Question

A uniform disk of mass m is released from rest from the rim of a fixed hemispherical bowl so that it rolls along the surface. If the rim of the hemispherical is kept horizontal, find the normal force exerted by the bowl on the disk when it reaches the bottom of the bowl.

Answer 1

Total normal force at bottom $=mg+\frac{m{v}^2}{R-r}$

The potential energy at the initial position gets converted to kinetic energy and rotational energy.

$\therefore mg(R-r)=\frac{1}{2}I{w}^2+\frac{1}{2}m{v}^2\Rightarrow mg(R-r)=\frac{1}{2}\times \frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2\Rightarrow \frac{3}{4}m{v}^2=mg(R-r)\Rightarrow v^2=\frac{4}{3}g(R-r)$

$\therefore$ Total normal force

$=mg+\frac{m\left(\frac{4}{3}\right)g(R-r)}{(R-r)}=mg+\frac{4}{3}mg=\frac{7}{3}mg$