Question

What should be the value of coefficient of static friction between the tyre and the road, when a car travelling at speed of $60km{h}^{-1}$ makes a level turn of radius $40\text{m}$?

• A. $0.5$
• B. $0.66$
• C. $0.71$
• D. $0.80$

Answer 1

The correct option is C $0.71$

Applying equation for frictional force as centripetal force at the circular turn gives,
$f=\frac{m{v}^2}{r}$
Here, $f$ is magnitude of frictional force.
$\therefore$ $\mu mg=\frac{m{v}^2}{r}$
$\mu =\frac{v^2}{rg}=\frac{{\left(\frac{60\times 5}{18}\right)}^2}{40\times 9.8}=0.71$