Question

Physics ❤

A car takes a circular turn of radius 20 m it a speed of 54 km/hr . what is the least coefficient of friction between tyres and road that can prevent slide slipping?

Need Explaination!​

Answer 1

Answer :-

$\mapsto \boxed{ \sf \: \mu \: = 1.12} \:$

Explanation :-

Given that :

1. Speed of car (v) = 54 km/hr
2. Radius (r) = 20 m

Here normal contact force is -

$\to \sf \: f = \mu \: m \: g \:$

→ It is moving in circular path hence ,its acceleration is

$\to \sf \: a = r \: \alpha \: \\ \\ \bf \: here \: \: \alpha \: = angular \: acceleration \: \\ \bf \: therefore \\ \\ \to \boxed{ \sf a = \frac{ {v}^{2} }{r} } \\ \\ \mapsto \sf \: f = m \: a \\ \\ \mapsto \sf f = \frac{m {v}^{2} }{r}$

hence,

$\to \sf \mu \: mg \: = \frac{m {v}^{2} }{r} \\ \\ \to \mu \sf \: = \frac{ {v}^{2} }{r \: g}$

→ Take g = 10 m/s

Firstly convert speed from km/hr to m/s

$\to \sf \: v = \frac{5}{18} \times 54 \: \frac{m}{s} \\ \\ \to \sf \: v = 15 \: \frac{m}{s}$

hence coefficient of friction will

$\to \sf \: \mu = \frac{15 \times 15}{20 \times 10} \\ \\ \mapsto \: \mu \: = \frac{225}{200} \\ \\ \mapsto \boxed{ \sf \: \mu \: = 1.12}$