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?
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.