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Unit 3 Notes, Part 2: Centripetal (Center-Seeking) Acceleration and Uniform Circular Motion
Uniform Circular Motion describes the motion of an object traveling at a constant speed on a circular path. The direction is not
constant, so the object does not have a constant velocity. Therefore, it must be accelerating and there must be a force causing
this acceleration. Examples of objects that have a uniform circular motion are a penny on a record on a record player going
around in circle, satellites or planets in orbit, an airplane changing directions, a car moving around a curve.
For an object moving with constant speed to remain in a circular path there must be a force acting on it that is perpendicular to
its motion. Look at the diagram below and picture each velocity vector as the direction the object’s motion. If there was no
force acting on the object, the object would just continue in a straight line at that same speed. However, an object in uniform
circular motion doesn’t move in a straight line; it continually changes direction. That direction change is due to a force that
acts perpendicular at every single moment or point in time. Since the force is directed perpendicular to the object’s velocity, it
won’t change the speed of the object, only the direction. Hence the name “uniform” circular motion.
The force acting on the object
causes an acceleration in the same
direction as the force. Both the
force and the acceleration are
“center seeking” so objects moving
in a circle are referred to as being
in “Centripetal Motion” as
“centripetal” literally translates to
“center seeking”. Since the force
that the object is experiencing is
constantly changing direction, the
acceleration that the object is
experiencing is constantly
changing direction, too, so it is not
a constant acceleration. Due to
this you can’t use the four funky
equations to solve problems in this
unit (those equations are only for
constant acceleration). Also, since
the force is always directed
perpendicular to the velocity,
absolutely no work is ever done by the centripetal force (this statement will make sense in the next unit). So, here comes the
Physics…..
Objects undergoing UCM (Uniform Circular Motion) will have a constant speed (not velocity). In previous units you used
d
to calculate an object’s constant (average) speed. You will use a similar equation in this unit, just specific for
t
“circular” motion. Since the unit circle is based on π, the “d” in the “speed equation” will also be based on π. The equation
v
you will use to find out how fast an object is moving in UCM use the following equation:
v
2  r

T
where:
v=
speed (NOT velocity); units are m/s
r=
the radius of the circle ; unit is the meter
T=
the “Period” of the object; the period is the time (in seconds) of one complete revolution (around the circle).
Notice the proportionalities: If the Period (the time of one revolution) remains constant, a larger radius will lead to a greater
speed. Conversely, with a constant radius, the larger the Period (the time of one revolution), the smaller the speed.
Unit 3 Part 2--- Uniform Circular Motion Notes
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In order to calculate the acceleration that an object in UCM has, you would use the following centripetal acceleration equation:
ac 
v2
r
where:
ac =
the centripetal acceleration that an object in UCM has; the units are m/s/s (just like any acceleration)
v=
r=
the speed of the object in UCM; units are m/s
the radius of the circle; unit is the meter
Notice in the above equation that the acceleration is proportional to the speed squared. So, if the speed doubles, the
acceleration must have quadrupled. You will NOT be required to know the derivation of this particular formula. I don’t even
know it. And I don’t want to.
And…in order for an object to have an acceleration, there must be a net Force acting on the object to provide the accelation.
The net force must always be directed toward the center of the circle (in order for an object to remain in a circular path) and is
in the same direction as the acceleration. It is perpendicular to the speed of the object at every single moment. This net force
is always provided by something. For example, if you are swinging an airplane connected to a string in a perfectly horizontal
circle around your head, the Fc (centripetal force) only comes from the tension in the string. So, T = F c for this particular
scenrio. And, if you are driving a car around an unbanked curve on a horizontal level of ground, the Fc comes from the Ff
(static force of friction, where Ff = μN). The centripetal force (Fc) is simply the net force of all the forces that happen to be
acting on the object in the plane that is directed toward the center of the circle.
The equation that you will use that will make use of the net centripetal force is:
Fc  mac
Fc 
 2  r 
m

T 

Fc 
r
mv 2
r
2
Again, the centripetal force is simple the name of the NET force that is required to keep a mass (m) moving at speed (v) in a
circular path of radius ( r). It points in the same direction as the acceleration (toward the CENTER of the circular path) and is
always changing direction so it is NOT a constant force (it might or might not be be a constant magnitude, but never a constant
direction). Examples of forces that might contribute to the overall net centripetal force are the Gravitational force (mg),
friction, tension, and the normal force. Each situation will be a little different, so you will have to use your head when
determine what “force” to use. FYI: Always choose the direction of the centripetal acceleration as “positive” for centripetal
motion problems.
Problems:
1. A model airplane has mass of 0.90 Kg and moves on a 17m guideline at constant speed in a circle that is parallel to
the ground. Find the tension in the guideline for a speed of 19m/s.
m = 0.90 Kg
r = 17m
v = 19m/s
T= ?
you’re actually looking for the Fc (net force directed toward the center of the circle) which happens to be the
tension in this problem
Fc 
2.
mv 2
r
19.11 N
Compare the maximum speed at which a car can safely negotiate an unbanked turn of radius 50m in dry weather
(coefficient of static friction = 0.900).
v =?
r = 50m
μ = 0.9
Fc 
mv 2
r
Ff 
mv 2
r
mg 
mv 2
r
g 
v2
r
v  gr
21 m/s
Unit 3 Part 2--- Uniform Circular Motion Notes
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In the above problem the net force that is causing the centripetal motion is the friction force between the tires and the road.
3.
In an amusement park ride, a small child stands against the wall of a cylindrical room that is then made to rotate. The
floor drops downward (out from under his feet) and the child remains pinned against the wall. If the radius of the
device is 3.34 m and the relevant coefficient of friction between the child and the wall is 0.47, with what minimum
speed is the child moving if he is to remain pinned against the wall?
mg
r
N
Ff (= mg)
For a situation such as above, ALWAYS draw a FBD (free-body diagram). The person is standing as the room starts to spin.
All of a sudden the floor drops away and the only force that is holding him against the wall is the friction between him and the
wall. Since the friction is opposite the weight and the person isn’t accelerating in the vertical plane, the Ff must be the same
magnitude as the mg. But remember that friction is related to the Normal Force by the equation F f =  N. AND…in this
scenario the weight is NOT equal to the N .
r = 3.34m
μ = 0.47
v=?
Ff =  N
mv 2
Fc 
r
mv 2
N
r
mv 2


r
Ff
mv 2


r
mg
v2

 r
g
v
gr
u
In the above example, the N is the entire net Fc as that is the only force that is in the direction of the acceleration. You might
be tempted to call the Ff the NET force (Fc) as that is what we did in the last example; but in the last example it truly was the
friction that was directed toward the center of the circle. Remember, the Fc is the “net center seeking force”.
Vertical Uniform Circular Motion
FYI: Always choose the direction of the centripetal acceleration as “positive” for centripetal motion problems.
You know those little hot-rod tracks that the little cars can go around a vertical circle and not loose contact with the track at the
top (inside) of the track? That is an example of “Vertical Circular Motion”. In vertical motion, the problem will ask you
(usually) one of two things: (1) For “inside the track” vertical motion, what is the minimum speed that an object needs to have
in order to remain in contact with the track? And (2) For “outside the track” vertical motion, what is the maximum speed that
an object can have and still remain in contact with the surface? We’ll examine situation #1 first.
Inside of the “circle” motion:
mg
N
For a situation where the object is inside the loop, the object will have contact
with the track as long as there is a “normal force”. The moment the normal force
goes to “zero”, the object has lost contact with the track. In the picture to the left,
the normal force points down as the object is inside the track (so it’s upside down
on the top of the track). So, in this example, BOTH the normal and the weight
point in the same direction…which is down.
So, in order to find the minimum “v” the object must have in order to remain in
contact with the track, set-up a “Fnet” equation for the object (but with UCM
equations ).
Unit 3 Part 2--- Uniform Circular Motion Notes
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mg + N = mac
mg + N =
mv 2
r
The minimum speed is right when N goes to zero (there is no Normal Force if the car isn’t in contact with the surface). So,
to find that velocity, set N = 0N and the equation becomes:
mg 

mv 2
r
g
v2
r
v  gr
Summary of the above: For an object moving inside of a circular track at the top of the track, the net centripetal (Fc)
force is the sum of the weight and the normal force (mg and N are in the same direction).
If the object is inside of the track but at the bottom of the circular track, the
weight is opposite the normal force so the centripetal force (F c) would be the
difference between the two. If the object is moving with some speed in this circle,
it must have some acceleration and net force toward the center of the circle. This
makes it so mg will not equal N at the bottom of the circle…the N must be greater
than the weight at the bottom of the circle. It’s a lot like the elevator problems. (+
dxn toward the acceleration).
mg
N
N - mg = mac
N - mg =
mv 2
r
Outside of the “circle” motion
mg
For an object moving outside of the circle but on top of the loop (like a skier on
top of a mountain), if they are moving too fast they will loose contact with the
surface of the loop. To determine the maximum speed that an object can have
without losing contact with the surface, write the Fnet equation for the object just
like you did in the previous example (+ dxn toward the acceleration).
N
mg – N = mac
mg – N =
mv 2
r
The maximum speed is right when N goes to zero (there is no Normal Force if you
aren’t in contact with the surface). So, to find that velocity, set N = 0N and the
equation becomes:
mg 
mv 2
r
g
v2
r
v  gr
Do NOT try to memorize these equations. You will need to understand how to derive them on the
AP exam as they will NOT be given to you.
Satellites (in Uniform Circular Motion)
Every satellite (and the space station, the space shuttle, the moon, etc) is kept on its circular path by a net force that is directed
toward the center of its circular radius. This net force is referred to as the centripetal force (Fc) and is simply equal to the
object’s weight (i.e., the “gravitational force” that the earth pulls on the object) at that particular height above the earth’s
surface. There are no other forces acting on an object in orbit. Remember, “g” fluctuates depending on how far the object is
from the earth’s surface so an object’s weight will fluctuate at that height, too (review “Universal Gravitation” if you need to).
Sooo, you can’t simply use “9.8 m/s/s” for “g” for these problems.
Unit 3 Part 2--- Uniform Circular Motion Notes
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Every object in orbit around the earth (or any other planet) is simply in a state of free-fall. The only reason why these objects
don’t “fall” back to the surface of the earth is the speed the object has in its orbit. Together, the speed of any orbiting object in
conjunction with the pull of the gravitational force on it, is such that they cause the object to “fall” at the same rate the earth
curves-out from underneath it. So, since it falls at the same rate the ground essentially moves out from under it, it will never
reach the “ground” unless it slows down (which satellites or that old Russian space station eventually will do/did). So, what
you need to get out of that last statement is that there is only ONE speed that a satellite (or any other orbiting body) can have to
stay in orbit with a “fixed” radius. Follow the logic in the following equalities:
in orbit: “ Fnet = mg”
=
GmM
r2
=
m  ac =
m  v2
r
All of the above statements are equal to each other. So, setting two of the above equations equal to each other in the following
equality (by themselves) will lead to the following equation:
GmM
r2
=
m  v2
r

GM
2
=v
r
v

GM
r
where m is the mass of the satellite, M is the mass of the particular planet the “satellite” is orbiting around, G is a constant that
will never, ever change, and r is the radius of the particular planet the satellite orbits around. Below are some constants that
will be given to you on all exams that will make your life easier:
G= 6.67 E -11 N m2/ kg2
rEarth = 6.38 E6 m
MEarth = 5.98 E 24 kg
The above “derived” equation will NOT be given to you on the AP exam….as a matter of fact, the only equations of this unit
that WILL be given to you are:
GmM
Fg =
r2
and
v2
ac 
r
You will need to understand how to derive all the equations in this unit except the two immediately above. It’s really not that
hard if you follow the logic….
Sometimes a problem will ask you to determine the Period of a satellite (the time of one complete revolution). The only
equation that has Period in it is the equation for the speed of an object in uniform circular motion. But from this equation, you
can link all the equations together in the following way:
v
d
t

Fc  m
vc 
v2
r
=
2  r
v2
and a c 
and Fc  mac = mg
T
r
2
 2  r 
m

 T 
r
=
GmM
r2
=
m  ac
=
m  v2
r
=
 2  r 
m

 T 
r
2
=
Whenever I solve problems with satellites, I am never 100% certain which equation(s) I will use to solve the problem until I
write down the whole equality that governs satellites. Once I write down everything I know, I can usually pick out the path
that I want to take in a few seconds. You need to get in the practice of doing this. Write down what you know….Remember:
That is how you will get points on the AP exam. In addition, if you do as I suggest, that is probably how you, too, will
eventually “see” the solution.
In the above equalities, the mass of the satellite is 99% of the time irrelevant and will cancel out. In addition, make sure you
remember that you might just be tempted to make a shortcut in some problems by sticking 9.8 m/s/s in for “g”. But…you can’t
Unit 3 Part 2--- Uniform Circular Motion Notes
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do that for satellite problems as “g” won’t be 9.8 m/s/s. But, for all satellites, the “g” will be the same value as the ac …and
you might have to determine that depending on whatever each problem asks of you.
Artificial Gravity
To live in outer space (not just inhabit the space station for a few months but to actually “live” there), we would need to have a
way to provide an artificial gravity (g) so we could go about our “normal” lives (such as standing, sleeping, eating, going to the
bathroom, etc). If a space station was rotated about its axis (geometric center) and you were rotating along with it, you would
be accelerating at the same rate (your centripetal acceleration would be the same as the station’s). If you were on a space
station that was being rotated about its axis, you could eventually reach a circular speed that would make it so the floor
provided a normal force that would equal your true weight (mg). At whatever speed a rotating station can produce a normal
force that equals your true weight (mg), that is the speed that is necessary to provide an artificial gravity equal to 9.8 m/s/s.
 2  r 
m

2
GmM
mv
T 


N  mg  mac 

r2
r
r
2
So, for example, if you wanted to know how fast a space station needs to rotate to provide an artificial gravity of 9.8 m/s/s
(“g”), you might choose to go about determining that the following way:
At that “g”, your weight would equal the normal force the floor exerts back on you. So….
N  mg

mg 
mv 2
r

v  gr
, where “g” would be 9.8 m/s/s.
Or, if you just want to determine what speed (v) is necessary, for example, to produce an artificial g on a space station with a
radius of r that equals Earth’s g, you could simply use g = ac = v2/r. It’s not that hard, it’s just weird to think about.
Again, I can NOT emphasize enough the importance of understanding (not memorizing, but actually understanding) how all
the equations in this unit are connected and then using Newtons’s three laws (really, Law #2 will be used the most) and your
“knowns” and “unknown(s)” to figure out which equation(s) to use. Do not memorize…understand…and write all your
equations down so you can figure out which one to use.
Unit 3 Part 2--- Uniform Circular Motion Notes
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