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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-7 System Of Particles And Rotational Motion



Q.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.

 

 

Answer:-

(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 Li = 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, LT = (I1 + I2) ω
Using the law of conservation of angular momentum, we have:
Li = LT
I1ω1 + I2ω2 = (I1 + I2
∴ ω = (I1ω1 + I2ω2) / (I1I2)



(b) Kinetic energy of disc I, E1 = (1/2) I1ω12
Kinetic energy of disc II, E1 = (1/2) I2ω22
Total initial kinetic energy, Ei = (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 Ef:  =  (1/2) ( I1 + I2) ω2
=  (1/2) ( I1 + I2) [ (I1ω1 + I2ω2) / (I1I2) ]2
= (1/2) (I1ω1 + I2ω2)2 / (I1 + I2)
EiEf
= (1/2) ( I1ω12 + I2ω22) – (1/2) (I1ω1 + I2ω2)2 / (I1 + I2)
Solving the equation, we get
= I1 I21 – ω2)2 / 2(I1 + I2)
All the quantities on RHS are positive
EiEf > 0
Ei > Ef
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.