1. Circular Motion and Gravitation
Chapter 7
Circular Motion
and
Gravitation

2. Centripetal Acceleration
An object traveling in a circle, even though it moves with a constant speed, will have an acceleration
The centripetal acceleration is due to the change in the direction of the velocity

3. Centripetal Acceleration, cont.
Centripetal refers to “center-seeking”
The direction of the velocity changes
The acceleration is directed toward the center of the circle of motion

4. Centripetal Acceleration, cont.
a = Δv (eq. I) Δt
By similar triangles
Δv = Δs
v r
therefore
Δv = Δs v r
Sub into eq. I
a = Δs v = v2
r Δt r
Since Δs = v

5. Centripetal Acceleration and Centripetal Force
The centripetal acceleration can also be related to the angular velocity
ac = vt2 Fc = mac r
Therefore, Fc = mvt2

6. Forces Causing Centripetal Acceleration
Newton’s Second Law says that the centripetal acceleration is accompanied by a force
F = maC
F stands for any force that keeps an object following a circular path
Tension in a string
Gravity
Force of friction

7. Level Curves
Friction is the force that produces the centripetal acceleration
Can find the frictional force, µ, v
Fc= mac = mv2 r
Fc =f !!!!!!
f = μn = μmg = mv2
Therefore,
Example: The coefficient of friction between
The tires of a car and the road is 0.6. What is
the smallest circle that the car will be able to
Negotiate at 60mph (26.8 m/s)?

8. Vertical Circle Consider the forces at the top of the circle
If car just makes it over the top then the forces exerted by track are zero.
Ftrack = 0
The minimum force at the top of the circle to continue circular motion is mg
Therefore
mg = mv2 r
v gr

9. Newton’s Law of Universal Gravitation (1687)
Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

10. Law of Gravitation, cont.
G is the constant of universal gravitational
G = x N m²/kg²
This is an example of an inverse square law
Example: find the magnitude and direction of the gravitational force between the sun and the earth.

11. Escape Speed (see p. 206 in text)
If an object is launched from the earth with a high enough velocity it will escape the earth’s gravity.
The escape speed is the speed needed for an object to soar off into space and not return
For the earth, vesc is about 11.2 km/s
Note, v is independent of the mass of the object

12. Satellite Law Motion GMm = mv2 r2 r r = GM v2
Assuming a circular orbit is a good approximation
Fg = Fc F
GMm = mv2
r r
r = GM v2

13. Kepler’s Laws Based on observations made by Tycho Brahe
Newton later demonstrated that these laws were consequences of the gravitational force between any two objects together with Newton’s laws of motion

14. Kepler’s Laws:
All planets move in elliptical orbits with the Sun at one of the focal points.
A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals.
The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.

15. Kepler’s First Law
All planets move in elliptical orbits with the Sun at one focus.
Any object bound to another by an inverse square law will move in an elliptical path
Second focus is empty

16. Kepler’s Second Law
A line drawn from the Sun to any planet will sweep out equal areas in equal times
Area from A to B and C to D are the same

17. 5.5 Satellites in Circular OrbitsGM

18. Kepler’s Third Law
The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.
T = circumference of orbit
orbital speed
For orbit around the Sun, KS = 2.97x10-19 s2/m3
K is independent of the mass of the planet
K = 42 GM
Example: A planet is in orbit 109 meters from the center of the sun. Calculate its orbital period and velocity.
Ms=1.991 x 1030 kg

19. 5-9 Kepler’s Laws and Newton's Synthesis
The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun.

20. Gravitational Force Chapter 7 Orbiting objects are in free fall.
Section 2 Newton’s Law of Universal Gravitation
Chapter 7
Gravitational Force
Orbiting objects are in free fall.
To see how this idea is true, we can use a thought experiment that Newton developed. Consider a cannon sitting on a high mountaintop.
Each successive cannonball has a greater initial speed, so the horizontal distance that the ball travels increases. If the initial speed is great enough, the curvature of Earth will cause the cannonball to continue falling without ever landing.

21. Torque The force can be resolved into its x- and y-components
The x-component, F cos Φ, produces 0 torque
The y-component, F sin Φ, produces a non-zero torque

22. Mechanical Equilibrium
In this case, the First Condition of Equilibrium is satisfied
The Second Condition is not satisfied
Both forces would produce clockwise rotations

23. Kepler’s Third Law
The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.
T = circumference of orbit
orbital speed
For orbit around the Sun, KS = 2.97x10-19 s2/m3
K is independent of the mass of the planet
K = 42 GM
Example: A planet is in orbit 109 meters from the center of the sun. Calculate its orbital period and velocity.
Ms=1.991 x 1030 kg

24. Simple Machines Chapter 7
Section 4 Torque and Simple Machines
Chapter 7
Simple Machines
A machine is any device that transmits or modifies force, usually by changing the force applied to an object.
All machines are combinations or modifications of six fundamental types of machines, called simple machines.
These six simple machines are the lever, pulley, inclined plane, wheel and axle, wedge, and screw.

25. Simple Machines, continued
Section 4 Torque and Simple Machines
Chapter 7
Simple Machines, continued
Because the purpose of a simple machine is to change the direction or magnitude of an input force, a useful way of characterizing a simple machine is to compare the output and input force.
This ratio is called mechanical advantage.
If friction is disregarded, mechanical advantage can also be expressed in terms of input and output distance.

26. Simple Machines, continued
Section 4 Torque and Simple Machines
Chapter 7
Simple Machines, continued
The diagrams show two examples of a trunk being loaded onto a truck.
In the first example, a force (F1) of 360 N moves the trunk through a distance (d1) of 1.0 m. This requires 360 N•m of work.
In the second example, a lesser force (F2) of only 120 N would be needed (ignoring friction), but the trunk must be pushed a greater distance (d2) of 3.0 m. This also requires 360 N•m of work.

27. Simple Machines, continued
Section 4 Torque and Simple Machines
Chapter 7
Simple Machines, continued
The simple machines we have considered so far are ideal, frictionless machines.
Real machines, however, are not frictionless. Some of the input energy is dissipated as sound or heat.
The efficiency of a machine is the ratio of useful work output to work input.
The efficiency of an ideal (frictionless) machine is 1, or 100 percent.
The efficiency of real machines is always less than 1.

28. Horizontal Circle Find ac of yoyo
The horizontal component of the tension causes the centripetal acceleration

29. 5-9 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.