Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc has angular velocities and . They are brought into contact face to face coinciding the axis of rotation. The expression fo

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Solution

Q.
Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc has angular velocities $ω_{1}$ and $ω_{2}$ . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is [2017]

1 $\frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$

2 $I {(ω_{1} - ω_{2})}^{2}$

3 $\frac{1}{8} I {(ω_{1} - ω_{2})}^{2}$

4 $\frac{1}{2} I {(ω_{1} + ω_{2})}^{2}$

Ans.
(1)

Initial angular momentum $= I ω_{1} + I ω_{2}$

Let $ω$ be angular speed of the combined system.

Final angular momentum = $2 I ω$

$∴$   According to conservation of angular momentum

$I ω_{1} + I ω_{2} = 2 I ω or ω = \frac{ω_{1} + ω_{2}}{2}$

Initial rotational kinetic energy,

$E = \frac{1}{2} I (ω_{1}^{2} + ω_{2}^{2})$

Final rotational kinetic energy

   $E_{f} = \frac{1}{2} (2 I) ω^{2} = \frac{1}{2} (2 I) {(\frac{ω_{1} + ω_{2}}{2})}^{2} = \frac{1}{4} I {(ω_{1} + ω_{2})}^{2}$

$∴$   Loss of energy $Δ E = E_{i} - E_{f}$

$= \frac{I}{2} (ω_{1}^{2} + ω_{2}^{2}) - \frac{I}{4} (ω_{1}^{2} + ω_{2}^{2} + 2 ω_{1} ω_{2})$

   $= \frac{I}{4} [ω_{1}^{2} + ω_{2}^{2} - 2 ω_{1} ω_{2}] = \frac{I}{4} {(ω_{1} - ω_{2})}^{2}$

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