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 starts rotating about PQ . Find the linear speed of the particle as it reaches its lowest position.

Question

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 starts rotating about PQ . Find the linear speed of the particle as it reaches its lowest position., - 2gR, 3gR, - 4gR, 5gR

MARKS App · Accepted Answer

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 'conservation of mechanical energy', 1 2 I ω 2 = mg AA ′ + mgCC ′ = mg 2AD + 2CD i.e., 1 2 × 15 8 mR 2 ω 2 = 2 mg R + 1 4 R + 1 4 R , i.e., ω = 4 g 5R so v A ′ = r ω = R + 1 4 R × 4 g 5R = 5gR

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