A disc is rotating with ω =10rad/s about a fixed central axis which is perpendicular to its plane. The disc has a mass =2kg & radius =10cm. A small particle of mass 100gm is put slowly on the disc’s outer circumference. There is sufficient friction between the disc and the particle. What will be the new angular velocity of the system?

Correct Answer: 9.09 rad/s
The friction acting between the two surfaces will ensure that the particle stays on the disc and doesn’t fly out. The angular momentum of the system will be conserved because there is no external torque. Moment of inertia of disc = MR2/2 = 2*0.01/2 = 0.01kgm2. Moment of inertia with particle on disc = 0.01 + mR2 = 0.01+0.1*0.01 = 0.011kgm2. ∴ I1w1 = I2w2 ∴ 0.01*10 = 0.011*w2 ∴ w2 = 9.09 rad/s.