# What forces are applied as a car exits a curve with constant speed?

Imagine a car taking a curve with constant speed. Assuming no friction or air resistance all we would have would be centripetal force. As the car's front wheels straighten to exit the curve the centripetal force gets smaller and smaller. Yet, the car's velocity's direction follows the wheels' direction. So there has to be some force making that happen, right?

What force is that? Where does it come from? How is it calculated?

When a car turns, the wheels are set to rotate in a direction other than forward. The normal rolling action of the wheel will cause the car to turn. This is the path of least friction - in the absence of friction (such as on ice), the car would continue to go forward.

When the car has started to turn, the occupants and other loose things would continue to head straight forward, under the inertia. But friction and physical restraints would cause these to stop their motion. (Excess motion could cause toppling).

Centrifugal force is not a real thing. It is simply an indicated force required to keep something in circular motion. In saying that centrifugal force is matched by gravity (in an orbit), the centrifugal force is a condition needed to be applied to keep the object in orbit, and the gravity is the substance of the condition.

• Sure, it makes sense that the car would follow that path of least friction. But for that to happen the car needs to rotate, which would indicate that there needs to be some sort of torque on the car, right? – Mark May 9 '15 at 8:35
• It's actually friction on the tyres which cause the rotation. This produces a force at an angle to the motion, which causes the turning. The force is transmitted to the car through a series of structual stresses. This is what makes tyres blow, for example. – wendy.krieger May 9 '15 at 9:17
• So, as the tyre are rotating to the right, there will be friction force to the right (perpendicular to the wheel) as it undergoes rotation? – Mark May 9 '15 at 9:28
• You should understand that when the wheel is turned, there is some resistance to the forward motion, and this resistance comes from a sideways push of the ground on the tyre. This is where the wear comes. You then have this force pushing at an angle to the car which cause it to turn. It does not happen on ice, and burnouts are caused by excessive friction. – wendy.krieger May 9 '15 at 22:30
• +1 I would only add that in a car it takes a certain amount of time to move the steering wheel when transitioning from straight line to circular arc and back. In that time, the car is actually traveling in a spiral, so modern highways and railways all have spiral segments leading into and out of curves. (I programmed those on a 360 model 40!) – Mike Dunlavey Dec 1 '15 at 1:26

You may have inadvertently begun to answer your own question. The centripetal force decreases as the car decreases it's rate of turning to exit the curve.

If you think about it, it is actually really intuitive. The car wants to go in a straight line, and if you are driving on ice and the wheels stop turning and you no longer have friction on the wheels, the car will slide in a straight line approximately tangent to the curve it was just following. The only thing keeping the car on a curve is the centripetal force, so when that force decreases, the cars velocity starts to straighten out until there is no longer a centripetal force and you are travelling in a straight line.

In other words, there is no force causing the exiting of a curve, it is precisely the lack of force.

• I'm not sure. As you said, without centripetal force the car would slide tangent to the curve it was following. So the velocity would still be tangent to the curve as the wheel straightens, but the wheel would not be tangent anymore. – Mark May 9 '15 at 6:40
• I don't understand what you are trying to ask. Could you please elaborate? – user01520 May 9 '15 at 10:27