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

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

Further more as \(\alpha_{1} - \frac{dv}{dt} - \frac{dv}{ds} \cdot \frac{ds}{dt} - v \frac{dv}{dt}\) …(ii)

(By chain rule)

Which in light of equation (i) i.e. \(\nu \quad s_{1/2} \sqrt{\frac{2a}{m}}\) 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 \(a = \sqrt{a_{R}^{2} + a_{t}^{2}} = \sqrt{\left|\frac{2as^{2}}{r \times R}\right|^{2} + \left|\frac{2as}{m}\right|^{2}}\)

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

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