1. Circular Motion

2. Circular Motion Definition:
Uniform Circular motion is the motion of an object traveling at a constant (uniform) speed on a circular path

3. Axis – the straight line around which rotation takes place
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

4. 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

5. 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

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.

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

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

9. 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

10. 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 = ?
This is also called “Tangential Speed”
v = 2πr/T

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.

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 vt, CAN the cars still have an acceleration?

13. 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.

14. 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
The acceleration is still a vector qty, and will always point toward the center of the circle.
ac = v2/r

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 Apply the expression for centripetal acceleration.

16. ac = vt2/r Practice Problem 1
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

17. Answer:19.7m/s

18. ac = vt2/r 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/s2

19. Answer = 89 m/s2

20. Velocity Practice From Packet
Classwork: (Holt: Physics) pg 236 Practice A

21. Velocity Practice From Packet
7 m/s
0.26 m/s
49 m/s

22. Practice A Centripetal Acceleration
2.5m/s
11m/s
1.5m/s2
58.4m

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/s2
Turn 2: 48 m/s2

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?

25. Centripetal force is necessary for circular motion.
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?

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

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??

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.

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

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 (vt),
and radius (r)

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

32. Practice B pg 238

33. Practice B pg 238
1) 29.6kg
2) 40m
3) 40N
4) 35m/s

34. Can you identify the different sources of centripital force?
Watch Video

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.

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 ½ way is 0.5g

37. F = ma Fc = mac or Fc = mv2/r Centripetal Force
The force equation will not change:
However, remember this is now Fc, so…
F = ma
Fc = mac or Fc = mv2/r

38. Fc = mv2/r 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 track’s radius?
What are you given? needed?
r = 40m

39. Answer: 40m

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.

41. At the top We should see that the Fnet = Fc + Fg OR
Ften = (mv2)/r + mg

42. At the bottom We should see that the Fnet = Fc - Fg OR
Ften = (mv2)/r - mg

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

44. Answer:
A) 143N
B) 153N

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/s2
Turn 2: 48 m/s2

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

47. What is gravity?
Is there gravity in space?
Why do astronauts float?
What keeps the moon from flying off in space?

48. 6.1 Gravitational Force and Field
6.1.1 State Newton’s universal law of gravitation.
Watch Veritesium Videos 1, 2, 3

49. Two cars are parked 3m away from each other
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?

50. 6.1 Gravitational Force and Field
6.1.1 State Newton’s universal law of gravitation.
Galileo ( ) – g = 9.81m/s2, even with different masses.
David Scott – feather and hammer dropped on the moon, Apollo 15
Isaac Newton( ) –
Idea about a cannon ball that never hit the ground.
Orbit period of the moon – 27.3days
Radius of moons orbit – RM = x 108m, RE = x 106m
mid-1600s Earth’s and Moon’s masses had been determined
MM = 7.35 x 1022kg
ME = 5.98 x 1024kg)

51. 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 = m/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

52. 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

53. 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 Nm2kg -2

54. 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

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

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 Moon’s gravitational field strength at its surface.
Answer: 1.63 N/kg

57. 6.1 Gravitational Force and Field
6.1.3 Determine the gravitational field due to one or more point masses Derive an expression for gravitational field strength at the surface of a planet, assuming that all its mass is concentrated at its Center Solve problems involving gravitational forces and fields.

58. 6.1 Gravitational Force and Field
6.1.3, ,
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

59. 6.1 Gravitational Force and Field
6.1.3, ,
Problem 3
Use the following data to calculate the gravitational field strength at the surface of the Earth.
RE = x 106 m
ME = 5.98 x 1024 kg
Answer: 9.80 N/kg

60. 6.1 Gravitational Force and Field
6.1.3, ,
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

61. 6.1 Gravitational Force and Field
6.1.3, ,
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= 24º as seen on the diagram.

62. Motion in space

63. 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.

64. Motion in space
About 300BC, Greek named Aristarchus had a theory that the Earth revolved around the sun.
No one accepted his idea.

65. Motion in space Claudius Ptolemy
Around 200 AD, developed extremely complex theory
Planets and Sun, traveled in small circles called epicycles.

66. Motion in space Claudius Ptolemy
Planets and Sun, also traveled in larger circular orbits with Earth at the center.
Didn’t explain all observations.

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.

68. 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.

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?

70. Tychonian system

71. 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.

72. 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.

73. 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

74. Kepler’s Laws of Planetary Motion
First Law:
The amount of "flattening" of the ellipse is termed the eccentricity.

75. 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

76. 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

77. 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.

78. Kepler’s Laws of Planetary Motion
Third Law:
Originally T12 = r13
T r23
Restated T2 = (4π2/Gm) r3

79. 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

80. 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

81. Practice makes perfect!
Pg 251, Practice D