Question

A turn of radius 20 m is banked for the vehicles going at a speed of 36 $km{h}^{-1}$. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speed of a vehicle so that it neither slips down nor skids up?

Answer 1

Angle of banking for designed speed,

$tan\theta =\frac{v_0^2}{Rg}$ (i)

$V=36km{h}^{-1}=10m{s}^{-1}$

$\Rightarrow tan\theta -\frac{v_0^2}{Rg}=\frac{{10}^2}{20\times 10}=\frac{1}{2}$

The vehicle may have the tendency to slide up or down depending on the speed of the vehicle. If speed of the vehicle is more it has tendency so slide up and visa-versa.

For speed greater than designed speed: The vehicle has the tendency to slide. Friction will act downward.

In vertical direction $\sum F_y=0.Ncos\theta -\mu Nsin\theta =mg$

In horizontal direction $\sum F_r=\frac{m{v}_{max}^2}{R}$

$Nsin\theta +\mu Ncos\theta =\frac{m{v}_{max}^2}{R}$ (ii)

$Ncos\theta -\mu Nsin\theta =mg$ (iii)

From (ii) and (iii)

$\frac{N(sin\theta +\mu cos\theta )}{N(cos\theta -\mu sin\theta )}=\frac{\frac{m{v}_{max}^2}{R}}{mg}\Rightarrow \frac{(sin\theta +\mu cos\theta )}{cos\theta -\mu sin\theta }=\frac{v_{max}^2}{Rg}$

$\Rightarrow \left(\frac{tan\theta +\mu }{1-\mu tan\theta }\right)-\frac{v_{max}^2}{Rg}$ (v)

$\Rightarrow \left(\frac{tan\theta +\mu }{1-\mu tan\theta }\right)=\left(\frac{0.5+0.4}{1-0.4\times 0.5}\right)=\frac{v_{max}^2}{20\times 10}$

$\Rightarrow v_{max}=15m{s}^{-1}$

For speed less than designed speed: The vehicle has tendency to slide down friction will act upward.

In horizontal direction $\sum F_r=\frac{m{v}_{max}^2}{R}$

$N^{\prime }sin\theta -\mu N^{\prime }cos\theta =\frac{m{v}^{\prime 2}}{R}$ (iv)

From (v) and (vi), we get $\frac{(sin\theta -\mu cos\theta )}{cos\theta +\mu sin\theta }=\frac{v^{\prime 2}}{Rg}$

$\left(\frac{tan\theta -\mu }{1+\mu tan\theta }\right)=\frac{v^{\prime 2}}{Rg}\Rightarrow v_{min}=10\sqrt{\frac{1}{6}}m{s}^{-1}$