MHT CET · Physics · Rotational Motion

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 $ω.$ 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 $\frac{I ω^{2}}{2}$
B $\frac{I ω^{2}}{4}$
C $\frac{I ω^{2}}{6}$
D $\frac{I ω^{2}}{8}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{I ω^{2}}{4}$

Step-by-step Solution

Detailed explanation

From conservation of angular momentum

I_{1} ω_{1} = I_{2} ω_{2}

I ω = 2 I ω_{2}

ω_{2} = \frac{ω}{2}

New

KE = \frac{1}{2} . (2 I) {(\frac{ω}{2})}^{2} = \frac{I ω^{2}}{4}

Loss in

KE = \frac{1}{2} I ω^{2} - \frac{I ω^{2}}{4} = \frac{I ω^{2}}{4}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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