Question

A car is moving with a speed of $60\text{m/s}$ on a curved road banked at an angle of ${45}^{\circ }$. If the coefficient of static friction between the wheels of the vehicle and the road is $0.4$, what should be the minimum radius of turn for the car so that it does not skid? Take $g=10{\text{m/s}}^2$.

• A. $175.5\text{m}$
• B. $154.3\text{m}$
• C. $170\text{m}$
• D. $200\text{m}$

Answer 1

The correct option is B $154.3\text{m}$

At minimum radius , $f_{max}$ will be acting down the incline, to prevent upward skidding of vehicle.
$F_{\text{centripetal}}=\frac{m{v}^2}{r}$
So if radius will be minimum then maximum $F_{\text{centripetal}}$ will be required, hence limiting value of friction will act i.e $f_{max}$


$N\sin \theta +f_{max}\cos \theta =\frac{m{v}^2}{r}....\left(i\right)$
where $f_{max}={\mu }_{s}N....\left(ii\right)$

and $N\cos \theta -f_{max}\sin \theta =mg...\left(iii\right)$

From Eq. ($i$), ($ii$), ($iii$) :-

$\frac{\sin \theta +{\mu }_{s}N\cos \theta }{\cos \theta -{\mu }_{s}N\sin \theta }=\frac{v^2}{rg}$
$\Rightarrow v^2=\left(\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }\right)rg$
$\therefore v=\sqrt{\frac{rg(\tan \theta +{\mu }_s)}{1-{\mu }_s\tan \theta }}\text{m/s}$

Putting $\theta ={45}^{\circ }$ and ${\mu }_s=0.4$:
${60}^2=\left(\frac{0.4+\tan {45}^{\circ }}{1-0.4\times \tan {45}^{\circ }}\right)\times r\times 10$
$\Rightarrow 3600=\frac{1.4}{0.6}\times r\times 10$
$\therefore r=\frac{3600\times 0.6}{1.4\times 10}=154.3\text{m}$
Here, radius is the minimum radius for safe travelling on the banked road.