According to given problem \(\frac{1}{2} m v^{2} - \omega s^{2}\) \(\rightarrow v \quad s \sqrt{\frac{2 \pi}{m}}\)

So \(0_r = -\frac{v^2}{R} - \frac{2 \omega s^2}{r R}\) …(i)

Further more as \(\alpha_{t} - \frac{d v}{d t} - \frac{d v}{d s} \cdot \frac{d s}{d t} - v \frac{d r}{d s}\) …(ii)

(By chain rule)

Which in light of equation (i) i.e. \(\nu_s \sqrt{\frac{2a}{m}}\) yields

\(a_r = \left| \begin{matrix} s & i2a \\ l & m \end{matrix} \right| \left| \begin{matrix} i2a \\ l & m \end{matrix} \right| = \frac{2as}{m}\) … (iii)

So that \(a = \sqrt{a_{R}^{2} + a_{t}^{2}} = \sqrt{\left|\frac{2as^{2} \cdot t}{rr\beta}\right| + \left|\frac{2as}{m}\right|^{2}}\)

Hence \(a = \frac{2 \alpha s}{r} \sqrt{1 + \left[\frac{s}{R}\right]^2}\)

\ \ \(F = ma = 2as \sqrt{1 + (s/R)^2}\)