Question

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 ω ₁ 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 1

Given, the moment of inertia of two discs are I 1 and I 2 respectively. The angular speed of two discs are ω 1 and ω 2 respectively.

a)

Let L 1 and L 2 be the angular momentum of the first and second disc respectively.

The angular momentum of first disc is given as,

L 1 = I 1 ω 1

The angular momentum of first disc is given as,

L 2 = I 2 ω 2

The total initial angular momentum is given as,

L i = L 1 + L 2 = I 1 ω 1 + I 2 ω 2 (1)

The moment of inertia of the two disc system is given as,

I= I 1 + I 2

The final angular momentum of two disc system is given as,

L f =Iω =( I 1 + I 2 )ω (2)

Where, ω is the angular speed of the two disc system.

From the principle of conservation of angular momentum,

L i = L f

By substituting the values from equation (1) and (2) in the above equation, we get

I 1 ω 1 + I 2 ω 2 =( I 1 + I 2 )ω ω= I 1 ω 1 + I 2 ω 2 ( I 1 + I 2 )

Thus, the angular speed of two disc system is I 1 ω 1 + I 2 ω 2 ( I 1 + I 2 ) .

b)

Let E 1 and E 2 be the kinetic energy of the first and second disc respectively.

The kinetic energy of first disc is given as,

E 1 = 1 2 I 1 ω 1 2

The kinetic energy of second disc is given as,

E 2 = 1 2 I 2 ω 2 2

The total initial kinetic energy is given as,

E i = 1 2 I 1 ω 1 2 + 1 2 I 1 ω 1 2

The final kinetic energy is given as,

E f = 1 2 I ω 2 = 1 2 ( I 1 + I 2 ) ω 2

By substituting the value of ω in the above equation, we get

E f = 1 2 ( I 1 + I 2 ) [ I 1 ω 1 + I 2 ω 2 ( I 1 + I 2 ) ] 2 = 1 2 ( I 1 ω 1 + I 2 ω 2 ) 2 I 1 + I 2

The difference between the initial and final kinetic energy is given as,

E i − E f = 1 2 ( I 1 + I 2 ) ω 2 − 1 2 ( I 1 ω 1 + I 2 ω 2 ) 2 I 1 + I 2 = 1 2 I 1 ω 1 2 + 1 2 I 2 ω 2 2 − 1 2 I 1 2 ω 1 2 ( I 1 + I 2 ) − 1 2 I 2 2 ω 2 2 ( I 1 + I 2 ) − 1 2 2 I 1 I 2 ω 1 ω 2 ( I 1 + I 2 ) = 1 ( I 1 + I 2 ) [ 1 2 I 1 ω 1 2 + 1 2 I 1 I 2 ω 1 2 + 1 2 I 1 I 2 ω 2 2 + 1 2 I 2 ω 2 2 − 1 2 I 1 ω 1 2 − 1 2 I 2 ω 2 2 1 2 I 1 2 ω 1 2 ( I 1 + I 2 ) − I 1 I 2 ω 1 ω 2 ] = I 1 I 2 2( I 1 + I 2 ) [ ω 1 2 + ω 2 2 −2 ω 1 ω 2 ]

Further solve the above expression.

E i − E f = I 1 I 2 ( ω 1 − ω 2 ) 2 2( I 1 + I 2 )

In the above equation, all the quantities on the right side are positive therefore,

E i − E f >0 E i > E f

This loss of kinetic energy is due to the frictional force produces when two disc come in contact.