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.

Explanation

Here, total initial angualr momentum of the two discs `L_(1) = I_(1) omega_(1) + I_(2) omega_(2)`
Under the given conditions, moment of inertia of the two disc system `= (I_(1) + I_(2))`
If `omega` is angular speed of the combined system, the final angular momentum of the system `L_(2) = (I_(1) + I_(2)) omega`
As no external torque is involved in this exercise, therefore, `L_(2) = L_(1)`
`(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))` ..(i)
(b) Initial K.E. of two discs `E_(1) = (1)/(2)I_(1)omega_(1)^(2) + (1)/(2)I_(2)omega_(2)^(2)`
Final K.E. of the system `E_(2) = (1)/(2)(I_(1) + I_(2)) omega^(2)`
using (i), `E_(2) = (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)))`
Now, `E_(1) - E_(2) = (1)/(2)I_(1)omega_(1)^(2) + (1)/(2)I_(2)omega_(2)^(2) - ((I_(1)omega_(1) + I_(2)omega_(2))^(2))/(2(I_(1) + I_(2)))` which on simplification, gives us
`E_(1) - E_(2) = (I_(1)I_(2)(omega_(1) - omega_(2))^(2))/(2(I_(1) + I_(2)))`,
which is a positive quantity (i.e `gt 0`), involving a perfect square.
`:. E_(1) - E_(2) gt 0 or E_(1) gt E_(2) or E_(2) lt E_(1)`
Hence there occurs a loss of K.E. in the process. Loss of `K.E. = E_(1) - E_(2)`. This loss must be due to friction in the contact of the two discs.
Note that angular momentum is conserved as torque due to friction is only an internal torque.