Question

A car moves uniformly along a horizontal sine curve $y=a\sin \left(\frac{x}{\alpha }\right)$, where $x$ and $\alpha$ are certain constants. The coefficient of friction between the wheels and the road is equal to $k$. The velocity at which the car ride without sliding is $v\le \alpha \sqrt{\frac{kg}{na}}$ . What is the value of n?

Answer 1

Since the car follows a curve, so the maximum velocity at which it can ride without sliding at the point of minimum radius of curvature is the sought velocity and obviously in this case the static friction between the car and the road is limiting.
Hence from the equation $F_n=m{w}_n$
$kmg\le \frac{m{v}^2}{R}$ or $v\le \sqrt{kRg}$
so $v_{max}=\sqrt{k{R}_{min}g}$. (1)
We know that, radius of curvature for a curve at any point (x,y) is given as,
$R=\left|\frac{{\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^{\frac{3}{2}}}{\frac{\left(d^{2}y\right)}{d{x}^2}}\right|$ (2)
For the given curve,
$\frac{dy}{dx}=\frac{a}{\alpha }\cos \left(\frac{x}{\alpha }\right)$ and $\frac{d^{2}y}{d{x}^2}=\frac{-a}{{\alpha }^2}\sin \frac{x}{\alpha }$
Substituting this value in (2) we get,
$R=\frac{{\left[1+\left(\frac{a^2}{{\alpha }^2}\right){\cos }^2\left(\frac{x}{\alpha }\right)\right]}^{\frac{3}{2}}}{\left(\frac{a}{{\alpha }^2}\right)\sin \left(\frac{x}{\alpha }\right)}$
For the minimum $R$, $\frac{x}{\alpha }=\frac{\pi }{2}$
and therefore, corresponding radius of curvature
$R_{min}=\frac{{\alpha }^2}{a}$ (3)
Hence from (1) and (2)
$v_{max}=\alpha \sqrt{\frac{kg}{a}}$

$\therefore n=1$