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: Understanding the Concept:
No external torque acts, so angular momentum is conserved. The second ring starts at rest and then moves with the first, so energy is lost as friction heat.

Step 2: Find the final angular velocity:
\(I\omega=(I+I)\omega'\), so \(\omega'=\dfrac\omega2\).

Step 3: Find the kinetic energies:
Initial: \(K_i=\dfrac12I\omega^2\). Final: \(K_f=\dfrac12(2I)\left(\dfrac\omega2\right)^2=\dfrac14I\omega^2\).

Step 4: Find the loss:
\(K_i-K_f=\dfrac12I\omega^2-\dfrac14I\omega^2=\dfrac14I\omega^2\). Option C.

Step 5: Why the other options are wrong.
\(I\omega^2\) and \(\dfrac{I\omega^2}{2}\) are bigger than or equal to the initial energy, which cannot be lost. \(\dfrac{I\omega^2}{8}\) comes from halving the correct loss.

Final Answer:
The loss in kinetic energy is I omega^2 / 4. \[ \boxed{\text{(C) }\dfrac{I\omega^2}{4}} \]