Circular Motion in Practice: Cyclist, Level Road, and Banked Road

Building directly on the general result of Section 4.11 -- that circular motion at constant speed always requires a net inward (centripetal) force of magnitude $mv^2/r$, supplied by whatever real force fits the situation -- this section works through the three practical examples of circular motion named explicitly in WBCHSE's syllabus.

1. A cyclist negotiating a turn. A cyclist riding around a curve leans their bicycle inward, toward the centre of the turn, rather than staying upright. Leaning shifts the combined centre of gravity of rider and bicycle inward and lets the (still vertical) normal reaction from the road, together with friction, provide the necessary centre-directed net force while keeping the whole system in rotational balance (so that it does not simply topple over sideways). For a cyclist leaning at angle $\theta$ to the vertical while turning at speed $v$ on a curve of radius $r$, balancing the torques (about the point where the tyres meet the road) leads to exactly the same relation used for banking in part (c) below:

   $\tan\theta = \frac{v^2}{rg}$

2. A vehicle on a level (unbanked) circular road. Here the road surface is flat, so the normal reaction $N$ on the vehicle is purely vertical and can supply no horizontal, centre-directed component at all. The entire centripetal force must then be supplied by friction between the tyres and the road, acting horizontally, toward the centre of the curve:

   $f = \frac{mv^2}{r}$

   Since friction cannot exceed its limiting value, $f_{\max} = \mu_s N = \mu_s mg$ (taking $N = mg$ on a level road), there is a strict maximum safe speed for taking a given unbanked curve without skidding:

   $\frac{mv_{\max}^2}{r} = \mu_s mg \quad \Longrightarrow \quad v_{\max} = \sqrt{\mu_s\, r g}$

   Beyond this speed, the friction available is simply not enough to keep the vehicle turning along the curve, and it skids outward.
3. A vehicle on a banked road. Banking tilts the entire road surface, at some angle $\theta$ to the horizontal, so that the normal reaction $N$ -- now perpendicular to the tilted surface rather than purely vertical -- itself has a horizontal component pointing toward the centre of the curve, which can contribute to (or, ideally, entirely supply) the centripetal force, easing the burden on friction and allowing higher, safer speeds around the same curve. Resolving $N$ into a vertical component $N\cos\theta$ (balancing the vehicle's weight $mg$) and a horizontal component $N\sin\theta$ (supplying the centripetal force) gives, for the ideal banking case where friction is entirely absent (or simply not needed):

   $N\cos\theta = mg \qquad N\sin\theta = \frac{mv^2}{r}$

   Dividing the second equation by the first eliminates both $N$ and $m$ from the result entirely: $\tan\theta = \frac{v^2}{rg}$ …