QJEEAdv17Phfc09: JEE Advanced 2017 Physics - Class 11 JEEAdv17Phfc09 Solution: The total kinetic energy of the ring is

Let P be the contact point of the finger and the ring. The point P revolves with an angular velocity $\omega$ ₀ in a circle of radius r centred at the point O (see figure). The contact point P, the centre O and the centre of the ring C are in a straight line because P is common to both the circles (i.e., circular trajectory and circular ring). Thus, the line CP will have same angular velocity as the line OP i.e., angular velocity of the line CP (or the ring) is $\omega$ ₀.


The velocity of the finger at the point P is $\omega$ ₀r tangential to the circle in which it revolves. The ring rolls without slipping. Thus, velocity of the ring at the point P is also $\omega$ ₀r tangential to the circle in which the finger revolves. Let velocity of C be v_cm. Since P and C lie on the ring (a rigid body), the velocities of C and P are related by ${\overrightarrow v _C} = {\overrightarrow v _P} + \overrightarrow \omega \times \overrightarrow {PC}$ which gives

v_cm = $\omega$ ₀(R $-$ r).

The ring rotates with an angular velocity $\omega$ = $\omega$ ₀ and its centre of mass translates with a velocity v_cm = $\omega$ ₀(R $-$ r). Thus, kinetic energy of the ring is given by

$K = {1 \over 2}Mv_{cm}^2 + {1 \over 2}{I_{cm}}{\omega ^2}$

$= {1 \over 2}M\omega _0^2{(R - r)^2} + {1 \over 2}M{R^2}\omega _0^2$

$= {1 \over 2}M\omega _0^2[{(R - r)^2} + {R^2}]$ .