|animation by Behrooz Mostafavi
This is the first of two pages on this
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
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
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
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
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. 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
Fnet,y = Tcosθ - mg
Applying the second law,
max = Tsinθ
0 = Tcosθ - mg.
Rearranging the second equation,
max = Tsinθ
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,
ax/g = tanθ.
We can now substitute ax = v²/r, where r is the
radius of the circular path, and solve for v.
v²/rg = tanθ
v = (rgtanθ)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?
Go on to additional examples.