Mass of a disc is non-uniformly distributed over the area of a thin disc of radius R and as a result the distance of the centre of mass of the disc from its centre O is found to be R/4. Initially the disc is at rest. Now it is made to roll without slipping with an initial velocity of its centre ${\mathrm{V}}_0.$ What is the maximum value of ${\mathrm{V}}_0$ for which the disc will not loose contact with the surface ?

Initially the centre of mass of the disc is at the lower most position C₁. The disc may loose contact with the surface when the centre of mass of mass is at the highest position C₂. Angular velocity of the disc in Figure (2) is $\mathrm{\omega }=\mathrm{v}/\mathrm{R}.$ Velocity of C₂ relative to O is $\mathrm{\omega }.\frac{\mathrm{R}}{4}=\frac{\mathrm{V}}{4}$. Vertical component of acceleration of C₂ relative to O is $\frac{{\left(\mathrm{V}/4\right)}^2}{\mathrm{R}/4}=\frac{{\mathrm{V}}^2}{4\mathrm{R}}$  (directed towards O).

Now,   $\mathrm{mg}-\mathrm{N}=\mathrm{m}\left(\frac{{\mathrm{V}}^2}{4\mathrm{R}}\right)\Rightarrow \mathrm{N}=\left(\mathrm{mg}-\frac{{\mathrm{mV}}^2}{4\mathrm{R}}\right).$

The disc will not loose contact with the surface if

$\mathrm{N}>0\Rightarrow \mathrm{mg}-\frac{{\mathrm{mV}}^2}{4\mathrm{R}}>0\Rightarrow {\mathrm{V}}^2<4\mathrm{gR}.....\left(1\right)$

Conserving energy, $\frac{1}{2}{\mathrm{mV}}^2+\mathrm{m}.\mathrm{g}.\left(2\times \frac{\mathrm{R}}{4}\right)=\frac{1}{2}{\mathrm{mV}}_0^2$

$\Rightarrow {\mathrm{V}}^2={\mathrm{V}}_0^2-\mathrm{gR} ....\left(2\right)$

From (1) and (2), ${\mathrm{V}}_0^2-\mathrm{gR}<4\mathrm{gR}\Rightarrow {\mathrm{V}}_0<\sqrt{5\mathrm{gR}}$

$\therefore {\left({\mathrm{V}}_0\right)}_{\max } = \sqrt{5\mathrm{gR}}$