$\text{ For circular motion, the radial acceleration, }{a}_r\text{ is given by }\frac{v^2}{R}\text{, and the tangential }$

$\text{ acceleration, }{a}_t\text{ is given by }\frac{dv}{dt}$

$\text{ Since these accelerations are equal, so }\frac{v^2}{R}=\frac{dv}{dt}$

$\text{ Hence, }\left(\frac{\mathrm{R}}{{\mathrm{v}}^2}\right)\mathrm{dv}=\mathrm{dt}$

$\int \frac{\mathrm{R}}{{\mathrm{v}}^2}\mathrm{dv}=\int \mathrm{dt}$

Upon integration

$\mathrm{t}=\mathrm{R}\left[\frac{\mathrm{v}-{\mathrm{v}}_0}{{\mathrm{v}}_0\cdot \mathrm{v}}\right]$

$\text{ Now, }v=\frac{ds}{dt}\text{, }$

where, s is the length of the arc covered by the particle as it moves in the circle

$\text{ Therefore, }t=\frac{R}{v_0}-\frac{R}{\frac{d}{dt}}$

$t=\frac{R}{v_0}-R\cdot \frac{dt}{ds}$

$\mathrm{tds}=\frac{\mathrm{R}}{{\mathrm{v}}_0}-\mathrm{Rdt}$

$\mathrm{ds}\left(\frac{\mathrm{R}}{{\mathrm{v}}_0}-\mathrm{t}\right)=\mathrm{Rdt}$

Upon integrating, and putting all initial condition

$\frac{\mathrm{s}}{\mathrm{R}}=\frac{\ln R}{{\mathrm{v}}_0}-\ln \left(\frac{\mathrm{R}}{{\mathrm{v}}_0}-\mathrm{t}\right)=\ln \left(\frac{\mathrm{R}}{\mathrm{R}-{\mathrm{v}}_0\mathrm{t}}\right)$

For complete revolution, putting s=2πR, t=T

$\frac{2\pi R}{R}=\ln \left(\frac{R}{R-v_{0}T}\right)$

$\mathrm{T}=\frac{\mathrm{R}}{{\mathrm{V}}_0}\left(1-{\mathrm{e}}^{-2\pi }\right)$