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

• Correct answer
  A

  $\frac{14}{15}$
• B

  $\frac{4}{5}$
• C

  $\frac{2}{5}$
• D

  $\frac{7}{15}$

Solution

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 $\frac{L_{\text {sphere }}}{L_{\text {cylinder }}}=\frac{\frac{7}{5} m R^2 \omega}{\frac{3}{2} m R^2 \omega} \Rightarrow \frac{L_1}{L_2}=\frac{14}{15}$

Hence, the correct answer is option: 'A'.