Question

A 1500-\rm kg car drives at 30 \rm m/s around a flat circular track 300 \rm m in diameter. What are the magnitude and direction of the net force on the car? 9000N

Since the net force is equivalent to the force of static friction, your answer to Part D is the magnitude f_s. Based on this value, what is the minimum coefficient mu_s of static friction between the road and the car?

Answer 1

Concepts and reason

The main concepts used to solve the problem are Centripetal force, weight, normal force, friction force, and Newton’s second law.

Initially, calculate the radius by using the relation between diameter and radius. Later, Use the expression of centripetal force to calculate the net force on the car. Finally, use the expression of frictional force to calculate the coefficient of static friction by using the force calculated in the previous step.

Fundamentals

The relation between the radius and the diameter is expressed as follows:

$r = \frac{d}{2}$

Here, r is the radius and d is the diameter.

The expression for the centripetal force is expressed as follows:

${F_{\rm{C}}} = \frac{{m{v^2}}}{r}$

Here, m is the mass, v is the velocity, r is the radius of the path, and ${F_{\rm{C}}}$ is the centripetal force.

The expression to calculate the frictional force is expressed as follows:

$F = \mu mg$

Here, F is the frictional force, m is the mass, g is the acceleration due to gravity, and $\mu$ is the coefficient of static friction.

Calculate the magnitude and direction of the net force on the car.

The relation between the radius and the diameter is expressed as follows:

$r = \frac{d}{2}$

Here, r is the radius and d is the diameter.

Substitute 300 m for d in expression $r = \frac{d}{2}$ .

$\begin{array}{c}\\r = \frac{{300\,{\rm{m}}}}{2}\\\\ = 150\,{\rm{m}}\\\end{array}$

The expression for the centripetal force is expressed as follows:

${F_{\rm{C}}} = \frac{{m{v^2}}}{r}$

Here, m is the mass, v is the velocity, r is the radius of the path, and ${F_{\rm{C}}}$ is the centripetal force.

Substitute 1500 kg for m, $30\,\,{\rm{m/s}}$ for $v$ , and 150 m for r in expression ${F_{\rm{C}}} = \frac{{m{v^2}}}{r}$ .

$\begin{array}{c}\\{F_{\rm{C}}} = \frac{{\left( {1500\,{\rm{kg}}} \right){{\left( {30\,{\rm{m/s}}} \right)}^2}}}{{150\,{\rm{m}}}}\\\\ = 9000\,{\rm{N}}\\\end{array}$

Calculate the minimum coefficient of static friction between the road and the car.

The expression to calculate the frictional force is expressed as follows:

$F = \mu mg$

Here, F is the frictional force, m is the mass, g is the acceleration due to gravity, and $\mu$ is the coefficient of static friction.

Substitute $9000\,{\rm{N}}$ for F, 1500 kg for m , $9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for g in expression $F = \mu mg$ .

$\begin{array}{c}\\9000\,{\rm{N}} = \mu \left( {1500\,{\rm{kg}}} \right)9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\\\\mu = \frac{{9000\,{\rm{N}}}}{{\left( {1500\,{\rm{kg}}} \right)9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}\\\\ = 0.6122\\\end{array}$

Ans:

The magnitude and direction of the net force on the car is $9000\,{\rm{N}}$ and towards the center.