Rotational Motion

JEE Main level, single correct, medium

269 of 280

Question 269 of 280

A thick-walled hollow sphere has outside radius R0R_0. It rolls down an incline without slipping and its speed at the bottom is v0v_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 5v0/45 v_0 / 4. The radius of gyration of the hollow sphere about an axis through its centre is

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Answer: (B) 3R0/43 R_0 / 4

When body rolls dawn on inclined plane with velocity V0\mathrm{V}_0 at bottom then body has both rotational and translational kinetic energy.
Therefore, by law of conservation of energy,
P.E.=K.Etrans +K.Erotational =12mV02+12Iω2=12mV02+12mk2V02R02… (i) [∵I=mk2,ω=VR0]\begin{aligned} & P . E .=K . E_{\text {trans }}+K . E_{\text {rotational }} \\ & =\frac{1}{2} m V_0^2+\frac{1}{2} I \omega^2 \\ & =\frac{1}{2} m V_0^2+\frac{1}{2} m k^2 \frac{V_0^2}{R_0^2} \quad \ldots \text { (i) } \\ & {\left[\because \mathrm{I}=m k^2, \omega=\frac{V}{R_0}\right]} \end{aligned}
When body is sliding down then body has only translatory motion.
∴P.E.=K.Etrans =12m(54v0)2\begin{aligned} \therefore P . E . & =K . E_{\text {trans }} \\ & =\frac{1}{2} m\left(\frac{5}{4} v_0\right)^2 \end{aligned}
Dividing (i) by (ii) we get
P.E.P.E.=12mv02[1+K2R02]12×2516×mV02=2516=1+K2R02⇒K2R02=916 or, K=34R0.\begin{aligned} & \frac{P . E .}{P . E .}=\frac{\frac{1}{2} m v_0^2\left[1+\frac{K^2}{R_0^2}\right]}{\frac{1}{2} \times \frac{25}{16} \times m V_0^2} \\ & =\frac{25}{16}=1+\frac{K^2}{R_0^2} \Rightarrow \frac{K^2}{R_0^2}=\frac{9}{16} \\ & \text { or, } K=\frac{3}{4} R_0 . \end{aligned}

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