Let I1‌and‌I2 be the moments of inertia of two discs have angular speed ω1‌and‌ω2. When they are brought in contact, the M.I. of the two discs system will be I1+I2.
Let ω = angular speed of the combined system.
(a) ∴ total initial angular momentum of the two discs,
L1=I1ω1+I2ω2
Total final angular momentum of the combined system,
L2=(I1+I2)ω
From the law of conservation of angular momentum,
L2=L1‌or‌ (I1+I2)‌ω=I1ω1+I2ω2
or ω=

I1ω1+I2ω2

I1+I2

...(i)
(b) Initial K.E. of the two disc,
K1=

1

2

I1ω12+

1

2

I2ω22...(ii)
Final K.E. of the combined system,
K2=

1

2

(I1+I2)ω2...(iii)
∴ From (i) and (iii), we get
K2=

1

2

(I1+I2)(

I1ω1+I2ω2

I1+I2

)2
K2=

1

2

(I1ω1+I2ω2)2

I1+I22

...(iv)
Eqn. (ii) – eqn. (iv) gives,
=

I1I2

2(I1+I2 )

(ω12+ω22−2ω1ω2)
=

I1I2

2(I1+I2)

(ω1−ω2 )2
which is a positive quantity i.e. >0
Hence, K1–K2>0orK1>K2
or K2<K1 i.e. rotational K.E. of the combined system is less than the sum of the initial energies of the two discs.
Hence there occurs a loss of K.E. on combining the two discs and is the dissipation of energy because of the frictional forces between the faces of the two discs. These forces bring about a common angular speed of the two discs on combining. This however is an internal loss and angular momentum remains conserved.