Circular Motion

Imagine that a particle is subject to a force of constant magnitude but whose direction may change. The particle's acceleration at any instant would be in the direction of the force at that instant. The change in the particle's velocity over a very short time would be a vector in the direction of the average acceleration. The new velocity at the end of this tiny time interval would be the vector sum of the original velocity and the change in velocity. The displacement of the particle during the little time slice would be given by the average velocity times the Dt. Now suppose that the changing direction of the force was such that the force was always perpendicular to the velocity. The Central Force display illustrates this situation.

Notice that in this example that the force bends the path of the particle into a circle and that the force vector and therefore the acceleration always points toward the center of that circular path. The magnitude of the velocity along the path remains constant. Under these conditions the particle is said to be undergoing uniform circular motion where "uniform" means the speed of the particle is constant. We have evidently caught this system in a delicate balance where in each Dt the force deflects the particle just enough from the trajectory it would have followed, a straight line in the direction of the velocity, that it ends up on a circular path. The question now is what must be the relationship among the acceleration, velocity and radius of the circle for us to get this nice result.

Here we are going to work some tricks that you might leave you thinking, "There is no way I would have thought of this on my own!". What we are going to do is a typical physicist's ploy of looking around for any relationship among the variables in which we are interested. Then seeing if there is any logic that leads to the relationship we want. One of the things that makes this seem like magic is that we do not show you all the false leads and dead ends that were tried before this line of reasoning presented itself. The other thing is, this business gets easier with experience. Having worked out a few of these connections helps in working out new ones.

This sort of thing, by the way, drives us mathematicians crazy. We like things to follow absolutely one step after the other so that we are driven inevitably to the correct solution. This business of jumping in sort of in the middle of a problem with some idea what the answer is going to be and using a mixture of physics, logic, geometry and intuition to get a result that then may be tested by experiment is really a physicist thing.

Using the image at the left, taken from the
Central Force display, Take a good look at the little
red/yellow/blue triangle made up of original velocity, change in
velocity and new velocity vectors. Now compare that to the figure
made up of the two gray radius lines and the arc included between
them. Except for the fact that the second figure has a curved
line for one side the two are similar triangles, meaning that the
angles in the two triangles are the same. By taking Dt sufficiently small, the effect
of the curvature may be made negligible.

Now the thing about similar triangles is that the ratios of corresponding sides are equal. So the ratio of the yellow side over the red side in the small triangle is equal to the ratio of the arc length over the radius in the larger figure. The length of the yellow side is the magnitude of the change in velocity, |Dv|. The length of the red side is the magnitude of the velocity, |v|. The length of the arc is the magnitude of the velocity times the time increment |v|*Dt. And the length of the gray line is just the radius of the circle, r. So we get the following relationship.

|Dv| /
|v| = |v| * Dt / r .

We were interested in the relationship among acceleration, velocity and radius which gave us this nice circular motion. The magnitude of the acceleration,|a|, is |Dv| / Dt so let's divide both sides of the preceding equation by Dt. Then to get acceleration by itself on one side of the equation, multiply both sides by |v|. These maneuvers get us this relationship,

|a| = |v|^{2} / r ,

which ties together the acceleration, velocity and radius as we
set out to do. Any time we find a particle in uniform circular
motion it has an acceleration of magnitude |v|An acceleration of the sort we have been talking about, one
that points toward the center of the circular motion of a
particle, is called centripetal (center seeking) acceleration.
Any particle whose direction is changing is undergoing a
centripetal acceleration of magnitude |v|^{2} / r where r
is the radius of
curvature of the particle's path. The direction of the
centripetal acceleration is along the radius of curvature.

Now
let's imagine a particle whose path is curved but not
circular. We know that one component of the acceleration must be
|v|^{2} / r in the direction of r, where r is the radius
of curvature of the path. This component of the acceleration
contributes only to the change in direction of the particle since
it is perpendicular to the path and therefore can not affect the
speed of the particle along the path. If the total acceleration
includes a component tangent to the path then the speed of the
particle is affected. The Curved Path display illustrates these
acceleration components. In the Curved
Path display you will see the path of a particle which is
moving along the x axis at constant speed and subject to a force
toward the x axis proportional to the y displacement.

Now let's go back and look at circular motion, but not uniform circular motion. Consider a weight attached to a rod of negligible mass which is suspended from a pivot so the rod and weight could swing freely in the vertical plane. The Pendulum Accelerations display shows you such an arrangement.

There is a lot to learn from this little pendulum
display. Perhaps the first lesson is that the physics of everyday
objects like a pendulum can get pretty messy, and we haven't
even got to the friction part of the story yet. The second lesson
is that we must pay careful attention to what we are really
seeing. Because of the path and speed of the pendulum weight, we
know that the radial and tangential components of its
acceleration are as displayed. If they were any different the
weight would have some other motion. What this display does not
show you is the actual forces which result in these *net*
accelerations. The only forces acting on the pendulum weight are
gravity and the force applied by the rod. Somehow these must
always add up to the total acceleration times the mass of the
pendulum weight, in accordance with Newton's second law.

Near the bottom of the swing, the tension in the rod must be sufficient to both support the weight and curve its path into a circle when it is moving its fastest. Near the 180 degree position, the rod will actually go into compression, supporting the weight when its speed is near zero. You will encounter many interesting problems based on a pendulum like this. For example, what would have to be the speed of the pendulum at the top of its loop for the force exerted by the rod to be zero? Try to work this one out based on what you now understand.

In the next lesson in this course we will introduce the ideas of work and energy.