Guide 6-3a. Solving Circular Motion Problems animation by Behrooz Mostafavi

This is the first of two web pages on this subject.

Circular motion is one of those topics in physics that we can easily experience ourselves.  Riding roller coasters, tilt-a-whirls, giant swings, the Starship 2000 and other such rides are a fun way to experience circular motion.  It's common nowadays for physics classes to take field trips to amusement parks and measure accelerations on the rides.  There are other more common but hopefully less thrilling circular motion experiences such as going around curves in a car, going up and down hills, swinging, and so on.  What these experiences all have in common, other than the fact that they involve circular motion, is that you experience them firsthand from a non-inertial frame of reference.  As you move in a circle, your frame of reference is accelerating.  Even if you're moving at constant speed, your velocity is changing, because you're changing directions.  Since your reference frame is non-inertial, Newton's Laws don't apply!  This means that your experiences of circular motion don't really help you in solving circular motion problems from inertial frames of reference.  You need to train yourself to talk about these situations in a different way than you're used to.

For example, when you ride on a merry-go-round (especially a fast one) you feel like you're being pushed outward.  That's your point of view as a non-inertial observer.  Someone on the ground (an inertial frame of reference) who knows Newton's Laws realizes that a net force is required to make you move in a circular path.  Otherwise, you'd just move in a straight line at constant velocity.  The net force is pulling you inward away from the straight-line path you would follow if the force were not present.  This leads us to the first important thing that you need to remember about circular motion.

Circular motion is the result of a net force pulling the moving object toward the center of the circular path.

The next thing to realize is that the inwardly-directed force is due to the influence of those you've already studied, namely, friction, normal, weight, and tension.  Thus, when you identify on a force diagram those forces that give rise to circular motion, you use the same words you've used before.  This brings us to the second important thing that you need to remember about circular motion.

The forces that give rise to circular motion are friction, normal, weight, and tension.

Note that we haven't used the word centripetal up until now.  That's because centripetal is simply a name we give to net force when it results in circular motion.  Centripetal force is simply net force.  You never show net force on a force diagram.  That's because net force is simply the vector sum of the individual forces that act on an object.  The same applies to centripetal force.  Thus, we come to the third important thing.

Centripetal force, like net force, is never drawn on a force diagram.

Like the last item, this next item isn't anything new, but it's worth restating.  In solving a net force problem involving circular motion, select the direction of positive displacement to be in the direction of the acceleration.  In the specific case of circular motion,

Select the direction of positive displacement to be toward the center of the circular path.

Granted, the direction selected will change as the object moves.  However, the net force analysis is done at one instant of time.  If the object is moving at constant speed in a circle, then the analysis applies equally well to any instant of time.  There's one other thing you need in order to be ready to solve net force problems involving circular motion.  When an object experiences a net force, hence an acceleration, toward the center of a circular path, the magnitude of that acceleration may be calculated using v²/r.  Thus,

The magnitude of centripetal acceleration is given by the square of the magnitude of the velocity of the object divided by the radius of its circular path.

Keeping these 5 things in mind, let's look at some common situations that involve circular motion. These situations are the ones you will typically encounter. You need to be able to recognize them when you see them.

The Horizontal Swing (the most common circular motion problem)

The following problem will show up again and again in different disguises, but the physics will always be the same.  So we present it to you first.

Consider the case of a ball being swung in a horizontal circle at the end of a string as shown to the right. The point P is a fixed support.  The acceleration is toward the center of the circle C.  An overhead view of the path is shown in Figure A below.  The velocity vector is tangent to the path, and the acceleration vector points to the center of the path.  Figure B shows the forces from two perspectives.  From an overhead perspective, we see that the force responsible for the ball's circular motion is a tension force.  We have to be careful, though, to examine the forces from the side view in order to get the full story.  The tension force has both horizontal and vertical components, since the string makes an angle θ with the vertical.  The vertical component of the tension force balances the weight if the ball remains at a constant height.  The horizontal component of the tension force is responsible for the circular motion.  Now let's look at the net force equations.

 Figure A Figure B

Note that +x is horizontal, because the plane of the circular path is horizontal.

Fnet,x = Tsinθ  (We use the sine function, because we're interested in the component of the tension force opposite θ, as defined in the diagram.)
Fnet,y = Tcosθ - mg

Applying the second law,

max = Tsinq
0 = Tcosθ - mg.

Rearranging the second equation,

max = Tsinq
mg  = Tcosθ

The reason for the rearrangement is to show that if we divide the upper equation by the lower, the mass and tension divide out, leaving

ax/g = tanq.

We can now substitute ax = v²/r, where r is the radius of the circular path, and solve for v.

v²/rg = tanθ

v = (rgtanq)½

A caution before we go on to another problem:  It's never a good idea to memorize a solution such as the one above.  You need to be able to analyze such situations by starting with a force diagram and from that setting up net force equations which you can then solve.  Of course, it helps to have experience with the algebra specific to this situation.  That's what you get by doing practice problems.

When you read section 6-5, you encountered a problem that leads to the same solution as the one above.  What problem was it?