A circular disc of mass M and radius R is rotating about its axis with angular speed ω ₁ . If another stationary disc having radius R/2 and same mass M is droped co-axially on to the rotating disc. Gradually both discs attain constant angular speed ω ₂ the energy lost in the process is p% of the initial energy. Value of p is __________.

Solution

Physics sir · 4 steps

No outside torque acts, so angular momentum is shared; the friction between the discs eats the kinetic energy.

1. Moments of inertia. For a disc about its axis, I = 1/2mr². The first disc has I₁ = 1/2MR². The second has I₂ = 1/2M(R/2)² = 1/8MR², so together I₁ + I₂ = 5/8MR².
2. Conserve angular momentum. Friction between the discs is internal, so I₁ω₁ = (I₁ + I₂)ω₂. This gives ω₂ = (1/2)/(5/8) ω₁ = 4/5ω₁.

   K_f/Kᵢ = (1/2(I₁+I₂)ω₂²)/(1/2I₁ω₁²) = I₁/(I₁+I₂) = (1/2)/(5/8) = 4/5

3. Energy lost. The final energy is 4/5 of the initial energy. The fraction lost is 1 - 4/5 = 1/5, which is 20%.
4. So. p = 20.

The common slip: 80. Reporting the fraction of energy that remains instead of the fraction lost gives 80, not 20.

The idea to keep. When a stationary body joins a spinning one with no outside torque, K_f/Kᵢ = I₁/(I₁+I₂). The fraction lost is I₂/(I₁+I₂).

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