What is the maximum speed with which a 1200-kg car can round a turn of radius 90.0 m on a flat road if the coefficient of friction between tires and road is 0.65? Is this result independent of the mass of the car?

The car takes a turn around a frictional path. The car will experience a maximum frictional value when it achieves a maximum speed. It may cause slipping at that time.In this problem, equate the frictional force and centripetal force to determine the maximum speed of the car.

The given data can be listed below as,

• The radius of curve is, $r=90\mathrm{m}$.
• The mass of the car is, $m=1200\mathrm{kg}$.
• The coefficient of friction between the tires and the road is, $\mu =0.65$.
• The acceleration due to gravity is, $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

The forces on the car can be shown as,

Here, mgis the weight of the car, $F_{\mathrm{fr}}$ is the frictional force, $F_{\mathrm{N}}$ is the normal force,m is the mass of the car.

Apply the equilibrium equation conditions in the vertical direction from the above figure. The car is not speeding up, so the net vertical forces can be equated to zero. This can be expressed as,

$\begin{array}{c}\sum F_y=0\\ F_{\mathrm{N}}-mg=0\\ F_{\mathrm{N}}=mg\end{array}$

From the above figure, the car is represented as it's coming out from the page. The round turn center is shown in the rightward direction on the page. The static frictional force will cause the car to turn around the curved path. Therefore, equate the static frictional force and the centripetal force of the given car.

$\begin{array}{c}{F}_{\mathrm{c}}=F_{\mathrm{fr}}\\ m{a}_{\mathrm{c}}={\mu }_s{F}_{\mathrm{N}}\\ \frac{m{v}^2}{r}={\mu }_{s}mg\end{array}$

role="math" localid="1646112139869" $v^2=r{\mu }_{s}g\dots \left(i\right)$

Here,v is the car’s speed, $F_{\mathrm{c}}$is the centripetal force, $a_{\mathrm{c}}$is the centripetal acceleration of the car, ${\mu }_{\mathrm{s}}$is the coefficient of static friction.

Substitute the values in the above equation.

$\begin{array}{c}{\left(v\right)}^2=90\mathrm{m}\times 0.65\times 9.81\mathrm{m}/{\mathrm{s}}^2\\ v=\sqrt{90\mathrm{m}\times 0.65\times 9.81\mathrm{m}/{\mathrm{s}}^2}\\ v=\sqrt{573.885}\mathrm{m}/\mathrm{s}\\ v=23.956\mathrm{m}/\mathrm{s}\end{array}$

Thus, the speed of the car is $23.956\mathrm{m}/\mathrm{s}$.

The above speed equation (i) of the car is not dependent on the mass. It only depends on the coefficient of static friction, radius of the curved path, the acceleration due to gravity.

Thus, the above result is independent of the mass of the car.