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 speed `omega_(1) and omega_(2)` are brought into contact face to face with their axes of rotation coincident . What is the angular speed of the two-disc system ?

Explanation

Let `I_(1) and I_(2)` be moment of inertia of discs and let the common angular speed be `omega`
By angular momentum conservation, Initial angular momentum of two discs = final angular momentum of combined system
`I_(1) omega_(1)+I_(2) omega_(2)=(I_(1)+I_(2)) omega`
`:. omega=(I_(1) omega_(1)+I_(2) omega_(2))/(I_(1)+I_(2))`
(b) Rotational K.E. before contact
`=(1)/(2)I_(1) omega_(1)^(2)+(1)/(2)I_(2) omega_(2)^(2)`
Rotational K.E, after contact `=(1)/(2)(I_(1)+I_(2)) omega^(2)`
`=(1)/(2)(I_(1)+I_(2))[(I_(1) omega_(1)+I_(2) omega_(2))/(I_(1)+I_(2))]^(2)`
Loss of energy
`E_(1)-E_(2)=(1)/(2)I_(1)omega_(1)^(2)+(1)/(2)I_(2) omega_(2)^(2)-((1)/(2)(I_(1) omega_(1)+I_(2))^(2))/(I_(1)+I_(2))`
On solving we get
`E_(1)=E_(2)=(I_(1)I_(2)(omega_(1)-omega_(2))^(2))/(2(I_(1)+I_(2)))` (=positive quantity)
`:.` There is loss in kinetic energy in this process. Because some work is done against friction.