A rotating body has angular momentum 'L'. If its frequency of rotation is halved and rotational kinetic…

A rotating body has angular momentum 'L'. If its frequency of rotation is halved and rotational kinetic energy is doubled, its angular momentum becomes
  1. $2 \mathrm{~L}$
  2. $\frac{\text { L }}{4}$
  3. $4 \mathrm{~L}$
  4. $\frac{\mathrm{L}}{2}$

Solution

Kinetic energy $k=\frac{1}{2} I \omega^{2}$ $\therefore \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{I}_{2} \mathrm{w}_{2}^{2}}{\mathrm{I}_{1} \omega_{1}^{2}}$ $\therefore 2=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}\left(\frac{\mathrm{l}}{2}\right)^{2} \quad \therefore \frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=8$ Angular momentum $\quad \mathrm{L}=\mathrm{I} \omega$ $\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}=\frac{\mathrm{I}_{2} \omega_{2}}{\mathrm{I}_{1} \omega_{1}}=8 \times \frac{1}{2}=4$ $\therefore \mathrm{L}_{2}=4 \mathrm{~L}_{1}=4 \mathrm{~L} \quad$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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