Question 206 of 280

A spherical shell of $1kg$ 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 $\dfrac{a}{3}{R}^{2}\omega$. The value of $a$ will be

A$2$B$3$C$5$D$4$

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Answer: (C) $5$

Angular momentum of a rotating and translating body is given by,
${L}_{O}=mvR+I\omega =mvR+\dfrac{2}{3}M{R}^{2}\omega$
For rolling, $v=\omega R$
$\Rightarrow {L}_{O}=m\omega {R}^{2}\left(1+\dfrac{2}{3}\right)=\dfrac{5}{3}m\omega {R}^{2}$
Therefore, the value of $a=5$.