Two discs of moments of inertia l_(1) and I_(2) about their respective axes (normal to the disc and passing through the centre), and rotating with angular speedomega_(1) and omega_(2) are brought intocontact face to face with their axes of rotation coincident, (i) What is the angular speed the two-disc system? (i) 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)ne omega_(2).

Two discs of moments of inertia l_(1) and I_(2) about their respective

1.	Two discs of moments of inertia l_(1) and I_(2) about their respective axes (normal to the disc and passing through the centre), and rotating with angular speedomega_(1) and omega_(2) are brought intocontact face to face with their axes of rotation coincident, (i) What is the angular speed the two-disc system? (i) 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)ne omega_(2).
Answer» Solution :Let `omega` be the angular speed of the TWO-dise system. Then by conservation of angular momentum `(I_(1)+I_(2)) omega=I_(1)+I_(2) omega_(2) or omega=(I_(1) omega_(1)+I_(2) omega_(2))/(I_(1)I_(2))` (ii) INITIAL K.E. of the two dises `K_(1)+(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 in `K.E. =K_(1)-K_(2)=(1)/(2)(I_(1)omega_(1)^(2)+1_(2) omega_(2)^(2))-(1)/(2(I_(1)+I_(2)))(1_(1)omega_(3)+I_(2)omega_(2))^(2)` `=(I_(1)I_(2))/(2(I_(1)+I_(2))) (omega_(1)-omega_(2))^(2)=a` positive quantity `[ :. omega_(1) ne omega_(2)]` Hence there is a loss of rotational K.E. which appears as heat. When the two discs are BROUGHT together work is done against friction between the two discs