Question:

A cyclist speeding at 18 Kmh^-1 on a level road takes a sharp turn of radius 3 m without reducing the speed.The coefficient of friction between the tyre and the road is 0.1.Will the cyclist slip while taking the turn?

Solution: Cyclist sharp turn

Centripetal Force:

Centripetal force acts inwards pulling the cyclist towards the centre of the turn.It depends on the cyclist’s speed and the turn radius.

A higher speed(v) or smaller turn radius will lead to a greater centripetal force.

Frictional force:

Frictional force acts outwards, between the tires and the road,providing the necessary centripetal force to keep the cyclist from slipping.It depends on the coefficient of friction and the normal force(the cyclist’s weight pressing the road).

(A higher coefficient of friction(mu) or a greater normal force(N) will result in a larger frictional force).

Calculate cyclist’s speed from Kmh^-1 to ms^-1

Cyclist speeding at 18 Kmh^-1

$${18 \ \frac{Km}{h}=18 \ \frac{(1000 \ m)} {(3600 \ s)}}$$

$${18 \ \frac{(5)}{(18)} \ \frac{m}{s}}$$

Cyclist speeding at 5 ms^-1

Centripetal force:

Apply centripetal force formula $${f_c=\frac{mv^2}{r}}$$

Speed of the cycle should not exceed

$${\sqrt {\mu rg}}$$

Coefficient of friction $${\mu=0.1}$$

Radius of the sharp turn

$${r=3 \ m}$$

Acceleration due to gravity

$${9.8 \ \frac{m}{s^2}}$$

we can calculate the maximum velocity

$${\sqrt{\mu rg}=\sqrt{(0.1)(3)(9.8)}=1.7}$$

The cyclist’s maximum safe speed for the turn is 1.7 ms^-1 , which is equal to 6.1 Kmh^-1

Will the cyclist slip?

Based on force:

For the cyclist to not slip, the frictional force must be greater than or equal to the centripetal force.

Maximum frictional force is

$${f_f=\mu_s mg = 0.1mg}$$

$${f_f=0.98}$$

Centripetal force is

$${f_c=\frac{mv^2}{r}=\frac{25m}{3}}$$

$${f_c=8.33}$$

Frictional force(f_f) is less than the centripetal force(f_c).

Therefore, the available frictional force is insufficient to provide the necessary centrietal force.

Based on speed

Here speed of the cyclist is exceeding the safe limit. Any speed above the above said 6.1 Kmh^-1 would increase the centripetal force and risk exceeding the frictional force, leading to a slip.

Hence the cyclist will slip while taking the turn.

For a safe turn:

• The cyclist should reduce his speed before entering the turn.
• If possible, take a wider turn with a larger radius.