Question

What is the maximum speed of car taking a turn of radius 16m on a level road if the coefficient of friction between the tyres and the road is 0.4

Answer 1

Given :

▪ Radius of circular path = 16m

▪ Co-efficient of friction = 0.4

To Find :

▪ Maximum speed of car while taking a circular turn.

Formula :

↗ A vehicle taking a circular turn on a level road. If u is the coefficient of friction b/w tyres and road, then the maximum velocity with which the vehicle can safely take a circular turn at radius r is given by

$\bigstar\bf\:v_{max}=\sqrt{\mu rg}$

Calculation :

$\dashrightarrow\sf\:v_{max}=\sqrt{\mu rg}\\ \\ \dashrightarrow\sf\:v_{max}=\sqrt{0.4\times 16 \times 10}\\ \\ \dashrightarrow\sf\:v_{max}=\sqrt{64}\\ \\ \dashrightarrow\underline{\boxed{\bf{\gray{v_{max}=8\:ms^{-1}}}}}\:\bigstar$

Answer 2

Given ,

• Radius of circular road (r) = 16 m
• Coefficient of friction b/w tyres and road $\sf (\mu)$= 0.4

We know that , the maximum speed with which a vehicle can negotiate a turn without skidding is given by

$\large \sf \fbox{v_{m} = \sqrt{ \mu gr \: } }$

Note : Vm is independent of mass of vehicle

Thus ,

$\sf \Rightarrow v_{m} = \sqrt{0.4 \times 10 \times 16 } \\ \\\sf \Rightarrow v_{m} = \sqrt{64} \\ \\\sf \Rightarrow v_{m} = 8 \: \: m/s$

$\therefore \sf \bold{ \underline{The \: maximum \: speed \: is \: 8 \: m/s</p><p> \: \: } }$