Question

A car travelling on a level road with a curve of radius $32\text{m}$. The coefficient of static friction between the road and tyre of car is $0.2$. What speed will put the car on verge of sliding? $(g=10{\text{m/s}}^2)$

• A. $6\text{m/s}$
• B. $13\text{m/s}$
• C. $8\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is C $8\text{m/s}$


The necessary centripetal force for the car will be provided by frictional force acting towards centre of circular curve.
To avoid slipping, friction will act at it's limiting value $f_{max}$.
From FBD of car:
$f_{max}=\frac{m{v}^2}{r}$....................$i$
$N=mg$.............................$ii$
$f_{max}={\mu }_{s}N={\mu }_{s}mg$.................$iii$
Car will start sliding when; $v=v_{max}$
$\therefore \frac{m{v}_{max}^2}{r}={\mu }_{s}mg$
$\Rightarrow v_{max}=\sqrt{{\mu }_{s}rg}$
putting ${\mu }_s=0.2,r=32\text{m}$
$\therefore v_{max}=\sqrt{0.2\times 32\times 10}=8\text{m/s}$