Is there an alternative to solving the equation for a bead on a hoop with friction? 
Given a circular horizontal loop of radius R and coefficient of friction $\mu$, and a bead which slides along the loop with some velocity, determine the initial velocity $v_o$, that should be given to the bead so that it makes exactly one full turn before stopping. 

I believe the set up looks something like this:

I will work in cylindrical polar coordinates (r,$\phi$,z). The normal force N can be written as:
$$N(\phi) \hat{r} +N_z \hat{z}$$
I arrive at the following set of equations:
\begin{align}
F_r &= -m r \dot{\phi } (t)^2 = -N _r (\phi) \\
F_\phi &= m r \ddot{\phi } (t) = -f = -\mu \lvert N\rvert \\
F_z &= ma_z = N_z-mg
\end{align}
Immediately it can be deduced that $a_z=0$ so $N_z=mg$.
I use the first equation to express $\lvert N \rvert$ as follows:
$$\lvert N \rvert = \sqrt{\bigl(mr\dot{\phi}(t)^2\bigr)^2+(mg)^2}=m\sqrt{r^2\dot{\phi}(t)^4+g^2}$$
I sub $\lvert N\rvert$ into the second equation and arrive at:
$$r\ddot{\phi}(t) = -\mu \sqrt{r^2\dot{\phi}(t)^4+g^2}$$
I'm not sure how to work with this differential equation. In an ideal world I would have the initial conditions be $\phi (0) = 0$ and $\dot{\phi}(0)=\frac{v_o}{R}$ and there would be an analytic solution. I would then chose $v_o$ so that the time at which $\phi (t) = 2\pi$ is the time at which $\dot{\phi}(t)=0$ which ensures the single turn condition. Is there some alternative procedure that would be less complicated?
 A: Maybe this is one way to go, Assuming one side of bead touching the loop as centrifugal force pushes it outwards.
Viewing from frame of the table, the Normal force on bead would be equal to its centripetal force;  

$ N = \frac{mv^2}{r}$

Also velocity at any moment can be found out by integrating   

$ \frac{dv}{dt} = \frac{\mu v^2}{r}$

Find work done by frictional force over one loop using :

$ W = \int N dx$

Put the value for velocity at any moment in equation for normal and value of normal in equation of work, and integrate from $0$ to $2\pi r$. Since no work is being done against gravity or other forces they need not be included. Equate this work done with initially provided kinetic energy

$ W = \frac{m {v_0}^2}{2} $

And solve for $v_0$.
A: In fact,the the given ODE can be easily solved even without computer assistance.
$$r\ddot{\phi}(t) = -\mu \sqrt{r^2\dot{\phi}(t)^4+g^2}$$
This can be converted to,
$$r\dot{\phi}(t)\frac{d\dot{\phi(t)}}{d\phi} = -\mu \sqrt{r^2\dot{\phi}(t)^4+g^2}$$
Separating variables we have,
$$\frac{\dot{\phi}(t)d\dot{\phi}(t)}{\sqrt{\dot{\phi}(t)^4+\frac{g^2}{r^2}}}=-\mu d\phi$$
Now let $u=\dot{\phi}(t)^2$,therefore $du=2\dot{\phi}(t)d\dot{\phi}(t)$.Using this substitution in our ODE in have,
$$ \frac{du}{2\sqrt{u^2+\frac{g^2}{r^2}}}=-\mu d\phi$$
This can be easily integrated using $\int \frac{1}{\sqrt{x^2+a^2}}=\ln(x+\sqrt{x^2+a^2})+c$ with the given limits.
