In this chapter we aim to treat a certain class of problems where the force or forces acting on an object change in time, but in a different way than we've treated in past chapters.  Now the force will be constantly changing in direction while it's magnitude may or may not also change.  [br][br]If there is a force on an object that remains orthogonal to the velocity of the object, it will result only in a change in the direction of the velocity while the magnitude (speed) remains constant.  Furthermore, if that orthogonal force itself remains constant in magnitude, circular motion will result.[br][br]While the motion of planets in orbits and cars going around bends in roads is not always along a circular path, at any moment there is always an effective radius of curvature of any curved path. In mathematics this is referred to as an osculating or kissing circle. [br][br]In this sense, if we understand circular motion, we can treat other cases of curvilinear motion by just using the radius of the osculating circle as the radius in our equations of circular motion.  An example of an osculating circle on a sinuous path is shown below.

In the next section we will do all the mathematical derivations of the acceleration due to circular motion, but first let's discuss the basic results and properties since I think it'll help motivate the derivation.  [br][br]The term [i]centripetal [/i]comes from the Latin term centripetus which means center-seeking.  It entered English vocabulary in the 18[sup]th[/sup] century.  In the context of physics it is not at first clear what exactly is doing the "seeking" of the center.  The idea is that the acceleration vector is always pointing toward the center of a path. So the object is always "seeking" to accelerate toward the center of the path.[br][br][b]When objects turn while traveling at constant speed, the acceleration is always directed toward the center of curvature of the path. [/b] In this sense you always know at least the direction of the acceleration from the outset of solving such a problem.  Further, it turns out that the acceleration vector always equals [math]\vec{a}_c=-\frac{v^2}{r}\hat{r}[/math] in such cases where the expression contains the radius of the path and the speed of travel along the path. The negative r-hat part just means radially inward toward the center of the path.[br][br]The term [b]centripetal force[/b] is used to describe the force that causes a centripetal acceleration.  What's confusing about this force is that it doesn't exist.  I don't mean to say that there is no force pulling toward the center of curvature, but rather that we are not witnessing a new force.  The force or forces pulling toward the center of curvature could be many things.  For a planet in orbit it is the force of gravity - not some new or strange force called the centripetal.  For a rock being spun around from a string the centripetal force is the tension or at least one component of it - again, not some strange new centripetal force.  In some cases the centripetal force is no single force acting alone.  It might be, as we'll see in an example shortly, the vector sum of forces like the gravitational force and the normal force.[br][br]Since we know the direction of the acceleration from the outset - and its magnitude if we know speed and path radius, then we also know the direction of the net force.  Why?  Because Newton's second law says so.[br][br]

On a level road, a car is only able to make a turn because there is friction between the tires and the road.  The maximum magnitude of the friction force depends on the coefficient of friction [math]\mu[/math] and on the normal force acting on the vehicle.  The friction force is the only net force acting on the vehicle if we assume the speed is constant (which we will) because in the vertical direction the downward gravitational force is cancelled by the upward normal force.  The friction force acts perpendicular to the velocity, toward the center of the turn. We will derive this fact mathematically in the next section, but for now just realize that the centripetal acceleration acts in that direction and that therefore the friction (being the cause of that acceleration) must also act in that direction.  Since friction is the sole cause of the centripetal acceleration in this example, using Newton's second law we can use friction as the sum of the forces and put a subscript on the acceleration to remind ourselves that it's a centripetal acceleration:  [math]F_f = ma_c.[/math] [br][br]Since [math]F_f = \mu mg[/math] and [math]a_c=\frac{v^2}{r},[/math] we can combine these and find [math]\mu mg=ma_c=m\frac{v^2}{r}.[/math]  This simplifies to [math]\mu g=\frac{v^2}{r}.[/math]Take note that all terms here are scalar, meaning that the gravitational field and velocity terms are magnitudes. Obviously we will look rigorously at the vector side of this expression in this chapter.[br]  [br]In any case, it is always a good idea to take a look at a results that you calculate and ask yourself if they make sense.  Sometimes it will be obvious, sometimes you might learn something.  Other times you might see that you messed up the algebra.  In this case, does it make sense that [math]v[/math] should rise with [math]\mu?[/math]  This means that you can round a turn faster if the friction coefficient is larger. Does it make sense that at the same speed that if you decrease [math]r[/math] that you must either have control of gravity and increase it, or will need stickier tires (bigger [math]\mu[/math])?  In other words, small r mean a sharper turn. Of course that requires more friction. Why is [math]g[/math] in the equation?  Do you see that a car couldn't corner on level ground without gravity?  Why do you suppose that this is true?  I won't answer this except to ask you to look at the friction expression.[br]