91Ó°ÊÓ

When the car of mass m moves in a circular path having radius rwith a constant speed v, it requires an amount of force that keeps the car moving in the circular motion. This force acts on the car act in the circle center direction, and it is termed the centripetal force. Mathematically, it is written as:

$F_{\mathrm{R}}=\frac{m{v}^2}{r}$

• The inclination of the steep hill of the road is, $\mathrm{θ}=18°$.
• The speed of the car is, $v=95\mathrm{km}/\mathrm{h}=\left(\frac{95\mathrm{km}}{1\mathrm{h}}\right)×\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)×\left(\frac{1\mathrm{h}}{60\min }\right)×\left(\frac{1\min }{60\min }\right)=26.39 \mathrm{m}/\mathrm{s}$.

The free-body diagram of the car is shown below:

It can be seen from the figure that the centripetal force $F_{\mathrm{R}}$is provided to the car by the difference of horizontal component of weight $mg$of the car and normal force $F_{\mathrm{N}}$acting on the car, then the equation can be expressed as,

role="math" localid="1646118286857" $\begin{array}{rcl}{F}_{\mathrm{R}}& =& mg\cos \mathrm{θ}-F_{\mathrm{N}}\\ \frac{m{v}^2}{r}& =& mg\cos \mathrm{θ}-F_{\mathrm{N}}...\left(i\right)\end{array}$

Using (i), the expression of the radius will be:

$r=\frac{m{v}^2}{mg\cos \mathrm{θ}-F_{\mathrm{N}}}$

When the car is just about to leave the road, then the normal force will be zero.

Therefore, the above expression becomes:

$r_{\mathrm{m}}=\frac{v^2}{g\cos \mathrm{θ}}$

So, the minimum radius will be:

$\begin{array}{c}{r}_{\mathrm{m}}=\frac{{\left(26.39\mathrm{m}/\mathrm{s}\right)}^2}{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\cos 18°\right)}\\ =74.8\mathrm{m}\\ ≈75\mathrm{m}\end{array}$

Thus, the minimum radius with which transition should be rounded is approximately 75 m.