A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0 is v₀, the time taken to complete the first revolution is

Given that  ${\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{t}}$

$∴{\mathrm{Rω}}^2=\mathrm{RÎ\pm }\text{ or }{\mathrm{ω}}^2=\mathrm{Î\pm }=\frac{\mathrm{dω}}{\mathrm{dt}}\text{ or }\frac{\mathrm{dω}}{{\mathrm{ω}}^2}=\mathrm{dt}$

Integrating this expression, we get

$\begin{array}{l}{∫}_{{\mathrm{ω}}_0}^{\mathrm{ω}} \frac{\mathrm{dω}}{{\mathrm{ω}}^2}={∫}_0^{\mathrm{t}} \mathrm{dt}\text{ or }\mathrm{ω}=\frac{{\mathrm{ω}}_0}{\left(1−{\mathrm{ω}}_0\mathrm{t}\right)}\\ ∴\frac{\mathrm{dθ}}{\mathrm{dt}}=\frac{{\mathrm{ω}}_0}{\left(1−{\mathrm{ω}}_0\mathrm{t}\right)}\text{ or }\mathrm{dθ}=\frac{{\mathrm{ω}}_0\mathrm{dt}}{\left(1−{\mathrm{ω}}_0\mathrm{t}\right)}\end{array}$

Further integrating, we get

${∫}_0^{2\mathrm{π}} \mathrm{dθ}={∫}_0^{\mathrm{T}} \frac{{\mathrm{ω}}_0\mathrm{dt}}{\left(1−{\mathrm{ω}}_0\mathrm{t}\right)}$

Solving, we get $\left(1−{\mathrm{ω}}_0\mathrm{T}\right)={\mathrm{e}}^{−2\mathrm{π}}$

$\mathrm{or} \mathrm{T}=\frac{1}{{\mathrm{ω}}_0}\left(1−{\mathrm{e}}^{−2\mathrm{π}}\right)=\frac{\mathrm{R}}{{\mathrm{v}}_0}\left(1−{\mathrm{e}}^{−2\mathrm{π}}\right)$