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 is $5v/4$. What is the radius of gyration of the sphere?

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

Answer 1

The correct option is C $3R/4$

Case 1: Pure rolling i-e, $v=Rw$

Work-energy theorem, $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2$ $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I\frac{v^2}{R^2}$ .................(1)

Case 2: No friction, thus there is only translational motion of the sphere.

Work-energy theorem, $mgh=\frac{1}{2}m(\frac{5v}{4}{)}^2$ .................(2)

From (1) & (2), $\frac{1}{2}m{v}^2\times \frac{25}{16}=\frac{1}{2}m{v}^2+\frac{1}{2}I\frac{v^2}{R^2}$

$⟹I=\frac{9}{16}m{R}^2$

Let K be the radius of gyration.

$m{K}^2=I=\frac{9}{16}m{R}^2⟹K=\frac{3}{4}R$