1. Bending of cyclist at circular path: We know that without centripetal force, the path of an object can never be circular. When a cyclist reaches a turn, its path becomes circular. Therefore he requires a centripetal force. To obtain the centripetal force, he bends towards the centre of the turn. On doing so, the force of friction between the tyres and the road provides the required centripetal force and cyclist safely crosses the turn.
The forces acting the cyclist at turn in fig. Cyclist crosses turn in equilibrium of clockwise torque due to mg about point A and anticlockwise torque due to force of friction fs about centre of gravity (C.G.)
∴ In equilibrium,
Momentum of Force Ff about C.G. = Moment of force mg about A.
As we have considered here that required centripetal force\((\frac{mv^2}{r})\) is provided by force of friction Ff. Therefore for safe passing through the turn,
Ff ≥ Fc
or µs R = µs mg ≥ Fc
or µs \(mg\geq\frac{mv^2}{r}\)
For optimum speed on turn.
Optimum speed or maximum safe speed on turn can be calculated with the help of above equation (2).
Cyclist slows down the speed at the turn: When a cyclist reaches a circular turn, it requires a centripetal force because without centripetal force circular motion is impossible. To obtain this centripetal force, he bends towards the centre of the turn and O be the angle of inclination with vertical,
then it can be proved, that
\(\tan\theta = \frac{v^2}{rg}\)
Since the values of r and g are constant for a turn, therefore
tanθ ∝ v2
Thus it is clear, for greater value of v, θ will also be greater and for high value of θ, the possibility of slipping the cycle will increase. This is in ω to avoid this mis-happening, cyclist shows down his speed on the turn.
2. Circular turn of road: At the turn, the roads are, generally, made of circular shape. Therefore the vehicles require centripetal force on turn and this required centripetal force is provided to vehicles by the force of friction between the tyres of the vehicles and the road. But if the friction force is not sufficient, the road is made banked at the turn, due to which the required centripetal force is obtained by following three types:
(i) By friction force or motion of a car on leveled road: Suppose, m be the mass of the vehicle, r be the radius of the turn and u be the velocity of the vehicle on turn, then the forces acting on the vehicle are shown in the fig. and these forces are:
(i) Weight of the car (mg)
(ii) Normal reaction (R)
(iii) Force of friction (f)
For balancing of vertical forces,
For circular motion, the centripetal force is provided by the friction force, therefore,
f ≤ µsR or \(\frac{mv^2}{r}\) ≤ µs mg
or v2 ≤ µs rg
\(\therefore\ \leq \sqrt{\mu_srg} \ ...(2)\)
Here the velocity does not depend on the mass of the vehicle. Therefore the optimum speed, i.e., maximum safe speed on the turn,
\(v_{max} = \sqrt{\mu rg}\)
If the vehicle moves on the turn with more than this safe speed, then either the vehicle will overturn or it will slip.
(ii) By banking of road: In above first part we have studied that if vehicle passes through curved horizontal road on turn, then friction provides the required centripetal force and maximum safe velocity on turn is \(V_{max} = \sqrt{\mu _s rg}\). If velocity of the vehicle is more than this velocity, the vehicle will slip. To solve this problem, the roads are made banked on the turn so that banking of road may also provide centripetal force. Here we are considering only the banking of road, not the friction.
When a vehicle passes over banked road at turn with optimum velocity v, then there is no friction between the tyres of the vehicle and road. In this case the required centripetal force is applied by horizontal component of reaction (R1 + R2) and its vertical component balances the weight of the vehicle.
