For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### Uniform Circular Motion

- For an object to maintain the same speed while moving in a circle, its velocity must constantly be changing-the object has an acceleration.
- The formula for acceleration if you have uniform circular motion (the same speed throughout the circle):
| →a| = | →v| ^{2}r. - The acceleration vector always points from the object to the center of the circle.
- The velocity vector is always tangential to the circle.
- Remember: in the above equation, those are the
__magnitudes__of the vectors, since the direction of both acceleration and velocity must constantly be changing. - A revolution is one complete circuit of the circle.
- The circumference of a circle is 2πr and the speed is |→v|, so the time it takes for one revolution is
T = 2πr | →v| .

### 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
- Centripetal Acceleration
- Centripetal Acceleration of a Rock Being Twirled Around on a String
- Looking Closer: Instantaneous Velocity and Tangential Velocity
- Magnitude of Acceleration
- Centripetal Acceleration Formula
- You Say You Want a Revolution
- Example 1: Centripetal Acceleration of a Rock
- Example 2: Magnitude of a Car's Acceleration While Turning
- Example 3: Speed of a Point on the Edge of a US Quarter

- Intro 0:00
- Centripetal Acceleration 1:21
- Centripetal Acceleration of a Rock Being Twirled Around on a String
- Looking Closer: Instantaneous Velocity and Tangential Velocity
- Magnitude of Acceleration
- Centripetal Acceleration Formula
- You Say You Want a Revolution 6:11
- What is a Revolution?
- How Long Does it Take to Complete One Revolution Around the Circle?
- Example 1: Centripetal Acceleration of a Rock 7:40
- Example 2: Magnitude of a Car's Acceleration While Turning 9:20
- Example 3: Speed of a Point on the Edge of a US Quarter 13:10

### High School Physics Online Course

I. Motion | ||
---|---|---|

Math Review | 16:49 | |

One Dimensional Kinematics | 26:02 | |

Multi-Dimensional Kinematics | 29:59 | |

Frames of Reference | 18:36 | |

Uniform Circular Motion | 16:34 | |

II. Force | ||

Newton's 1st Law | 12:37 | |

Newton's 2nd Law: Introduction | 27:05 | |

Newton's 2nd Law: Multiple Dimensions | 27:47 | |

Newton's 2nd Law: Advanced Examples | 42:05 | |

Newton's Third Law | 16:47 | |

Friction | 50:11 | |

Force & Uniform Circular Motion | 26:45 | |

III. Energy | ||

Work | 28:34 | |

Energy: Kinetic | 39:07 | |

Energy: Gravitational Potential | 28:10 | |

Energy: Elastic Potential | 44:16 | |

Power & Simple Machines | 28:54 | |

IV. Momentum | ||

Center of Mass | 36:55 | |

Linear Momentum | 22:50 | |

Collisions & Linear Momentum | 40:55 | |

V. Gravity | ||

Gravity & Orbits | 34:53 | |

VI. Waves | ||

Intro to Waves | 35:35 | |

Waves, Cont. | 52:57 | |

Sound | 36:24 | |

Light | 19:38 | |

VII. Thermodynamics | ||

Fluids | 42:52 | |

Intro to Temperature & Heat | 34:06 | |

Change Due to Heat | 44:03 | |

Thermodynamics | 27:30 | |

VIII. Electricity | ||

Electric Force & Charge | 41:35 | |

Electric Fields & Potential | 34:44 | |

Electric Current | 29:12 | |

Electric Circuits | 52:02 | |

IX. Magnetism | ||

Magnetism | 25:47 |

### Transcription: Uniform Circular Motion

*Hi, welcome back to educator.com, Today we are going to be talking about uniform circular motion.*0000

*We have not really talked about what does it mean to have a uniform circular motion.*0005

*Uniform circular motion just means that you have got something traveling at a uniform rate, or a constant rate.*0010

*If you have got something move at a constant speed uniformly in a circle, say we have got a rock on a string, moving around in a circle at a constant speed.*0016

*The rock has a constant speed, but what does it mean about the velocity, does it have a constant velocity?*0030

*It might be going at say, 10 m/s at this point, and at this point, and at this point, and at this point, but will the velocity remain the same?*0040

*Well, that does not quite make sense.*0050

*Because we have got something that is going around in a circle, that means at this point that is actually moving something like this, and at this point it is moving something like this, and at this point it is moving something like this, so it is actually changing as it goes around.*0051

*Speed remains the same, but its velocity actually changes.*0064

*Otherwise, if its velocity were not changing, it would just continue to go off in a straight line.*0067

*It would not be able to make a circle, it would go off in a straight line if it did not have a changing velocity.*0071

*So, its velocity cannot be constant even though speed is constant.*0077

*What exactly is that change in ' v'?*0081

*We do not know quite yet what the change in v is precisely, but let us consider two opposite cases.*0084

*What about the string has the rock at the very top, and the string has the rock at the very bottom?*0090

*If you have ever swung something around by its end and let go at a certain point, like a slinky or a yoyo, and you are spinning it around in a circle and let go (like here), you might notice that it does not go off like this, it goes off like this.*0099

*That is because, as it curves around the circle, for it to be moving in that circle at that moment in time, it has to be going off directly this way.*0118

*At the bottom it has to be going off directly this way.*0126

*If you cut the string at that specific moment, the rock would fly off in a tangential path, a tangent is something that barely brushes the edge and goes off straight.*0131

*It is parallel to the curve at that one point.*0143

*The instantaneous velocity is the tangent at that point on the circle.*0146

*Wherever it is, it is going to be tangential.*0151

*What about if we look at a bunch of really close positions?*0155

*If we looked at a rock moving just a little bit each time, what is going to happen then is that we are going to have, is these little lines are only going to change by a bit each time.*0158

*So they are only sort of like, tick down only a teensy bit every time as they move their way around the circle.*0168

*What does that mean f we are doing that?*0174

*With each little step forward, the instantaneous velocity, (the instantaneous velocity is this part right here, that moment if we were to take a freeze frame what would its velocity be),*0179

*the instantaneous velocity has to be tugged down a bit with each one of these freeze frames.*0190

*Otherwise it is going to move off in a straight line, it is just going to turtle off away from the circle.*0194

*So to keep it in a circle, we have to have something pulling it in.*0199

*What do we have pulling it in?*0202

*We have got the string pulling it in.*0203

*So we are pulling it in with the string.*0205

*The string is going to keep it tangential, keep the velocity tangential by giving us an acceleration pulling directly in.*0208

*So we need something to pull this velocity directly in to the circle by just a little bit each time.*0214

*Wherever we go, we have an acceleration pulling in to keep that velocity tangential.*0220

*So velocity always tangential, and acceleration directly in for us to be able to maintain the circle.*0227

*What exactly would be the size or the magnitude of the acceleration, the size of a is the same as the magnitude of a.*0234

*The magnitude is going to depend on v.*0245

*If we have got a circle, and something is moving around very very quickly, we are going to need a very fast, very large acceleration to pull that in, to keep it in a circle continuously.*0247

*On the other hand, if we have got something that is making a very slow circle, we do not need a very fast acceleration at all.*0258

*We can just have something to pull, tug lightly to be able to keep it in a circle because it is moving so slowly.*0266

*The magnitude of a very clearly depends on how fast the object is traveling.*0271

*High speed require a big acceleration to be able to keep it in the circle.*0276

*Also, the radius is going to have an effect.*0279

*Think about this: You have two circles, one like this, and one like this.*0282

*At constant velocity, this acceleration is going to have to be able to pull a lot harder to be a able to keep it in the circle.*0287

*This one, it has got all this time.*0295

*It has got his time because it takes so much longer to make its way around the circle.*0296

*In the time it takes for this object with constant velocity this far, this one will manage to wrap its whole way around the circle.*0301

*With this one, it has got the time, it can take its time pulling that thing in, so the acceleration does not have to be as large.*0307

*That reasoning which we just talked about brings us to this formula.*0314

*We cannot derive it explicitly because that takes some calculus, and we do not have that yet.*0321

*But we can at least understand intuitively why this comes about.*0324

*Your acceleration, and you do not have the precise acceleration because the acceleration will always be changing, remember, the acceleration always points in.*0328

*It pulls in to the circle.*0336

*So, that reasoning leads us to this formula.*0338

*The magnitude of the acceleration = (speed) ^{2}/radius, which lets us have an understanding that the velocity has an effect, a larger velocity is going to cause a large acceleration, and a larger radius is going t o cause a smaller acceleration.*0341

*That gives us the centripetal acceleration.*0359

*Centripetal means 'to the centre', so centripetal acceleration is based off this, this formula right here.*0362

*It is an important formula for all things in uniform circular motion.*0368

*What about a revolution, what does it mean to make a revolution?*0371

*A revolution is really simple, it is just the time that it takes for an object to make its way one time around the circle.*0376

*A 360 degree turn is a single revolution of an object or if you have gotten into trigonometry, in terms of radians it would be 2 π radians. (2×π)*0384

*We are going to be using degrees in this course for most part, but 360 degrees or 2 π radians, these are all names for the same thing, which is just a single circle.*0395

*A revolution is complete when a point on the edge has made all the way around the circle.*0411

*For this point, for it to make all the way around the circle, what does that mean?*0415

*That means it has managed to travel the entire distance of this circle.*0419

*We know its speed, we know what the distance on a circle is, we know how long the track is, it is 2 π times the radius (2 π r), or π times diameter, so distance/speed is going to give us the time that it took.*0424

*Time for a revolution = circumference / speed = (2 π r) / speed*0446

*Let us take some examples.*0459

*We have got a rock, twirled around at 8 m/s on a rope of length 1 m .*0462

*What is the rock's acceleration?*0478

*We just use our formula, magnitude of the acceleration = (speed) ^{2} / (radius of the circle) = (8^{2}) / 1 = 64 m/s/s , is the acceleration.*0480

*That is the magnitude of the acceleration, remember, depending on where we are, the acceleration is going to be constantly be changing, it is always going to be pulling in.*0505

*We know that magnitude is going to be 64 m/s/s .*0512

*The specific components of that vector will change, we could solve that if we knew what angle it was at, we could solve for what it was.*0516

*But we know it is always going to 64 m/s/s pointing to the centre of the circle, from the point that we are looking at.*0523

*How long will it take the object to complete one revolution?*0530

*We know what the velocity is, we have the information to figure out what the circumference is, so from here we just use our formula.*0532

*So, the time for a revolution = (2 π r) / (speed) = (2 π × 1m) / (8 m/s) = 0.79 seconds, it is how long it will take to complete a revolution.*0540

*Here is another example, we got a car driving at a constant speed of 20 m/s .*0560

*It is driving on some road, and it is currently driving to the East.*0566

*It is going to go through a turn , and it is going to come out going to the North.*0569

*If the car is driving at a constant speed throughout the turn, and it goes through a quarter circle turn, that has a radius of 150 m, (part of an imaginary circle if we were to complete this quarter circle),*0576

*...so this is a really big slow turn, what will be the acceleration that the car will experience during the turn?*0590

*It is going to experience it as soon as it enters the circle, and it is always going to be pointing in, to the middle.*0597

*What will that be?, use our formula.*0613

*The size of the acceleration, the magnitude = (speed) ^{2} / radius = (20)^{2} / (150m) , it is a good idea to keep your units in it, because it gives us a chance to double check, if units are wrong in the end, you missed something!,*0615

*Sometimes, when the problems are easier, when you know for sure what you are doing, it is OK to omit it, but in general you are encouraged to keep your units because it helps you to understand what you are doing with your work.*0655

*So, (20 m/s) ^{2} / (150 m) = (400 m^{2}/s^{2}) / 150 m = 400/150 = 2.67 m/s/s .*0674

*m/s ^{2} is also the normal, acceptable form for acceleration, I just prefer m/s/s because it gives you a better idea of what is happening.*0714

*Now we have got our answer, 2.67 m/s/s is the amount that the car is being pulled in to the centre of the turn at all times.*0720

*Keep in mind, that is a pretty fair acceleration, if gravity = 9.8 m/s/s, 2.67 m/s/s/ is a pretty considerable amount.*0729

*This person is pulling about more than a quarter of a 'g' is pulling them in, quarter of a 'g' being quarter of a 'g' acceleration, 'g' being 9.8 m/s/s.*0740

*So, that is a fair amount of acceleration being pulled in, which is why when you are turning in a car you experience that feeling of being pulled somewhere.*0750

*We will talk about later on, you do not feel like you are being pulled in to the centre of the car, you feel like you are being pulled in to the centre of the circle.*0757

*If you feel like you are being pulled out, that is the difference between centripetal and centrifugal, and we will explain why you will have that experience.*0766

*But, now you have an understanding of why there is that acceleration taking place , and a little bit later in the course, when we get to talking about uniform circular motion and force, you will get the chance to understand why it is that feels like you are being pulled against the car.*0777

*You are being pushed up against the door, but you will understand why that is later.*0785

*Example 3, final example.*0789

*We have got a US quarter on a turn table.*0793

*The quarter is spinning at 1776 rpm, which is revolutions per minute.*0795

*It has a diameter of 24.26 mm.*0804

*What will be the speed of a point on the edge of the quarter in m/s?*0813

*To get this in m/s, we need to get everything in S.I. units.*0818

*Right now, none of our stuff is in S.I. units.*0821

*We need to work towards that.*0823

*Starting off, 1776 rpm, what would that be if we were to get this in seconds?*0825

*We divide that by 60, 1776/60 =29.6 rev. per second.*0833

*But that is not equal to 'T'.*0846

*T is equal to that number flipped, because T = 1/29.6 = number of seconds to the revolution.*0848

*So, it takes (1/29.6) seconds per revolution.*0861

*It makes a lot of sense, if we had 29.6 revolutions in a second, then if we want to see how many seconds it takes to do that, we are going to have to flip it, take the reciprocal, 1/29.6 = 0.0338 s.*0868

*That is how long it takes to give us a single revolution.*0885

*What is the diameter, if we want that in S.I. units?*0888

*Diameter = 24.26 mm = 0.02426 m , (move once to cm, another to dm, another to m.)*0893

*Now we have got the information we need to solve.*0912

*T = (diameter × π) / speed = D × π / speed , we do not know speed, but we do know T and D and π is just a constant, so we can solve this now.*0914

*Remember that velocity will always be tangential to the point we are looking at, speed is the only thing we can solve that will be truly constant.*0931

*speed = (0.02426 m × π) / 0.0338 = 2.255 m/s, is the speed that a point on the edge is moving at. (Once again, I encourage you to keep the units.)*0944

*And that tells us the answer.*0987

*Hope you enjoyed this, hope you learned something.*0988

*See you later when you come back to educator.com for the next lesson.*0990

1 answer

Last reply by: Professor Selhorst-Jones

Wed Aug 31, 2016 5:54 PM

Post by Claire yang on August 31, 2016

What is magnitude?