A particle executes uniform circular motion with angular momentum 'L'. Its rotational kinetic energy becomes half, when the angular frequency is doubled. Its new angular momentum is

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Aicharya · Accepted Answer

$\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}, \quad \mathrm{~L}=\mathrm{I} \omega$ $\mathrm{K}^{\prime}=\frac{1}{2} \mathrm{I}^{\prime} \omega^{\prime 2}, \quad \omega^{\prime}=2 \omega$ $	herefore \mathrm{K}^{\prime}=\frac{1}{2} \mathrm{I}^{\prime}(2 \omega)^{2}, \quad \mathrm{~K}^{\prime}=\frac{\mathrm{K}}{2}$ $	herefore \frac{1}{2}\left(\frac{1}{2} \mathrm{I} \omega^{2}ight)=\frac{1}{2} \mathrm{I}^{\prime}(2 \omega)^{2}$ $	herefore \mathrm{I}^{\prime}=\frac{\mathrm{I}}{8}$ $\mathrm{~L}^{\prime}=\mathrm{I}^{\prime} \omega^{\prime}=\frac{\mathrm{I}}{8} 	imes 2 \omega=\frac{\mathrm{I} \omega}{4}=\frac{\mathrm{L}}{4}$

A particle executes uniform circular motion with angular momentum 'L'. Its rotational kinetic energy becomes…

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