Two disc of moments of inertia `I_(1)andI_(2)` about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds `omega_(1)andomega_(2)` 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 `omega_(1)neomega_(2)`.

Explanation

Let the common angular velocity of the system is `omega`.
(a) Yes, the law of conservation of angular momentum applies. Because, there is no net external torque on the system of the two discs.
(b) By conservation of angular momentum,
`L_(t)=L_(i)`
`therefore Iomega=I_(1)omega_(1)+I_(2)omega_(2)`
`therefore omega=(I_(1)omega_(1)+I_(2)omega_(2))/(I)=(I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(1))(because I=I_(1)+I_(2))`
(c ) `K_(t)=(1)/(2)I_(1)omega^(2)+(1)/(2)I_(2)omega^(2)`
`=(1)/(2)(I_(1)+I_(2))((I_(1)omega_(1)+I_(2)omega_(2))^(2))/((I_(1)+I_(2))^(2))=(1)/(2)((I_(1)omega_(1)+I_(2)omega_(2))^(2))/((I_(1)+I_(2))^(2))`
`K_(i)=(1)/(2)(I_(1)omega_(1)^(2)+I_(2)omega_(2)^(2))`
`DeltaK=K_(t)-K_(i)=-(I_(1)I_(2))/(2(I_(1)+I_(2)))(omega_(1)-omega_(2))^(2)lt0`
(d) Hence, there is loss in KE of the system. The loss in kinetic energy is mainly due to the work against the friction between the two discs.