(a) Let I₁  and I₂ be the moment of inertia of the two turntables respectively.

Let  ω₁  and ω₂ be the angular speed of the two turntables respectively.

So, we can say,

Angular momentum of turntable 1, L₁ = I₁ω₁

Angular momentum of turntable 2, L₂ = I₂ω₂

So, the total initial angular momentum will be,

L_i = I₁ω₁ + I₂ω₂

Considering when the two turntables are combined together:

Moment of inertia of the two turntable systems, I = I₁ + I₂

Let the angular speed of the system be ω

So, the final angular momentum will be,

L_T = (I₁ + I₂) ω

According to the principle of conservation of angular momentum, we can write,

L_i = L_T

I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω

As a result,

ω = (I₁ω₁ + I₂ω₂) / (I₁ + I₂) . . . . . . . . . .  ( 1 )

(b) Kinetic energy of turntable 1 has the expression,

K.E₁ = ( 1/2 ) I₁ω₁²

Kinetic energy of turntable 2 has the expression,

K.E₁ = ( 1/2 ) I₂ω₂²

The total initial kinetic energy of the turntables will be,

KE_I = (1/2) ( I₁ω₁² + I₂ω₂²)

When combined together, their moments of inertia add up.

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

The angular speed of the system is ω

Final kinetic energy can be calculated as,

KE_F :  =  (1/2) ( I₁ + I₂) ω²Using the value of ω from  (1)

=  (1/2) ( I₁ + I₂) [ (I₁ω₁ + I₂ω₂) / (I₁ + I₂) ]²

KE_F = (1/2) (I₁ω₁ + I₂ω₂)² / (I₁ + I₂)

Now, KE_I – KE_F

= ( I₁ω₁² + I₂ω₂²) (1/2) – [ (1/2) (I₁ω₁ + I₂ω₂)² / (I₁ + I₂) ]

On solving the above equation, we get :

= I₁ I₂ (ω₁ – ω₂)² / 2(I₁ + I₂)

As (ω₁ – ω₂)²  will only yield a positive quantity and I₁  and I₂  are both positive, the RHS will be positive.

This means KE_I – KE_F > 0

Or, KE_I > KE_F

When the two turntables came into contact, some of the kinetic energy was lost as a result of the friction forces.