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
$\therefore L_0=L_{\text{linear motion }}+L_{\text{rotational motion }}$
$=mvR+I\omega$
$=m{R}^2\omega +I\omega$
$=\left(m{R}^2+I\right)\omega$
$[\therefore v=R\omega ]$ 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
${\left(L_{\text{sphere }}\right)}_{\text{about }}\cdot 0^{\prime }=\left(m{R}^2+\frac{1}{2}m{R}^2\right)\omega =\frac{3}{2}m{R}^2\omega$
So, ratio of $\frac{L_{\text{sphere }}}{L_{\text{cylinder }}}=\frac{\frac{7}{5}m{R}^2\omega }{\frac{3}{2}m{R}^2\omega }$
$\Rightarrow \frac{L_1}{L_2}=\frac{14}{15}$