Question

Two discs of moments of inertia `I_(1) and I_(2)` about their respective axes (normal to the disc and passing through the centre) and rotating with angular speeds `omega_(1) and omega_(2)` are brought into contact face to face with their axes of rotation coincident.
(a) Does the law of conservation of angular momentum apply to the situation ? Why ?
(b) Find the angular speed of the two-disc system.
(c ) Calculate the loss in kinetic energy of the system in the process.
(d) Account for this loss.

Answer 1

(a) Yes, the law of conservation of angular applies to the situation. This is because no external torque in involved in bringing the two discs into contact face to face. External forces, gravitation and normal reaction, act through the axis of rotation producing no torque.
(b) If `omega` is angular speed of the two disc system, then from the principle of conservation of angular momentum,
`(I_(1) + I_(2)) omega = I_(1) omega_(1) + I_(2)omega_(2)`
`omega = (I_(1)omega_(1) + I_(2)omega_(2))/(I_(1) + I_(2))`
(c) Initial `KE` of two discs, `E_(i) = (1)/(2)I_(1)omega_(1)^(2) + (1)/(2)I_(2)omega_(2)^(2)`
Final `KE` of the system, `E_(f) = (1)/(2)(I_(1) + I_(2)) omega_(2)`
Using (i), `E_(f) = (1)/(2)(I_(1) + I_(2)) ((I_(1)omega_(1) + I_(2)omega_(2))^(2))/((I_(1) + I_(2))^(2)) = ((I_(1)omega_(1) + I_(2)omega_(2))^(2))/(2(I_(1) + I_(2)))`
`DeltaE = E_(f) - E_(i) = ((I_(1)omega_(1) + I_(2)omega_(2))^(2))/(2(I_(1) + I_(2))) - (I_(1)omega_(1)^(2) + I_(2)omega_(2)^(2))/(2)`
`=(I_(1)^(2)omega_(1)^(2) + I_(2)^(2)omega_(2)^(2) + 2I_(1)omega_(2) - I_(1)^(2)omega_(1)^(2) - I_(1)I_(2)omega_(2)^(2) - I_(2)I_(1)omega_(1)^(2) - I_(2)^(2)omega_(2)^(2))/(2(I_(1) + I_(2)))`
`= (-I_(1)I_(2))/(2(I_(1) + I_(2))) (omega_(2) - omega_(2))^(2)` which is a negative quantity. Hence, some `KE` is always lost.
(d) The loss in `KE` is due to work done against friction between the two discs.