Question
A spherical shell of $1 \,kg$ mass and radius $R$ is rolling with angular speed $\omega$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin $O$ is $\frac{a}{3} R^{2} \omega$. The value of a will be ..............

Answer

$L _{0}=$ angular momentum of shell about $O$.

As shell is rolling

$\text { so } V _{ cm }=\omega R$

$L _{0}= mV _{ cm } R + I \omega$

$=1 \times \omega R \times R +\frac{2}{3} R ^{2} \omega$

$=\frac{5}{3} R ^{2} \omega$

$\text { so } a =5$

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