal Kinetic energy of rotation. If $L_P$ and $L_Q$ be their angular momenta respectively then
  1. $\mathrm{L}_{\mathrm{Q}}=0$
  2. $\mathrm{L}_{\mathrm{Q}}=\mathrm{L}_{\mathrm{P}}$
  3. $\mathrm{L}_{\mathrm{Q}} < \mathrm{L}_{\mathrm{P}}$
  4. $\mathrm{L}_{\mathrm{Q}}>\mathrm{L}_{\mathrm{P}}$

Solution

$\begin{aligned} & \text { Kinetic energy }=\frac{\mathrm{L}_{\mathrm{P}}^2}{2 \mathrm{I}_{\mathrm{P}}}=\frac{\mathrm{L}_{\mathrm{Q}}^2}{\mathrm{LI}_{\mathrm{Q}}} \\ & \therefore \frac{\mathrm{L}_{\mathrm{Q}}^2}{\mathrm{~L}_{\mathrm{P}}^2}=\frac{\mathrm{I}_{\mathrm{Q}}}{\mathrm{I}_{\mathrm{P}}} \\ & \because \mathrm{I}_{\mathrm{Q}}>\mathrm{I}_{\mathrm{P}} \text { we get } \mathrm{L}_{\mathrm{Q}}>\mathrm{L}_{\mathrm{P}}\end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

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