(i) When a bicyclist takes a turn along an unbanked road, the force of friction \(\vec{Fs}\) provides the centripetal force; the normal reaction of the road \(\vec{N}\) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, f_{s}·h, due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by and the weight g, mg·a = f_{s}·h_{1}.
(ii) Since the force of friction provides the centripetal force,
,\(f_s=\frac{mv^2}{r}\)
If the cyclist leans from the vertical by an angle q, the angle between N⃗ and F⃗ as shown in fig (b).
\(\frac{f_s}{N}\ = \frac{mv^2/r}{mg}=\frac{v^2}{gr}\)
Hence, the cyclist must lean by an angle
θ =\(\tan^{-1}\frac{v^2}{gr}\)
(iii) When a car takes a turn along a level road, apart from the risk of skidding off outward, it also has a tendency to roll outward due to an outward torque about the centre of gravity due to the friction force. But a car is an extended object with four wheels, So, when the inner wheels just pet lifted above the ground, it can be counterbalanced by a restoring torque of the couple formed by the normal reaction (on the outer wheels) and the weight.