Question

On a highway, the maximum speed of vehicles is $50\text{m/s}$. The material of the road provides a coefficient of static friction ${\mu }_s=0.35$. If the highway has a circular turn with a radius of curvature $90\text{m}$, then what should be the banking angle $\left(\theta \right)$ for this turn to ensure safe driving ?

• A. ${\tan }^{-1}\left(2.24\right)$
• B. ${\tan }^{-1}\left(1.23\right)$
• C. ${\sin }^{-1}\left(2.26\right)$
• D. ${\cos }^{-1}\left(1.26\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(1.23\right)$

For maximum safe speed $v=50\text{m/s}$, the vehicle will have a tendency to skid up the incline, hence friction
($f_{max}=\mu N$) will act down the incline to ensure no slipping.


From equation of dynamics:
$N\sin \theta +f_{max}\cos \theta =\frac{m{v_{max}}^2}{r}....\left(i\right)$
where $f_{max}=\mu N.......\left(ii\right)$
$N\cos \theta -f_{max}\sin \theta =mg.......\left(iii\right)$

Dividing ($i$) by ($iii$):
$\frac{\sin \theta +{\mu }_s\cos \theta }{\cos \theta -{\mu }_s\sin \theta }=\frac{v_{max}^2}{rg}$
$\Rightarrow \frac{\tan \theta +{\mu }_s}{1-{\mu }_s\tan \theta }=\frac{v_{max}^2}{rg}$
Dividing by $\cos \theta$ on L.H.S and solving for $\tan \theta$ gives:-
$\tan \theta =\frac{\left(\frac{v_{max}^2}{gr}\right)-{\mu }_s}{1+\left(\frac{{\mu }_s{v}_{max}^2}{gr}\right)}$
$=\frac{\frac{{50}^2}{10\times 90}-0.35}{1+\frac{0.35\times {50}^2}{10\times 90}}=1.23$
or, $\theta ={\tan }^{-1}\left(1.23\right)$