Question

A thick-walled hollow sphere has outside radius $R_0$. It rolls down an incline without slipping and its speed at the bottom is $v_0$. Now the incline is waxed, so that it is practically frictionless and the sphere is observed to slide down (without any rolling). Its speed at the bottom is observed to be $5v_0/4$. The radius of gyration of the hollow sphere about an axis through its centre is

• A. $3R_0/2$
• B. $3R_0/4$
• C. $9R_0/16$
• D. $3R_0$

Answer 1

Answer: (B)

When body rolls dawn on inclined plane with velocity $V_0$ at bottom then body has both rotational and translational kinetic energy.
Therefore, by law of conservation of energy,
$P.K = K.E_{traos} + K.T._{rotational}$
$= \frac{1}{2}mV^{2}_{0} +\frac{1}{2}I\omega^{2}$
$= \frac{1}{2}mV^{2}_{0} +\frac{1}{2}mk^{2} \frac{V^{2}_{0}}{R^{2}_{0}} \quad\ldots\left(i\right)$
$\left[\because I = mk^{2}, \omega =\frac{V}{R_{0}} \right]$
When body is sliding down then body has only translatory motion.
$\therefore P.E. = K.E_{trans}$
$= \frac{1}{2}m\left(\frac{5}{4}v_{0}\right)^{2}\quad\quad...\left(ii\right)$
Dividing $\left(i\right)$ by $\left(ii\right)$ we get
$\frac{P.E.}{P.E.} = \frac{\frac{1}{2}mv^{2}_{0}\left[1+\frac{K^{2}}{R^{2}_{0}}\right]}{\frac{1}{2}\times\frac{25}{16}\times mv^{2}_{0}}$
$= \frac{25}{16} 1+\frac{K^{2}}{R^{2}_{0}} \Rightarrow \frac{K^{2}}{R^{2}_{0}} = \frac{9}{16}$
or, $K= \frac{3}{4}R_{0}.$