Two discs of same moment of inertia ( I ) are rotating in same sense about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω 1 and ω 2 . They are brought into contact face to face coinciding the axis of rotation. The expression for loss in…

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Two discs of same moment of inertia ( I ) are rotating in same sense about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω 1 and ω 2 . They are brought into contact face to face coinciding the axis of rotation. The expression for loss in energy during this process is

Aicharya · Accepted Answer

The initial angular momentum of two discs is $I_{1} ω_{1} + I_{2} ω_{2} = I (ω_{1} + ω_{2})$ .

The final angular momentum of system is $(I_{1} + I_{2}) ω = 2 I ω$

From the conservation of angular momentum, $ω = \frac{ω_{1} + ω_{2}}{2}$ .

Initial kinetic energy, $K E_{i} = \frac{1}{2} I (ω_{1}^{2} + ω_{2}^{2})$

and final kinetic energy, $K E_{f} = \frac{1}{2} (2 I) \frac{{(ω_{1} + ω_{2})}^{2}}{4}$

Loss in kinetic energy is $Δ K E = \frac{1}{2} \frac{I_{1} I_{2}}{I_{1} + I_{2}} {(ω_{1} - ω_{2})}^{2}$
$= \frac{1}{2} \frac{I^{2}}{(2 I)} {(ω_{1} - ω_{2})}^{2}$
$= \frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$

Two discs of same moment of inertia ( I ) are rotating in same sense about their regular axis passing…

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