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 ${}^{\prime }{\omega }^{\prime }$. 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 is 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

The correct option is B $\frac{I{\omega }^2}{4}$

Initial angular momentum, $L_1=I\omega$

$K_1=\frac{1}{2}I{\omega }^2$

When, second ring is put on it and they start rotating with same angular speed, say ${\omega }_o$.

Now, final angular momentum, $L_2=2I{\omega }_o$

Conservation of angular momentum:-

$L_1=L_2$

$⟹I\omega =2I{\omega }_o$

$⟹{\omega }_o=\frac{\omega }{2}$

$K_2=2\times \frac{1}{2}I{\omega }_o^2=I{\omega }_o^2$

$K_2=\frac{1}{4}I{\omega }^2$

$⟹K_1-K_2=\frac{1}{4}I{\omega }^2$

Hence, answer is option-(B).