Question

The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is $I$ . It is rotating with angular velocity $ω.$ Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss in kinetic energy is

• A. $\frac{I \omega ^{2}}{2}$
• B. $\frac{I \omega ^{2}}{4}$
• C. $\frac{I \omega ^{2}}{6}$
• D. $\frac{I \omega ^{2}}{8}$

Answer 1

Answer: (B)

From conservation of angular momentum
$I_{1} \omega _{1} = I_{2} \omega _{2}$
$I \omega = 2 I \omega _{2}$
$\omega _{2} = \frac{\omega }{2}$
New $KE=\frac{1}{2}.\left(\right. 2 I \left.\right)\left(\right. \frac{\omega }{2} \left.\right)^{2}=\frac{I \left(\omega \right)^{2}}{4}$
Loss in $KE=\frac{1}{2}I\omega ^{2}-\frac{I \omega ^{2}}{4}=\frac{I \omega ^{2}}{4}$