1. Physics: Principles with Applications, 6th edition
Lecture PowerPoints
Chapter 5
Physics: Principles with Applications, 6th edition
Giancoli
© 2005 Pearson Prentice Hall
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2. Circular Motion
Gravitation

3. 5-1 Kinematics of Uniform Circular Motion
Uniform circular motion: motion in a circle of constant radius at constant speed
Instantaneous velocity is always tangent to circle.

4. 5-1 Kinematics of Uniform Circular Motion
This acceleration is called the centripetal, or radial, acceleration, and it points towards the center of the circle.

5. 5-2 Dynamics of Uniform Circular Motion
For an object to be in uniform circular motion, there must be a net force acting on it.
We already know the acceleration, so can immediately write the force:
(5-1)

6. 5-2 Dynamics of Uniform Circular Motion
We can see that the force must be inward by thinking about a ball on a string:

7. 5-2 Dynamics of Uniform Circular Motion
Net forces that can cause centripetal acceleration
Friction of tires on road
Tension of cord
Gravity keeping object in orbit
Source: physicsclassroom.com

8. 5-2 Dynamics of Uniform Circular Motion
Net forces that can cause centripetal acceleration
Normal force
Multimedia: Roller Coaster G forces
Source: physicsclassroom.com

9. 5-2 Dynamics of Uniform Circular Motion
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.

10. 5-1 Kinematics of Uniform Circular Motion
Frequency (f): Number of revolutions per second
Period (T): Time it takes for one revolution
T = 1 / f
Frequency: units are s-1
Period: units are s

11. 5-2 Dynamics of Uniform Circular Motion
Formulas for Circular Motion
Where R is radius of circle, T is period, v is velocity, m is mass
Average speed = 2πR / T
Acceleration = v2 / R
Substitute 2πR / T for v
Acceleration = (2πR/T)2 / R = 4π2R / T2 ; velocity not required
Fnet r= mar
Substitute mv2/R for ar
Fnetr = mv2 / R
Substitute 4π2R / T2 for ar
Fnet r= m4π2R / T2 ; velocity not required

12. Circular Motion
Examples

13. Circular Motion
Examples
Force on revolving ball (horizontal)

14. Circular Motion Examples Force on revolving ball (vertical)
Calculate FT at bottom of arc
At bottom, mg is downward, FT2 upward
FT2 - mg = m ((v2)2 / r)
FT2 = m ((v2)2 / r) + mg
= kg (6.56 m/s2) / 1.10 m
+ (0.150 kg)(9.8 m/s2)
= 7.34 N
(continued)

15. Circular Motion Examples Force on revolving ball (vertical)
0.150 kg ball at end of 1.10 m cord swung vertically
Calculate maximum speed at top of arc so ball doesn’t fall
At top, forces are mg and FT1
downward
FT + mg = m ((v1)2 / r)
v = √gr = √(9.80 m/s2 1.10m)
v = 3.28 m/s

16. Circular Motion Examples Tetherball
Acceleration is toward pole, not top
So, net force is toward pole
FTy – mg = 0
FTx is centripetal acceleration, towards center

17. Circular Motion Examples
Centripetal acceleration = aR = v2 / r ; / 9.8 m/s2 to find ‘g’s
Particle revolves in circle of circumference = 2πr = 2π(0.0600m) = m /revolution
5 x 104 rpm x 60 s / min = 833 rev / s
T = 1 / (833 rev / s) = 1.2 x 10-3 s / rev
v = 2πr / T = 3.14 x 102 m/s
aR = v2 / r = (3.14 x 102 m/s)2 /.0600 m
= 1.64 x 106 m/s2
‘g’s = 1.64 x 106 m/s2 / 9.8 m/s2 = 1.67 x 105

18. 5-3 Highway Curves, Banked and Unbanked
When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.

19. 5-3 Highway Curves, Banked and Unbanked
If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.

20. Circular Motion Examples Forces are mg, FN and horizontal FF
FN = mg = 9800 N
Net horizontal force to keep car moving in circle
(continued)

21. Circular Motion Examples Skidding on a curve Dry pavement
Maximum static friction of tires (μS = 0.60)
5900N > 3900N so car stays on road
Icy road
Maximum static friction of tires (μS = 0.250)
2500N < 3900N so car skids off of road

22. Circular Motion Examples
Car moving along horizontal circle, so aR is horizontal.
FN sin(θ) = mv2 / r
FN cos(θ) - mg = 0 (no vert. motion)
FN = mg / cos(θ) [FN > mg]
Substitute for FN in first equation
sin(θ) (mg / cos(θ)) = mv2 / r
or mg x tan(θ) = mv2 / r
tan(θ) = v2 / rg =
(14 m/s)2 / (50m)(9.8 m/s2) = 0.40
Θ = 22°

23. 5-6 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.

24. 5-6 Newton’s Law of Universal Gravitation
The gravitational force on you is one-half of a Third Law pair: the Earth exerts a downward force on you, and you exert an upward force on the Earth.
When there is such a disparity in masses, the reaction force is undetectable, but for bodies more equal in mass it can be significant.

25. 5-6 Newton’s Law of Universal Gravitation
Therefore, the gravitational force must be proportional to both masses.
By observing planetary orbits, Newton also 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
(5-4)

26. 5-6 Newton’s Law of Universal Gravitation
The magnitude of the gravitational constant G can be measured in the laboratory.
This is the Cavendish experiment.

27. 5-7 Gravity Near the Earth’s Surface; Geophysical Applications
Now we can relate the gravitational constant to the local acceleration of gravity. We know that, on the surface of the Earth:
Solving for g gives:
Now, knowing g and the radius of the Earth, the mass of the Earth can be calculated:
(5-5)

28. 5-7 Gravity Near the Earth’s Surface; Geophysical Applications
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.

29. 5-8 Satellites and “Weightlessness”
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.
Velocity at which satellite escapes from Earth’s gravity called escape velocity

30. Escape Velocities
Source:

31. 5-8 Satellites and “Weightlessness”
The satellite is kept in orbit by its speed – it is continually falling, but the Earth curves from underneath it.

32. 5-8 Satellites and “Weightlessness”
Objects in orbit are said to experience weightlessness. They do have a gravitational force acting on them, though!
The satellite and all its contents are in free fall, so there is no normal force. This is what leads to the experience of weightlessness.

33. 5-8 Satellites and “Weightlessness”
More properly, this effect is called apparent weightlessness, because the gravitational force still exists. It can be experienced on Earth as well, but only briefly:

34. Newton’s Law of Universal Gravitation Examples
Since satellite at r=2, FG is ¼ as great
FG = ¼ mg
= ¼ (2000 kg) (9.80 m/s2)
= 4900 N

35. Newton’s Law of Universal Gravitation Examples
a) Period of satellite must be 1 day
Only force on satellite is gravity, so
G mSat x mE / r2 = mSat v2 / r ; unknowns are v and r
But, period of satellite = period of rotation of Earth = 24 h
24 h x 3600 s / 1 h = 86,400 s
Speed of satellite v = 2πr / T (same speed at Earth)
Substitute: G x mE / r2 = (2πr)2 / r T2
r = 4.23 x 107 m = 42,300 km
Radius of Earth = 6380 km, so satellite orbits at 36,000 km

36. Newton’s Law of Universal Gravitation Examples
Geosynchronous satellite
b) v = √ G mE
= √ (6.67x10-11 x Nm2/kg2) (5.98 x 1024 kg) / 4.23 x 107 m
= 3070 m/s

37. Newton’s Law of Universal Gravitation Examples

38. 5-9 Kepler’s Laws and Newton's Synthesis
Kepler's First and Second Laws (FYI)
Orbits of planets are ellipses with Sun at one focus
Planets sweep out equal areas in equal times (velocity not constant)

39. 5-9 Kepler’s Laws and Newton's Synthesis
Deriving Kepler's Third Law
Since the centripetal force equals the
force of gravity for an orbiting body
M is mass of Sun,
m is mass of planet
We know the formulas for centripetal acceleration and velocity so
Substituting for ac in original formula
divide out m
(If the units of R are AU (astronomical units) and
the units of time are Earth years, then R3 = T2)

40. 5-9 Kepler’s Laws and Newton's Synthesis
Kepler’s 3rd Law: The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun.

41. Gravitational field
Gravitational field exists for every object that has mass
Field is responsible for gravitational force on objects
Field is force per unit mass
Magnitude of Earth’s gravitational field at any point above Earth’s surface is G ME / r2
At Earth’s surface, gravitation field = g

42. Gravitational field
Since gravity is an attractive only force, field lines point toward center of Earth (center of mass)

43. Summary
An object moving in a circle at constant speed is in uniform circular motion.
It has a centripetal acceleration
There is a centripetal force given by
The centripetal force may be provided by friction, gravity, tension, the normal force, or others.
Period: Time required for one revolution
Frequency: Revolutions per second

44. Summary of Chapter 5 Newton’s law of universal gravitation:
Satellites are able to stay in Earth orbit because of their large tangential speed.
Objects in freefall are said to be weightless because there is no normal force, but there is still gravitational force

45. Summary of Chapter 5
Kepler’s Third Law states that the square of the period of revolution is proportional to the cube of the mean distance from the Sun

46. Summary of Chapter 5
Multimedia – Circular Motion