3.5 Notes – Special Case 2: Circular Motion
Q: What determines whether or not something will travel in a circular path?
A: Whether or not there is a force that pulls it into a circular path.
Q: What is centripetal motion? Is there a true “Newtonian” force associated
with centripetal motion?
A: Center-seeking, thus, circular motion. There IS a force pulling the object
in. The evidence is two-fold:
1. The object accelerates due to this force by constantly changing its
direction.
2. There is a definitive action-reaction pair—force source pulls on object;
object pulls on force source.
Q: What is centrifugal motion? Is there a true “Newtonian” force associated
with centrifugal motion?
A: Outward-seeking motion. There is NOT a force pulling the object outward.
The evidence is that when the force source is “cut,” the object does not
continue to fly outward, but rather, flies off tangent to the circular motion at
that point. Thus, there is no true Newtonian counterpart to the force, or no
action-reaction pair.
Q: What is centrifugal motion, REALLY?
A: The object initially flies outward due to the application of an initiation
force. Once that happens, the object is set in motion, and thus continues
moving due to its own INERTIA. If it is somehow tethered (by a rope or cord,
etc.), or stopped by the wall of a container, it will, from that point forward, be
undergoing centripetal force. Again, the centrifugal force is not really a force
at all; just the inertia of the object.
Consider a pendulum bob swinging back and forth.
Q: Is it undergoing centripetal motion? How do you know?
A: Even though it’s only part of a circle, if traced all the way around (3600)
it’s still a circular path, so YES, it’s undergoing centripetal motion.
Q: What is Newton’s 2nd Law of Motion (N2)? What does it really mean?
A: Newton’s 2nd Law of Motion is F is proportional to m••a. In other words, if
you apply an outside force to an object, it will accelerate in direct proportion to
the force applied, and in inverse proportion to its own mass.
With regard to N2, If you apply an outward force to an object:
1. You will accelerate it by changing its velocity, either by changing its
a. Magnitude
b. Direction
2. How much it accelerates will be in direct proportion to the force you apply,
and in inverse proportion to the mass of the object itself. In other words, if
you apply a force of 1N to a 1 kg object, in the absence of friction, you will then
accelerate it by 1 m/s/s. If you keep the force the same, but then double the
mass, because the object is now morr massive, obviously the amount to which
you can now accelerate it is going to be lessened. Specifically, all other things
considered equal, it will be cut in half.
With regard to the pendulum, If you trace out its path, you will see that it
gains speed as it falls (due to gravity), maxing out at the bottom of the curve,
and then starts to lose speed as it ascends up the other “half” of the curve.
Q: What would you get, graphically, algebraically, and conceptually, if you
align v1 with v3? (Arrange them tail to tail, smaller subscript on the bottom,
and make sure magnitudes and slopes are maintained ☺.)
A: You would get ∆v, which is the same as acceleration.
Q: What would you get if you align v3 with v5 tail to tail? What does this
vector look like? WHY does it look that way? Should it look that way?
A: It is in the direction of the centripetal source of the motion, thus, straight
up toward the center of the implied circle. It should look like this because that
is the point at which it stops curving down and starts curving up. For all of the
other points, only the component of acceleration that is simultaneously in the
direction of the motion contributes to the continuance of that motion, so YES, it
should look that way. ☺
So, for diagram (d) above, the ∆v, or a, goes straight up. In diagram (e), the
resultant acceleration vector points inward (in agreement with the continuance
of circular motion). But, in determining ∆v for the motion depicted in
traveling from v5 to v7, the resultant acceleration vector points to the left,
meaning in will then be pulling “backward” to perpetuate circular motion in
that direction.
Uniform Circular Motion
DEF: Uniform Circular Motion = the speed of an object traveling in a circular
path stays constant.
For the above diagram, Assume the following:
1. The position and velocity vectors are accurately drawn for a particle
undergoing uniform circular motion.
2. The change in the angle that occurs between the velocity at one instant,
and the new velocity that occurs at the next instant will be the same as
the change in the angle that occurs between the position at one instant,
and the new position that occurs at the next instant. This is because the
velocity and position vectors have to remain mutually perpendicular for
there to be uniform circular motion. If not, you would get a speeding
up-slowing down effect, which would then affect your change in
position, accordingly.
3. The maintenance of ∆θ means that you have isosceles triangles for both
the formation of ∆v and ∆r vectors.
4. Sharpen your brain for this one: To find the direction of the acceleration
vector, look at two velocity vectors, and consider their resultant ∆v.
a. The sum of the interior angles for ANY triangle is 1800.
b. Because the triangles are isosceles, the base angles are equal.
c. As ∆t → 0, ∆θ also → 0. This means you are “pulling the triangle
out of existence,” and the two base angle are, thus, approaching
900 each. This also means ∆v is ⊥ to v, the velocity at that point.
d. Drawing ∆v, which is acceleration, means pointing it in the
centripetal direction.
e. Since the two triangles are similar (by geometric definition, AA
postulate), if you divide both sides by ∆t, you get ∆v = v
∆r

∆t
r ∆t
f. Take the limit as ∆t → 0, and the left side looks like the magnitude
of the instantaneous acceleration, and the right side looks like the
magnitude of the instantaneous velocity. This gives us
g. The motion of the particle can be described in terms of the period,
T, where the particle travels 2π
πr during one period. So v = 2π
πr / T
STUDY EX 3-14. White-board The exercise under the “remarks” section. ☺