A particle moving in a circular path has an angular momentum of $L$. If the frequency of rotation is halved, then its angular momentum becomes

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Aicharya · Accepted Answer

Angular momentum of particle is given by:
$\mathrm{L}=\mathrm{mr}^{2} \omega=2 \pi \mathrm{mr}^{2} \mathrm{f} \quad[\because \mathrm{W}=2 \pi \mathrm{f}]$
If frequency is halved then, $\mathrm{L}^{\prime}=\mathrm{mr}^{2} \frac{\omega}{2}=\pi \mathrm{mr}^{2} \mathrm{f} \therefore \mathrm{L}^{\prime}=\frac{\mathrm{L}}{2}$ .

A particle moving in a circular path has an angular momentum of $L$. If the frequency of rotation is halved,…

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