A rotating body has angular momentum ' $L$ '. If its frequency is doubled and kinetic energy is halved, its…

A rotating body has angular momentum ' $L$ '. If its frequency is doubled and kinetic energy is halved, its angular momentum will be
  1. $\frac{\mathrm{L}}{4}$
  2. $\frac{\mathrm{L}}{2}$
  3. 2 L
  4. 4 L

Solution

Angular momentum of a particle performing UCM $\mathrm{L}=\mathrm{I} \omega...(i)$
Kinetic energy, $\mathrm{k}=\frac{1}{2} \mathrm{I} \omega^2$...(ii) $\therefore \quad \mathrm{L}=\frac{2 \mathrm{k}}{\omega}$ $\ldots[$ From (i) and (ii)] $\begin{aligned} & \therefore \quad \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{k}_1}{\mathrm{k}_2} \times \frac{\omega_2}{\omega_1} \\ & \frac{\mathrm{~L}}{\mathrm{~L}_2}=\frac{1}{\frac{1}{2}} \times \frac{2}{1} \\ & \therefore \quad \mathrm{~L}_2=\frac{\mathrm{L}}{4} \end{aligned}$ ...(given) ~

Asked in: MHT CET 2024 (04 May Shift 1)

Practice more Rotational Motion questions on Aicharya