Question

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 tothe same fixed point along its path. The ratio $\frac{ L _{1}}{ L _{2}}$ is

• A. $\frac{14}{15}$
• B. $\frac{4}{5}$
• C. $\frac{2}{5}$
• D. $\frac{7}{15}$

Answer 1

Answer: (A)

We take origin as fixed point.

Now, angular momentum about origin $0=$ angular momentum due to linear motion of centre of mass $+$ angular momentum due to rotation of body about its' centre of mass
$\therefore L_{0}=L_{\text {linear motion }}+L_{\text {rotational motion }}$
$=m v R+I \omega$
$=m R^{2} \omega+I \omega$
$=\left(m R^{2}+I\right) \omega$
$[\therefore v=R \omega]$ For solid sphere,
$I=\frac{2}{5} m R^{2}$ and for solid cylinder, $I=\frac{1}{2} m R^{2}$
So, $\left(L_{\text {sphere }}\right)_{\text {about 'o' }}=\left(m R^{2}+\frac{2}{5} m R^{2}\right) \omega=\frac{7}{5} m R^{2} \omega$ and
$\left(L_{\text {sphere }}\right)_{\text {about }} \cdot 0'=\left(m R^{2}+\frac{1}{2} m R^{2}\right) \omega=\frac{3}{2} m R^{2} \omega$
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} $
$\Rightarrow \frac{L_{1}}{L_{2}}=\frac{14}{15}$