Question

A cyclist speeding at 18km/hr on a level road takes a sharp circular turn of radius 3m without
reducing the speed. The coefficient of static friction between the tyres and road is 0.1
Will the cyclist slip while taking the turn? Explain.

Answer 1

Given:

Speed of cyclist = 18 km/h = 5 m/s

Radius of sharp circular turn taken by cyclist (R) = 3 m

Coefficient of static friction between tyres and road $\sf (\mu_s)$ = 0.1

To Explain:

Will the cyclist slip while taking the turn

Explanation:

Maximum safe speed of cyclist while taking sharp circular turn:

$\boxed{ \bf{v_{max} = \sqrt{\mu_s Rg}}}$

By substituting values we get:

$\rm \implies v_{max} = \sqrt{0.1 \times 3 \times 10} \\ \\ \rm \implies v_{max} = \sqrt{3} \: m {s}^{ - 1} \\ \\ \rm \implies v_{max} = 1.732 \: m {s}^{ - 1}$

By comparing maximum safe speed and speed of cyclist we can conclude that speed of cyclist is more than maximum safe speed. So, The cyclist will slip while taking turn.

Answer 2

Given : A cyclist speeding at 18km/hr on a level road takes a sharp circular turn of radius 3m without reducing the speed. The coefficient of static friction between the tyres and road is 0.1.

　

To find : Will the cyclist slip while taking the turn?

　

Using formula :

★ Maximum velocity = Static friction × Radius of sharp circular × Gravity.

　

Calculations :

→ Maximum velocity = √0.1 × 3 × 9.81

→ Maximum velocity = √2.943

→ Maximum velocity = 1.71551741 m/s

　

Therefore, maximum velocity is equal to 1.71551741 m/s.

　

• Required final answer :

The cyclists will slip down while turning because the speed of cyclists is more than the speed of maximum safe speed.