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

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

TestHub · Accepted Answer

Initial angular momentum of the system is $L_{i} = I ω$

Initial kinetic energy is $K_{i} = \frac{1}{2} I ω^{2}$

When an identical ring is placed on it, the new moment of inertia becomes $I_{f} = I + I = 2 I$

By conservation of angular momentum, $I_{f} ω_{f} = I_{i} ω_{i}$

$2 I ω_{f} = I ω \Rightarrow ω_{f} = \frac{ω}{2}$

Final kinetic energy is $K_{f} = \frac{1}{2} (2 I) (\frac{ω}{2})^{2} = \frac{1}{4} I ω^{2}$

Loss in kinetic energy is $Δ K = K_{i} - K_{f} = \frac{1}{2} I ω^{2} - \frac{1}{4} I ω^{2} = \frac{I ω ^{2}}{4}$

Answer: $\frac{I ω ^{2}}{4}$

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