A particle performs uniform circular motion w | vedclass

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Question

$\mathrm{KE}=\frac{1}{2} \mathrm{I} \omega^{2}$

$\Rightarrow \frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\righ

Vedclass · Accepted Answer

$L/4$

Vedclass · Answer

$\mathrm{KE}=\frac{1}{2} \mathrm{I} \omega^{2}$

$\Rightarrow \frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}ight)\left(\frac{\omega_{1}}{\omega_{2}}ight)^{2} \Rightarrow 2=\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}ight)\left(\frac{1}{2}ight)^{2} \Rightarrow \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=8$

Also, $\mathrm{L}^{2}=2 \mathrm{I}(\mathrm{KE})$

$\left(\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}ight)^{2}=\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}ight)\left(\frac{\mathrm{K}_{1}}{\mathrm{K}_{2}}ight)=8 	imes 2=16$

$\Rightarrow \frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}=4 \quad \Rightarrow \mathrm{L}_{2}=\frac{\mathrm{L}_{1}}{4}$

A particle performs uniform circular motion with an angular momentum $L.$ If the angular frequency of the particle is doubled and kinetic energy is halved, its angular momentum becomes

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