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

Answer 1

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

Case I: $$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}m{k}^2{\omega }^2$$
$$\because V=\omega R$$
$$\therefore mgh=\frac{1}{2}m{v}^2+\frac{1}{2}m{k}^2\frac{v^2}{R^2}$$

Case II: $$mgh=\frac{1}{2}m\times {\left(\frac{5v}{4}\right)}^2$$ $$\frac{1}{2}m{k}^2\frac{v^2}{R^2}+\frac{1}{2}m{v}^2=\frac{1}{2}m\times \frac{25v^2}{16}$$ $$\frac{K^2}{R^2}+1=\frac{25}{16}\Rightarrow k=\frac{3R}{4}$$