For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

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## Table of Contents

## Transcription

## Related Books

### Uniform Circular Motion

- The distance around a circular path is the circumference.
- Frequency is the number of revolutions or cycles which occur each second. The symbol is f, and the units are 1/s, or Hertz (Hz).
- Period, the inverse of frequency, is the amount of time it takes for each revolution or cycle. The symbol for period is T, and units are seconds (s).
- Objects moving in uniform circular motion (circular motion at constant speed) are accelerating because their direction is always changing, causing their velocity to continuously change.
- The direction of this acceleration is toward the center of the circular path. This type of acceleration is known as a centripetal acceleration (center-seeking). The force causing a centripetal acceleration is known as a centripetal force.

### Uniform Circular Motion

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Uniform Circular Motion
- Distance Around the Circle for Objects Traveling in a Circular Path at Constant Speed
- Average Speed for Objects Traveling in a Circular Path at Constant Speed
- Frequency
- Period
- Frequency and Period
- Example I: Race Car
- Example II: Toy Train
- Example III: Round-A-Bout
- Example III: Part A - Period of the Motion
- Example III: Part B- Frequency of the Motion
- Example III: Part C- Speed at Which Alan Revolves
- Uniform Circular Motion
- Direction of Centripetal Acceleration
- Magnitude of Centripetal Acceleration
- Example IV: Car on a Track
- Centripetal Force
- Calculating Centripetal Force
- Example V: Acceleration
- Example VI: Direction of Centripetal Acceleration
- Example VII: Loss of Centripetal Force
- Example VIII: Bucket in Horizontal Circle
- Example IX: Bucket in Vertical Circle
- Example X: Demon Drop

- Intro 0:00
- Objectives 0:08
- Uniform Circular Motion 0:42
- Distance Around the Circle for Objects Traveling in a Circular Path at Constant Speed
- Average Speed for Objects Traveling in a Circular Path at Constant Speed
- Frequency 1:42
- Definition of Frequency
- Symbol of Frequency
- Units of Frequency
- Period 2:04
- Period
- Frequency and Period 2:19
- Frequency and Period
- Example I: Race Car 2:32
- Example II: Toy Train 3:22
- Example III: Round-A-Bout 4:07
- Example III: Part A - Period of the Motion
- Example III: Part B- Frequency of the Motion
- Example III: Part C- Speed at Which Alan Revolves
- Uniform Circular Motion 5:28
- Is an Object Undergoing Uniform Circular Motion Accelerating?
- Direction of Centripetal Acceleration 6:21
- Direction of Centripetal Acceleration
- Magnitude of Centripetal Acceleration 8:23
- Magnitude of Centripetal Acceleration
- Example IV: Car on a Track 8:39
- Centripetal Force 10:14
- Centripetal Force
- Calculating Centripetal Force 11:47
- Calculating Centripetal Force
- Example V: Acceleration 12:41
- Example VI: Direction of Centripetal Acceleration 13:44
- Example VII: Loss of Centripetal Force 14:03
- Example VIII: Bucket in Horizontal Circle 14:44
- Example IX: Bucket in Vertical Circle 15:24
- Example X: Demon Drop 17:38
- Example X: Question 1
- Example X: Question 2
- Example X: Question 3
- Example X: Question 4

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: Uniform Circular Motion

*Hello, everyone, and welcome back to www.educator.com.*0000

*I'm Dan Fullerton and in this lesson we are going to talk about uniform circular motion.*0004

*Our objectives include calculating the speed of an object traveling in a circular path or portion of the circular path.*0008

*Calculating the period and frequency for objects moving in circles at constant speed.*0015

*Explaining the acceleration of an object moving in a circle at constant speed.*0020

*Solving problems involving calculations of centripetal acceleration.*0024

*Defining centripetal force and recognizing that is not a special kind of force*0030

*but that is provided by forces such as tension, gravity, and friction.*0034

*Solving problems involving calculations of centripetal force.*0038

*Uniform circular motion, objects travel a circular path at constant speed, when they do that we call that uniform circular motion.*0043

*The distance around the circle is its circumference.*0052

*If there is a circle and we call the distance from the center to the edge of the radius, then circumference is 2π r,*0055

*the distance around the outside.*0063

*If instead, you define a diameter through the center point across a circle as your diameter then circumference is just π × D.*0066

*The average speed formula that we talked about from our kinematic section still applies here.*0077

*Average speed is distance traveled / time but for something moving in a circle once around the circle,*0082

*it would be 2 π r the circumference ÷ the time it takes to go once around the circle.*0087

*If it is once around the circle, you can say that is 2 π r ÷ period, where the period is the time for one complete cycle or revolution.*0092

*Frequency is the number of revolutions or cycles which occur each second.*0103

*It gets the symbol f and the units are 1/s, which are also known as a Hertz abbreviated capital Hz.*0107

*Frequency is the number of cycles per seconds or the number of revolutions per second.*0117

*Corresponding to frequency, we have the period that is the time it takes for one complete revolution or cycle.*0123

*Its symbol is T and the units are seconds.*0129

*T period time for one cycle is the time for one revolution.*0133

*Frequency in period are closely related.*0138

*The frequency is 1/ the period and the period is 1/ the frequency.*0142

*1 When you know one, you can easily find the other.*0146

*A couple of examples here starting off by talking about a race car.*0151

*The combined mass of a race car and its driver is 600 kg.*0155

*Traveling at constant speed the car completes one lap around a circular track of radius 160m in 36 s, calculate the speed of the car.*0159

*We will start with their circle.*0168

*A radius of 160 m, average velocity is distance travel ÷ time.*0170

*Its distance traveled is once around so that is a circumference ÷ time or 2 π × the radius 160 m/36 s which is 27.9 m/s .*0177

*A very simple example but probably worth getting a good solid foundation in the basics before we move further.*0195

*Taking a look at a toy train, a 500g toy train completes 10 laps of its circular track in 00:01:40.*0202

*If the diameter of the track is 1m, find the trains period T and its frequency f.*0210

*Period is going to be, it takes it 100s or 1min and 40s to do 10 laps.*0219

*Period being the time for 1 lap or 1 revolution is just going to be 10 s.*0227

*Once we know that, the frequency becomes 1/ the period or 1/10s which could be 0.1 Hz.*0234

*Let us take a look at a more detailed example but still pretty straightforward.*0246

*Allen makes 38 complete revolutions on the playground around about 30s.*0250

*If the radius of the roundabout is 1 m, determine the period of the motion.*0256

*The frequency of the motion, the speed at which Allen revolves, and how sick he is when he is out.*0260

*A, period of the motion, the period it takes him 30s to do 38 revolutions, the period is 0.789s.*0267

*The frequency is just 1/ the period which would be 1/0.789s or 1.27 Hz and the speed at which Allen revolves.*0284

*Average speed is distance travel ÷ time.*0303

*The distance he travels is 2 π × the radius 1m, he does that 38 times.*0306

*To do that 38 times, it takes him 30s so the speed would be 7.96 m/s.*0315

*A little more on uniform circular motion.*0330

*Is an object undergoing uniform circular motion accelerating?*0332

*We draw our circle for something moving at constant speed and is it accelerating?*0337

*It is kind of a trick question.*0346

*If we draw the velocity along the path, notice that it is changing direction depending on where it is at.*0348

*And because velocity is a vector, it has direction and acceleration is change in velocity.*0355

*It has a vector, because you have a change in direction, you have a change in velocity.*0364

*Therefore, yes you are accelerating.*0368

*Yes, it is accelerating due to that change in direction even though the speed is staying the same.*0371

*How do you determine the direction of that centripetal acceleration?*0380

*Here, we have a look at an object moving counterclockwise around the circle.*0386

*At this point, it is instantaneous velocity is up and a little bit later its velocity is in that direction.*0391

*If we want to find its acceleration, that will be the change in velocity ÷ time which will be final velocity - initial velocity ÷ time.*0398

*Velocity final - velocity initial that is the same as final velocity + negative initial velocity.*0412

*Let us see if we can draw it to see the direction of the change in velocity.*0422

*Our final velocity, this vector here in green I’m going to try and reproduce here.*0427

*It is sliding over but something like that.*0432

*If our initial velocity is up, negative initial velocity must be down from that point.*0435

*Something like that, where this is VF this is – VI.*0447

*To add those vectors, we draw a line from the starting point of the first to the ending point of our last.*0452

*That is going to be the direction of our acceleration vector.*0460

*If we look here by these, if we draw that same vector here, we are pointing toward the center of the circle.*0464

*That is why I refer to it as a centripetal acceleration.*0470

*Centripetal acceleration, oftentimes abbreviated ac because centripetal means center seeking.*0474

*It always points towards the center of the circle, even though it is moving in a constant speed the object going*0490

*in uniform circular motion is constantly accelerating towards the center of the circle.*0496

*How do you find the magnitude of acceleration?*0503

*The magnitude is straightforward.*0506

*That is the square of the speed ÷ the radius of the circle, AC =V² /r.*0508

*You got to know that one.*0515

*Let us do another example here.*0518

*Miranda drives a car clockwise around a circular track of radius 30m.*0521

*It is a circle that you get the idea 30m.*0528

*She completes 10 laps around the track in 2 minutes.*0530

*Find Miranda’s total distance traveled in the average speed and centripetal acceleration.*0534

*Distance traveled is going to be the distance around the circle × 10 laps or 2 π × the radius 30m × 10 laps which will be about 1885m.*0540

*To find her average speed, average speed is distance over time.*0561

*That will be 1885m and it took her 2 minutes or 120s to do that, which is 15.7 m/s.*0567

*Centripetal acceleration, that is going to be the square of speed ÷ the radius which will be 15.7m/s² ÷ 30m or 8.22 m/s².*0581

*Let us take a look now at what causes the centripetal acceleration.*0607

*We have acceleration, we must have a force.*0611

*We are going to call that a centripetal force.*0614

*If an object is traveling in a circle, it is accelerating towards the center of the circle, we have to find that.*0617

*For an object to accelerate there must be a net force.*0622

*This is what we call a centripetal force, it is a center seeking force.*0625

*Centripetal points towards the center in order to cause a center seeking acceleration.*0630

*When we talked about net force in the x direction, caused an acceleration*0637

*in the x direction or net force in the y direction cause an acceleration in the y direction,*0642

*we can also note now the net force towards the center of a circle causes an acceleration towards the center of the circle.*0650

*A part to note here, as centripetal force is not a new force.*0660

*It is just a label placed on an existing force when it is directed to the center of circle.*0665

*Tensions can cause a centripetal force, gravity, even friction, can provide us centripetal force.*0670

*A centripetal force is not any magical new force that just comes into existence because something was in a circle.*0676

*Something has a force towards the center of the circle that causes it to move in a circle.*0682

*We will call that forces centripetal force.*0687

*It is just a more generic label for any force going towards the center of the circle.*0689

*As such, you do not want to label anything FC or F centripetal on a free body diagram.*0693

*Be more specific, write down what it is that is causing that force.*0700

*Calculating centripetal force.*0706

*If net force is mass × acceleration, net force for the center of the circle is mass × acceleration towards the center of circle.*0709

*We know that centripetal acceleration is V² /r.*0720

*We can write net force must be equal to MV² /r, that easy.*0726

*If you want a look at units, the units of force are going to be kilogram m² / s² mass × speed² ÷ distance which is going to be kilogram m/s²,*0737

*which is our definition of a Newton, the unit of force.*0754

*Let us take another example.*0761

*If the car is accelerating, is its speed increasing?*0762

*That depends.*0767

*If you think about that, we could have a car traveling with some velocity to the right and it could be accelerating to the right.*0767

*If that is the case, yes its speed is going to increase.*0780

*On the other hand, we can have a car with some velocity to the right and acceleration to the left.*0785

*Is its speed increasing?*0797

*No, it is decreasing.*0800

*Or we could take a look at a car traveling in a circular path with some centripetal acceleration*0802

*that is accelerating but it is moving at constant speed.*0811

*If the car is accelerating is its speed increasing?*0815

*Perhaps, it could be but not necessarily.*0819

*In the diagram below, the car travels clockwise a constant speed in a horizontal circle.*0825

*At the position shown in the diagram which air indicates the direction of the centripetal acceleration of the cart.*0830

*It is got to be A toward the center of the circle.*0837

*Here we have a ball attached to a spring to a string move at constant speed in a horizontal circular path.*0844

*The target is located near the path of the ball is shown.*0850

*At which point along the balls path should the string be released removing the centripetal force if the ball is to hit the target?*0854

*If you want the ball to hit the target, he would release it at the point when it is instantaneous velocity*0863

*is tension to the circle which would be right at B.*0868

*The moment you get rid of that centripetal force, it no longer has a centripetal acceleration.*0873

*Therefore, it travels in the straight line by Newton’s first law.*0878

*Let us take a look at the bucket swung in a horizontal circle.*0885

*The diagram shows a 5 kg bucket of water swung a horizontal circle of radius 0.7m at the constant speed of 2 m/s.*0888

*What is the magnitude of the centripetal force on the bucket of water?*0896

*The centripetal force, the magnitude of the centripetal force will be MV² /r or 5 kg × its speed 2m /s² ÷ 0.7m which is 28.6 N.*0900

*What happens if we do this in a vertical circle?*0924

*The diagram now shows the same bucket, same radius, constant speed 3 m/s,*0927

*find the magnitude of tension of a string at the top of the circle and at the bottom of the circle.*0931

*Let us draw our free body diagram for the top and we will start with that as our analysis.*0937

*We have a tension pulling down, strings can only pull, and we have the weight MG.*0943

*When we write Newton’s second law, I am going to write Newton’s second law in this centripetal direction for the center of the circle.*0950

*Towards the center the circle is down so we have T + MG and that must equal mass × acceleration or MV² /r.*0957

*Which implies then the tension must be MV² /r - MG which will be 5 kg × 3 m/s² ÷ the radius 0.7m - 5 × 10 or T =14.3 N.*0968

*Let us do the same analysis down here when the bucket is at the bottom of the circle.*0992

*At the bottom, our free body diagram looks a little different.*1000

*Our tension is pointing up, gravity still pulls down, and we will write Newton’s second law equation*1005

*f net C = T is pointing towards the center of the circle so that is positive.*1011

*Gravity MG is pointing away from the center of a circle so that is negative.*1017

*T - MG =MV² /r.*1021

*When we solve for tension, T =MV² /r + MG which is going to be 5 × 3² / 0.7 + 5 × 10.*1026

*Therefore, the tension now in the string is 114.3 N.*1043

*Quite a difference.*1053

*Let us take a look at the demon drop problem.*1058

*Put up this after an amusement park ride at an amusement park that used to visit when I was a kid.*1061

*The diagram shows at the top of you the 65 kg student here at point A on an amusement park ride.*1069

*The ride spins a student in a horizontal circle of radius 2 1/2 m at the constant speed of 8.6 m /s.*1076

*While that is happening, the floor is lowered and the student remains against the wall without falling to the floor.*1083

*There is no floor there but the student remains against the wall.*1089

*First off, draw the direction of the centripetal acceleration of the student on the diagram.*1092

*Centripetal acceleration, center seeking, nice easy towards the center of the circle.*1097

*There we go for that part.*1102

*For part 2, determine the centripetal acceleration of the student and the centripetal force acting on the student.*1106

*Centripetal acceleration is V² /r so that is just going to be 8.6 m/s² ÷ our radius is 2.5m which will be 29.6 m/s².*1113

*If we want the centripetal force, the centripetal force is going to be MAC which is 65kg × centripetal acceleration 29.6 m/s² or 1924 N.*1133

*Some force is pushing the student or pulling the student toward the center of the circle with 1924 N of force.*1153

*Number 3, what force keeps the student from sliding to the floor?*1163

*Let us draw a picture.*1169

*Here is our student against the wall, is it spinning in the direction kind of like that.*1170

*If I draw a free body diagram from the side, we have weight down, normal force from the wall, and we must have force of friction.*1177

*What keeps the student from sliding to the floor?*1190

*It must be the force of friction.*1193

*How do you get a formula for force of friction?*1195

*Remember, force of friction is μ × the normal force.*1197

*You must have a big normal force and that is what is causing the centripetal force is the normal force.*1200

*Force of friction keeps a student from sliding.*1209

*What is the minimum coefficient of friction between the student and the wall required to keep the student from sliding down the wall?*1214

*In this case, in order to not slide down the wall, force of friction and the weight of the student have to be balanced.*1221

*Force of friction must equal the weight of the student.*1227

*We have just said force of friction is μ × the normal force.*1231

*μ × the normal force must equal MG which implies then that μ must equal MG / the normal force.*1237

*We also know for all of this to work that the normal force is providing the centripetal force which must be equal to mac.*1247

*We can write that μ =MG / mac which is just G /centripetal acceleration or 10/ centripetal acceleration was 29.6.*1259

*We find that we have a μ of 0.34, that is the coefficient of friction that we need in order to keep the student from sliding down the wall.*1275

*Hopefully, that is a pretty good refresher of fundamentals on uniform circular motion.*1287

*Thanks so much for joining us at www.educator.com and make it a great day everyone.*1292

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