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 figure. The disc can rotate freely about a 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 the point A at its highest position. It is then allowed to fall, so that it starts rotation about PQ. Find the linear speed of the particle as it reaches its lowest position.

Answer 1

Initial and final positions are shown below. Decrease in potential energy of mass $=mg\{2\times \frac{5R}{4}\}=\frac{5mgR}{2}$ Decrease in potential energy of disc $-mg\{2\times \frac{R}{4}\}=\frac{mgR}{2}$ Therefore, total decrease in potential energy of system $=\frac{5mgR}{2}+\frac{mgR}{2}=3mgR$ Gain in kinetic energy of system $=\frac{1}{2}l{\omega }^2$ where I = moment of inertia of system (disc + mass) about axis PQ = moment of inertia of disc + moment of inertia of mass $=\{\frac{m{R}^2}{4}+m(\frac{R}{4}{)}^2$ $I=\frac{15m{R}^2}{8}$ From conservation of mechanical energy, Decrease in potential energy = Gain in kinetic energy Therefore, linear speed of particle at its lowest point $v=\left(\frac{5R}{4}\right)\omega =\frac{5R}{4}\sqrt{\frac{16g}{5R}}$ $v=\sqrt{5gR}$