Question

A sphere rolls without slipping on a rough horizontal surface with centre of mass having constant speed $v_{0}$ . If mass of the sphere is $m$ and its radius is $R$ , then what is the angular momentum of the sphere about the point of contact?

• A. $\frac{5}{2}mv_{0}R$
• B. $\frac{7}{5}mv_{0}R$
• C. $\frac{3}{5}mv_{0}R$
• D. $\frac{1}{2}mv_{0}R$

Answer 1

Answer: (B)

Assume the angular velocity of the sphere is $\omega $ , as the sphere is rolling without slipping
$\omega =\frac{v_{0}}{R}$
Now Angular momentum about the lowest point is the sum of angular momentum of center of mass and the angular momentum due to rotation of the sphere
$\Rightarrow mv_{0}R+I\omega =mv_{0}R+\frac{2}{5}mR^{2}\frac{v_{0}}{R}=\frac{7}{5}mv_{0}R$