Question

A curve has a radius of $50\text{metres}$ and a banking angle of ${15}^{\circ }$. What is the critical speed (the speed for which no friction is required between the car's tyres and the surface) for a car to travel safely on this curve ? Take $g=10{\text{m/s}}^2$.

• A. $16\text{m/s}$
• B. $9-\sqrt{2}\text{m/s}$
• C. $10\sqrt{5(2-\sqrt{3})}\text{m/s}$
• D. $\sqrt{15}\text{m/s}$

Answer 1

The correct option is C $10\sqrt{5(2-\sqrt{3})}\text{m/s}$

Here, radius of curve, $r=50\text{m}$
Banking angle, $\theta ={15}^{\circ }$
$g=10{\text{m/s}}^2$

To meet the condition of the critical speed $v$, the necessary centripetal force must be provided by the horizontal component of the normal reaction ($N\sin \theta$)

From the free-body diagram for the car:-

$F_{\text{centripetal}}=N\sin \theta .......i$

For the vertical equilibrium of the car:
$N\cos \theta =mg$
From equation of dynamics towards the centre of horizontal circle:
$N\sin \theta =\frac{m{v}^2}{r}$.........$ii$

On dividing Eq. $ii$ and $i$, we get
$\Rightarrow \tan \theta =\frac{v^2}{rg}$
$\therefore v^2=rg\tan \theta$

$\therefore v=\sqrt{rg\tan \theta }=\sqrt{50\times 10\times \tan {15}^{\circ }}$
Using formula $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A.\tan B}$

$\tan {15}^{\circ }=\frac{\tan {45}^{\circ }-\tan {30}^{\circ }}{1+\left(\tan {45}^{\circ }\right)\left(\tan {30}^{\circ }\right)}=2-\sqrt{3}$

$\therefore v=10\sqrt{5(2-\sqrt{3})}\text{m/s}$

If the car has above speed then it can safely overcome the curve without any friction coming into play.