Circular Motion
Circular Motion

Definition:
 Uniform Circular motion is the motion of an
object traveling at a constant (uniform)
speed on a circular path
2.4.1Draw a vector diagram to illustrate that the
acceleration of a particle moving with constant speed
in a circle is directed towards the center of the circle.
Axis – the straight line around which
rotation takes place
 Rotation – a spin around an internal axis.
i.e.: a carnival ride or record (big CD)
 Revolution – a spin around an external axis.
i.e.: the Earth around the sun

2.4.1Draw a vector diagram to illustrate that the
acceleration of a particle moving with constant speed in
a circle is directed towards the center of the circle.

Speeds for objects in a straight line are
called linear (or tangential) speeds,
 Linear speeds are a rate at which an object
covers a certain distance (v =d/t)
 Ex. Unit – m/s , km/hr , mph

Can’t express speeds of rotation with a
linear speed,
 b/c objects at different points on the rotating
object have different linear speeds

Rotational speed
 Expresses the rate at which an object rotates
through a portion of a circle ( an angle)
 Ex. Unit --- RPM’s
Are all people on Earth moving at
the same speed??


Earth is rotating about an axis through its
poles
So that means we are all moving since we
are all on the Earth.
Below, a record spinning on a axis through its
center (black dot)

Faster linear speed, Star or Smiley??
Smiley, travels a greater distance for each
Full spin.

Faster rotational speed, Star or smiley??
 Both the same, b/c entire record is rotating at the same rate

Are some of us moving with a greater
LINEAR SPEED than others??
 Yes, closer to the Equator, the faster you are
moving…. Closer to poles, the slower you are
moving

Are some of us moving with a greater
ROTATIONAL SPEED than others??
 No, all people on earth have same rotational
speed, because Earth is spinning at the same
rate everywhere
Velocity was…
v = d/t
 Distance is now the
circumference of the circle
(2πr)
 Period (T) is the time it
takes for one revolution.
 So… Speed = ?

v = 2πr/T
Velocity was…
v = d/t
 Distance is now the
circumference of the circle
(2πr)
 Period (T) is the time it
takes for one revolution.
 So… Speed = ?

v = 2πr/T

This is also called
“Tangential Speed”
Check Your Neighbor…
 If
a meter stick supported at the 0-cm
mark swings like a pendulum from your
fingers (look at demo), how fast at any
given moment is the 100-cm mark
compared to the 50-cm mark? It takes 2
seconds to make one complete rotation.
2.4.2Apply the expression for centripetal
acceleration.
 Think
about a Ferris wheel.
 The cars in on the Ferris
wheel are in uniform circular
motion.
 Even though they have a
constant vt, CAN the cars still
have an acceleration?
2.4.2Apply the expression for centripetal
acceleration.
 This
is due to what defines acceleration:
a = vf – vi
tf - ti
 Because velocity is a vector, acceleration
can be changed by the magnitude or
direction of the velocity.
2.4.2 Apply the expression for centripetal
acceleration.

Well, velocity has changed, so centripetal
acceleration (ac) will be a little different too
 Centripetal acceleration = (tangential
speed)2 / radius of circular path
ac = v2/r

The acceleration is still a vector qty, and
will always point toward the center of the
circle.
2.4.1Draw a vector diagram to illustrate that the
acceleration of a particle moving with constant speed in
a circle is directed towards the center of the circle.
2.4.2 Apply the expression for centripetal acceleration.
Practice Problem 1
2
ac = vt /r
 A test car moves at a constant speed around a
circular track. If the car is 48.2m from the track’s
center and has a centripetal acceleration of 8.05
m/s2, what is the car’s tangential speed?

What are you given?
 Needed?
 Vt = 19.7 m/s

Answer:19.7m/s
Practice Problem 2
2
ac = vt /r
 The cylindrical tub of a washing machine has a
radius of 34 cm. During the spin cycle, the wall of
the tub rotates with a tangential speed of 5.5 m/s.
Calculate the centripetal acceleration of the clothes
sitting against the tub.
 What is given? needed?
 A = 89 m/s2

Answer = 89 m/s2

Velocity Practice From Packet
 Classwork: (Holt: Physics) pg 236 Practice A

Velocity Practice From Packet
1) 7 m/s
2) 0.26 m/s
3) 49 m/s
Practice A Centripetal Acceleration
1) 2.5m/s
2) 11m/s
3) 1.5m/s2
4) 58.4m
Centripetal Acceleration

A bobsled travels at 34 m/s
and goes around two turns
in the track as seen here.
What is the acceleration of
the sled in each turn?
Turn 1: 35 m/s2
Turn 2: 48 m/s2
Centripetal Force
 Any
force that causes an object to follow
a circular path
 Watch the demo. (spinning cup of water)
– What provided the Centripetal Force on the
cup?
– On the water?
 Do
you know how your washing machine
works?
2.4.3 Identify the force producing circular
motion in various situations.

Centripetal force is
necessary for circular
motion.
 What would happen if the
string attached to the cup
broke?
2.4.3 Identify the force producing circular
motion in various situations.

When driving in a circle, in what direction ?
is a force acting on you?
 Pushing you outward from the circle, or
inward?
 If you are swinging a yo-yo in a circle, and the
string breaks…. What path does the yo – yo
take??


Ans. -- Inwards, toward the center of the
circle
Ans -- yo- yo goes in a path tangent to
the circle
2.4.3 Identify the force producing circular
motion in various situations.

HOWEVER, People commonly think there
is a force pushing you out from the circle
 Feels like you are being pushed outward
 Example ….. The Rotor- amusement park
ride, a centrifuge, CD on your dashboard
moving to the right when your turning left
 Why is this??
The
Rotor
People Stand with backs against
wall of a large cylinder, cylinder
then starts spinning, and people
are seemingly pushed against
the wall, then floor drops, and
people are stuck against the wall.
http://www.youtube.com/watch?
v=uz_DkRs92pM
So why is there no Force pushing you out
from the circle??


A force does not cause this…… your
INERTIA does!!
Inertia makes you want to stay in a
straight line, and by going in a circle,
you are fighting your own inertia
 This is how Rotor works, and why CD on
dashboard happens

The only actual force acting on you is the
Centripetal Force
2.4.3 Identify the force producing circular
motion in various situations.
Centripetal means “centerSeeking”
 Force pushes you toward the
center of the circle
 Is the force that keeps you
moving in a circle, and
keeps your inertia from
taking you in a straight line

Centripetal Force is affected
by.. Mass (m),
linear speed (vt),
and radius (r)
Centripetal Force

Inertia wants to take objects in a tangent
line, to the circular path
 Inertia is why you feel like your being pushed
outward
– This outward pushing is sometimes called the
Centrifugal Force
actually a force, is only inertia
 Every object that moves in circular motion
must experience a centripetal force from
somewhere

but it is not

Practice B pg 238

Practice B pg 238
 1) 29.6kg
 2) 40m
 3) 40N
 4) 35m/s
Can you identify the different
sources of centripital force?

Watch Video
2.4.3 Identify the force producing circular
motion in various situations.
By definition – the net force that is directed
toward the center of the circle.
 This force is provided by a frame of reference
 Consider a ladybug in a spinning can
 The apparent centrifugal force on the lady bug is
only an effect, not an
interaction.

Simulated Gravity
A
spinning wheel can provide a
“gravity” to occupants within it.
 If multiple rings are built, the gravity
would vary depending on distance
from center.
 Outer edge 1g, then ½ way is 0.5g
Centripetal Force

The force equation will not change:
F = ma

However, remember this is now Fc, so…
Fc = mac or Fc =
2
mv /r
Practice Problem 3
Fc =


2
mv /r
A bicyclist is riding at a angular speed of
13.2m/s around a circular track. The
magnitude of the centripetal force is 377N,
and the combined mass of the bicycle and
rider is 86.5kg. What is the track’s radius?
What are you given? needed?
 r = 40m

Answer: 40m
Vertical Circular Motion

If an object is suspened on the end of a cord
and is rotated in vertical circle what forces
are acting on it?
 Lets watch Video #2
 Lets draw an FBD.
At the top

We should see that the Fnet = Fc + Fg

OR

Ften = (mv2)/r + mg
At the bottom

We should see that the Fnet = Fc - Fg

OR

Ften = (mv2)/r - mg
Practice Problem 4

A 0.5kg mass, suspended on the end of a
light cord, 1.2m long, is rotated in a vertical
circle at a constant speed such that one
revolution is completed in 0.4s. Calculate
the tension in the cord when the weight is:
– A) at the top of the circle
– B) at the bottom of the circle

Answer:
 A) 143N
 B) 153N
Centripetal Acceleration

A bobsled travels at 34 m/s
and goes around two turns
in the track as seen here.
What is the acceleration of
the sled in each turn?
Turn 1: 35 m/s2
Turn 2: 48 m/s2
TOPIC 6.1: Gravitational
Fields and Forces
These notes were typed in association
with Physics for use with the IB
Diploma Programme by Michael
Dickinson

What is gravity?
 Is there gravity in space?
 Why do astronauts float?
 What keeps the moon from flying off in
space?
6.1 Gravitational Force and Field
6.1.1 State Newton’s universal law of gravitation.
 Watch Veritesium Videos 1, 2, 3
http://www.youtube.com/watch?v=mezkHBPLZ4
A&list=PL772556F1EFC4D01C
http://www.youtube.com/watch?v=zN6kCa6xi9k&
list=PL772556F1EFC4D01C
http://www.youtube.com/watch?v=iQOHRKKNN
LQ&list=PL772556F1EFC4D01C

Two cars are parked 3m away from each
other. One car has a mass of 1500kg while
the other has a mass of 2000kg. What is the
gravitational force between them?
6.1 Gravitational Force and Field
6.1.1 State Newton’s universal law of gravitation.
 Galileo (1564-1642) – g = 9.81m/s2, even with
different masses.
 David Scott – feather and hammer dropped on the
moon, Apollo 15
 Isaac Newton(1643-1727) –
– Idea about a cannon ball that never hit the ground.
– Orbit period of the moon – 27.3days
– Radius of moons orbit – RM = 3.844 x 108m, RE = 6.378
x 106m

mid-1600s Earth’s and Moon’s masses had been
determined
– MM = 7.35 x 1022kg
– ME = 5.98 x 1024kg)
6.1 Gravitational Force and Field
6.1.1 State Newton’s universal law of gravitation.
 From all that data Newton calculated the
centripetal acceleration on the moon due to the
earth’s gravitational attraction.
ac= v2/r= (2πr/T)2 x 1/ r = 0.00272m/s2
 Compared this with the “g”
 3600 times lower.
 Concluded gravitational force of attraction is
inversely proportional to the square of the
distance between the centers of the two object.
 FG ∝ 1/r2
6.1 Gravitational Force and Field
6.1.1 State Newton’s universal law of gravitation.
 Also concluded that the gravitational force was
proportional to the product of the two masses.
 FG ∝ m1m2

Combined = FG ∝ m1m2/r2
6.1 Gravitational Force and Field
6.1.1 State Newton’s universal law of gravitation.
IB Equation and Formula
 Newton’s Law of Universal Gravitation –
every object attracts every other object with
a force that is proportional to the product of
the two masses and inversely proportional to
the square of the distance between them.
 F = G(m1m2/ r2)
 Universal law of gravitation
– G = 6.67 x 10-11 Nm2kg -2
6.1 Gravitational Force and Field
6.1.1 State Newton’s universal law of gravitation.
Practice 1
 Calculate the gravitational force of attraction
between you and the person sitting next to you!
Assume your mass is 65kg, their mass is 55kg
and the distance is 2m
 Answer 5.9x10-8N
6.1 Gravitational Force and Field
6.1.2 Define gravitational field strength.
 Gravitational field is like a “force field” that
exist around every object.
 It is dependent on the mass of an object. So
larger mass means a larger field. Smaller mass
means smaller field.
IB Definition and Formula
 Gravitational field strenghth – the force per
unit mass acting on mass in a gravitational
field
 g = F/m
6.1 Gravitational Force and Field
6.1.2 Define gravitational field strength.
Practice 2
 A 2.45kg object feels a gravitational force of
4.0N at the surface of the Moon. Calculate the
Moon’s gravitational field strength at its surface.
 Answer: 1.63 N/kg
6.1 Gravitational Force and Field
6.1.3 Determine the gravitational field due to one
or more point masses.
6.1.4 Derive an expression for gravitational field
strength at the surface of a planet, assuming that all
its mass is concentrated at its Center.
6.1.5 Solve problems involving gravitational forces
and fields.
6.1 Gravitational Force and Field
6.1.3, 6.1.4, 6.1.5
 Gravitational Field strength is a vector. This
means we must use the rules of vector addition!
 Given that g = F/m and F = G(Mm)/r2
– g = gravitational field strength
– F = gravitational force
– M = mass of the planet
– m = mass of object in gravitational field

Substitute for F and you get:
g = G(Mm)/r2m = GM/r2
6.1 Gravitational Force and Field
6.1.3, 6.1.4, 6.1.5
Problem 3
 Use the following data to calculate the
gravitational field strength at the surface of the
Earth.
 RE = 6.378 x 106 m
 ME = 5.98 x 1024 kg
 Answer: 9.80 N/kg
6.1 Gravitational Force and Field
6.1.3, 6.1.4, 6.1.5
Problem 4
 If we take a look at the Earth-Moon system it becomes
apparent that there must be a point somewhere between
them where the gravitational field strength is zero.
Meaning a mass, a spacecraft, is not pulled in either
direction at that point. Calculate the position of this point
– distance, x.
 Earth mass ME = 5.98 x 1024 kg
 Moon mass MM = 7.35 x 1022 kg
 Earth Moon distance 3.84 x 108 m
Hint: this will be gravitational field strengths are equal
Answer: 88.9% of the distance from the Earth to the Moon
or 3.41 x 108m
6.1 Gravitational Force and Field
6.1.3, 6.1.4, 6.1.5
Problem 5
 The Sun, Earth and Moon, at a particular moment in
time are perpendicular to each other. (See board for
diagram). Calculate the gravitation field strength
and direction at the moon’s position in space, due to
the Earth and Sun.
 Dist. for Sun to Moon – RS = 1.49x1011m
 Sun Mass – MS = 1.99x1030
 Answer: g= 0.0066N/kg @ 24º as seen on the
diagram.
Motion in space
Motion in space
 Planetary
motion has
been studied as long as
people have looked into
the skies.
 Most people believed
that the Earth was the
center of the universe.
Motion in space
 About
300BC, Greek
named Aristarchus had a
theory that the Earth
revolved around the sun.
 No one accepted his
idea.
Motion in space
 Claudius
Ptolemy
 Around 200 AD,
developed extremely
complex theory
 Planets and Sun, traveled
in small circles called
epicycles.
Motion in space
 Claudius
Ptolemy
 Planets and Sun, also
traveled in larger circular
orbits with Earth at the
center.
 Didn’t explain all
observations.
Motion in space
 Polish
astronomer Nicolaus Copernicus
 1543 published On the Revolutions of the
Heavenly Spheres
 Proposed all planets orbited the sun in
perfect circles.
Motion in space
 Late
1500’s Tycho Brahe
 Technology had progressed
 Made very precise observations.
 Brahe’s data didn’t agree with
Copernicus’s model.
 Developed the Tychonian system.
Tychonian system
 Sun
and moon revolved around
Earth
 Other
planets revolved around Sun
 Argued that if Earth moved you
could see the change in our
position relative to the stars –
parallax. Can we?
Tychonian system
Motion in space
 Early
1600’s, Johannes Kepler
 Worked for Tycho Brahe.
 Used Copernicus’s theory and Brahe’s data
 Figure’s out how the two relate
 Developed three laws of planetary motion.
Kepler’s Laws of Planetary Motion
First Law: Each planet travels in an elliptical
orbit around the sun, and the sun is one of
the focal points.
Kepler’s Laws of Planetary Motion
First Law:
–
–
For an ellipse there are two points called foci
sum of the distances to the foci from any point on the
ellipse is a constant
Kepler’s Laws of Planetary Motion
First Law:
–
The amount of "flattening" of the ellipse is termed the
eccentricity.
Kepler’s Laws of Planetary Motion
Second Law: An imaginary line drawn from
the sun to any planet sweeps out equal
areas in equal time intervals
Kepler’s Laws of Planetary Motion
Second Law:
–
–
Perihelion & Aphelion
the planet moves fastest when it is near perihelion
and slowest when it is near aphelion
Kepler’s Laws of Planetary Motion
Third Law: The square of a planet’s orbital
period (T2) is proportional to the cube of
the average distance (r3) between the
planet and the sun.
 Applies to satellites orbiting the Earth,
including our moon.
Kepler’s Laws of Planetary Motion
Third Law:
 Originally T12 = r13
T22
r23

Restated
T2 = (4π2/Gm) r3
Kepler’s Laws of Planetary Motion




Period of an object in circular orbit
T2 = (4π2/Gm) r3
Or T = 2π√(r3/Gm)

Speed of an object in circular orbit
υt = √(Gm/r)

where m is mass of central object
Let’s try one
A
spacecraft, Magellan, took pictures of the
planet Venus. On the spacecraft’s flight it
traveled at a mean altitude of 361km. If the
orbit had been circular, what would it’s
period and speed have been?
 Mass of Venus is 4.87x1024m
 Radius of Venus is 6.05x106m
Answer: Period = 5.66x103s, speed = 7.12x103m/x
Practice makes perfect!
 Pg
251, Practice D