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Circular Motion
Speed/Velocity in a Circle
Consider an object moving in a circle
around a specific origin. The DISTANCE the
object covers in ONE REVOLUTION is
called the CIRCUMFERENCE. The TIME
that it takes to cover this distance is called
the PERIOD.
scircle 
d 2r

T
T
Speed is the MAGNITUDE of the
velocity. And while the speed may be
constant, the VELOCITY is NOT. Since
velocity is a vector with BOTH
magnitude AND direction, we see that
the direction of the velocity is ALWAYS
changing.
We call this velocity, TANGENTIAL velocity as its
direction is draw TANGENT to the circle.
Centripetal Acceleration
Suppose we had a circle with angle, , between 2
radaii. You may recall:
s
r
s  arc length in meters

v
v
v
vo

vo
s v
 
r
v
s  vt
vt v

r
v
v 2 v

 ac
r
t
ac  centripetal acceleration
Centripetal means “center seeking” so that means that the
acceleration points towards the CENTER of the circle
Drawing the Directions correctly
So for an object traveling in a
counter-clockwise path. The
velocity would be drawn
TANGENT to the circle and the
acceleration would be drawn
TOWARDS the CENTER.
To find the MAGNITUDES of
each we have:
2r
vc 
T
2
v
ac 
r
Circular Motion and N.S.L
2
Recall that according to
Newton’s Second Law,
the acceleration is
directly proportional to
the Force. If this is true:
v
FNET  ma ac 
r
2
mv
FNET  Fc 
r
Fc  Centripetal Force
Since the acceleration and the force are directly
related, the force must ALSO point towards the
center. This is called CENTRIPETAL FORCE.
NOTE: The centripetal force is a NET FORCE. It
could be represented by one or more forces. So
NEVER draw it in an F.B.D.
Examples
2r
vc 
T
The blade of a windshield wiper moves
through an angle of 90 degrees in 0.28
seconds. The tip of the blade moves on
the arc of a circle that has a radius of
0.76m. What is the magnitude of the
centripetal acceleration of the tip of the
blade?
2 (.76)
vc 
 4.26 m / s
(.28 * 4)
v 2 (4.26) 2
ac  
 23.92 m / s 2
r
0.76
Examples
Top view
What is the minimum coefficient of static friction
necessary to allow a penny to rotate along a 33
1/3 rpm record (diameter= 0.300 m), when
the penny is placed at the outer edge of the
record?
F f  Fc
FN
mg
Side view
Ff
mv 2
FN 
r
mv 2
mg 
r
v2

rg
rev 1 min
33.3
*
 0.555 rev
sec
min 60 sec
1sec
 1.80 sec
T
rev
0.555 rev
2r 2 (0.15)

 0.524 m / s
T
1.80
v2
(0.524) 2
 
 0.187
rg (0.15)(9.8)
vc 
A 1500-kg car is traveling at 24 m/s
through a flat 100-meter radius turn.



How large is the frictional force required to
keep the car moving in its circular path?
What is the correspondingly minimal
coefficient of friction between the road
surface and the car's tires?
How would this coefficient of friction be
changed if a 3000-kg pickup truck were
traveling through the same curve?
Answers
Ff = Fc
Ff = m(v2/r)
Ff = (1500)(242/100)
Ff = 8640 FN
Ff = μFN
Ff = μ(mg)
8640 = μ(1500)(9.81)
0.587
Vertical Circle


Unlike horizontal circular motion, in vertical
circular motion the speed, as well as the
direction of the object, is constantly changing.
Gravity is constantly either speeding up the
object as it falls, or slowing the object down
as it rises.
We will begin by looking at two special
positions which are usually analyzed in
problems: the very top of a vertical circle and
the very bottom of the circle.
Now, consider an example of a person riding a
roller coaster through a circular section of the track,
a "loop-the-loop”
Let's look at the formulas needed to
calculate the normal force, N, exerted
on a object traveling on the inside
surface of a vertical circle as it passes
through the bottom and through the
top of the ride
While at the bottom
At the top
net force to the center = N - mg
N - mg = m(v2/r)
N = m(v2/r) + mg
net force to the center = N + mg
N + mg = m(v2/r)
N = m(v2/r) - mg
While driving to work you pass over a "crest" in
the road that has a radius of 30 meters.
How fast would
you need to be
traveling to
experience
apparent
"weightlessness"
while passing
over the crest?
If we let the
normal approach 0
to represent
apparent
weightlessness,
then
Examples
The maximum tension that a 0.50
m string can tolerate is 14 N. A
0.25-kg ball attached to this
string is being whirled in a
vertical circle. What is the
maximum speed the ball can
have (a) the top of the circle,
(b)at the bottom of the circle?
mv 2
FNET  Fc  mac 
r
mv 2
T  mg 
 r (T  mg )  mv 2
r
r (T  mg )
0.5(14  (0.25)(9.8))
v

m
0.25
v  5.74 m / s
T
mg
Examples
mv 2
FNET  Fc  mac 
r
mv 2
T  mg 
 r (T  mg )  mv 2
r
r (T  mg )
0.5(14  (0.25)(9.8))
v

m
0.25
v  4.81 m / s
At the bottom?
T
mg
Conical Pendulums

Our next example is also an object on the end of
string but this time it is a conical pendulum.
Notice, that its path also tracks out a horizontal
circle in which gravity is always perpendicular to
the object's path.
T cos θ is balanced by the
object's weight, mg. It is T sin θ
that is the unbalanced central
force that is supplying the
centripetal force necessary to
keep the block moving in its
circular path: T sin θ = Fc = mac.
Suppose a 75-gram ball is being whirled as a
conical pendulum by a child. The ball is attached
to a 50-cm string and tracks out a horizontal circle
with a radius of 40 cm.




What is the measure of angle θ?
What is the tension in the rope?
How fast is the ball traveling as it swings?
What is the period of the stopper?
Answers
T cos θ = mg
T cos (53º) = (0.075)(9.81)
T = (0.075)(9.81)/(0.602)
T = 1.22 N
r = L sin θ
0.40 = 0.50 sin θ
θ = 53º
T sin θ = m(v2/r)
1.22(sin 53º) = (0.075)(v2/0.40)
v2 = 5.20
v = 2.28 m/sec
v = 2πr/T
2.28 = 2π(0.40)/T
T = 1.1s
Banked Curves

If instead, the curve is banked then there is a
critical speed at which the coefficient of
friction can equal zero and the car still travel
through the curve without slipping out of its
circular path.

At this critical speed, there is no need for any
friction between the car and the road's
surface. If the speed of the car were to
exceed vcritical then the car would drift up the
incline. If the speed of the car is less than
vcritical then the car would slip down the
incline.
Free body diagram of the forces acting on the car would
show weight and a normal. Since the car is not sliding
down the bank of the incline, but is instead traveling
across the incline, components of the normal are
examined.
N sin θ is the unbalanced central force;
that is, N sin θ = Fc = mac. This component
of the normal is supplying the centripetal
force necessary to keep the car moving
through the banked curve
Dividing the equations
N sin θ = mv2/r
N cos θ = mg
Solving for v produces the desired result yields the equation.
tan θ = v2/rg.
vcritical = √ (rg tan θ).
At a local NASCAR racetrack, cars travel through
a 316-meter radius curve that is banked at 31º



At what speed would the race cars be
traveling if they wanted to pass through this
curve frictionlessly?
Does your answer to the previous question
depend on the mass of the car?
If the cars actually travel through this turn in
excess of 195 mph, or 87 m/sec, what would
supply the additional centripetal force?
Answers
No, in the derivation for our
equation for vcritical, the mass of
the race car cancelled out.
friction
Newton’s Law of Gravitation
What causes YOU to be pulled down? THE EARTH….or
more specifically…the EARTH’S MASS. Anything that
has MASS has a gravitational pull towards it.
Fg Mm
What the proportionality above is
saying is that for there to be a
FORCE DUE TO GRAVITY on
something there must be at least 2
masses involved, where one is
larger than the other.
N.L.o.G.
As you move AWAY from the earth, your
DISTANCE increases and your FORCE DUE
TO GRAVITY decreases. This is a special
INVERSE relationship called an InverseSquare.
1
Fg  2
r
The “r” stands for SEPARATION DISTANCE
and is the distance between the CENTERS OF
MASS of the 2 objects. We us the symbol “r”
as it symbolizes the radius. Gravitation is
closely related to circular motion as you will
discover later.
N.L.o.G – Putting it all together
m1m2
r2
G  constant of proportion ality
Fg 
G  Universal Gravitatio nal Constant
G  6.67 x10
Fg  G
 27
Nm 2
m1m2
r2
Fg  mg  Use this when you are on the earth
Fg  G
m1m2
 Use this when you are LEAVING th e earth
r2
kg 2
Try this!
Let’s set the 2 equations equal to each other since they BOTH
represent your weight or force due to gravity
Mm
r2
M
g G 2
r
M  Mass of the Earth  5.97 x10 24  kg
mg  G
r  radius of the Earth  6.37 x10 6  m
SOLVE FOR g!
(6.67 x1027 )(5.97 x1024 )
2
g

9
.
81
m
/
s
(6.37 x106 ) 2
Examples
Venus rotates slowly about its
axis, the period being 243
days. The mass of Venus is
4.87 x 1024 kg. Determine the
radius for a synchronous
satellite in orbit around
Venus. (assume circular
orbit)
Mm mv2
Fg  Fc
G 2 
r
r
GM
2r
 v 2 vc 
r
T
Fg
2
GM 4 2 r 2
GMT 2
GMT
3

r 
r 3
2
2
r
T
4
4 2
11
24
7 2
(
6
.
67
x
10
)(
4
.
87
x
10
)(
2
.
1
x
10
)
9 m
1.54x10
r3

4 2