System of particles and Rotational motion NCERT Solutions Class 11 Physics - Solved Exercise Question 7.25

Question 7.25:

Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω₁ and ω₂ are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω₁ ≠ ω₂.

Solution:
(a) Moment of inertia of disc I = I₁

Angular speed of disc I = ω₁

Moment of inertia of disc I = I₂
Angular speed of disc I = ω₂
Angular momentum of disc I, L₁ = I₁ω₁
Angular momentum of disc II, 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.
Total final angular momentum, L_T = (I₁ + I₂) ω
Using the law of conservation of angular momentum, we have:
L_i = L_T
I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω
∴ ω = (I₁ω₁ + I₂ω₂) / (I₁ +I₂)

(b) Kinetic energy of disc I, E₁ = (1/2) I₁ω₁²
Kinetic energy of disc I, E₁ = (1/2) I₂ω₂²
Total initial kinetic energy, E_i = (1/2) ( I₁ω₁² + I₂ω₂²)
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
Final kinetic energy E_f:  =  (1/2) ( I₁ + I₂) ω²
=  (1/2) ( I₁ + I₂) [ (I₁ω₁ + I₂ω₂) / (I₁ +I₂) ]²
= (1/2) (I₁ω₁ + I₂ω₂)² / (I₁ +I₂)
∴ E_i - E_f
= (1/2) ( I₁ω₁² + I₂ω₂²) - (1/2) (I₁ω₁ + I₂ω₂)² / (I₁ +I₂)
Solving the equation, we get
= I₁ I₂ (ω₁ - ω₂)² / 2(I₁ + I₂)
All the quantities on RHS are positive
∴ E_i - E_f > 0
E_i > E_f
The loss of KE can be attributed to the frictional force that comes into play when the two discs come in contact with each other.