Definition: Uniform Circular motion is the motion of an object
traveling at a constant (uniform) speed on a circular path
Slide 3
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
Slide 4
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
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.
Slide 5
Cant 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
--- RPMs
Slide 6
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.
Slide 7
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
Slide 8
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
Slide 9
Velocity was v = d/t Distance is now the circumference of the
circle (2r) Period (T) is the time it takes for one revolution. So
Speed = ? v = 2r/T
Slide 10
Velocity was v = d/t Distance is now the circumference of the
circle (2r) Period (T) is the time it takes for one revolution. So
Speed = ? This is also called Tangential Speed v = 2r/T
Slide 11
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.
Slide 12
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 v t, CAN
the cars still have an acceleration?
Slide 13
2.4.2Apply the expression for centripetal acceleration. This is
due to what defines acceleration: a = v f v i t f - t i Because
velocity is a vector, acceleration can be changed by the magnitude
or direction of the velocity.
Slide 14
2.4.2 Apply the expression for centripetal acceleration. Well,
velocity has changed, so centripetal acceleration (a c ) will be a
little different too Centripetal acceleration = (tangential speed)
2 / radius of circular path The acceleration is still a vector qty,
and will always point toward the center of the circle. a c = v 2
/r
Slide 15
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.
Slide 16
Practice Problem 1 A test car moves at a constant speed around
a circular track. If the car is 48.2m from the tracks center and
has a centripetal acceleration of 8.05 m/s 2, what is the cars
tangential speed? What are you given? Needed? V t = 19.7 m/s a c =
v t 2 /r
Slide 17
Answer:19.7m/s
Slide 18
Practice Problem 2 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/s 2 a c = v t 2 /r
Slide 19
Answer = 89 m/s 2
Slide 20
Velocity Practice From Packet Classwork: (Holt: Physics) pg 236
Practice A
Slide 21
Velocity Practice From Packet 1) 7 m/s 2) 0.26 m/s 3) 49
m/s
Slide 22
Practice A Centripetal Acceleration 1) 2.5m/s 2) 11m/s 3)
1.5m/s 2 4) 58.4m
Slide 23
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/s 2 Turn 2: 48
m/s 2
Slide 24
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?
Slide 25
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?
Slide 26
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
?
Slide 27
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??
Slide 28
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
Slide 29
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
Slide 30
2.4.3 Identify the force producing circular motion in various
situations. Centripetal means center- Seeking 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 (v t
), and radius (r)
Slide 31
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 but it is not actually a force, is only inertia
Every object that moves in circular motion must experience a
centripetal force from somewhere
Slide 32
Practice B pg 238
Slide 33
1) 29.6kg 2) 40m 3) 40N 4) 35m/s
Slide 34
Can you identify the different sources of centripital force?
Watch Video
Slide 35
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.
Slide 36
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 way is
0.5g
Slide 37
Centripetal Force The force equation will not change: However,
remember this is now F c, so F = ma F c = ma c or F c = mv 2
/r
Slide 38
Practice Problem 3 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 tracks radius? What are you given? needed? r =
40m F c = mv 2 /r
Slide 39
Answer: 40m
Slide 40
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.
Slide 41
At the top We should see that the F net = F c + F g OR F ten =
(mv 2 )/r + mg
Slide 42
At the bottom We should see that the F net = F c - F g OR F ten
= (mv 2 )/r - mg
Slide 43
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
Slide 44
Answer: A) 143N B) 153N
Slide 45
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/s 2 Turn 2: 48
m/s 2
Slide 46
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
Slide 47
What is gravity? Is there gravity in space? Why do astronauts
float? What keeps the moon from flying off in space?
Slide 48
6.1 Gravitational Force and Field 6.1.1 State Newtons 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
Slide 49
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?
Slide 50
6.1 Gravitational Force and Field 6.1.1 State Newtons universal
law of gravitation. Galileo (1564-1642) g = 9.81m/s 2, 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 R M = 3.844 x 10 8 m, R E = 6.378 x 10 6 m mid-1600s
Earths and Moons masses had been determined M M = 7.35 x 10 22 kg M
E = 5.98 x 10 24 kg)
Slide 51
6.1 Gravitational Force and Field 6.1.1 State Newtons universal
law of gravitation. From all that data Newton calculated the
centripetal acceleration on the moon due to the earths
gravitational attraction. ac= v 2 /r= (2r/T) 2 x 1/ r = 0.00272m/s
2 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. F G
1/r 2
Slide 52
6.1 Gravitational Force and Field 6.1.1 State Newtons universal
law of gravitation. Also concluded that the gravitational force was
proportional to the product of the two masses. F G m 1 m 2 Combined
= F G m 1 m 2 /r 2
Slide 53
6.1 Gravitational Force and Field 6.1.1 State Newtons universal
law of gravitation. IB Equation and Formula Newtons 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(m 1 m 2 / r 2 ) Universal law of gravitation G = 6.67 x 10
-11 Nm 2 kg -2
Slide 54
6.1 Gravitational Force and Field 6.1.1 State Newtons 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 -8 N
Slide 55
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
Slide 56
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 Moons
gravitational field strength at its surface. Answer: 1.63 N/kg
Slide 57
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.
Slide 58
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)/r 2
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)/r 2 m = GM/r 2
Slide 59
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. R E = 6.378 x 10 6 m M E =
5.98 x 10 24 kg Answer: 9.80 N/kg
Slide 60
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 M E = 5.98 x 10 24
kg Moon mass M M = 7.35 x 10 22 kg Earth Moon distance 3.84 x 10 8
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
10 8 m
Slide 61
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 moons position in
space, due to the Earth and Sun. Dist. for Sun to Moon R S =
1.49x10 11 m Sun Mass M S = 1.99x10 30 Answer: g= 0.0066N/kg @ 24
as seen on the diagram.
Slide 62
Motion in space
Slide 63
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.
Slide 64
Motion in space About 300BC, Greek named Aristarchus had a
theory that the Earth revolved around the sun. No one accepted his
idea.
Slide 65
Motion in space Claudius Ptolemy Around 200 AD, developed
extremely complex theory Planets and Sun, traveled in small circles
called epicycles.
Slide 66
Motion in space Claudius Ptolemy Planets and Sun, also traveled
in larger circular orbits with Earth at the center. Didnt explain
all observations.
Slide 67
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.
Slide 68
Motion in space Late 1500s Tycho Brahe Technology had
progressed Made very precise observations. Brahes data didnt agree
with Copernicuss model. Developed the Tychonian system.
Slide 69
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?
Slide 70
Tychonian system
Slide 71
Motion in space Early 1600s, Johannes Kepler Worked for Tycho
Brahe. Used Copernicuss theory and Brahes data Figures out how the
two relate Developed three laws of planetary motion.
Slide 72
Keplers 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.
Slide 73
Keplers 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
Slide 74
Keplers Laws of Planetary Motion First Law: The amount of
"flattening" of the ellipse is termed the eccentricity.
Slide 75
Keplers Laws of Planetary Motion Second Law: An imaginary line
drawn from the sun to any planet sweeps out equal areas in equal
time intervals
Slide 76
Keplers Laws of Planetary Motion Second Law: Perihelion &
Aphelion the planet moves fastest when it is near perihelion and
slowest when it is near aphelion
Slide 77
Keplers Laws of Planetary Motion Third Law: The square of a
planets orbital period (T 2 ) is proportional to the cube of the
average distance (r 3 ) between the planet and the sun. Applies to
satellites orbiting the Earth, including our moon.
Slide 78
Keplers Laws of Planetary Motion Third Law: Originally T 1 2 =
r 1 3 T 2 2 r 2 3 Restated T 2 = (4 2 /Gm) r 3
Slide 79
Keplers Laws of Planetary Motion Period of an object in
circular orbit T 2 = (4 2 /Gm) r 3 Or T = 2(r 3 /Gm) Speed of an
object in circular orbit t = (Gm/r) where m is mass of central
object
Slide 80
Lets try one A spacecraft, Magellan, took pictures of the
planet Venus. On the spacecrafts flight it traveled at a mean
altitude of 361km. If the orbit had been circular, what would its
period and speed have been? Mass of Venus is 4.87x10 24 m Radius of
Venus is 6.05x10 6 m Answer: Period = 5.66x10 3 s, speed = 7.12x10
3 m/x