Two bodies rotate with kinetic energies ' $E_1$ ' and ' $E_2$ '. Moment of inertia about their axis of…

Two bodies rotate with kinetic energies ' $E_1$ ' and ' $E_2$ '. Moment of inertia about their axis of rotation are ' $I_1$ ' and ' $I_2$ '. If $I_1=\frac{I_2}{3}$ and $\mathrm{E}_1=27 \mathrm{E}_2$, then the ratio of angular momenta ' $\mathrm{L}_1$ ' to ' $\mathrm{L}_2$ ' is
  1. $1: 3$
  2. $3: 1$
  3. $1: 1$
  4. $2: 1$

Solution

$\begin{aligned} & \mathrm{E}_1=\frac{1}{2} \mathrm{I}_1 \omega_1^2 \\ & \mathrm{E}_2=\frac{1}{2} \mathrm{I}_2 \omega_2^2=\frac{1}{2}\left(3 \mathrm{I}_1\right) \omega_2^2=\frac{3}{2} \mathrm{I}_1 \omega_2^2 \quad\left[\because \mathrm{I}_2=3 \mathrm{I}_1\right] \\ & \mathrm{E}_1=27 \mathrm{E}_2 \\ & \frac{1}{2} \mathrm{I}_1 \omega_1^2=\frac{81}{2} \mathrm{I}_1 \omega_2^2 \\ & \omega_1=9 \omega_2 \\ & \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{I}_1 \omega_1}{\mathrm{I}_2 \omega_2}=\frac{\mathrm{I}_1 \times 9 \omega_2}{3 \mathrm{I}_1 \times \omega_2}=3 \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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