A uniform circular disc has radius R and mass m . A particle, also of mass m , is fixed at a point A on the edge of the disc as shown in the diagram. 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 st

Step-by-step solution

As moment of inertia of a disc about a diameter is 1 2 1 2 mR 2 , the moment of inertia of the disc about the chord PQ by 'theorem of parallel axes' will be I D PQ = 1 4 mR 2 + m 1 4 R 2 = 5 16 mR 2 and as particle of mass m is at a distance [R + (R/4) = (5/4)R] from PQ, the moment of inertia of the system about PQ I = I D PQ + I P PQ = 5 16 mR 2 + m 5 4 R 2 = 1 5 8 mR 2 Now if ω is the angular speed of the system when A reaches the lowest point A' on rotation about the axis PQ, by 'conservati