Question

A track (circular) is designed for car passing through a point with speed $72km/hr$ & radius equals to $40m$. If coefficient of friction between tyres & road is $\mu$$=0.6$, find maximum speed at which the car can turn?

• A. $72km/hr$
• B. $108km/hr$
• C. $180km/hr$
• D. $144km/hr$

Answer 1

The correct option is D $144km/hr$

If a car can pass at an optimum speed of 72 kmph in absence of friction, we can calculate the angle of banking as
$tan\theta =\frac{v^2}{rg}=\frac{20\text{x}20}{40\text{x}10}=1$

Hence, $\theta ={45}^o$

At max speed vehicle is about to skid & $f=\mu N$
$\frac{v_{max}^2}{rg}=\frac{\mu +tan\theta }{1-\mu tan\theta }$

$\frac{v_{max}^2}{\left(40\right)\left(10\right)}=\frac{0.6+1}{1-0.6}=\frac{1.6}{0.4}$

$v_{max}^2=4\times 400v_{max=40m/s}$

$v_{max}=40m/s=40\times \frac{18}{5}km/hr=144km/hr$