Question

A thick walled hollow sphere has outer radius $R$. It rolls down an inclined plane without slipping and its speed at bottom is $v_0$. Now the incline is waxed so that the friction becomes zero. The sphere is observed to slide down without rolling and the speed now is $\left(5\frac{v_0}{4}\right)$. The radius of gyration of the hollow sphere about the axis through its centre is

• A. $\frac{3R}{4}$
• B. $\frac{R}{2}$
• C. $\frac{R}{4}$
• D. $\frac{4}{5}R$

Answer 1

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

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 $Mgh=\frac{1}{2}M{v}_0^2+\frac{1}{2}I{\omega }_0^2=\frac{1}{2}M{v}_0^2+\frac{1}{2}I\frac{v_0^2}{r^2}$
In case of no rotation, the potential energy converts to gain in linear kinetic energy only.
Thus, $Mgh=\frac{1}{2}M{\left(\frac{5v_0}{4}\right)}^2$

Thus, $Mgh=\frac{1}{2}M{v}_0^2+\frac{1}{2}I{\omega }_0^2=\frac{1}{2}M{\left(\frac{5v_0}{4}\right)}^2\Rightarrow I=\frac{9M{R}^2}{16}$
$\frac{9M{R}^2}{16}=M{k}^2\Rightarrow k=\frac{3R}{4}$
$\therefore \left(a\right)$ is the correct option.