Explanation:
Given, diameter of circle, \(d = 50\,cm\)
\( = 50 \times {10^{ - 2}}m\)
and frequency , \(f = 2Hz\)
The acceleration of particle in a uniform circular motion can be given as
\(a = {\omega ^2}x\)
where, \(\omega = \) angular frequency \( = 2\pi f\), and \(x = \) distance from centre \( = \frac{d}{2}\).
\( \Rightarrow a = 4{\pi ^2}{f^2} \times \frac{d}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
\( \Rightarrow a = 4{\pi ^2} \times 4 \times \frac{{50 \times {{10}^{ - 2}}}}{2} = 4{\pi ^2}\)