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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

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 _______ \(\mathrm{J}\).

  1. A \(20\)
  2. B \(22\)
  3. C \(24\)
  4. D \(30\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(24\)

Step-by-step Solution

Detailed explanation

\(I_1 \omega_1+I_2 \omega_2=\left(I_1+I_2\right) \omega_0 \text { (C.O.A.M.) }\) \(\text { gives } \omega_0=8 \mathrm{rad} / \mathrm{s}\) \(E_1=\frac{1}{2} I_1 \omega_1^2+\frac{1}{2} I_2 \omega_2^2=216 J\) \(E_2=\frac{1}{2}\left(I_1+I_2\right) \omega_0^2=192 J\)…