Minimum distance around a track

[better361]

Homework Statement

A motorbike starts from rest at time t=0 and begins to accelerate around a circular track of radius 10m. Eventually, at time t=t1 the motorbike reaches the maximum velocity possible without slipping off the track. What's the minimum length in meters the motorbike must travel between t=0 and t=t1?

Homework Equations

mv^2/r=Force of friction in radial direction
m*dv/dt=Force of friction in tangential direction

The Attempt at a Solution

Here is the solution that was given but I did not understand it.
Call β the the angle formed by friction with respect to the radial axis by taking the arctan of (friction in tangential/friction in radial). So, mv^2/r=F*sinβ. Differentiating w.r.t, the equation is now (2mv/r)*dv/dt=F*cosβ*dβ/dt. The second equation becomes, with the substitution of β ,
m*dv/dt=F*cosβ. Dividing these two equations, it now becomes 2v/r=dβ/dt. v/r is the rate of change of the angle,θ, formed from the traveled distance(the arclength) over the radius. Making this substitution and canceling the differential dt, the equation is now dθ=dβ/2.
Then, as β goes from 0 to pi/2, the distance traveled is s=θ*r=pi/4*r=7.85m.

Is there another way to solve this problem? The part that I am still confused with is how to realize to take the derivative and divide the equations.

 

[TSny]

Hello.
This same problem was discussed a couple of months ago here:
https://www.physicsforums.com/showthread.php?t=690914

If you want to wade through all of that you will discover another way to get to the solution. But, in my opinion, the solution you have provided is much nicer than the solution given there.

I don't have an answer as to how one sees the "trick" of doing it the way shown in your solution. I guess it's a combination of insight, experience, and playing around with the basic equations.