Question

A curve of radius 78 m is banked for a design speed of 85 km/h. If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve? [Hint. Consider the direction of the friction force when the car goes too slow or too fast.]

Short answer

The car can safely make the curve if its speed ranges from $16.42\mathrm{m}/{\mathrm{s}}^2$to $31.75\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Step 1. Understanding the concept of force acting on an object in circular motion

Any object having mass m moving in a circular path consists of radius r with a constant speed v.Then the car needs a net amount of force that continues to keep the car moving in the circular motion.This force can be termed the centripetal force.

The direction of this forceis directed towards the center of the circle and the magnitude of the force is written as:

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

Step 2. Identification of the given information

• The radius of the banked curve is, r = 78 m.
• The speed of the car for which road is banked is,

$v=85\mathrm{km}/\mathrm{h}=\left(\frac{85\mathrm{km}}{1\mathrm{h}}\right)\times \left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\times \left(\frac{1\mathrm{h}}{60\min }\right)\times \left(\frac{1\min }{60\min }\right)=23.61\mathrm{m}/\mathrm{s}$.

• The coefficient of the static friction is, ${\mu }_{\mathrm{s}}=0.30$.

Step 3. Determination of banking angle of the road

Since the road is designed to travel with the speed v, thus no frictional force is required to bank the curve with this speed. Therefore, the banking angle of a road banked for cars travelling with velocity v is:

$\tan \theta =\frac{v^2}{rg}$

On substituting the given values, you will get:

$\begin{array}{c}\tan \theta =\frac{{(23.61\mathrm{m}/\mathrm{s})}^2}{\left(78\mathrm{m}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\\ =0.73\end{array}$

Thus, the banking angle is:

$\begin{array}{c}\theta ={\tan }^{-1}\left(0.73\right)\\ =36.1°\end{array}$

Step 4. Determination of the highest speed of the car

If the car travels with the speed larger than the v, then more centripetal force will be required which is supplied by the force of friction (f) acting downward along the incline as shown in the free-body diagram below:

Thus, the forces acting on the car are. normal force acting perpendicular to the incline (N), weight (mg) of the car acting downwards and static frictional force $\left(f={\mu }_{s}N\right)$acting downward along the incline.

On applying Newton’s second law in the vertical direction, you get:

role="math" localid="1646127747080" $\begin{array}{c}\sum F_{\mathrm{y}}=0\\ N\cos \theta -mg-f\sin \theta =0\\ N\cos \theta -mg-{\mu }_{\mathrm{s}}N\sin \theta =0\\ N=\frac{mg}{\cos \theta -{\mu }_s\sin \theta }...\left(i\right)\end{array}$

On applying Newton’s second law in the horizontal direction, you get:

role="math" localid="1646127792318" $\begin{array}{c}\sum F_{\mathrm{x}}=F_{\mathrm{R}}\\ N\sin \theta +f\cos \theta =\frac{m{v\text{'}}^2}{r}\\ N\sin \theta +{\mu }_{\mathrm{s}}N\cos \theta =\frac{m{v\text{'}}^2}{r}\\ N=\frac{m{v\text{'}}^2}{r\left(\sin \theta +{\mu }_s\cos \theta \right)}...\left(ii\right)\end{array}$

On equating expressions (i) and (ii), the highest speed at which car can safely make the curve is calculated as:

$\begin{array}{c}\frac{mg}{\cos \theta -{\mu }_s\sin \theta }=\frac{m{v\text{'}}^2}{r\left(\sin \theta +{\mu }_s\cos \theta \right)}\\ v\text{'}=\sqrt{\frac{rg\left(\sin \theta +{\mu }_s\cos \theta \right)}{\cos \theta -{\mu }_s\sin \theta }}\\ =\sqrt{\frac{rg\left(\tan \theta +{\mu }_s\right)}{1-{\mu }_s\tan \theta }}\end{array}$

On putting the given values, we get:

$\begin{array}{c}v\text{'}=\sqrt{\frac{\left(78\mathrm{m}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.73+0.30\right)}{1-\left(0.30\right)\left(0.73\right)}}\\ =\sqrt{1008.11{\mathrm{m}}^2/{\mathrm{s}}^2}\\ =31.75\mathrm{m}/{\mathrm{s}}^2\end{array}$

Step 5. Determination of the lowest speed of the car

If the car travels with the speed smaller than the v, then less centripetal force will be required. This decrease in centripetal force is compensated by the force of friction (f) acting upward along the incline as shown in the free-body diagram below:

On applying Newton’s second law in the vertical direction, we get:

$\begin{array}{c}\sum F_{\mathrm{y}}=0\\ N\cos \theta -mg+f\sin \theta =0\\ N\cos \theta -mg+{\mu }_{\mathrm{s}}N\sin \theta =0\\ N=\frac{mg}{\cos \theta +{\mu }_s\sin \theta }...\left(iii\right)\end{array}$

On applying Newton’s second law in the horizontal direction, we get:

$\begin{array}{c}\sum F_{\mathrm{x}}=F_{\mathrm{R}}\\ N\sin \theta -f\cos \theta =\frac{m{v\text{'}}^2}{r}\\ N\sin \theta -{\mu }_{\mathrm{s}}N\cos \theta =\frac{m{v\text{'}}^2}{r}\end{array}$

$N=\frac{m{v\text{'}}^2}{r\left(\sin \theta -{\mu }_s\cos \theta \right)}$ ...(iv)

On equating expressions (iii) and (iv), the lowest speed at which car can safely make the curve is calculated as:

$\begin{array}{c}\frac{mg}{\cos \theta +{\mu }_s\sin \theta }=\frac{m{v\text{'}}^2}{r\left(\sin \theta -{\mu }_s\cos \theta \right)}\\ v\text{'}=\sqrt{\frac{rg\left(\sin \theta -{\mu }_s\cos \theta \right)}{\cos \theta +{\mu }_s\sin \theta }}\\ =\sqrt{\frac{rg\left(\tan \theta -{\mu }_s\right)}{1+{\mu }_s\tan \theta }}\end{array}$

On putting the given values, you get:

$\begin{array}{c}v\text{'}=\sqrt{\frac{\left(78\mathrm{m}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.73-0.30\right)}{1+\left(0.30\right)\left(0.73\right)}}\\ =\sqrt{269.64{\mathrm{m}}^2/{\mathrm{s}}^2}\\ =16.42\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the car can safely make the curve if the speed of the car lies between the range $16.42\mathrm{m}/{\mathrm{s}}^2\le v\text{'}\le 31.75\mathrm{m}/{\mathrm{s}}^2$.