Question

A circular racing track of radius $300\text{m}$ is banked at an angle of ${17}^{\circ }$ with the horizontal. If the coefficient of static friction $\left({\mu }_s\right)$ between the wheels of the racing car and the road is $0.2$, what is the maximum permissible speed to avoid slipping? Take $g=10{\text{m/s}}^2$. Use the given data: $\tan {17}^{\circ }=0.30,\sin {17}^{\circ }=0.29,\cos {17}^{\circ }=0.96$ and $\frac{0.50}{0.94}\simeq 0.5$

• A. $10\sqrt{15}\text{m/s}$
• B. $3\sqrt{35}\text{m/s}$
• C. $5\sqrt{15}\text{m/s}$
• D. $50\text{m/s}$

Answer 1

The correct option is A $10\sqrt{15}\text{m/s}$

For the maximum permissible speed $\left(v\right)$, static friction will act at it's maximum value $f_{max}$.

Horizontal component of friction and normal reaction will provide the necessary centripetal force for the car. $f_{max}={\mu }_{s}N$ ,will act down the inclined plane to prevent slipping of the car.



Applying the equation of dynamics towards the centre of circular path:
$N\sin \theta +f_{max}\cos \theta =m{a}_c=\frac{m{v}^2}{r}$
$N(\sin \theta +{\mu }_s\cos \theta )=\frac{m{v}^2}{r}...\left(i\right)$

For the vertical equilibrium:
$N\cos \theta =f_{max}\sin \theta +mg$
$N(\cos \theta -{\mu }_s\sin \theta )=mg...\left(ii\right)$

Dividing Eq. $\left(i\right)$ and $\left(ii\right)$:
$\frac{\sin \theta +{\mu }_s\cos \theta }{\cos \theta -{\mu }_s\sin \theta }=\frac{v^2}{rg}$
$\Rightarrow \frac{\tan \theta +{\mu }_s}{1-{\mu }_s\tan \theta }=\frac{v^2}{rg}$
$\Rightarrow v=\sqrt{\frac{rg(\tan \theta +{\mu }_s)}{1-{\mu }_s\tan \theta }}...\left(iii\right)$

Putting the given values in (iii) $r=300\text{m},\tan {17}^{\circ }=0.30,{\mu }_s=0.20$
$\Rightarrow v=\sqrt{\frac{300\times 10(0.30+0.20)}{1-(0.20\times 0.30)}}$
$v=\sqrt{3000\times \frac{0.50}{0.94}}=\sqrt{3000\times 0.5}$
$\therefore v=10\sqrt{15}\text{m/s}$
This is the maximum permissible speed on the track.