Given:

Moment of inertia of disc 1 = I₁

Angular speed of disc 1 = ω₁

∴ Angular momentum of disc 1 , L₁ = I₁ω₁

Angular speed of disc 2 = ω₂

∴ Angular momentum of disc 2, L₂ = I₂ω₂

⇒ Total initial angular momentum, L_i = I₁ω₁ + I₂ω₂

When the two discs are joined together, their moments of inertia get added up.

Moment of inertia of the system of two discs, I = I₁ + I₂

Let ω be the angular speed of the system.

Let total final angular momentum, L_f = (I₁ + I₂) × ω_f

Using the law of conservation of angular momentum, we have:

L_i = L_f

⇒ I₁ω₁ + I₂ω₂ = (I₁ + I₂) × ω_f

⇒ ω_f = (I₁ω₁ + I₂ω₂) / (I₁ + I₂) …(i)

(b) Kinetic energy of disc I, E₁ = 1/2 I₁ω₁²

Kinetic energy of disc II, E₂ = 1/2 I₂ω₂²

⇒ Total initial kinetic energy, E_i = 1/2 I₁ω₁² + 1/2 I₂ω₂² …(ii)

When the discs are joined, their moments of inertia get added up.

Moment of inertia of the system, I = I₁ + I₂

Angular speed of the system = ω_f

Final kinetic energy,

E_f = 1/2 (I₁ + I₂)ω_f²

Substituting the value of ω from equation (i)

⇒ E_f = 1/2 (I₁ + I₂) {(I₁ω₁ + I₂ω₂) / (I₁ + I₂)}² …(iii)

From (ii) and (iii)

E_i- E_f = [1/2 I₁ω₁² + 1/2 I₂ω₂²] - 1/2 (I₁ + I₂) {(I₁ω₁ + I₂ω₂) / (I₁ + I₂)}²

By solving the equation we get,

E_i- E_f = I₁I₂(ω₁-ω₂)²/ 2(I₁ + I₂)

Since all Quantities on RHS are positive so we conclude that,

E_i-E_f> 0

i.e. E_i> E_f

The loss can be attributed to the frictional forces that arise when discs come into contact with each other.