A bead on a circle

We will measure the distance $x$ along the circular path of the bead. The normal force is the centripetal force, $mv^2/R$ , and the equation of motion along the circular path is

$m\frac{dv}{dt}= -\mu m \frac{v^2}{R}$

We will use $\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$ , and we get

$\frac{dv}{dx}= -v \frac{\mu}{R}$

The solution of this differential equation is $v=v_0 e^{-\mu x/R}$ , satisfying the initial condition of $v=v_0$ at $x=0$ . After the first circle is completed we have $x=2\pi R$ and we get

$\frac {v'}{v_0}= e^{-2\pi\mu}=0.2846$