Question

The moment of inertia of a ring about an axis passing through its 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 rings are rotating about the same axis then loss in kinetic energy is

• A. \(Iω^2\)
• B. \(\frac{Iω^2}{2}\)
• C. \(\frac{Iω^2}{4}\)
• D. \(\frac{Iω^2}{8}\)

Hint:

Angular momentum is conserved, but kinetic energy is not.

Answer 1

The Correct Option is C

Step 1: Use the angular momentum form of kinetic energy:
$K=\dfrac{L^2}{2I}$. The angular momentum $L=I\omega$ stays the same.

Step 2: Compare initial and final:
$K_i=\dfrac{L^2}{2I}$ and $K_f=\dfrac{L^2}{2(2I)}=\dfrac{L^2}{4I}$.

Step 3: Take the difference:
$K_i-K_f=\dfrac{L^2}{4I}=\dfrac{I^2\omega^2}{4I}=\dfrac{I\omega^2}{4}$. Option C.

Final Answer:
Angular momentum conservation halves omega, so half of the energy is lost. \[ \boxed{\text{(C) }\dfrac{I\omega^2}{4}} \]