# Cars on Curves

One of the most common uses of the phenomenon of centripetal acceleration occurs in the design of highways and affects the way a car turns in a curve. When the car is on a level road and turning, all motion is in the horizontal plane. The forces are:

We can see from the above that the net force in the y direction is zero, and that the normal force is equal to the weight of the car. In the x-direction, or horizontal axis, the force acting on the car is the friction between the road and the tires. This is the centripetal force. The passengers feel like they are being pushed outward, but there is no centrifugal (outward) force, which is a common misconception. Inertia directs the passengers in a straight line and the car pushes them around the turn. Hopefully, the coffee that was left on the dashboard will also make the turn. But probably not. The direction of the net force continually changes so that it is always directed towards the center.

Here is an interesting phenomenon: The friction between the tires and the pavement is static friction. The rolling of the tires is always bringing new tire in contact with the pavement and the coefficient of friction is relatively large. But, if something causes the wheels to lock, then the friction immediately becomes kinetic and the coefficient drops rapidly, causing a loss of control of the car. This is why we have anti-lock brakes.

When a car is on a banked curve (that is, the road is not level but is on an angle with the horizontal), there are now forces in both
the x and y direction. The important thing to remember here is that *the centripetal force is still directed towards the center of the circle, and not down the slope as might be expected*. The
forces look like:

The banking of curves are designed to reduce the chance of skidding because the normal force, which is perpendicular to the road will have a component toward the center of the circle reducing the dependence of friction. The normal force is resolved into horizontal and vertical components. The centripetal acceleration is horizontal and not parallel to the slope of the road. For a given bank angle, there is one speed for which no friction is required where the horizontal component of the normal force toward the center of the curve is equal to the force required to give a vehicle its centripetal acceleration, which is expressed through this formula:

*F _{N}
sin*

*q*

*= mv2 /r*

Using this information, the formula at which a road should be so that no friction is required can be determined. The formula for a banking angle is:

# *tanq
= v*^{2} /rg

^{2}/rg

The banking angle of a road is chosen so that the condition holds for a particular speed, known as the “design speed”.

**Sample Problem:**

A truck exits the highway on an off ramp. The banking angle of the off ramp is 25° and has a radius of 60m. What is the maximum speed the truck can be traveling at in order to make the turn safely?

*tan**q =* *v ^{2} /rg*

tan25° = v^{2} / (60m) (9.8m/s^{2})

.47 = v^{2} / 588

v^{2} = 1251.06

v= √1251.06

v =35.37m/s

(I've only included the numbers here. You should try this problem on your own - draw a picture, include units with your equations to see how they work out.)