algebra topology said:
$$ v\frac{\mathrm{d} v}{\mathrm{d} \theta } + gRcos \theta + \mu gRcos \theta = \mu v^2 $$
It looks very different to what I got. But I'm not an expert on this. I'll share my attempt.
I solved the left half of the problem. The equations that describe the other half should be equivalent. It's just a matter of getting the velocity of the bead at the end of the left loop and using it as the initial condition for the next loop.
Here is the diagram I used.
Using Newton we know that:
$$\sum \vec{F}=m\ddot{\vec{r}} \rightarrow m\vec{g}+\vec{F_f}+\vec{N}=m\ddot{\vec{r}} \tag {1}$$
(I prefer to express them all being positive and allow trigonometry to give me the right signs)
Since we're dealing with a circular path, it makes sense to use polar coordinates. Therefore, I'll express the cartesian coordinates ##x## and ##y## in terms of ##r## and ##\theta##. Take into account that ##r## is constant so its derivative with respect to time will be 0.
$$x=r\cos\theta \tag{2}$$
$$\dot{x}=-r\dot{\theta}\sin\theta \tag{3}$$
$$\ddot{x}=-r(\ddot{\theta}\sin\theta+\dot{\theta}^2\cos\theta) \tag{4}$$
$$y=r\sin\theta \tag{5}$$
$$\dot{y}=r\dot{\theta}\cos\theta \tag{6}$$
$$\ddot{y}=r(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta) \tag{7}$$
I'll also need to express the horizontal and vertical components of the forces. The weight is clear, it's always pointing down and with the same magnitude. For the other forces, we will need ##\theta##.
$$N_x=N\cos(\theta+\pi)=-N\cos\theta \tag{8}$$
$$N_y=N\sin(\theta+\pi)=-N\sin\theta \tag{9}$$
$$F_{fx}=F_f\cos(\theta +\pi /2)=-F_f\sin(\theta)=-\mu N\sin\theta \tag{10}$$
$$F_{fy}=F_f\sin(\theta +\pi /2)=F_f\cos(\theta)=\mu N\cos\theta \tag{11}$$
Finally, applying Newton to the horizontal and vertical directions results in:
$$\sum F_x=m\ddot{x}\rightarrow N_x+F_{fx}=m\ddot{x}$$
$$-N\cos\theta-\mu N\sin\theta=-mr(\ddot{\theta}\sin\theta+\dot{\theta}^2\cos\theta) \tag{12}$$
$$\sum F_y=m\ddot{y}\rightarrow -mg +N_y+F_{fy}=m\ddot{y}$$
$$-mg-N\sin\theta+\mu N\cos\theta=mr(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta) \tag{13}$$
I believe from ##(12)## and ##(13)## it should be possible to obtain the expressions for ##N## and ##\theta##.
I believe we can take advantage of the fact that, since we know it's a circular path, the centripetal acceleration is defined.
$$a_n = \dot{\theta}^2r \rightarrow m\dot{\theta}^2r = - N - mg \cos(\theta-\pi /2) \tag{14}$$
Although I have some doubts about ##14##. I guess one possibility would be to solve it both ways and see if the methods are equivalent or not. I can't do that though because those differential equations are too hard for me now.
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