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A solid sphere and a solid cylinder, each of mass $M$ and radius $R$ are rolling with a linear speed on a flat surface without slipping. Let $L_1$ be magnitude of the angular momentum of the sphere with respect to a fixed point along the path of the sphere. Likewise $L_2$ be the magnitude of angular momentum of the cylinder with respect to the same fixed point along its path. The ratio $\frac{L_1}{L_2}$ is
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The correct answer is:
$\frac{14}{15}$
We take origin as fixed point.
Now, angular momentum about origin $O=$ angular momentum due to linear motion of centre of mass + angular momentum due to rotation of body about its' centre of mass
$\therefore \quad L_O=L_{\text {linear motion }}+L_{\text {rotational motion }}$
$=m v R+I \omega$
$=m R^2 \omega+I \omega \quad[\therefore v=R \omega]$
$=\left(m R^2+I\right) \omega$
For solid sphere, $I=\frac{2}{5} m R^2$
and for solid cylinder, $I=\frac{1}{2} m R^2$
So, $\left(L_{\text {sphere }}\right)_{\text {about ' } O \text { ' }}=\left(m R^2+\frac{2}{5} m R^2\right) \omega=\frac{7}{5} m R^2 \omega$
and $\left(L_{\text {sphere }}\right)_{\text {about ' } O}=\left(m R^2+\frac{1}{2} m R^2\right) \omega=\frac{3}{2} m R^2 \omega$
So, ratio of $\frac{L_{\text {sphere }}}{L_{\text {cylinder }}}=\frac{\frac{7}{5} m R^2 \omega}{\frac{3}{2} m R^2 \omega} \Rightarrow \frac{L_1}{L_2}=\frac{14}{15}$
Now, angular momentum about origin $O=$ angular momentum due to linear motion of centre of mass + angular momentum due to rotation of body about its' centre of mass
$\therefore \quad L_O=L_{\text {linear motion }}+L_{\text {rotational motion }}$
$=m v R+I \omega$
$=m R^2 \omega+I \omega \quad[\therefore v=R \omega]$
$=\left(m R^2+I\right) \omega$
For solid sphere, $I=\frac{2}{5} m R^2$
and for solid cylinder, $I=\frac{1}{2} m R^2$
So, $\left(L_{\text {sphere }}\right)_{\text {about ' } O \text { ' }}=\left(m R^2+\frac{2}{5} m R^2\right) \omega=\frac{7}{5} m R^2 \omega$
and $\left(L_{\text {sphere }}\right)_{\text {about ' } O}=\left(m R^2+\frac{1}{2} m R^2\right) \omega=\frac{3}{2} m R^2 \omega$
So, ratio of $\frac{L_{\text {sphere }}}{L_{\text {cylinder }}}=\frac{\frac{7}{5} m R^2 \omega}{\frac{3}{2} m R^2 \omega} \Rightarrow \frac{L_1}{L_2}=\frac{14}{15}$
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