A thick - walled hollow sphere has outside radius R_0. It rolls d… - Vidyadip

MCQ

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 $5{v_0}/4$. 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

When body rolls dawn on inclined plane with velocity $V_0$ at bottom then body has both rotational and translational kinetic energy.
Thereforce, by law of conservation of energy,
$P.E\, = \,K.{E_\text{ trans}} + K.{E_\text{ rotational}}$
$= \frac{1}{2}mV_0^2 + \frac{1}{2}I{\omega ^2}$
$= \frac{1}{2}mV_0^2 + \frac{1}{2}m{k^2}\frac{{V_0^2}}{{R_0^2}},...\left( i \right)$
$\left[ {I = m{k^2},\omega = \frac{V}{{{R_0}}}} \right]$
When body is sliding down then body has only translatory motion.
$\therefore P.E = K.{E_\text{ trans}}$
$ = \frac{1}{2}m{\left( {\frac{5}{4}{v_0}} \right)^2}...\left( {ii} \right)$
Dividing $(i)$ by $(ii)$ we get
$\frac{{P.E}}{{P.E}} = \frac{{\frac{1}{2}mv_0^2\left[ {1 + \frac{{{K^2}}}{{R_0^2}}} \right]}}{{\frac{1}{2} \times \frac{{25}}{{16}} \times mV_0^2}}$
$= \frac{{25}}{{16}} = 1 + \frac{{{K^2}}}{{R_0^2}} $
$\Rightarrow \frac{{{K^2}}}{{R_0^2}} = \frac{9}{{16}}$
or $,K = \frac{3}{4}\,{R_{0.}}$

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