Question 496: Centre of Mass and Rotational Motion

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 $\omega$. 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

Accepted Answer

Step 1: Initial kinetic energy
Only first ring is rotating:
$ KE_	ext{initial} = 	frac{1}{2} I \omega^2 $
Step 2: Conservation of angular momentum
System is isolated (no external torque about axis).
Let final angular velocity = $\omega_f$.
Initial angular momentum:
$ L_	ext{initial} = I \omega $
Final total moment of inertia = $I + I = 2I$.
So,
$ L_	ext{final} = (2I)\,\omega_f $
By conservation:
$ I \omega = 2I \,\omega_f \;\;\Rightarrow\;\; \omega_f = \frac{\omega}{2} $
Step 3: Final kinetic energy
$ KE_	ext{final} = 	frac{1}{2} (2I) \left(\frac{\omega}{2}ight)^2 = I \cdot \frac{\omega^2}{4} = \frac{I \omega^2}{4} $
Step 4: Loss in kinetic energy
$ \Delta KE = KE_	ext{initial} - KE_	ext{final} = \frac{1}{2} I \omega^2 - \frac{1}{4} I \omega^2 = \frac{1}{4} I \omega^2 $
Ã¢ÂÂ Final Answer:
$ \boxed{	frac{I \omega^2}{4}} $
That corresponds to
Option C
.

Answer

$\frac{\mathrm{I} \omega^2}{16}$

Answer

$\frac{\mathrm{I} \omega^2}{8}$

Answer

$\frac{\mathrm{I} \omega^2}{4}$

Answer

$\frac{\mathrm{I} \omega^2}{2}$

Centre of Mass and Rotational Motion

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