JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2024
- Hard
- Numerical
Two discs of moment of inertia \(I_1=4 \mathrm{~kg} \mathrm{~m}^2\) 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 axe of rotation coincident. The loss in kinetic energy of the system in the process is _______ \(\mathrm{J}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(24\)
Step-by-step Solution
Detailed explanation
\(I_1 \omega_1+I_2 \omega_2=\left(I_1+I_2\right) \omega_0 \text { (C.O.A.M.) }\)
\(\text { gives } \omega_0=8 \mathrm{rad} / \mathrm{s}\)
\(E_1=\frac{1}{2} I_1 \omega_1^2+\frac{1}{2} I_2 \omega_2^2=216 J\)
\(E_2=\frac{1}{2}\left(I_1+I_2\right) \omega_0^2=192 J\)
\(\therefore \Delta E=24 J\)
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