Question

A race track banked at an angle of ${37}^{\circ }$ has a coefficient of static friction $0.4$. What is the range of the safe velocities for the cars travelling on the track, for a turn of radius $40\text{m}$? Take $g=10{\text{m/s}}^2$.

• A. $9\text{m/s}\text{to}12\text{m/s}$
• B. $10\text{m/s}\text{to}20\text{m/s}$
• C. $\sqrt{188.46}\text{m/s}\text{to}\sqrt{243.32}\text{m/s}$
• D. $\sqrt{107.69}\text{m/s}$ to $\sqrt{657.14}\text{m/s}$

Answer 1

The correct option is D $\sqrt{107.69}\text{m/s}$ to $\sqrt{657.14}\text{m/s}$

For case of $v_{min}$, car will have a tendency to slip down. Hence, friction $f_{max}$ will act up the plane.


Applying the equations of circular dynamics:
$N\sin \theta -f_{max}\cos \theta =\frac{m{v_{min}}^2}{r}...\left(i\right)$
where $f_{max}={\mu }_{s}N....\left(ii\right)$

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

From Eq. ($i$), ($ii$), ($iii$) :-
$v_{min}=\sqrt{\frac{rg(\tan \theta -{\mu }_s)}{1+{\mu }_s\tan \theta }}\text{m/s}$
$=\sqrt{\left(\frac{\tan {37}^{\circ }-0.4}{1+0.4\times \tan {37}^{\circ }}\right)40\times 10}$
$=\sqrt{107.69}\text{m/s}$

For the case of $v_{max}$, car will have a tendency to skid up the incline. Hence, friction will act down the plane:


Simillarly, solving the equations of circular dynamics, we get:
$v_{max}=\sqrt{\left(\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }\right)rg}$
$v_{max}=\sqrt{\left(\frac{0.4+\tan {37}^{\circ }}{1-0.4\times \tan {37}^{\circ }}\right)40\times 10}$
$v_{max}=\sqrt{657.14}\text{m/s}$

Hence the range of safe velocities on the track is $\sqrt{188.46}\text{m/s}$ to $\sqrt{657.14}\text{m/s}$
$\therefore$ option (D) is correct.