Question

A $1500kg$ car drives around a flat $200-m$-diameter circular track at $25m/s$. What are the magnitude and direction of the net force on the car? What causes this force?

Short answer

The net force is $F_{\text{net}}=9.4\times {10}^3\mathrm{N}$ and its direction is toward the center.

The static friction causes this force.

Step-by-step solution

Given Information

A $1500kg$ car drives around a flat $200-m$-diameter circular track at $25m/s$.

Explanation

The car's velocity is tangent to the circle of motion, and its acceleration is called the centripetal acceleration and it points toward the center of the circle. This acceleration is given by equation (8.4) in the form

$\overrightarrow{a}=\frac{v^2}{r}$

Where $r$ is the radius of the circle and $v$ is the velocity of the car. At a uniform circular motion, the velocity vector has a tangential component and the acceleration vector has a radial component. From Newton 's first law, the car has a net force exerted on it as it doesn't move with a constant velocity. So, using the expression of the acceleration from equation (1), we get the net force by

$\overrightarrow{F_{\text{net}}}=m\overrightarrow{a}=\frac{m{v}^2}{r}$

Explanation

This force has a direction toward the center of the circular motion. The car moves in a circle with a diameter $d=200\mathrm{m}$, so we can find its radius $r$by

$r=\frac{d}{2}=\frac{200\mathrm{m}}{2}=100\mathrm{m}$

Now, we plug the values for $m,v$and $r$into equation (2) to get $F_{\text{net}}$

$F_{\text{net}}=\frac{m{v}^2}{r}$

$=\frac{(1500\mathrm{kg})(25\mathrm{m}/\mathrm{s}{)}^2}{100\mathrm{m}}$

$=9.375\times {10}^3\mathrm{N}$

$=9.4\times {10}^3\mathrm{N}$

The car moves on a road in a circle motion, there is no tension force between the car and the center of the circle, also there is no normal force, so the static friction causes this force.

Final Answer

The net force is $F_{\text{net}}=9.4\times {10}^3\mathrm{N}$ and its direction is toward the center.

The static friction causes this force