A stone tied to the end of a string of $$1\,m$$  long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in $$44\,s,$$  what is the magnitude and direction of acceleration of the stone?

Solution :
Since, speed is constant throughout the motion, so, it is a uniform circular motion. Therefore, its radial acceleration is given by
$$\eqalign{ & {a_r} = r{\omega ^2} \cr & = r{\left( {\frac{{2\pi n}}{t}} \right)^2} = r \times \frac{{4{\pi ^2}{n^2}}}{{{t^2}}} \cr & = \frac{{1 \times 4 \times {\pi ^2} \times {{\left( {22} \right)}^2}}}{{{{\left( {44} \right)}^2}}} \cr & = {\pi ^2}\,m/{s^2} \cr} $$
This acceleration is directed along radius of circle.