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Topic: Angular momentum and its conservation· Rotational Motion

Two discs of same moment of inertia rotating 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 e…

Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities \(\omega_1\) and \(\omega_2\). They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is :

1. \(\frac14 I(\omega_1-\omega_2)^2\)

2. \(I(\omega_1-\omega_2)^2\)

3. \(\frac18 I(\omega_1-\omega_2)^2\)

4. \(\frac12 I(\omega_1+\omega_2)^2\)

1 answer

Best answer

Answer: 1. \(\frac14 I(\omega_1-\omega_2)^2\)

Key idea: angular momentum is conserved when the discs join.

\(I\omega_1+I\omega_2=2I\omega\Rightarrow\omega=\dfrac{\omega_1+\omega_2}{2}\).

Loss \(=\tfrac12I\omega_1^2+\tfrac12I\omega_2^2-\tfrac12(2I)\omega^2=\tfrac12I(\omega_1^2+\omega_2^2)-\tfrac14I(\omega_1+\omega_2)^2=\tfrac14I(\omega_1-\omega_2)^2\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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