The two uniform discs rotate separately on parallel axles. The upper disc (radius a and momentum of inertia ${\mathrm{I}}_1$) is given an angular velocity ${\mathrm{\omega }}_0$ and the lower disc of (radius b and momentum of inertia ${\mathrm{I}}_2$) is at rest. Now the two discs are moved together so that their rims touch. Final angular velocity of the upper disc is

detailed solution

Correct option is A

The two discs exert equal and opposite forces on each other when in contact. The torque due to these forces changes the angular momentum of each disc. From angular impulse-angular momentum theorem, we havefa∆t = I1(ω0-ω1)----------(i)and fb∆t = I2ω2---------(ii)From eqns. (i) and (ii), we get ab = I1(ω0-ω1)I2ω2------(iii)When slipping ceases between the discs, the contact points of the two discs have the same linear velocity,i.e.,aω1 = bω2-----(iv)On substituting ω2 in eqn. (iii) we getω1 = (I1ω0)[I1+(a2I2b2)]