Chapter 6, Problem 85A

To determine

The banking angle of the road for the given cases.

Answer to Problem 85A

The banking angle of the road is $\theta =34.9°$ when the speed limit is $v=15.7\text{m/s}$ .

The banking angle of the road is $\theta =48.9°$ when the speed limit is $v=20.1\text{m/s}$ .

Explanation of Solution

Given:

Radius of curve is 36 m.

Initially, the speed limit is $v=15.7\text{m/s}$

Then, speed limit is increased to $v=20.1\text{m/s}$

Formula used:

The expression for the centripetal force acting on an object moving in a circular path is given by,

  $F_{\text{c}}=\frac{m{v}^2}{r}$

Here, $m$ is the mass of the object, $v$ is the velocity of the object, and $r$ is the radius of the circular path.

The expression for the gravitational force acting on the body is,

  $F_{\text{g}}=ma$

Here, $m$ is the mass of the object and $a$ is the acceleration of the object.

Calculation:

Consider the road is banked at an angle $\theta$ .

Draw the free body diagram of the forces acting on the banked road as shown below.

  

Consider the gravitational force acting on the car is denoted by $F_{\text{g}}$ .

Consider the normal force acting on the car is denoted by $F_{\text{N}}$ .

Consider the centripetal force acting on the car is denoted by $F_{\text{C}}$ .

For Equilibrium of forces in the vertical direction,

  $\begin{array}{l}{F}_{\text{N}}\cos \theta =F_{\text{g}}\\ F_{\text{N}}\cos \theta =mg\\ F_{\text{N}}=\left(\frac{mg}{\cos \theta }\right)\end{array}$

For the banking angle to be $\theta$ , the friction between the road and the car’s tyre should be zero.

Hence, the horizontal component of the normal force should be equal to the centripetal force acting on the car.

  $\begin{array}{l}{F}_{\text{N}}\sin \theta =F_{\text{C}}\\ \left(\frac{mg}{\cos \theta }\right)\sin \theta =\frac{m{v}^2}{r}\\ g\tan \theta =\frac{v^2}{r}\\ \theta ={\tan }^{-1}\left(\frac{v^2}{gr}\right)\end{array}$

Take the value of acceleration due to gravity of the Earth as $g=9.80{\text{m/s}}^2$ .

The radius of the curve is $r=36\text{m}$ .

Consider the speed limit corresponding to the radius of the curve is $v=15.7\text{m/s}$ .

The banking angle of the road is,

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{v^2}{gr}\right)\\ \theta ={\tan }^{-1}\left(\frac{{15.7}^2}{9.80\times 36}\right)\\ \theta ={\tan }^{-1}\left(0.69866\right)\\ \theta =34.9°\end{array}$

Consider the speed limit corresponding to the radius of the curve is $v=20.1\text{m/s}$ .

The banking angle of the road is,

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{v^2}{gr}\right)\\ \theta ={\tan }^{-1}\left(\frac{{20.1}^2}{9.80\times 36}\right)\\ \theta ={\tan }^{-1}\left(0.69866\right)\\ \theta =48.9°\end{array}$

Conclusion:

Therefore, the banking angle of the road is $\theta =34.9°$ when the speed limit is $v=15.7\text{m/s}$ .The banking angle of the road is $\theta =48.9°$ when the speed limit is $v=20.1\text{m/s}$ .