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 at 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 rotating about PQ. Find the linear speed of the particle (in m/s) as it reaches the lowest position. (R=0.5m and g=10m/${\mathrm{s}}^2$)

Applying the theorem of parallel axis, the moment of inertia of the disc about PQ is

${\left({\mathrm{I}}_{\mathrm{disc}}\right)}_{\mathrm{PQ}} = \left[\frac{{\mathrm{mR}}^2}{4} + \mathrm{m}{\left(\frac{\mathrm{R}}{4}\right)}^2\right] = \frac{5}{16}{\mathrm{mR}}^2 \left(1\right)$

The moment of inertia of mass m about PQ is

${\left({\mathrm{I}}_{\mathrm{particle}}\right)}_{\mathrm{PQ}} = \mathrm{m}{\left(\mathrm{R}+ \frac{\mathrm{R}}{4}\right)}^2 = {\left(\frac{5\mathrm{R}}{4}\right)}^2 = \frac{25{\mathrm{mR}}^2}{16} \left(2\right)$

So, the moment of inertia of the system (disc + particle) about PQ is

$\mathrm{I} = \frac{5}{16}{\mathrm{mR}}^2 + \frac{25}{16}{\mathrm{mR}}^2 = \frac{15{\mathrm{mR}}^2}{8} \left(3\right)$

Now, we shall calculate the loss in potential energy when the particle reaches in
lowest position. The lowest position of particle is shown in figure.

Loss in potential energy of the particle is

$\mathrm{mg}\left[\left(\mathrm{R} + \frac{\mathrm{R}}{4}\right) + \left(\mathrm{R}+ \frac{\mathrm{R}}{4}\right)\right] = \frac{5\mathrm{mgR}}{2}$

Total loss in P.E. = 3 mgR .
Now, $\omega$ be the angular velocity of the disc at the lowest position. Thus,

$\frac{1}{2}{\mathrm{I\omega }}^2 = 3\mathrm{mgR} \Rightarrow \mathrm{\omega } = \sqrt{\left(\frac{16\mathrm{g}}{5\mathrm{R}}\right)}$

Hence, the linear speed of the particle is given by

$\mathrm{v} = \left(\mathrm{R} + \frac{\mathrm{R}}{4}\right)\mathrm{\omega } = \frac{5\mathrm{R}}{4}\mathrm{X} \sqrt{\frac{16\mathrm{g}}{5\mathrm{R}}} = 5\mathrm{m}/\mathrm{s}$