Circular Motion## Newton's Second Law

## applied to a

## Flat Curve

What provides thecentripetal forcenecessary for a car to make a curve on aflatroad? What happens if that force isnotpresent?

Gravity pulls down on the car with its weight,

w= mg. The level road pushesupwith a normal forcenandhorizontallywith a friction forceF_{frict}.This is the force of

staticfriction! As the car moves and the tires rotate, the tires aremomentarilyat rest with respect to the road. Otherwise, the tiresskid!

Remember, the friction force can be

anyvalue from zero up to a maximum of F_{s}=_{s}n when the car is just on the verge of sliding. We will consider this case, when the car is just on the verge of sliding. This meansF

_{frict}= F_{s}=_{s}nThe diagram at the right shows all these forces. There is no vertical component to the acceleration so we find that

n = m g

which means

F

_{net}= f_{s}=_{s}m g = m v^{2}/r

_{s}g = v^{2}/r

_{s}= v^{2}/g rThat is, we must have a coefficient of static friction of

_{s}= v^{2}/g r to provide the friction force to allow a car, traveling at speed v, to make it around a flat curve of radius r.Or, we might find the speed in terms of this coefficient of static friction,

v

^{2}=_{s}g rv = SQRT[

_{s}g r]This is the maximum speed that a curve of radius r can be taken when the coefficient of static friction between tires and pavement is

_{s}.If the velocity increases, the radius r will also increase! This means the car will not follow the intended curve and may run off the road entirely!

What must be the coefficient of friction between the tires and the level roadway to allow a car to make a curve of radius r = 350 m at a speed of 80 km/h?

Table of Contents(c) Doug Davis, 2001; all rights reserved