Question

Detennine the angle to be made with the vertical by a two wheeler rider while turning on a horizontal track.

Answer 1

When a bicyclist takes a turn along an unbanked road, the force of friction `vec f_s` provides the centripetal force; the normal reaction of the road `vec N` is vertically up. If the biker does not lean inward, the friction force will provide an uneven outward torque about the centre of gravity, f_s.h, tipping the rider 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 `vec N` and the weight `vec g`, mg.a = f_sh₁.

Since the force of friction provides the centripetal force,

Fs = `(mv^2)/r`

If the cyclist leans from the vertical by an angle θ, the angle between `vec N` and `vec F`.

tan θ = `f_s/N`

= `((mv^2)/r)/(mg)`

= `v^2/(gr)`

When an automobile makes a turn on a flat road, it may slide or roll outward due to the friction force. However, an automobile is elongated and has four wheels. When the inner wheels lift off the ground, a restoring torque from the couple formed by the outer wheels and the weight can counter it.