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AP Physics 1 Chapter 5
Circular Motion, Newton’s Universal Law of Gravity,
and Kepler’s Laws
Centripetal force at work!
Hammer throw
Tangential Velocity and
Centripetal Acceleration
Roadway banking
Sitges Terramar in Spain
(60 degree bank)
Roadway Banking at an extreme
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 o 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

s v

r
v
s  vt
vt v

r
v

v
v
v
vo

vo
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
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r
vc 
T
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 
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)
Fg  Fc
GM
 v2
r
Fg
Mm mv2
G 2 
r
r
2r
vc 
T
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
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
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 decrease. 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
Fg  mg  Use this when you are on the earth
Fg  G
m1m2
 Use this when you are LEAVING th e earth
2
r
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
Kepler’s Laws
Testing Models


Geocentric (or Ptolemaic) means the Earth is at
the center and motionless.
Heliocentric (or Copernican) means the Sun is at
the center and motionless.

Scholars wanted to differentiate models by
comparing the predictions with precise
observations.

This originated the modern scientific method.
Kepler’s Work


Tycho Brahe led a team which
collected data on the position
of the planets (1580-1600 with
no telescopes).
Mathematician Johannes
Kepler was hired by Brahe to
analyze the data.




He took 20 years of data on
position and relative distance.
No calculus, no graph paper,
no log tables.
Both Ptolemy and Copernicus
were wrong.
He determined 3 laws of
planetary motion (1600-1630).
Kepler’s First Law

The orbit of a planet is an ellipse with the sun
at one focus.
A path connecting the two foci to the
ellipse always has the same length.
Orbital Description

An ellipse is described by two axes.



Long – semimajor (a)
Short – semiminor (b)
The area is ab (becomes r2 for a circle).
b
a
Orbital Speed

The centripetal force is due to gravity.


GMm/r2 = mv2/r
v2 = GM/r

Larger radius orbit means slower speed.

Within an ellipse larger distance also gives
slower speed.
Kepler’s Second Law

The line joining a planet and the sun sweeps
equal areas in equal time.
t
The planet moves
slowly here.
t
The planet moves
quickly here.
Orbital Period

The speed is related to the period in a circular
orbit.



v2 = GM/r
(2r/T)2 = GM/r
T2 = 42r3/GM

Larger radius orbit means longer period.

Within an ellipse, a larger semimajor axis also
gives a longer period.
Kepler’s Third Law

The square of a planet’s period is
proportional to the cube of the length of the
orbit’s semimajor axis.


T2/a3 = constant
The constant is the same for all objects orbiting
the Sun
direction of orbit
semimajor axis: a
The time for one orbit
is one period: T
Hyperbolic Orbits

Some satellites have so much speed that
gravity can’t hold them in an orbit.

These objects take a hyperbolic orbit that
never returns.