Vehicle Along a Horizontal Circular Track

Consider a car (treated as a particle for simplicity) taking a turn on a FLAT, horizontal, circular road of radius r. Three forces act on it: its weight mg, vertically downward; the normal reaction N from the road, vertically upward, which balances mg (so $N=mg$ here, since there is no vertical acceleration); and the force of STATIC friction $f_s$ between the tyres and the road surface. This is static (not kinetic) friction because the tyres are not sliding relative to the road -- they only tend to slip outward, and friction opposes that tendency. Since weight and normal reaction cancel vertically, the force of static friction alone is the (horizontal) resultant force, and being directed towards the centre of the track, it is entirely responsible for supplying the centripetal force: $f_s=\frac{mv^2}{r}$.

For a given track (r fixed) and a given vehicle (mg, and hence N, fixed), as the speed v increases, the REQUIRED friction $f_s$ also increases proportionally to $v^2$. But static friction cannot grow without bound -- it has an upper limit $f_{s,max}=\mu_sN$, where $\mu_s$ is the coefficient of static friction between the road and the tyres. This sets an upper limit on the safe turning speed: at the maximum possible speed $v_{s,max}$, friction is at its limiting value, so $\frac{mv_{s,max}^2}{r}=\mu_sN=\mu_smg$, giving $v_{s,max}=\sqrt{\mu_srg}$. Exceeding this speed means the required centripetal force exceeds what friction can supply, and the vehicle skids outward. In practice this dependence on friction is a serious limitation, because $\mu_s$ is rarely constant (it drops sharply on a wet, oily, sandy or otherwise non-uniform road surface) -- which is exactly the motivation, developed in section 1.3.3, for BANKING a curved road instead of relying on friction alone. It is also worth noting that in reality a four-wheeler's normal reaction is shared unequally among all four tyres (not equal, as the point-mass model assumes) and …