Chapter 5, Problem 16P

To determine

The minimum radius so that the car traveling at $\text{95}\text{km/h}$ will not leave the road.

Answer to Problem 16P

Solution:

  $\text{74}\text{.72}\text{m}$

Explanation of Solution

Given:

The angle of inclination of the hill $={18}^0$

The design speed of the car $=\text{95}\text{km/h}$

  $\begin{array}{l}\text{=}\frac{\text{95×1000 m}}{\text{60×60 s}}\\ \text{=26}\text{.39 m/s}\end{array}$

Formula Used:

Radial acceleration of the car $=\frac{v^2}{R}$

v : speed of the car

R : radius of curvature

Newton’s second law of motion

  ${\displaystyle \sum F}=ma$

Where, m is the mass and a is the acceleration.

Calculation:

Free body diagram and resolution of vector

  

Apply Newton’s second law:

  $\begin{array}{l}mg\cos \theta -N=\frac{m{v}^2}{R}\\ R=\frac{m{v}^2}{mg\cos \theta -N}\end{array}$

For minimum radius, the normal reaction should be zero.

  $\begin{array}{l}{R}_{\min }=\frac{m{v}^2}{mg\cos \theta }\\ R_{\min }=\frac{v^2}{g\cos \theta }\\ v=26.39\text{m/s}\\ g=9.8{\text{m/s}}^{\text{2}}\\ \theta ={18}^0\\ R_{\min }=\frac{{\left(26.39\right)}^2}{\left(9.8\right)\times \cos {18}^0}\\ =74.72\text{m}\end{array}$

Conclusion:

Hence, minimum radius of curvature to avoid jumpofthe car would be greater than equal 74.72m.