A body of mass $$10\,kg$$  is attached to a wire of radius $$3\,cm.$$  It’s breaking stress is $$4.8 \times {10^7}\,N{m^{ - 2}},$$    the area of cross-section of the wire is $${10^{ - 6}}{m^2}.$$   What is the maximum angular velocity with which it can be rotated in the horizontal circle?

Solution :
Given that $$\frac{F}{A} = 4.8 \times {10^7}N{m^{ - 2}}$$
$$\eqalign{ & \therefore F = 4.8 \times {10^7} \times A \cr & {\text{or}}\,\,\frac{{m{v^2}}}{r} = 4.8 \times {10^7} \times {10^{ - 6}} = 48 \cr & {\text{or}}\,\frac{{m{r^2}{\omega ^2}}}{r} = 48\,\,{\text{or}}\,\,{\omega ^2} = \frac{{48}}{{mr}} \cr & \omega = \sqrt {\left( {\frac{{48}}{{10 \times 0.3}}} \right)} = \sqrt {16} = 4\,rad/\sec \cr} $$