Question

A van is moving with a speed of 72 km h${}^{-1}$ on a level road, where the coefficient of friction between its tyres and road is 0.5. The minimum radius of curvature, the road must have for safe driving of van is (g = 10 m s${}^{-2}$).

• A. 80 m
• B. 40 m
• C. 20 m
• D. 4

Answer 1

The correct option is A 80 m

$\text{Step 1: F.B.D of Van}$ $\text{[Ref. Fig. 1]}$

$\text{Step 2: Newton's Law along Centripetal direction}$

Here, The only possible horizontal force is friction, Therefore, it will produce centripetal acceleration:

So, Applying Newtons Second Law towards center:

$\Sigma F_c=m{a}_c$

$\Rightarrow f=m\left(\frac{v^2}{R}\right)$ $....\left(1\right)$

Since there is no acceleration in vertical direction, Therefore, From figure 2:

$N=mg$

$\text{Step 2: Calculation of minimum radius}$

From equation $\left(1\right)$, For minimum radius, maximum friction should act.

$\therefore f=\mu N$

$=\mu mg$

So, Equation $\left(1\right)\Rightarrow$ $\mu mg=\frac{m{v}^2}{R_{min}}$

$\Rightarrow R_{min}=\frac{v^2}{\mu g}$

$\text{Step 4: Calculations}$

Given, Speed of van $v=72Km/h$ $=72\times \frac{5}{18}m/s$ $=20m/s$

and Coefficient of friction $\mu =0.5$

$R_{min}=\frac{20\times 20}{0.5\times 10}=80m$

So, Minimum radius will be $80m$

Hence, option A is correct

$\text{Alternnate Solution}$

Use formula

$V_{max}=\sqrt{Rg\left(\frac{\tan \theta +\mu }{1+\mu \tan \theta }\right)}$ Where $\theta$ is the angle of inclination of the road

Put $\mu =0.5,v=20m/s$

& $\theta =0^o$

Therefore, $R_{min}=\frac{v^2}{\mu g}$

$R_{min}=\frac{20\times 20}{0.5\times 10}=80m$