Question

A solid sphere of mass $m$ and radius $R$ rolls down an inclined plane from rest without slipping as shown in the figure. What will be the maximum velocity of the sphere when it will reach the ground?

• A. $\sqrt{\frac{10gh}{7}}$
• B. $\sqrt{\frac{gh}{7}}$
• C. $\sqrt{\frac{10gh}{3}}$
• D. $\sqrt{\frac{gh}{3}}$

Answer 1

The correct option is A $\sqrt{\frac{10gh}{7}}$

The loss in gravitational potential energy is equal to the gain in translational kinetic energy and rotational kinetic energy for the case of maximum velocity of sphere when it reaches ground.

$⟹mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$⟹mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\omega }^2$

From the no-slip condition:

$v=R\omega$

$⟹mgh=\frac{1}{2}m{v}^2+\frac{2}{10}m{v}^2$

$⟹v=\sqrt{\frac{10gh}{7}}$