Two discs of same moment of inertia rotating about their regular axis with angular velocities ω_1 and ω_2 . They are brought into contact face to face - Physics MCQ NEET (UG) 2017 2026 | ExamTest.live

hardMCQNEET (UG) 20172026Physics

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Two discs of same moment of inertia rotating about their regular axis 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 of energy during this process is:

Solution & Step-by-step Explanation

By conservation of angular momentum: I ω_{1} + I ω_{2} = (I + I) ω_{f} ⟹ ω_{f} = \frac{ω _{1} + ω _{2}}{2} .Initial Kinetic Energy: K_{i} = \frac{1}{2} I ω_{1}^{2} + \frac{1}{2} I ω_{2}^{2} .Final Kinetic Energy: K_{f} = \frac{1}{2} (2 I) ω_{f}^{2} = I (\frac{ω _{1} + ω _{2}}{2})^{2} = \frac{I ( ω _{1} + ω _{2} ) ^{2}}{4} .Loss in Energy Δ K = K_{i} - K_{f} : Δ K = \frac{1}{2} I (ω_{1}^{2} + ω_{2}^{2}) - \frac{I ( ω _{1} + ω _{2} ) ^{2}}{4} Δ K = \frac{2 I ω _{1}^{2} + 2 I ω _{2}^{2} - I ( ω _{1}^{2} + ω _{2}^{2} + 2 ω _{1} ω _{2} )}{4} Δ K = \frac{I ( ω _{1}^{2} + ω _{2}^{2} - 2 ω _{1} ω _{2} )}{4} = \frac{1}{4} I (ω_{1} - ω_{2})^{2}

Try it yourself before checking the explanation above.

Two discs of same moment of inertia rotating about their regular axis 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 of energy during this process is:

\frac{1}{4} I (ω_{1} - ω_{2})^{2}

I (ω_{1} - ω_{2})^{2}

\frac{1}{8} I (ω_{1} - ω_{2})^{2}

\frac{1}{2} I (ω_{1} + ω_{2})^{2}

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