1. Foundations of Physics
CPO Science
Foundations of Physics
Chapter 9
Unit 3, Chapter 8

2. 8.1 Linear and Angular Speed
Radius (m)
Circumference
(m)
C = 2 P r
Distance (m)
2 P r
Speed
(m/sec)
v = d t
Time (sec)

3. 8.1 Linear and Angular Speed
Radius (m)
Linear speed
(m/sec)
v = w r
Angular speed
(rad/sec)
*This formula is used in automobile speedometers based on a tire's radius.

4. *Students read Section 8.2 AFTER Investigation 8.2
8.2 Centripetal Force
Key Question:
Why does a roller coaster stay on a track upside down on a loop?
*Students read Section 8.2 AFTER Investigation 8.2

5. 8.2 Centripetal Force
We usually think of acceleration as a change in speed.
Because velocity includes both speed and direction, acceleration can also be a change in the direction of motion.

6. 8.2 Centripetal Force
Any force that causes an object to move in a circle is called a centripetal force.
A centripetal force is always perpendicular to an object’s motion, toward the center of the circle.

7. Fc = mv2 r 8.2 Centripetal Force Mass (kg) Linear speed (m/sec)
force (N)
Fc = mv2 r
Radius of path
(m)

9. 8.2 Calculate centripetal force
A 50-kilogram passenger on an amusement park ride stands with his back against the wall of a cylindrical room with radius of 3 m.
What is the centripetal force of the wall pressing into his back when the room spins and he is moving at 6 m/sec?
1) You are asked to find the centripetal force.
2) You are given the radius, mass, and linear speed.
3) The formula that applies is Fc = mv2 ÷ r.
4) Solve: Fc = (50 kg)(6 m/sec)2 ÷ (3 m) = 600 N

10. 8.2 Centripetal Acceleration
Acceleration is the rate at which an object’s velocity changes as the result of a force.
Centripetal acceleration is the acceleration of an object moving in a circle due to the centripetal force.

11. 8.2 Centripetal Acceleration
Speed
(m/sec)
Centripetal
acceleration (m/sec2)
ac = v2 r
Radius of path
(m)

12. 8.2 Calculate centripetal acceleration
1) You are asked for centripetal acceleration and a comparison with g (9.8 m/sec2).
2) You are given the linear speed and radius of the motion.
3) ac = v2 ÷ r
4) Solve:
ac = (10 m/sec)2 ÷ (50 m) = 2 m/sec2
The centripetal acceleration is about 20% or 1/5 that of gravity.
A motorcycle drives around a bend with a 50-meter radius at 10 m/sec.
Find the motor cycle’s centripetal acceleration and compare it with g, the acceleration of gravity.

13. 8.2 Centrifugal Force
We call an object’s tendency to resist a change in its motion its inertia.
An object moving in a circle is constantly changing its direction of motion.
Although the centripetal force pushes you toward the center of the circular path...
...it seems as if there also is a force pushing you to the outside. This apparent outward force is called centrifugal force.

14. 8.2 Centrifugal Force
Centrifugal force is not a true force exerted on your body.
It is simply your tendency to move in a straight line due to inertia.
This is easy to observe by twirling a small object at the end of a string.
When the string is released, the object flies off in a straight line tangent to the circle.

16. 8.3 Universal Gravitation and Orbital Motion
Key Question:
How strong is gravity in other places in the universe?
*Students read Section 8.3 AFTER Investigation 8.3

17. 8.3 Universal Gravitation and Orbital Motion
Sir Isaac Newton first deduced that the force responsible for making objects fall on Earth is the same force that keeps the moon in orbit.
This idea is known as the law of universal gravitation.
Gravitational force exists between all objects that have mass.
The strength of the gravitational force depends on the mass of the objects and the distance between them.

18. 8.3 Law of Universal Gravitation
Mass 1
Mass 2
Force (N)
F = m1m2 r2
Distance between
masses (m)

19. 8.3 Calculate gravitational force
The mass of the moon is × 1022 kg.
The radius of the moon is × 106 m.
1) You are asked to find a person’s weight on the moon.
2) You are given the radius and the masses.
3) The formula that applies is Fg = Gm1m2 ÷ r2
4) Solve:
Fg = (6.67 x N.m2/kg2) x (90 kg) (7.36 x 1022 kg) / (1.74 x 106 m)2 = 146 N
By comparison, on Earth the astronaut’s weight would be 90 kg x 9.8 m/s2 or 882 N.
The force of gravity on the moon is approximately one-sixth what it is on Earth.
Use the equation of universal gravitation to calculate the weight of a 90 kg astronaut on the surface of the moon.

20. 8.3 Orbital Motion
A satellite is an object that is bound by gravity to another object such as a planet or star.
If a satellite is launched above Earth at more than 8 kilometers per second, the orbit will be a noncircular ellipse.
A satellite in an elliptical orbit does not move at a constant speed.

21. Application: Satellite Motion