JEE Advance - Physics (2022 - Paper 2 Online - No. 15)
A flat surface of a thin uniform disk $A$ of radius $R$ is glued to a horizontal table. Another thin uniform disk $B$ of mass $M$ and with the same radius $R$ rolls without slipping on the circumference of $A$, as shown in the figure. A flat surface of $B$ also lies on the plane of the table. The center of mass of $B$ has fixed angular speed $\omega$ about the vertical axis passing through the center of $A$. The angular momentum of $B$ is $n M \omega R^{2}$ with respect to the center of $A$. Which of the following is the value of $n$ ?
2
5
$\frac{7}{2}$
$\frac{9}{2}$
Explanation
Angular momentum of $B$ with respect to center of $A$
$$ \begin{aligned} \vec{L} & =\vec{L}_{C M}+\vec{L}_{\text {Body about CM }} \\\\ & =M(2 R)^2 \omega \hat{k}+\frac{M R^2}{2}\left(\omega_{\text {body }}\right) \hat{k} \\\\ & =M(2 R)^2 \omega \hat{k}+\frac{M R^2}{2}(2 \omega) \hat{k} \\\\ & =5 M R^2 \omega \hat{k} \end{aligned} $$
Comparing the magnitude with $n M \omega R^2$
$$ n=5 $$
$$ \begin{aligned} \vec{L} & =\vec{L}_{C M}+\vec{L}_{\text {Body about CM }} \\\\ & =M(2 R)^2 \omega \hat{k}+\frac{M R^2}{2}\left(\omega_{\text {body }}\right) \hat{k} \\\\ & =M(2 R)^2 \omega \hat{k}+\frac{M R^2}{2}(2 \omega) \hat{k} \\\\ & =5 M R^2 \omega \hat{k} \end{aligned} $$
Comparing the magnitude with $n M \omega R^2$
$$ n=5 $$
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