1. Unit 3: Motion and Forces in 2 and 3 Dimensions
A) Motion in Circles
B) Centripetal Force
C) Universal Gravitation and Orbital Motion

2. Key Points Calculate angular speed in radians per second.
Calculate linear speed from angular speed and vice-versa.
Describe and calculate centripetal forces and accelerations.
Describe the relationship between the force of gravity and the masses and distance between objects.
Calculate the force of gravity when given masses and distance between two objects.
Describe why satellites remain in orbit around a planet.

3. Vocabulary rotate radian revolve orbit axis centripetal force
law of universal gravitation
circumference
linear speed
angular speed
centrifugal force
radian
orbit
centripetal force
centripetal acceleration
ellipse
satellite
angular displacement
gravitational constant

4. Vectors and Direction Key Question:
How do we describe circular motion?

5. Motion in Circles
We say an object rotates about its axis when the axis is part of the moving object.
A child revolves on a merry-go-round because he is external to the merry-go-round's axis.

6. Angular Speed
Angular speed is the rate at which an object rotates or revolves.
There are two ways to measure angular speed
number of turns per unit of time (rotations/minute)
change in angle per unit of time (deg/sec or rad/sec)

7. Angular Speed
For the purpose of angular speed, the radian is a better unit for angles.
One radian is approx degrees.
Radians are better for angular speed because a radian is a ratio of two lengths.

9. w = q t Angular Speed Angle turned (rad) Angular speed (rad/sec)
Time taken (sec)

10. Angular Speed Why does this formula work?
Angle turned (rad)
Angular speed
(rad/sec)
w = q t
Time taken (sec)
rad/s = rad s

11. Calculating angular speed
A bicycle wheel makes six turns in 2 seconds.
What is its angular speed in radians per second?
1) You are asked for the angular speed.
2) You are given turns and time.
3) There are 2π radians in one full turn. ω = θ/t
4) Solve: ω = (6 × 2π) ÷ (2 sec) = 18.8 rad/sec

12. Calculating angular speed
Known: Unknown:
6 rotations w (rad/s)
2 s
1 rotation = 2π rad
= 6 rot x 2 π rad
2 s rot
= 6 π rad/s
1) You are asked for the angular speed.
2) You are given turns and time.
3) There are 2π radians in one full turn. ω = θ/t
4) Solve: ω = (6 × 2π) ÷ (2 sec) = 18.8 rad/sec
w = q t

13. Linear and Angular Speed
A wheel rolling along the ground has both a linear speed and an angular speed.
A point at the edge of a wheel moves one circumference in each turn of the circle.

14. 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)

15. 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.

16. Linear and Angular Speed Why does this formula work?
v = w r
Radius (m)
Linear speed
(m/sec)
Angular speed
(rad/sec)
Circumference m s
2πr m ×
2π rad
w rad s =
v = r w m s
Radians in one circumference

17. Calculating linear from angular speed
Two children are spinning around on a merry-go-round.
1) You are asked for the children’s linear speeds.
2) You are given the angular speed of the merry-go-round and radius to each child.
3) The relationship that applies is v = ωr.
4) Solve:
For Siv: v = (1 rad/sec)(4 m) = 4 m/sec.
For Holly: v = (1 rad/sec)(2 m) = 2 m/sec.
Siv is standing 4 meters from the axis of rotation and Holly is standing 2 meters from the axis.
Calculate each child’s linear speed when the angular speed of the merry go-round is 1 rad/sec.

18. Calculating linear from angular speed
Known: Unknown:
= 1 rad/s v1
r1 = 2 m v2
r2 = 4 m
v = w r
1) You are asked for the children’s linear speeds.
2) You are given the angular speed of the merry-go-round and radius to each child.
3) The relationship that applies is v = ωr.
4) Solve:
For Siv: v = (1 rad/sec)(4 m) = 4 m/sec.
For Holly: v = (1 rad/sec)(2 m) = 2 m/sec.
v1 = w r1 = (1)(2) = 2 m/s
v2 = w r2 = (1)(4) = 4 m/s

19. Linear and Angular Speed and Displacement

20. Calculating angular from linear speed
A bicycle has wheels that are 70 cm in diameter (35 cm radius).
The bicycle is moving forward with a linear speed of 11 m/sec.
Assume the bicycle wheels are not slipping and calculate the angular speed of the wheels in RPM (rotations per minute).
1) You are asked for the angular speed in RPM.
2) You are given the linear speed and radius of the wheel.
3) v = ωr, 1 rotation = 2π radians
4) Solve: ω = v ÷ r = (11 m/sec) ÷ (.35 m) = 31.4 rad/sec.
Convert to rpm using dimensional analysis:
ω = 31.4 rad/sec x 60 sec/min x 1 rotation/2π radians = 300 rpm

21. Calculating angular from linear speed
Known: Unknown:
v = 11 m/s w in rot/min
r = 0.35 m
1 rot = 2π rad
1 min = 60 s
11 m/s =
11 m/s__
0.35 m/rad
= 31.4 rad s
Circumference
1) You are asked for the angular speed in RPM.
2) You are given the linear speed and radius of the wheel.
3) v = ωr, 1 rotation = 2π radians
4) Solve: ω = v ÷ r = (11 m/sec) ÷ (.35 m) = 31.4 rad/sec.
Convert to rpm using dimensional analysis:
ω = 31.4 rad/sec x 60 sec/min x 1 rotation/2π radians = 300 rpm
31.4 rad x 1 rot x s
s π rad min
× w
(0.35 m)(2π)
2 π rad
= 300 rot/min
= w
Radians in one circumference

22. Centripetal Force Key Question:
Why does a roller coaster stay on a track upside down on a loop?

23. 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.

24. 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.

25. Centripetal forces keep these children moving in a circular path.

26. Uniform Circular Motion
Uniform circular motion is motion along a circular path in which there is no change in speed, only a change in direction.
Constant velocity tangent to path. v Fc
Constant force toward center.
Question: Is there an outward force on the ball?

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

28. Examples of Centripetal Force
You are sitting on the seat next to the outside door. What is the direction of the resultant force on you as you turn? Is it away from center or toward center of the turn? Fc
Car going around a curve.
Force ON you is toward the center.

29. The centripetal force is exerted BY the door ON you. (Centrally)
Car Example Continued
The centripetal force is exerted BY the door ON you. (Centrally) F’
Reaction Fc
There is an outward force, but it does not act ON you. It is the reaction force exerted BY you ON the door. It affects only the door.

30. Centripetal Acceleration
Consider ball moving at constant speed v in a horizontal circle of radius R at end of string tied to peg on center of table. (Assume zero friction.) n Fc R v W
Force Fc and acceleration ac toward center. W = n

31. Deriving Central Acceleration
Consider initial velocity at A and final velocity at B: vf vf B R vo
-vo Dv R vo s A

32. Deriving Acceleration (Cont.)vf
-vo R vo Dv s
ac = Dv t
Definition: = Dv v s R
Similar Triangles
mass m = Dv t vs Rt
ac = = vv R
Centripetal acceleration:

33. Car Negotiating a Flat Turnv Fc
What is the direction of the force ON the car?
Ans. Toward Center
This central force is exerted BY the road ON the car.

34. Car Negotiating a Flat Turnv Fc
Is there also an outward force acting ON the car?
Ans. No, but the car does exert a outward reaction force ON the road.

35. Car Negotiating a Flat Turn
The centripetal force Fc is that of static friction fs: R v m Fc n
Fc = fs fs R mg
The central force FC and the friction force fs are not two different forces that are equal. There is just one force on the car. The nature of this central force is static friction.

36. Finding the maximum speed for negotiating a turn without slipping.
Fc = fs n mg fs R R v m Fc
The car is on the verge of slipping when FC is equal to the maximum force of static friction fs.
Fc =
mv2 R
fs = msmg
Fc = fs

37. Maximum speed without slipping (Cont.)
Fc = fs n mg fs R v m Fc
mv2 R
= msmg
v = msgR
Velocity v is maximum speed for no slipping.

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

40. Calculating 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

41. 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.

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

43. Calculating 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.

44. 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.

45. 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.

47. Universal Gravitation and Orbital Motion
Key Question:
How strong is gravity in other places in the universe?

48. 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.

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

50. Calculating 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.