A circular disc of moment of inertia $I_{i}$ is rotating in a horizontal plane, about its geometrical axis, with a constant angular speed $ω_{i}$ . Another disc of moment of inertia $I_{f}$ is dropped coaxially onto the rotating disc. Initially the second disc has zero angular speed. Eventually both the discs rotate with a constant angular speed $ω_{f}$ . The energy lost by the initially rotating disc due to friction is:

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$\frac{1}{2} \frac{I_{i} I_{f}}{(I_{i} + I_{f})} {ω_{i}}^{2}$

$\frac{1}{2} \frac{{I_{f}}^{2}}{(I_{i} + I_{f})} {ω_{i}}^{2}$

$\frac{1}{2} \frac{{I_{i}}^{2}}{(I_{t} + I_{f})} {ω_{i}}^{2}$

$\frac{1}{2} \frac{I_{f} - I_{i}}{(I_{f} + I_{i})} {ω_{i}}^{2}$

answer is D.

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Detailed Solution

Initial KE of system= $K E_{1}$ = $\frac{1}{2} I_{i} {ω_{i}}^{2} + \frac{1}{2} I_{f} 0^{2} = \frac{1}{2} I_{i} {ω_{i}}^{2}$

Conserving angular momentum, common angular velocity $ω_{f} = \frac{I_{i} ω_{i} + I_{f} 0}{I_{i} + I_{f}}$

Final KE= $K E_{2} = \frac{1}{2} [I_{i} + I_{f}] {ω_{f}}^{2}$ = $\frac{1}{2} (I_{i} + I_{f}) {[I_{i} ω_{i} / (I_{i} + I_{f})]}^{2} = \frac{I_{i} I_{f} {ω^{2}}_{i}}{2 (I_{i} + I_{f})}$

KE lost= $∆ K E = K E_{i} - K E_{f}$ = $\frac{1}{2} I_{i} {ω_{i}}^{2} - (\frac{1}{2} I_{i} I_{f} {ω_{i}}^{2}) / (I_{i} + I_{f}) = \frac{1}{2} \frac{I_{i} I_{f}}{{(I_{i} + I_{f})}^{2}} {ω_{i}}^{2}$