1. Circular Motion and Gravitation

2. Circular
Motion

3. Circular motion equations:
v = 2π r f

4. Remember these?

5. Circular motion equations:
Fc = mac
v = 2π r f

6. UNIFORM CIRCULAR MOTION
Uniform circular motion is motion in which there is no change in speed, only a change in direction.

7. The period T is the time for one complete revolution
The period T is the time for one complete revolution. So the linear speed can be found by dividing the period into the circumference:
Units: m/s
Another useful parameter in engineering problems is the rotational speed, expressed in revolutions per minute (rpm), revolutions per second (rev/s) or Hz (s-1). This quantity is called the frequency f of rotation and is given by the reciprocal of the period.
Units: s-1 = Hz

8. 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.
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.

9. The term centripetal means that the acceleration a is always directed toward the center. The velocity
and the acceleration
are not in the same direction; v points in the direction of motion which is tangential to the circle.
v and a are perpendicular at every point.

10. v and a are perpendicular at every point.

11. CENTRIPETAL ACCELERATION
An object experiencing uniform circular motion is continually accelerating. The position and velocity of a particle moving in a circular path of radius r are shown at two instants in the figure.
When the particle is at point A, its velocity is represented by vector v1. After a time interval t, its velocity is represented by the vector v2.

12. The acceleration is given
by:
as Δt becomes smaller and smaller, the chord length becomes equal to the arc length s = v Δt

13. This acceleration is called the centripetal, and it points toward the center of the circle.
Units: m/s2

14. CENTRIPETAL ACCELERATION
An object experiencing uniform circular motion is continually accelerating. The direction and velocity of a particle moving in a circular path of radius r are shown at two instants in the figure. The vectors are the same size because the velocity is constant but the changing direction means acceleration is occurring.

15. To calculate the centripetal acceleration, we will use the linear velocity and the radius of the circle
Or substituting for v
We get ac

16. Example 2: A ball is whirled at the end of a string in a horizontal circle 60 cm in radius at the rate of 1 revolution every 2 s. Find the ball's centripetal acceleration.
r = 0.6 m
T = 2 s
= 5.92 m/s2

17. Centripetal Acceleration
Suppose we had a circle with angle, q, between 2 radaii. You may recall: Dv v v vo q vo
Centripetal means “center seeking” so that means that the acceleration points towards the CENTER of the circle

18. 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:

19. Circular Motion and N.S.L
Recall that according to Newton’s Second Law, the acceleration is directly proportional to the Force. If this is true:
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.

20. Uniform circular motion is motion along a circular path in which there is no change in speed, only a change in direction. v
Constant velocity tangent to path. Fc
Constant force toward the center.

21. Uniform Circular Motion (Cont.)
The question of an outward force can be resolved by asking what happens when the string breaks!
Ball moves tangent to path, NOT outward as might be expected. v
When central force is removed, ball continues in straight line.
Centripetal force is needed to change direction.

22. 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?

23. 4.11 A 2 kg body is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete revolutions every second, determine its linear speed and its centripetal acceleration.
UCM
m = 2 kg
r = 2 m
f = 3 rev/s
= 0.33 s
= m/s
= 725 m/s2

24. 4.12 A ball is whirled at the end of a string in a horizontal circle
60 cm in radius at the rate of 1 revolution every 2 s. Find the ball's centripetal acceleration.
UCM
r = 0.6 m
T = 2 s
= 5.92 m/s2

25. CENTRIPETAL FORCE
The inward force necessary to maintain uniform circular motion is defined as centripetal force.
From Newton's Second Law, the centripetal force is given by:

26. Fc = mac CENTRIPETAL FORCE
 The inward force necessary to maintain uniform circular motion is defined as centripetal force. From Newton's
Second Law, the centripetal force is given by:
Fc = mac Or

27. We can see that the force must be inward by thinking about a ball on a string:

28. As a car makes a turn, the force
The centripetal force is not a 'special' kind of force. The centripetal force is provided by the force that keeps the object in a circle, this is called the centripetal force requirement.
4.13 a. A car makes a turn, what force is required to keep it in circular motion?
As a car makes a turn, the force
of friction acting upon the tires of the car provide the centripetal force required for circular motion.

29. Force of friction between tires and road.
Example 3: A 1000-kg car rounds a turn of radius 30 m at a velocity of 9 m/s
a. How much centripetal force is required?
m = 1000 kg
r = 30 m
v = 9 m/s
= 2700 N
b. Where does this force come from?
Force of friction between tires and road.

31. Figure 6.32
Two paths around a race track curve are shown. Race car drivers will take the inside path (called cutting the corner) whenever possible because it allows them to take the curve at the highest speed.

32. Examples Top view FN Ff mg Side view
What is the minimum coefficient of static friction necessary to allow a penny to rotate along a 33 1/3 rpm record (diameter= m), when
the penny is placed at the outer edge of the record?
Top view FN Ff mg
Side view

33.  b. A bucket of water is tied to a string and spun in a circle, what force is required to keep it in circular motion?
As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion.
c. The moon orbits the Earth, what force is required to keep it in circular motion?
As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.

34. Without a centripetal force, an object in motion continues along a straight-line path.
With a centripetal force, an object in motion will be accelerated and change its direction.
             

35. The Forbidden F-Word
When the subject of circular motion is discussed, it is not uncommon to hear mention of the word "centrifugal."
Centrifugal, not to be confused with centripetal, means away from the center or outward. The use of this word combined with the common sensation of an outward force when experiencing circular motion, often creates or reinforces a deadly student misconception.

36. The Forbidden F-Word
The deadly misconception, is the notion that objects in circular motion are experiencing an outward force.
"After all," a well-meaning student may think, "I can recall vividly the sensation of being thrown outward away from the center of the circle on that roller coaster ride. Therefore, circular motion must be characterized by an outward force."

37. Centrifugal Force is often used to describe why mud gets spun off a spinning tire, or water gets pushed out of the clothes during the spin dry cycle of your washer.  It is also used to describe why we tend to slide to the outer side of a car going around a curve. 
Let’s imagine that you are riding in Granny’s car going around a curve.  Sitting on your dashboard is a cassette tape.  As you go around the curve, the tape moves to outside edge of the car.  Because you don't want to blame it on ghosts, you say “centrifugal force pushed the tape across the dashboard.”
Wwrroonngg!! 

38. The animation shows both views at the same time
The animation shows both views at the same time.  The top window shows you the passenger's view of the car and the tape, while the shows you the bird's eye view.

40. The tape on the slippery dashboard does not have enough friction to act as a centripetal force, so in the absence of a centripetal force the tape follows straight line motion.  If the car you are riding in has the windows rolled down, then the tape will leave the car (or does the car leave the tape?) as it follows its straight line path.  If the windows are rolled up, then the window will deliver a centripetal force to the tape and keep it in a circular path.

41. There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome.
If the centripetal force vanishes, the object flies off tangent to the circle.

42. Spin Cycle on a Washer
How is the water removed from clothes during the spin cycle of a washer?
Think carefully before answering Does the centripetal force throw water off the clothes?
NO. Actually, it is the LACK of a force that allows the water to leave the clothes through holes in the circular wall of the rotating washer.

43. The Conical Pendulum
A conical pendulum consists of a mass m revolving in a horizontal circle of radius R at the end of a cord of length L.
T cos q q h T L R T q mg
T sin q
Note: The inward component of tension T sin q gives the needed central force.

44. Solve two equations to find angle q
Angle q and velocity v: q h T L R mg
T sin q
T cos q
Solve two equations to find angle q
mv2 R
T sin q =
tan q =
v 2 gR
T cos q = mg

45. 2. Recall formula for pendulum.
Example 5: A 2-kg mass swings in a horizontal circle at the end of a cord of length 10 m. What is the constant speed of the mass if the rope makes an angle of 300 with the vertical?
1. Draw & label sketch.
q = 300 q h T L R
2. Recall formula for pendulum.
Find: v = ?
3. To use this formula, we need to find R = ?
R = L sin 300 = (10 m)(0.5)
R = 5 m

46. Example 5(Cont.): Find v for q = 300h T L R
q = 300
R = 5 m
4. Use given info to find the velocity at 300.
R = 5 m
g = 9.8 m/s2
Solve for v = ?
v = 5.32 m/s

47. Example 5b: Now find the tension T in the cord if m = 2 kg, q = 300, and L = 10 m.mg
T sin q
T cos q
2 kg
SFy = 0: T cos q - mg = 0; T cos q = mg
T = = mg
cos q
(2 kg)(9.8 m/s2)
cos 300
T = 22.6 N

48. Example 6: Find the centripetal force Fc for the previous example.q h T L R mg
T sin q
T cos q
q = 300 Fc
2 kg
m = 2 kg; v = m/s; R = 5 m; T = 22.6 N
Fc =
mv2 R
or Fc = T sin 300
Fc = 11.3 N

49. 4.14 A 75 g toy airplane is fastened to one end of a 44 cm string, and the other end is held fixed at the ceiling The plane whirls in a horizontal circle. Find the speed of the plane and the tension of the string if the angle to the vertical is 30˚.
UCM FT
60º FG
L = 0.44 m
m = kg
= 0.22 m

50. L = 0.44 m
m = kg
r = 0.22 m FT
60º FG
= 1.1 m/s

51. = 0.85 N

52. 4. 15 In a Rotor ride at the amusement park the room radius is 4
4.15 In a Rotor ride at the amusement park the room radius is 4.6 m and the rotation frequency is 0.5 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?
UCM
r = 4.6 m
f = 0.5 rev/s
T = 2 s

53. r = 4.6 m
f = 0.5 rev/s
T = 2 s
= 0.22

54. MOTION IN A VERTICAL CIRCLE
When a body moves in a vertical circle at the end of a string, the tension FT in the string varies with the body's position. The centripetal force Fc on the body at any point is the vector sum of FT and the component of the body's weight toward the center of the circle.

55. At the highest point in the circular turn the centripetal force is given by:
At the lowest point in the circular turn the centripetal force is given by:

56. Top of Circle Fc = FT + Fg FT = Fc - Fg = 50 - 9.8 = 40.2 N
4.16 A string 0.5 m long is used to whirl a 1-kg stone in a vertical circle at a uniform velocity of 5 m/s.
a. What is the tension in the string when the stone is at the top of the circle
UCM
Top of Circle
Fc = FT + Fg
FT = Fc - Fg
r = 0.5 m
m = 1 kg
v = 5 m/s
= = 40.2 N

57. Bottom of Circle Fc = FT - Fg FT = Fc + Fg = 50 + 9.8 = 59.8 N
b. What is the tension in the string when the stone is at the bottom of the circle?
Bottom of Circle
Fc = FT - Fg
FT = Fc + Fg =
= 59.8 N

59. Apparent weight will be the normal force at the top:
Example 12: What is the apparent weight of a 60-kg person as she moves through the highest point when R = 45 m and the speed at that point is 6 m/s? n mg + R v
Apparent weight will be the normal force at the top:
mg - n =
mv 2 R
n = mg -
mv 2 R
n = 540 N

60. For Motion in Circle v AT TOP: mv2 R + T = - mg R mg T AT BOTTOM: mv2
The force on you is the least
AT BOTTOM: T mg +
T = mg
mv2 R
The force on you is the greatest

61. Summary: Ferris Wheel n n = mg - n = + mg AT TOP: mg + AT BOTTOM:
mv 2 R
n = mg v
n = mg -

62. The Loop-the-Loop v AT TOP: mv2 R + R mg AT BOTTOM: mv2 R + mg
FN = mg
mv2 R FN mg +
AT BOTTOM: FN mg +
FN = mg
mv2 R

63. The Ferris Wheel v mv2 R AT TOP: R + mv2 R mg AT BOTTOM: mv2 R + mg
mg - FN =
mv2 R
AT TOP: FN mg +
FN = mg -
mv2 R
AT BOTTOM: FN mg +
FN = mg
mv2 R

64. 4.17 A ball of mass M is attached to a string of length R and negligible mass. The ball moves clockwise in a vertical circle, as shown. When the ball is at point P, the string is horizontal. Point Q is at the bottom of the circle and point Z is at the top of the circle. Air resistance is negligible. Express all algebraic answers in terms of the given quantities and fundamental constants.
UCM

65. a. On the figures below, draw and label all the forces exerted on the ball when it is at points P and Q. FT FT FG FG

66. b. Derive an expression for vmin the minimum speed the ball can have at point Z without leaving the circular path.
The minimum speed will occur when there is no FT FG

67. c. The maximum tension the string can have without breaking is Tmax Derive an expression for vmax, the maximum speed the ball can have at point Q without breaking the string.

68. d. Suppose that the string breaks at the instant the ball is at point P. Describe the motion of the ball immediately after the string breaks.
The ball will go straight up i.e. tangential to point P

69. 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? T mg

70. Examples
At the bottom? T mg

71. Like a ball on a string, Satellites are an example of uniform circular motion.
There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius.
The gravitational pull of the Earth provides the centripetal force and acts like an invisible guideline for a satellite.

72. Fc = mac

73. r = 4.23x107 m from the Earth’s center
Example 4: A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Determine:
a. The height above the Earth’s surface such a satellite must orbit and
FUG = FC
mE = 5.98x1024 kg
RE = 6.38x106 m
T = 1 day = s
GMET2 = 4 π2r3
r = 4.23x107 m from the Earth’s center

74. height = r - rE = 4.23x107 - 6.38x106 m = 3.592x107 m = 3070 m/s
b. The satellite’s speed
= 3070 m/s

75. 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
1.54x109 m

76. Kepler’s Laws and Newton's Synthesis
Kepler’s laws describe planetary motion.
The orbit of each planet is an ellipse, with the Sun at one focus.

77. 2. An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times.

78. Figure 6.30
The shaded regions have equal areas. It takes equal times for m to go from A to B, from C to D, and from E to F. The mass m moves fastest when it is closest to M . Kepler’s second law was originally devised for planets orbiting the Sun, but it has broader validity.

79. As the planet is closest the sun, the planet is moving fastest and as the planet is farthest from the sun,it is moving slowest. Nonetheless, the imaginary line adjoining the center of the planet to the center of the sun sweeps out the same amount of area in each equal interval of time.

80. KEPLER’S THIRD LAW
"If T is the period and r is the length of the semi-major axis of a planet’s orbit, then the ratio T2/r3 is the same for all planets."

81. The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun: T2/r3

82. Kepler’s laws can be derived from Newton’s laws
Kepler’s laws can be derived from Newton’s laws. Irregularities in planetary motion led to the discovery of Neptune, and irregularities in stellar motion have led to the discovery of many planets outside our Solar System.

84. Newton’s Law of Universal Gravitation
If the force of gravity is being exerted on objects on Earth, what is the origin of that force?
Newton’s realization was that the force must come from the Earth.
He further realized that this force must be what keeps the Moon in its orbit.

85. Newton’s Law of Universal Gravitation
The gravitational force must be proportional to both masses. By observing planetary orbits, Newton concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses.
In its final form, the Law of Universal Gravitation reads:
Where:

86. The magnitude of the gravitational constant G can be measured .
This is the Cavendish experiment.

87. 4. 18 The mean distance from the Earth to the Sun is 1
4.18 The mean distance from the Earth to the Sun is 1.496x108 km and the period of its motion about the Sun is one year. The period of Jupiter’s motion around the Sun is years. Determine the mean distance from the Sun to Jupiter. KL
rE = 1.496x108 km
TE = 1 year
TJ = years
= 7.77x108 km

88. 4.19 Derive Kepler’s Third Law from Newton’s Law of Gravitation.
For a Planet 1 of mass m1 and the Sun of mass MS
ULG/KL
ΣF = ma = mac
substituting:
rearranging:
For a Planet 2 of mass m2 and the Sun of
mass MS:
therefore:

89. 4.20 What is the force of gravity acting on a 2000 kg spacecraft when it orbits two Earth radii from the Earth’s center above the Earth’s surface?
ULG
m1 = 2000 kg
ME = 5.98x1024 kg
rE = 6380x103 (2) =1.276x107 m
= 4899 N

90. 4.21 a. Derive the expression for g from the Law of Universal Gravitation.
Fg = FUG
ULG

91. b. Estimate the value of g on top of the Everest (8848 m) above the Earth’s surface.
mE = 5.98x1024 kg
RE = 6.38x106 m
RT = x106 = 6.388x106 m
= 9.77 m/s2

92. The acceleration due to gravity varies over the Earth’s surface due to altitude, local geology, and the shape of the Earth, which is not quite spherical.

93. Satellites
Satellites are routinely put into orbit around the Earth. The tangential speed must be high enough so that the satellite does not return to Earth, but not so high that it escapes Earth’s gravity altogether.

95. The satellite is kept in orbit by its speed – it is continually free falling, but the Earth curves from underneath it.

96. r = 4.23x107 m from the Earth’s center
4.22 A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Determine:
a. The height above the Earth’s surface such a satellite must orbit
UCM
mE = 5.98x1024 kg
RE = 6.38x106 m
T = 1 day = s
FUG = FC
GMET2 = 4 π2r3
r = 4.23x107 m from the Earth’s center

97. height = r - rE = 4.23x107 - 6.38x106 m = 3.592x107 m = 3070 m/s
b. The satellite’s speed
= 3070 m/s

98. Figure 6.24
The Moon causes ocean tides by attracting the water on the near side more than Earth, and by attracting Earth more than the water on the far side. The distances and sizes are not to scale. For this simplified representation of the Earth-Moon system, there are two high and two low tides per day at any location, because Earth rotates under the tidal bulge.

99. Figure 6.25
(a, b) Spring tides: The highest tides occur when Earth, the Moon, and the Sun are aligned. (c) Neap tide: The lowest tides occur when the Sun lies at 90º to the Earth-Moon alignment. Note that this figure is not drawn to scale.

100. Figure 6.26
A black hole is an object with such strong gravity that not even light can escape it. This black hole was created by the supernova of one star in a two-star system. The tidal forces created by the black hole are so great that it tears matter from the companion star. This matter is compressed and heated as it is sucked into the black hole, creating light and X-rays observable from Earth.

101. Figure 6.27
Astronauts experiencing weightlessness on board the International Space Station. (credit: NASA)