Physics

SPH4U-B

4
Circular Motion

Physics SPH4U-B

Lesson 4

Introduction
Transportation has undoubtedly advanced. Not only have vehicles changed, but the means by which people can get themselves to and from different locations has changed as well. Maps and compasses now take a backseat to the Global Positioning System (GPS). Access to the
GPS is now very widespread; 24 satellites above our atmosphere are used to send and receive information in order to accurately determine locations of objects to within 15 m of their actual position. If you have used a GPS, or even a mobile phone or satellite TV or radio, you have used some of the nearly 300 artificial satellites that exist above our atmosphere to send and receive information. Satellites are projectiles that, through our knowledge of force and energy, have been launched into orbit. Because the earth curves downward by approximately 5 metres over
8000 metres along the horizon, satellites launched with a horizontal speed of 8000 m/s can take orbit. At this speed, thanks to earth’s curved surface and a satellite’s constant horizontal velocity, the satellites, which are in free fall once projected, keep falling toward the earth, but always miss it. Thus, under the influence of gravity, satellites maintain motion in a circular pattern at a uniform speed.
Most satellites are launched using rockets that fall to the ocean when their fuel is spent.
Sometimes a satellite may require minor adjustments to its orbit. This is accomplished through the use of built-in rockets, called thrusters. Once placed in the proper orbit, a satellite can stay there for a long time, with its operation and location monitored by computers and human operators at a control centre on earth. Solar panels provide a source of power for some satellites.
A satellite keeps its solar panels facing the sun and its antennae ready to receive information. A satellite remains in orbit until its speed decreases and gravity pulls it down into the atmosphere, where it slows down, due to its collisions with air molecules in the atmosphere. As the satellite falls further down into the denser atmosphere, it compresses the air in front of it, causing the air to become so hot that the satellite burns up. In this lesson, you will learn more about the forces that contribute to the motion of a satellite and to other circular motions.

Planning Your Study
You may find this time grid helpful in planning when and how you will work through this lesson. Suggested Timing for This Lesson (hours)
Circular Motion
Investigation into Circular Motion

1

Problem Solving with Circular Motion

1

Frames of Reference

1

Key Questions

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What You Will Learn
After completing this lesson, you will be able to
•

describe the relationships between variables affecting uniform circular motion

•

identify and analyze the forces contributing to uniform circular motion

•

distinguish between inertial and non-inertial frames of reference, and real and fictitious
(apparent) forces

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Lesson 4

Circular Motion
Before you begin this lesson, find an object that you can safely tie to a string and swing around in a circle, for example, a yo-yo, or even a belt with a heavy buckle. A heavier object at the end of the rope will give you a better “feel” for the ideas in this lesson, but safety should be your primary concern. Make sure that you have lots of space in which to swing the object and put on some safety glasses or sports goggles, if you have them. Try to swing your object so that it stays on a constant horizontal circular path (as if you were getting ready to lasso an animal). After this, try to make a vertical circular motion. Notice the differences. Can you feel how the tug on your hand changes at different times, in the vertical cycle? That’s what this lesson is all about: the forces involved in creating uniform circular motion.

Uniform Circular Motion
Uniform circular motion is motion in which an object travels in a circle, maintaining a constant radius and constant speed. The following diagram shows the path of an object in uniform circular motion and indicates three specific points in its path. The vectors indicate the object’s instantaneous velocity. Notice that the length of each vector is the same, indicating that the magnitude of velocity at all locations is the same. The direction of the vectors indicates the direction of the instantaneous velocity at each location, that is, the direction in which the object would travel, if the string to which it is connected suddenly snapped. Therefore, even though the magnitude of the velocity of the ball is not changing, the direction of the velocity is constantly changing, indicating that there is uniform acceleration, despite the constant speed.
→

v

→

v

r

→

v

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Centripetal Acceleration
The type of acceleration that results in uniform circular motion is called centripetal
→
acceleration ( ac ), also known as radial acceleration (because it is directed along the radius of the path of motion). Centripetal acceleration is always directed toward the centre of the circular path created by the motion; therefore, it is always perpendicular to the direction of the instantaneous velocity.
→

v

→

ac

Consider an object going around a circle with a constant speed and a constant radius.
→

v

r

If this object, at this particular location in the circular path, had continued to travel in a straight line instead of being forced to maintain a circular path, then, rearranging the basic
→
→ equation for determining velocity, the displacement would have been equal to Δ d = v Δt .
→
→
Δd =
However, the displacement was not equal to v Δt because the object experienced acceleration in
→
→
Δd = order to stay on the circle. As a result, the object’s displacement is actually equal to v Δt , plus
1→
the displacement that occurred due to the uniform acceleration: ac Δt 2 . This is based on the
2
kinematics equation you know for finding the displacement of an object that is experiencing
→
→
1→
uniform acceleration: Δ d = vi Δt + a Δt 2. Notice that, in the diagram, the direction of the
2
displacement due to acceleration is in the same direction as the centripetal acceleration: toward the centre of the circle.

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Lesson 4

→

→

Δ d = v Δt r r

1→ x = ac Δt 2
2

By drawing in the various displacement vectors for this object, a right triangle is formed. Notice that the displacement due to the acceleration is being set as equal to x.
Applying the Pythagorean theorem:
→2

r 2 + Δ d = (r + x)2
→

→

Given that Δ d = v Δt , the equation can be rewritten as:
→

2

r 2 + v Δt = (r + x)2

Expanding and simplifying, you get:
→2

r 2 + v Δt 2 = r 2 + 2rx + x 2
→2

v Δt 2 = 2rx + x 2

Look at the diagram again. If the two object locations that had been chosen were much closer together, what would happen to x? Can you see that as the two points get closer, x will become
2
smaller? In fact, as the time interval of the change in position approaches zero, x will approach zero faster, becoming negligible. (This actually involves a little bit of calculus.)
→2

2

Given that2it=becomes negligible, you can eliminate x to get: r 2 + v Δt r 2 + 2rx + x 2
→2

v Δt 2 = 2rx + x 2

Solving for x, you get:
→2

v Δt 2 = 2rx
→2

x=

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v Δt 2
2r

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Recall that:
1→
x = ac Δt 2
2
Substituting this into the equation and simplifying, you get an equation for centripetal acceleration: →2

1→ 2 v Δt 2 ac Δt =
2r
2
→2

→

ac =

v r In the equations above, you have written the vectors as they are meant to be communicated, using arrows above the variables. While this is correct, it is a common convention, when working with uniform circular motion, to omit the arrows above the vector symbols, even though you are talking about vector quantities. This simply means that the magnitude of the vectors—not the direction—is being considered in the equations. This is an acceptable convention in uniform circular motion questions, since
•

the direction of centripetal acceleration is always toward the centre of the circle.

•

the direction of the instantaneous velocity is always tangent to the circle.

In previous math courses, you learned that the circumference of a circle is equal to 2πr.
Similarly, you may recall that the time required to complete one rotation is referred to as the period of revolution and is given the symbol T. Since the magnitude of velocity of an object is
Δd
, the magnitude of velocity in uniform circular motion can be calculated through the v= Δt
2πr
. equation: v =
T
Substituting this expression into the centripetal acceleration equation you have derived gives you: 2

v r 2
⎛ 2πr ⎞
⎜
⎝ T ⎟
⎠
ac = r 2
4π r ac = 2
T
ac =

1 where f is the frequency of rotation (that is, how often the f rotations happen, measured in hertz). Therefore, if you know the frequency (f) in a situation, you may use the equation:

You may also recall that T =

2

ac = 4π rf

2

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Lesson 4

Equation Summary v 2

ac = r You now have three equations for calculating centripetal acceleration:
2
⎛ 2πr ⎞
2
v
⎜
⎟
1. ac = ⎝ T ⎠ ac = r r 2
4π r
2. ac = 2
T

3.

2

ac = 4π rf

2

So, if an object is in uniform circular motion, you have the tools to analyze that motion.
However, as you learned in the last lesson, where there is acceleration, there must be force.
What causes uniform circular motion?

Centripetal Force
According to Newton’s second law, whenever there is acceleration, there must be a net force causing that acceleration. The net force that keeps an object moving in uniform circular motion is referred to as centripetal force. Centripetal force is not a new “type” of force; it is simply the sum of the combination of forces that contribute to the circular motion. In other words, it is just another name for net force in a situation where there is uniform circular motion. Regardless of which forces contribute to the centripetal force causing the object to follow a circular path, the direction of this net force is always aimed toward the centre of the circle, in the same direction as the centripetal acceleration.

Example 1
In this diagram of a ball that is rotating clockwise in a horizontal plane, determine the directions of the velocity, centripetal acceleration, and centripetal force at the location indicated. N

W

E

S

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Solution
•

The velocity is directed south (tangent to the circle and in the direction in which the ball would move if the string broke, at this instant).

•

The centripetal acceleration is always directed toward the centre of the circle, which, at this instant, happens to be west.

•

The centripetal force is always directed toward the centre of the circle, which, at this instant, happens to be west (the same as acceleration).

Support Question
Do not send your answers in for evaluation.
28.

In this diagram of a ball that is rotating clockwise within a vertical plane, determine the directions of the velocity, centripetal acceleration, and centripetal force, at the location indicated. Up

Left

Right

Down

There are Suggested Answers to Support Questions at the end of this unit.

Example 2
Draw FBDs for the following objects and state which forces contribute to the centripetal force:
a)

A satellite in orbit around the earth

b)

A car travelling around a (left-directed) turn on a horizontal road surface

c)

A ball on a string being whirled in a circle in the vertical plane, at the bottom of the circle

d)

A person on an amusement-park ride who is standing on a platform with her back against a wall and is being rotated in a horizontal circle

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Lesson 4

Solution
a)

→

Fg

The force of gravity provides the centripetal force.
b)

Back view:

→

FN

→

Fs
→

Fg

It is actually the force of static friction that enables a car to make a turn on a horizontal surface.
Just imagine if the surface were icy—the car might just go straight, without this static friction to prevent it from slipping.
c)
→

FT

→

Fg

d)

It is the tension force in the string and the force of gravity on the ball that provide the centripetal force.
→

FN

→

FN

wall on person

→

Fg

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Note that it is often a normal force that contributes to the centripetal force to keep an object in a “container” (vehicle, ride, and so on) during uniform circular motion.

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Support Question
29.

Draw FBDs for the following objects and state which forces contribute to the centripetal force:
a) A ball on a string that is being whirled in a circle on a frictionless horizontal surface
b) A pilot in a jet who is making a loop in the vertical plane and is now at the top of the loop (the pilot is upside down at this location)

Investigation into Circular Motion
Before moving on to look at problem solving in situations involving uniform circular motion, you are going to perform an online investigation, in order to take a closer look at the variables involved in circular motion and how they affect one another.
Go to your course page on the ILC website and, under Lesson 4, open the “Uniform Circular
Motion” simulation of a mass being whirled in a circle. Notice the following:
•

At the end of the string, there is a mass that is hanging down. This mass provides the centripetal force for this motion.

•

The radius of the circular path followed by the stopper remains constant.

•

The speed of the stopper is constant, although its direction is constantly changing.

•

The string is merely hanging; it is not “held” in place. It is only the uniform circular motion that is keeping it in place. The tension in the string is constant throughout, despite the bend. You could try to get a feel for this for yourself by hanging a string through a cylinder (for example, a paper towel roll), with some mass on either end. Try to rotate the top mass so that it attains and maintains uniform circular motion.

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Lesson 4

Support Questions
Hint: In the following questions, you will be using the abovementioned simulation to make observations while variables are being manipulated. One way of helping to verify your observations is through rearranging the appropriate equation to solve for the variable in question. For example, to determine how changing the radius affects frequency, rearrange the centripetal acceleration equation for frequency. If r appears in the numerator, it means that frequency is proportional to radius (as increasing radius causes frequency to increase). If r appears in the denominator, it means that frequency is inversely proportional to radius (as increasing radius causes frequency to decrease). Understanding such relationships helps with problem solving.
30.

Click and drag to change the radius of the circular path. When the radius increases, what happens to the
a) period of rotation?
b) frequency of rotation?
c) speed of the stopper?
d) centripetal acceleration?
e) centripetal force?

31.

Press “reset” and restart the simulation. This time, add more mass to the hanging mass.
As you add mass, you are actually increasing the tension in the string. As the hanging mass increases, what happens to the
a) period of rotation?
b) frequency of rotation?
c) speed of the stopper?
d) centripetal acceleration?
e) centripetal force?

32.

Press “reset” again, so that the two set-ups have the same motion. This time, increase the mass that is experiencing circular motion. As the mass of the object in motion increases, what happens to the
a) period of rotation?
b) frequency of rotation?
c) speed of the stopper?
d) centripetal acceleration?
e) centripetal force?

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Problem Solving with Circular Motion
Recall that you now know three equations for calculating centripetal acceleration. Because
→
→
Fnet = m a , all of the equations developed for centripetal acceleration are easily transformed into equations that can be used to calculate centripetal force by multiplying the centripetal acceleration by “mass.” Therefore, the three equations for calculating centripetal force (Fc ) are:
1.

Fc =

mv 2 r 2.

Fc =

4π 2mr
T2

4π 2mr
2
2
F =
Tc = 4π mrf
Fc
In uniform circular motion, it is a common convention to omit the arrows above the vector
4π 2 (0.250 you are m) symbols, even though kg)(2.00 talking about vector quantities.
T=
50.0 N
T = 0.628 s

3.

Example 1

A rope with a length of 2.00 m is used to swing a 0.250 kg object in a horizontal circular motion with a uniform speed of 20.0 m/s. What is the force of tension in the rope?

Solution
Given:
r = 2.00 m m = 0.250 kg v = 20.0 m/s
Required: FT
Analysis and solution:
→

FN

→

FT

→

Fg

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Lesson 4

While you are not told the vertical forces, you are told that the object is maintaining horizontal motion. Therefore, the vertical forces must cancel each other out. The centripetal force is provided solely by the tension force in the rope. mv 2 r (0.250 kg)(20.0 m/s)2
Fc =
2.00 m
Fc = 50.0 N
Fc =

Paraphrase: The force of tension in the rope is 50.0 N.

Example 2
Determine the period of rotation (that is, the time it takes for the ball to complete one entire circle) in the situation described in example 1 (the previous example).

Solution
Given:
r = 2.00 m m = 0.250 kg v = 20.0 m/s
Required: T
Analysis and solution:
Fc =
T=

4π 2mr
T2
4π 2mr
Fc

4π 2 (0.250 kg)(2.00 m)
T=
50.0 N
T = 0.628 s

Paraphrase: Therefore, the period of rotation is 0.628 s.

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Example 3
The same ball (0.250 kg) is now swung with uniform circular motion using the same rope
(2.00 m) so that it maintains the same speed (20.0 m/s) in a vertical plane. Determine the tension force in the rope at the highest point, and then again at the lowest point, in the circular path. Solution
Given:
r = 2.00 m m = 0.250 kg v = 20.0 m/s
Required: FT at the top of the circle, FT at the bottom of the circle
Analysis and solution:
At the top:
Let [down] be positive.
→

→

Fg + F T

Fc =

mv 2 r mv 2 r mv 2
+ mg = r 2 mv =
− mg r (0.250 kg)(20.0 m/s)2
=
− (0.250 kg)(9.8 m/s 2 )
2.00 m
= 47.6 N

FT + Fg =
FT
FT
FT
FT

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Lesson 4

At the bottom:
Let [up] be positive.

→

FT

→

Fg

Fc =

mv 2 r mv 2 r mv 2
− mg = r mv 2
=
+ mg r (0.250 kg)(20.0 m/s)2
=
+ (0.250 kg)(9.8 m/s 2 )
2.00 m
= 52.5 N

FT − Fg =
FT
FT
FT
FT

Paraphrase: Therefore, the force of tension in the rope is 47.6 N, when the ball is at the top of the circle, and 52.5 N, when the ball is at the bottom of the circle.
Note: For both calculations, the direction of acceleration was made the positive direction, as this is always easier.

Support Questions
33.

A 1200.0 kg car goes around a curve with a radius of 45.0 m. What is the maximum speed at which the car can safely make the turn, if the coefficient of static friction between the tires and the dry pavement is 0.500? Assume that the road is flat.

34.

An airplane is flying in a loop, following a circular path with a radius of 2.50 km in the vertical plane. What is the minimum speed that the airplane can maintain without being forced out of the circular path?

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Frames of Reference
Have you ever taken a ride on a roller coaster? If you have, you’ll know that, as it goes around sharp turns, you feel pushed against the sides of the roller coaster car. You may also feel this way in a car that is going around a sharp bend. For a moment, you feel as if a force is pushing you in the opposite direction to the bend, away from the centre of the circle. There is, however, no such force directed away from the centre of the circular path; just because your frame of reference is accelerating, you imagine one!

Inertial versus Non-inertial Frames of Reference
An inertial frame of reference is a frame of reference that is either stationary or moving at constant velocity; that is, not accelerating. Motion in inertial frames of reference can be explained using Newton’s laws of motion.
A non-inertial frame of reference is a frame of reference that is undergoing acceleration, that is, experiencing a net force. Motion with respect to non-inertial frames of reference cannot be explained using Newton’s laws of motion. Therefore, people often use “fictitious” forces—forces that are not real—to try to explain the motion.
Now, go back to the example where you are riding in the turning roller coaster car. The fictitious “push” toward the centre of the circle is referred to as a centrifugal force. A centrifugal force is a fictitious force that is used to try to explain the push on objects to maintain circular motion. However, there is really no such force. If you were in a car that was rounding a bend, your body, according to Newton’s first law, would tend to continue moving in a straight line.
Because the car and seat belt prevent this, you, like the car, make the turn. The force you feel is the normal force of the vehicle on you. This is easy to explain from earth’s frame of reference, but not so easy to explain from the frame of reference of the accelerating vehicle. If you had no idea of your motion with respect to some inertial frame (and here, you’re assuming that the car is an inertial frame), you would have no way to explain the “push” that is being exerted on you.
Another example of a fictitious force that has been discussed already is “apparent weight.” If you recall the elevator example you worked with, you “felt” as if you had more weight or less weight, depending on the acceleration of the elevator. Similarly, someone in free fall might say that they are “weightless,” when really they weigh the same as they did when they were standing on solid ground! It is difficult to explain these force sensations, with respect to an accelerating frame of reference.

Example
You are riding in a car when, suddenly, the air freshener hanging above you swings forward.
Is the car haunted? Explain your answer by constructing two FBDs—one in the car’s frame of reference, and the other, in the earth’s frame of reference.

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Lesson 4

In the car’s frame of reference:
→

FT
→

Ffictitious

→

Fg

In the earth’s frame of reference:
→

FT

→

Fg

Because you are in the car with the air freshener, your frame of reference accelerates with the car, which makes it seem as if there is a force acting on the air freshener. An observer standing on the street would see the car’s acceleration and would not need to include the fictitious force in the FBD.

Support Question
35.

Imagine that you are given a pail of water and are told that, somehow, you must make this pail turn upside down without the water falling out. No equipment is allowed other than you, the pail, and the water. Using the knowledge that you’ve gained in this lesson, how can you do this?
D
 o you need help? If so, go to your course page on the ILC website under Lesson 4, and view the short video clip called “Water in a Bucket” for some help and a demonstration.

Review: Frames of Reference
G
 o to your course page on the ILC website under Lesson 4, and view the video called “Frames of Reference,” which demonstrates the ideas in this lesson and reviews relative velocity and circular motion.

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Key Questions
Save your answers to the Key Questions.
When you have completed the unit, submit them to ILC for marking.
(24 marks)
12.

Does the label “centripetal force” ever appear in an FBD? Explain. (2 marks)

13.

Sometimes, road surfaces have banked curves. Use an FBD to explain how this helps cars to make turns more safely. (3 marks)

14.

A bus passenger has her laptop sitting on the flat seat beside her as the bus, travelling at
10.0 m/s, goes around a turn with a radius of 25.0 m. What minimum coefficient of static friction is necessary to keep the laptop from sliding? (5 marks)

15.

Keys with a combined mass of 0.100 kg are attached to a 0.25 m long string and swung in a circle in the vertical plane. (9 marks)
a) What is the slowest speed that the keys can swing and still maintain a circular path?
b) What is the tension in the string at the bottom of the circle?

16.

Do research to find out what artificial gravity is and how it is related to centripetal motion. Explain how artificial gravity could be created in a weightless environment and give a reason why we would want to do this. Give at least one source that you used for your research. (5 marks)

This is the last lesson in Unit 1. When you are finished, do the Reflection for Unit
1. Follow any other instructions you have received from ILC about submitting your coursework, then send it to ILC. A teacher will mark your work and you will receive your results online.

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