Question

A thick walled hollow sphere has outer radius $R$. It rolls down an inclined plane without slipping and its speed at the bottom is $v$. If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom will be $5v/4$. What is the radius of gyration of the sphere?

• A. $\frac{R}{\sqrt{2}}$
• B. $\frac{R}{2}$
• C. $\frac{3R}{4}$
• D. $\frac{\sqrt{3}R}{4}$

Answer 1

The correct option is C $\frac{3R}{4}$

The loss in gravitational potential energy in both the cases remains the same and equal to the total gain in kinetic energy.

In case of no slipping, the potential energy converts to gain in linear and rotational kinetic energy.

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

$=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{I{v}^2}{R^2}$

In case of no rotation, the potential energy converts to gain in linear kinetic energy only.

Thus $mgh=\frac{1}{2}m(\frac{5v}{4}{)}^2$

Thus $\frac{1}{2}m{v}^2+\frac{1}{2}\frac{I{v}^2}{R^2}=\frac{1}{2}m\frac{25v^2}{16}$

$⟹I=\frac{9}{16}m{R}^2$

$=m{k}^2$

where $k$ is the radius of gyration

$⟹k=\frac{3R}{4}$

Hence correct answer is option C.