What is the maximum velocity of a motorbike on a curved track?

[TSny]

This one:
\left(\frac{v_{max}^2}{R}\right)^2+(a_t)^2=(\mu g)^2
I think I am wrong about substituting $a_t=v(dv/ds)$ in this equation as it is only applicable for max acceleration.

You can always write $a_c^2 + a_t^2 = a^2$ at any time during the motion, not just at the time when you reach maximum speed. See what you get if you substitute $a_t = v dv/ds$ in this equation with $v$ not at max value.

 

[Saitama]

You can always write $a_c^2 + a_t^2 = a^2$ at any time during the motion, not just at the time when you reach maximum speed. See what you get if you substitute $a_t = v dv/ds$ in this equation.

I tried substituting it but I get a differential equation which is not solvable. :(

 

[TSny]

I tried substituting it but I get a differential equation which is not solvable. :(

What equation did you get?

 

[Saitama]

What equation did you get?

Sorry, I replied hastily before.

\left(\frac{v^2}{R}\right)^2+\left(v\frac{dv}{ds}\right)^2=(\mu g)^2
It can be rewriten as (I am substituting $\mu g=k$)
\frac{dv}{ds}=\frac{1}{vR}\sqrt{R^2k^2-v^4}
\frac{vR dv}{\sqrt{R^2k^2-v^4}}=dx

Integrating v from 0 to $v_{max}$ and s from 0 to $s_{min}$
\frac{R}{2}arctan\left(\frac{v_{max}^2}{\sqrt{R^2k^2-v^4}}\right)=s_{min}
What should I substitute from $v_{max}$?

 

[TSny]

At $v_{max}$ what is the direction of the total acceleration?

 

[Saitama]

At $v_{max}$ what is the direction of the total acceleration?

Towards the center? So $v_{max}=\sqrt{\mu gR}$?

 

[TSny]

Towards the center? So $v_{max}=\sqrt{\mu gR}$?

Yes.

 

[Saitama]

Yes.

Thank you TSny!

Substituting this value of $v_{max}$ in the final equation, I get $s_{min}=\pi R/4$.

 

[TSny]

I get $s_{min}=\pi R/4$.

I think that's right.