Two discs of moment of inertia \(\mathrm{I}_{1}=4 \mathrm{~kg} \mathrm{~m}^{2}\) and \(\mathrm{I}_{2}=2 \mathrm{~kg} \mathrm{~m}^{2}\) about their central axes & normal to their planes, rotating with angular speeds \(10 \mathrm{rad} / \mathrm{s}\) & \(4 \mathrm{rad} / \mathrm{s}\) respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is ______ J.

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Rotational Mechanics

Two discs of moment of inertia

I_{1} = 4 kg m^{2}

and

I_{2} = 2 kg m^{2}

about their central axes & normal to their planes, rotating with angular speeds

10 rad / s

4 rad / s

respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is ______ J.

Answer: 24

Solution

When the two discs are brought into contact face to face along a common axis, angular momentum is conserved (no external torque):

I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω_{0}

4 (10) + 2 (4) = 6 ω_{0} ⟹ ω_{0} = \frac{48}{6} = 8 rad/s

Initial kinetic energy:

E_{1} = \frac{1}{2} (4) (10)^{2} + \frac{1}{2} (2) (4)^{2} = 200 + 16 = 216 J

Final kinetic energy:

E_{2} = \frac{1}{2} (6) (8)^{2} = 192 J

Loss in kinetic energy:

Δ E = 216 - 192 = 24 J

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Integer answer

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JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Rotational Mechanics

Solution

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