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Announcements
• CAPA Set #6 due Friday at 10 pm
• CAPA Set #7 now available, due next Friday
• Next week in Section 
Assignment 4: Circular Motion & Gravity
• Finish reading all sections of Chapter 5
• Advanced reminder  Exam #2 on Tuesday, October 11
• Reminder about office hours …
Nagle (Monday 2-3 in office, Wednesday 1:45-3:45 pm help room)
Kinney (Thursday 4-5 pm help room)
Uzdensky (Tuesday 11am-noon help room)
If you find a missing clicker, please bring it to me or the Main Physics Office
Circular Motion – fixed radius and at constant speed |v|
Always accelerating due to
change in direction of velocity vector.
Centripetal acceleration inwards towards the circle
center with magnitude |a| = v2/r
“Wall-of-Death”
But don’t I feel out outward force?
This is a “fictitious” force, not real.
Clicker Question
Room Frequency BA
Consider the “Wall-of-Death”
Which diagram correctly shows the real forces on the rider?
“Centripetal force”:
a real force.
Fictitious force:
“centrifugal force”
– in the rider’s frame.
Centrifugal force (from Latin
centrum, meaning "center", and
fugere, meaning "to flee“)
Room Frequency BA
Clicker Question
What are the three forces #1, 2, 3?
1
2
3
A)
1 - gravity
2 - centrifugal force
3 – friction
B)
1 – friction
2 – normal force of the wall
3 – gravity
C)
1 - centripetal force
2 – normal force of the wall
3 – friction
D)
1 – friction
2 – centrifugal force
3 - gravity
Dynamics of Uniform Circular Motion
Choose a coordinate system:
Usually radial and tangential.
Tangential (T)
m
r
Radial (R)
For uniform motion, velocity in
the tangential direction is
constant, so
Σ FT = m aT = 0
In the radial direction:
Σ FR =m aR = mv2/r
For every case of uniform circular motion, there must
be a force directed towards the center.
We say there is a centripetal force. However, there is
always a specific force that is acting. There is no
“circle force”. Circular motion does not cause a force.
Ball circling
around tied to a Centripetal force  Tension Force
string.
Wall of Death
ride
Centripetal force  Normal Force
Race Car driving Centripetal force  Friction Force
in circle
Spinning Bucket of Water
The Earth circles the Sun at an average distance of
1 Astronomical Unit = 1.5 x 1011 meters in one year.
What is its orbital centripetal acceleration?
 2r 


T 


r
2
E
v
aradial 
r
r = 1 AU
aradial 
2

4 2 r
T2
4 (1.5 10 m)
2
11
(3.155  10 s)
7
2
aradial  0.006 m / s 2
Sometimes we quote accelerations relative to g (9.81 m/s2).
a radial  0.006 m / s 
2
1g
9.81m / s
2
 0.0006 g  0.06 % of g
Clicker Question
Room Frequency BA
The Earth circles the Sun at an average distance of
1 AU = 1.5 x 1011 m in 1 year.
E
r = 1 AU
aradial  0.006 m / s 2
What’s causing the centripetal acceleration?
τ = 365 days
A) The electrostatic force between the
Earth and Sun.
B) The tension in the string connecting the
Earth to the Sun.
C) The force of gravity between the Earth
and the Sun.
D) Depends on the time of day.
Newton’s Law of Universal Gravitation
Insight: what keeps the Moon in
orbit around the Earth and the
Earth in orbit around the Sun
is exactly the same thing
that causes an
“apple to fall from a tree”.
“Every particle in the
universe attracts every
other particle.”
Newton’s Law of Universal Gravitation
“Every particle in the universe attracts every other particle
with a force proportional to the product of their masses
and inversely proportional to the
square of the distance between them.
The force points along the line joining the two particles.”

m1m2
| F | gravity  G 2
r
G  6.67 x 10
11
2
Nm / kg
2
Universal Gravitation Verification
1687: Isaac Newton
published Gravity Theory
1798: Henry Cavendish
confirmed this formula
experimentally
1915: Albert Einstein’s
General Theory of
Relativity explained why
gravity behaves this way.

mm
| F | gravity  G 1 2 2
r
How Strong is Gravity?
m1 = 70 kg
m2 = 70 kg
r = 1 meter

2 
m1m2 
Nm
| F | gravity  G 2   6.67 10 11
 70 kg70 kg
2 
2

r
kg


1
m



7
| F | gravity  3.3 10 Newtons  7.5 10 8 Pounds
That is about 1/60th the weight of a single hair.
Who does the force act on?
m1 = 70 kg
m2 = 70 kg
r = 1 meter

7
| F | gravity  3.3 10 Newtons
Answer = Both
Person #1 exerts a force on Person #2
Person #2 exerts a force on Person #1
Extended Objects
Objects are extended in space.
Newton’s Law of Universal Gravitation is based on
computing the distance between two objects,
but which distance?
RCM
In fact, every part of object #1 exerts a gravitational attraction on
every part of object #2, and vice versa.
When adding all vector components, we treat the force as acting
between the “center of mass” of each object.
“Center of mass” for sphere = middle
Big G, Little g
Consider the force of gravity exerted by the Earth
with mass ME on a person of mass m on its surface?
RE

mM E
| F | gravity  G
2
RE




24

m
5
.
98

10
kg
11
2
2
| F | gravity  6.67 10 Nm / kg
2
6
6.37 10 m


2
| F | gravity  m  9.81m / s

| F | gravity  mg


Gravitational force on an object on the surface of the earth!

Room Frequency BA
Clicker Question
You are standing on the surface of
the earth.
The earth exerts a gravitational force
on you Fearth, and you exert a
gravitational force on the earth
Fperson.
Which of the following is correct:
A) Fearth > Fperson
B) Fearth < Fperson
C) Fearth = Fperson Newton’s Third Law
D) It’s not so simple, we need more
information.