A thin circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity ω. If another disc of same dimensions but of mass M / 2 is placed gently on the first disc co-axially, then the new angular velocity of the system is :

Solution

Physics sir · 3 steps

No outside torque, so Iω stays fixed: the extra disc raises I, so ω must drop.

1. Why L is conserved. The second disc is placed gently, so the only horizontal forces are the friction forces between the two discs. These forces are internal to the system. There is no external torque about the axis, so angular momentum is conserved.
2. Moments of inertia. For a disc about its central perpendicular axis, I = 1/2MR². The second disc adds 1/2 × M/2R² = 1/4MR². The total becomes 1/2MR² + 1/4MR² = 3/4MR².

   1/2MR² ω = 3/4MR² ω' ⇒ ω' = 2/3 ω

3. So. The new angular velocity is 2/3ω, which is option (d).

The idea to keep. When something is dropped gently onto a spinning body, use I₁ω₁ = I₂ω₂. Adding mass raises I, so ω must fall, never rise.

Answer from the NTA final answer key. Solution by PrepWiser.