Answer: \(v_{min} = \sqrt{\dfrac{rg}{\mu_s}}\); the road should be banked at \(\theta = \tan^{-1}(1.276) \approx 51.9^\circ\) (about \(52^\circ\)).
(i) Well of death: In a well of death a motorcyclist rides along a horizontal circle of radius \(r\) on the inner surface of a vertical cylindrical wall.
[Figure: vertical cylindrical wall; the rider moves in a horizontal circle; forces — weight mg vertically down, friction \(f_s\) vertically up along the wall, normal reaction N horizontally towards the axis.]
Forces on the rider (with vehicle) of mass \(m\) moving with speed \(v\):
- The normal reaction \(N\) of the wall is horizontal, towards the centre, and provides the centripetal force: \(N = \dfrac{mv^2}{r}\).
- The weight \(mg\) acts downwards and is balanced by the static friction \(f_s\) acting upwards along the wall: \(f_s = mg\).
Friction cannot exceed its limiting value: \(f_s \le \mu_s N\). Hence
\[mg \le \mu_s\frac{mv^2}{r} \Rightarrow v^2 \ge \frac{rg}{\mu_s}\]
The minimum speed needed is
\[\boxed{v_{min} = \sqrt{\frac{rg}{\mu_s}}}\]
It is independent of the mass. Below this speed the rider would slip down.
(ii) Numerical: \(r = 288\) m, \(v = 216\) km/h \(= 216\times\dfrac{5}{18} = 60\) m/s, \(g = 9.8\) m/s\(^2\).
For a banked road (design speed, no reliance on friction):
\[\tan\theta = \frac{v^2}{rg} = \frac{60^2}{288\times9.8} = \frac{3600}{2822.4} = 1.2755\]\[\theta = \tan^{-1}(1.2755) \approx 51.9^\circ \approx 51^\circ 54'\]
(If \(g\) is taken as 10 m/s\(^2\), \(\tan\theta = 1.25\) and \(\theta \approx 51.3^\circ\).)
