Chapter 7, Problem 92A

To determine

To Calculate: The smallest radius of a track.

Answer to Problem 92A

77.8 m

Explanation of Solution

Given:

Speed of car, $v=12\text{m/s}$

Coefficient of friction, $\mu =0.50$

Acceleration due to gravity on Mars, $g=3.\text{7}{\text{m/s}}^{\text{2}}$

Formula used:

Centripetal force:

  $F=\frac{m{v}^2}{r}$

Here, m is the mass, v is the velocity and ris the radius of the circular track.

Frictional force:

  $f=\mu mg$

Here, $\mu$ is the coefficient of friction, m is the mass and gis the gravitational acceleration.

Calculations:

The centripetal acceleration must be equal to the frictional force otherwise the car would skid.

  $\begin{array}{c}\mu mg=\frac{m{v}^2}{r}\\ r=\frac{v^2}{\mu g}\\ =\frac{{12}^2}{0.50\times 3.7}\\ =77.8\text{m}\end{array}$

Conclusion:

Thus, the smallest radius of a track is 77.8 m.