A uniform circular disc has radius R and mass m. A particle, also of mass m, is fixed at point A on the edge of the disc as shown in figure. The disc can rotate freely about a fixed horizontal chord PQ that is at a distance R/4 from the centre C of the disc. The line AC is perpendicular to PQ. Initially, the disc is held vertical with point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. If R = 0.5 m then, find the linear speed of the particle (in m/s) as it reaches its lowest position. (Take g = 10 m/s2)

The system (disc + particle) is rotating about fixed axis without friction, its mechanical energy should be conserved. According to conservation of energy, ΔK+ΔU=0⇒Kf−Ki+ΔU=0        …(i)During the fall, the disc-mass system gains rotational kinetic energy. This is at the expense of potential energy.Change in gravitational potential energy.ΔU=−mg2R+R4+mg2R4=−3mgRChange in kinetic energy ΔK=12IPQω2−0where IPQ=MI of the disc-mass system about PQIPQ=Idisc PQ+Imass PQ(I)PQ=Idisc PQ+Imass PQ=mR24+MR42+m5R42=15mR28ΔK=1215mR28ω2−0Finally substituting ΔU and ΔK in the equation (i),we have1215mR28ω2−0+(−3mgR)=0⇒ω=16g5RLet v be the velocity of mass m at the lowest point of rotation v=ωR+R4=16g5R×5R4=5gR=5×10×0.5=5m/s