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 $'\omega'$. 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)

According to the law of conservation of angular momentum,
$Iw=$ constant
Now, according to the question,
$I_{1} \omega_{1} =I_{2} \omega_{2} \,\,\, \text { or } \,\,\,\, Iw=(2l) \omega_{2} $
$\omega_{2} =\frac{\omega}{2}$
New kinetic energy $=\left[\frac{1}{2} I_{2} \omega_{2}^{2}\right]=\frac{1}{2}(2 I) \times\left(\frac{\omega}{2}\right)^{2}=\frac{lw^{2}}{4}$
Loss in kinetic energy $\left(K_{L}\right)=K_{i}-K_{j}$
$=\frac{1}{2} Iw^{2}-\frac{Iw^{2}}{4}=\frac{Iw^{2}}{4}$