Question

A solid sphere of mass $m$ and radius $R$, is rolling without slipping, along a horizontal surface. Find the total angular momentum of the sphere, about the point of contact. The speed of its centre of mass is $v_0$

• A. $L=\frac{12}{5}m{v}_{0}R$
• B. $L=2m{v}_{0}R$
• C. $L=\frac{2}{5}m{v}_{0}R$
• D. $L=\frac{7}{5}m{v}_{0}R$

Answer 1

The correct option is D $L=\frac{7}{5}m{v}_{0}R$

As we know,
$\overrightarrow{L}={\overrightarrow{L}}_{trans}+{\overrightarrow{L}}_{rot}=mv{r}_{\perp }+I\omega$

Here, about point $P,r_{\perp }=R,I_{com}=\frac{2}{5}m{R}^2$

$\therefore L=mv{r}_{\perp }+I_{com}.\omega$

For rolling without slipping,
$\omega =\frac{v}{R}$
$\therefore L=m{v}_{0}R+\frac{2}{5}m{R}^2\times \left(\frac{v_0}{R}\right)$

$=m{v}_{0}R+\frac{2m{v}_{0}R}{5}=\frac{7m{v}_{0}R}{5}$

Hence, $\left(D\right)$ is the correct answer.