For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

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### Centripetal Acceleration & Force

- Objects moving in a circular path are accelerating because their velocity is changing (their direction is changing, and velocity is a vector and includes direction), even if their speed is constant.
- An object in UCM is accelerating toward the center of the circle. This is known as a centripetal (center-seeking) acceleration. The force causing this acceleration is known as a centripetal force.
- The magnitude of centripetal acceleration is equal to the square of velocity divided by the radius.
- Don't label forces as centripetal forces on free body diagrams. A centripetal force is just a label put on a force to indicate it points toward the center of a circle. Instead, label the force on the free body diagram as specifically as possible (force of tension, force of gravity, etc.)

### Centripetal Acceleration & Force

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
- Direction of ac
- Magnitude of ac
- Centripetal Force
- Calculating Centripetal Force
- Example 1: Acceleration
- Example 2: Direction of ac
- Example 3: Loss of Centripetal Force
- Example 4: Velocity and Centripetal Force
- Example 5: Demon Drop
- Example 6: Centripetal Acceleration vs. Speed
- Example 7: Calculating ac
- Example 8: Running Back
- Example 9: Car at an Intersection
- Example 10: Bucket in Horizontal Circle
- Example 11: Bucket in Vertical Circle
- Example 12: Frictionless Banked Curve

- Intro 0:00
- Objectives 0:08
- Uniform Circular Motion 0:38
- Direction of ac 1:41
- Magnitude of ac 3:50
- Centripetal Force 4:08
- For an Object to Accelerate, There Must Be a Net Force
- Centripetal Force
- Calculating Centripetal Force 6:14
- Example 1: Acceleration 7:31
- Example 2: Direction of ac 8:53
- Example 3: Loss of Centripetal Force 9:19
- Example 4: Velocity and Centripetal Force 10:08
- Example 5: Demon Drop 10:55
- Example 6: Centripetal Acceleration vs. Speed 14:11
- Example 7: Calculating ac 15:03
- Example 8: Running Back 15:45
- Example 9: Car at an Intersection 17:15
- Example 10: Bucket in Horizontal Circle 18:40
- Example 11: Bucket in Vertical Circle 19:20
- Example 12: Frictionless Banked Curve 21:55

### AP Physics 1 & 2 Exam Online Course

### Transcription: Centripetal Acceleration & Force

*Hi, folks, and welcome back to Educator.com.*0000

*I am Dan Fullerton. In this lesson, we are going to talk about centripetal acceleration and force.*0003

*Our objectives are going to be to explain the acceleration of an object moving in a circle at a constant speed, to solve problems involving calculations of centripetal acceleration...*0008

*...to define centripetal force and recognize that it is not a special kind of force, but that it is provided by forces such as tension, gravity and friction -- something always causes the centripetal force.*0018

*Finally, of course, we are going to solve problems involving calculations of centripetal force.*0030

*So, let us get back to uniform circular motion.*0037

*The big question "Is an object that is undergoing uniform circular motion accelerating?"*0040

*If it is moving in a circle at constant speed, is it accelerating?*0045

*To answer that, we really need to understand very well what acceleration is.*0056

*Acceleration is a change in velocity and for an object going around the circle -- at some point there -- its velocity at that instant in time is tangent to the circle.*0062

*There its velocity is tangent to the circle and there it has a velocity tangent to the surface.*0072

*Notice that the direction of the velocity keeps changing.*0078

*Since the direction of velocity is changing -- yes, it is accelerating. Absolutely!*0082

*An object undergoing circular motion, even though it is moving at constant speed, is accelerating.*0095

*What about the direction of its acceleration then?*0102

*To do that, I am going to take you through sort of a kind of quasi-proof.*0105

*If acceleration is the change in velocity divided by time, that is the final velocity (vF) minus the initial velocity (vI) divided by time.*0112

*Or, vF - vI -- final velocity - initial velocity for an object going around a circle -- is just going to be the same as vF plus the opposite of vI.*0125

*So if we have an object going around a circle here, its initial velocity at a point in time -- tangent to the circle that way and an instant or two later that way.*0140

*Let us see if we cannot add up those vectors to see what we get.*0150

*vF is easy. We have already got that in here. It is pretty much in that direction.*0155

*vI, we have pointing up, so the opposite of vI would be pointing down.*0160

*To do that, I have to line them up tip to tail and draw it there, so that would be -vI.*0165

*The sum of those are vector addition and again, we go from the starting point of the first to the ending point of the last.*0173

*That must be the direction of vF + -vI which is the direction of the acceleration.*0181

*So where am I going to draw that?*0187

*Since I did that for these two points, I am going to draw that vector when it is right in between those two and we see if I draw it, the (a) points towards the center of the circle.*0189

*That is what we mean by centripetal acceleration.*0199

*'Centripetal' actually means center-seeking. It is always toward the center of the circle- Centripetal, center seeking.*0202

*For an object moving in a circle, it is accelerating even if it is moving at constant speed in the direction of that object's acceleration toward the center of the circle.*0218

*Now, the magnitude of centripetal acceleration -- we talked about very briefly in our previous lesson, but finding the magnitude is straightforward.*0230

*It is the square of the speed divided by the radius; Ac = V ^{2}/r.*0240

*Centripetal force is something that causes a centripetal acceleration.*0248

*If the object is traveling in a circle, we know it is accelerating toward the center of the circle, and for it to accelerate, there must be a net force.*0253

*Remember, net force equals mass times acceleration -- Newton's Second Law.*0259

*We call this force a centripetal force because it too is pointed toward the center of the circle. It is center-seeking.*0265

*So a net force toward the center of the circle causing a centripetal acceleration is a centripetal force, which we label Fc.*0274

*So we had F _{net} = MA and we talked about breaking that down into the x direction, F_{net}x = MAx.*0282

*Then we did that in the y direction, F _{net}y = MAy.*0290

*We could even do that for the centripetal direction, where the centripetal direction is just pointing toward the center of the circle -- F _{net}c = MAc.*0295

*It is important to note here -- a centripetal force is not some new magic force.*0306

*An object moving in a circle does not automatically have some magic force that causes it to move in a circle called the centripetal force.*0310

*Something else must be causing it to move in a circle, something like gravity or attention or somebody pushing on it.*0317

*You have to have something causing it to move in a circle.*0325

*Think of a car going around a track really fast in a circle.*0328

*As it is moving in a circle, something has to be applying a force toward the center of the circle. 0337 What is it that is doing that?*0332

*Try and imagine what the car would do if some of those forces went away.*0339

*If it is going around in a circle and it is a car -- if all of a sudden, a friction between the tires and the road goes away, the car goes careening off in a straight line; it cannot go in a circle.*0344

*For a car moving in a circle, what is causing the centripetal force is the force of friction.*0353

*You have to understand, you do not label something like centripetal force on a free body diagram (FBD).*0359

*A centripetal force is just a net force pointed toward the center of the circle.*0363

*It is a label we put on another force because it is pointed toward the center.*0368

*Calculating centripetal force -- We already said that the net force is equal to mass times acceleration, and since it is in the centripetal or toward the center of the circle direction, F _{net}c = MAc.*0374

*But we also just learned that the magnitude of centripetal acceleration is the speed ^{2} divided by the radius.*0388

*Therefore, F _{net} c = M × V^{2}/r.*0397

*If we want to check the units -- their dimensional analysis -- mass is in kg, velocity is in m/s that is squared divided by (r), which is in meters.*0408

*So I am going to have in the top kg-m ^{2} per m/s^{2}.*0420

*One of the meters, we will make a ratio of 1 and I will be left with kg-m/s ^{2} which is a Newton (N).*0428

*That makes sense. It is a force that we are after.*0436

*So F _{net} c = MV^{2}/r -- just combining Newton's Second Law with the magnitude of the centripetal acceleration.*0440

*If a car is accelerating, is its speed increasing?*0451

*That is another one of those thought questions.*0455

*Well, if we have a car over here and it is traveling with a velocity to the right and it is accelerating to the right, its speed is going to be increasing.*0459

*Acceleration and velocity in the same direction -- it is going to be speeding up. Absolutely, yes!*0475

*On the other hand, if we have a car and it has a velocity to the right but an acceleration vector pointing to the left, is it accelerating?*0482

*No, what is going to happen in the next instant in time -- that velocity is getting smaller.*0497

*For the accelerations pointing to the left, our next velocity is going to be that way.*0502

*Then it is eventually going to stop and then it is eventually going to be going the other direction.*0506

*So in this case, no, or we could also look at the situation of the car traveling in a circle.*0510

*Remember, the entire time it is traveling in a circle, it is accelerating toward the center of the circle even if it is doing so at constant speed.*0519

*Another sample question -- In the diagram below, a cart travels clockwise at constant speed in a horizontal circle.*0534

*At the position shown right here, which arrow indicates the direction of the centripetal acceleration of the cart?*0541

*Again, centripetal acceleration, a vector always points toward the center of the circle. The correct answer must be A.*0548

*A ball attached to a string is moved at constant speed in a horizontal circular path.*0560

*A target is located near the path of the ball, as shown in the diagram here.*0565

*At which point, along the ball's path, should the string be released, thereby removing the centripetal force, if the ball is to hit the target?*0569

*So the string's tension is what is causing the force to keep it moving in a circle.*0577

*Where do we let go of that so it hits the target?*0580

*Well, if the ball is moving this way, the moment that centripetal force goes away, there is no longer any net force on the ball.*0584

*It is going to travel in a straight line -- Newton's First Law.*0591

*There is no net force, no acceleration -- constant velocity, straight line.*0596

*I would let go at B where that line is now tangent to the target.*0600

*Another example -- A 1,000 kg car travels a constant speed of 20 m/s around a horizontal circular track.*0609

*Which diagram correctly represents the direction of the car's velocity and centripetal force at a particular moment?*0615

*Again, velocity at a particular moment is going to be tangent to the circle -- centripetal force -- centripetal must point toward the center of the circle.*0623

*It must be answer 1.*0633

*Again, since it is a car, what provides the centripetal force?*0635

*Well, if all of a sudden, the tires give way and you hit an oil slick, that car right here is going to keep going in a straight line.*0638

*It cannot turn anymore. So it must be the frictional force that is providing the centripetal force.*0644

*Here, we have an example with the Demon Drop, a popular amusement park ride.*0656

*Maybe you have been on it.*0659

*When you get on the ride, they have you stand up against the wall and they start to spin you.*0660

*They spin you faster and faster and faster and then the bottom -- the floor -- drops out.*0665

*You stay up there! You do not fall with it.*0670

*It is called different names at different amusement parks, but the diagram here shows a top view of this with a 65 kg student spinning.*0673

*Right now they are at point A. It is a radius of 2.5 m and a constant speed of 8.6 m/s.*0683

*The floor is lowered and the student remains against the wall without falling to the floor.*0688

*Draw the direction of the centripetal acceleration of the student on the diagram.*0694

*Centripetal acceleration -- by now, drawing these are probably getting pretty easy.*0699

*There we go -- centripetal acceleration -- toward the center of the circle.*0703

*Now, however, we want to determine the centripetal acceleration of the student and the centripetal force acting on the student.*0709

*The centripetal acceleration -- AC = V ^{2}/r; that is 8.6 m/s^{2} divided by the radius (2.5 m) or about 29.6 m/s^{2}.*0716

*How about the centripetal force acting on the student?*0737

*To find that -- F _{net}c = MAc. Mass is 65 kg. We just determined Ac was 29.6.*0740

*If we multiply those together and I come up with the centripetal force of about 1923N.*0755

*What causes that centripetal force?*0764

*Again, imagine we are looking down overhead.*0766

*Well, what is pushing in on the student must be the normal force, the perpendicular force, from the wall.*0769

*What force keeps the student from sliding to the floor?*0779

*Now we have to think a little bit more.*0782

*Let us draw a FBD for the student.*0785

*There is the wall -- while they are going around in this circle -- FBD -- there is our student.*0788

*Of course, we have their weight -- the force of gravity down and we have a force from the wall, the normal force -- perpendicular -- toward the center of the circle.*0797

*And if the student wants to slide down, he must have a force of friction pointing up.*0809

*It moves so quickly that we have a large normal force, therefore, we have a large frictional force.*0816

*So what force keeps the student from sliding to the floor?*0825

*That frictional force balances the weight to keep the student on the wall.*0827

*How do they do that?*0833

*They spin so fast you have a big normal force and since the force of friction is μ times the normal force -- if you have a very big normal force, you get a big force of friction.*0834

*Here, we have an example where we are looking at graphs to best represent the relationship between the magnitude of the centripetal acceleration and the speed of an object moving in a circle of constant radius.*0851

*The first thing I do when I see these sorts of problems is I try and find the relationship we are looking at, so I want a formula that has centripetal acceleration and speed in it -- relationship between the two.*0863

*I know that Ac = V ^{2}/r, where the variables I am interested in are speed on the x and centripetal acceleration.*0873

*What happens as V gets larger? Well, Ac gets larger.*0885

*Right away, we can get rid of those two.*0888

*Now is this a linear relationship? No, it is a squared relationship. Therefore, the correct answer must be 2.*0891

*Another one -- We have a half kg object moving in a horizontal circular path with a radius of one-quarter meter at a constant speed of 4 m/s.*0904

*What is the magnitude or size of the object's acceleration?*0913

*Ac = V ^{2}/r, so that is going to be 4 m/s^{2}/0.25 m -- 16/0.25 will be 64 m/s^{2}.*0916

*For something to be in uniform circular motion, it does not even have to go all the way around a circle, it just has to be traveling in a circular path.*0936

*So let us take a look at the example of a running back.*0945

*An 800N running back turns a corner in a circular path of radius 1 m at a velocity of 8 m/s. Find the running back's mass, centripetal acceleration and centripetal force.*0947

*To find the mass, let us start off -- we are given the weight of the running back -- mg = 800N. 0968 Therefore, the running back's mass is 800N/g -- 10 m/s ^{2} -- or about 80 kg.*0960

*To find the running back's centripetal acceleration -- Ac = v ^{2}/r.*0984

*The Velocity is 8 m/s ^{2}/radius (1 m) -- 64 m/s^{2}.*0991

*The centripetal force -- F _{net}c = MAc.*1006

*So, the mass of our running back (80 kg), the centripetal acceleration (64 m/s ^{2}) -- I come up with a net force, a centripetal force of 5,120N.*1014

*Now we have a car at an intersection.*1034

*A 1200 kg car traveling at a constant speed of 9 m/s turns at an intersection.*1038

*The car follows a horizontal circular path with a radius of 25 m to point P.*1045

*At point P, the car hits an area of ice and loses all frictional force on its tires.*1051

*What is the frictional force on the car before it reaches point P?*1058

*To answer that, we first have to realise that the frictional force is what is causing the centripetal force.*1062

*So the frictional force, F _{net}c, is the frictional force, which is MV^{2}/r.*1069

*Our mass is 1200 kg, our speed is 9 m/s ^{2} and our radius is 25 m, so I get a frictional force of 3,888N.*1078

*What path does the car follow on the ice?*1099

*Once it hits the ice, no longer does it have a centripetal force -- anything pushing into the circle is going to travel in a straight line.*1102

*It is going to keep going straight ahead, the same direction it was and it is going to follow path B.*1109

*One of my favorite demonstrations -- In the diagram, we have a 5 kg bucket of water that swung in a horizontal circle of radius 0.7 m at a constant speed of 2 m/s.*1120

*What is the magnitude of the centripetal force on the bucket of water?*1132

*F _{net}c = MAc, which is MV^{2}/r.*1138

*Our mass is 5 kg, our speed is 2 m/s ^{2}/radius (0.7 m) or 28.6N.*1145

*Now we have the 5 kg bucket of water swung in a vertical circle of radius 0.7 m.*1161

*Now it is being swung vertically with a speed of 3 m/s.*1167

*What is the magnitude of the tension on the string at the top of the circle and at the bottom of the circle?*1171

*Let us start at the top and what I am going to do is I am going to draw the FBD for the bucket when it is at the top of the circle.*1176

*There is my bucket and forces acting on it -- we have the tension in the string and of course we have the weight of the bucket, mg.*1184

*So Newton's Second Law in the centripetal direction, F _{net}c, which is all the forces now pointing toward the center of the circle is going to be t + mg and F_{net}c is always equal to MV^{2}/r.*1192

*So if I want to solve for the tension, tension is going to be mv ^{2}/r - mg.*1209

*The mass is 5, speed is 3 ^{2} divided by the radius (0.7) minus the mass (5) times g (10).*1217

*So at the top of the circle, I come up with a tension of about 14.3N on the string.*1229

*Let us try the same thing at the bottom of the circle.*1237

*At the bottom of the circle, the FBD is a little bit different.*1243

*When the bucket is down here, now we have gravity still pulling down but the tension is pulling up.*1246

*So now the net force in the centripetal direction, we have (t) toward the center of the circle and mg away from the center of the circle, so that is -mg.*1255

*Again, always equal to MAc or MV ^{2}/r.*1265

*Therefore, tension = mg + mv ^{2}/r or tension = 5 kg × 10 m/s^{2} + 5 × speed (3^{2})/radius (0.7)...*1270

*...so our tension now is 114.3N, which is probably what you expect.*1289

*Try this one sometime -- take a bucket of water, tie a string to it or even hold it in your hand -- use your hand instead of the string -- and swing it up and down.*1297

*You will feel a lot more force on your arm when it is at the bottom of the vertical circle than when it is at the top.*1305

*One last problem here -- It is actually possible for a car to turn on a banked curve without friction if the speed of the car and the angle of the bank are just right.*1314

*This is a terrific physics application for folks who live in icy areas, especially when the roads freeze over, they try and design the banks at angles so that the cars do not need a whole lot of friction.*1325

*Let us determine the required speed of the car for a given bank angle.*1336

*I am going to start off by drawing my banked angle and on here, we will put our car and I am going to look at it from behind again.*1340

*There is the license plate and we have some angle θ to our incline.*1352

*So we are going to start off with our FBD.*1358

*Even though the car is on a banked angle, because it is turning in a circle, that means that the motion of the car, the acceleration, is toward the center of the circle.*1361

*That way we are not going to tilt our axis. We are going to keep the acceleration going straight toward the center of the circle.*1372

*So when we draw our FBD in this case, I am going to...*1379

*... put my y axis, my x axis -- here's the car and we of course have mg down -- its weight -- and the normal force, but the normal force is here at an angle and that angle there is θ.*1384

*So when we draw our psuedo free body diagram (P-FBD), now we have to break up the normal force in the components.*1401

*There is my x, there is my y, and we still have mg down.*1410

*Now, if we want the x component of the normal force, that is going to be the opposite side of this triangle.*1418

*So the x component here is going to be Fn -- opposite side -- sin θ and the y component again is going to be the adjacent side, Fn cos θ.*1425

*Once I have that -- my P-FBD -- I can come back here and I can write my Newton's Second law equation, F ^{net}c equals...*1445

*Well, I am going to replace F ^{net}c with all of the different forces I have pointing toward the center of the circle.*1455

*In that case, it is Fn sin θ.*1460

*So Fn sin θ points toward the center of the circle and I know that must be equal to MV ^{2}/r.*1463

*Let us do my F ^{net}y equation over here.*1478

*Net force in the y direction is going to be Fn cos θ minus mg and all of that must equal 0because the car is not going to spontaneously go flying up off the bank or down into it, not in the situation here where we are saying that it is just right that it is going to stay on the banked curve.*1481

*So I can say that the normal force, cos sin θ, must equal mg or the normal force equal mg/cos sin θ.*1500

*Now let us go back over here.*1513

*With that information, knowing that Fn = mg/cos θ, I am going to replace normal force here with mg/cos θ.*1514

*So I have mg/cos θ and I still have the sin θ here, times the sin θ and all of that equals MV ^{2}/r.*1527

*We can do a simplification here and divide both sides by m.*1542

*By the way, sin θ/cos sin θ -- that is the tangent function.*1546

*Therefore, g × the tangent of θ = V ^{2}/r or to solve for the velocity, V^{2} = g(r) tangent of θ, or to get V by itself, take the square root, V = the square root of g(r) tangent of θ.*1552

*I do not need any friction whatsoever when I have the situation where the velocity is equal to the acceleration due to gravity times the radius of the curve times the tangent of the bank angle of the curve -- square root.*1576

*Hope that gets you a good start on centripetal force and centripetal acceleration.*1589

*Thanks for watching Educator.com and make it a great day!*1593

1 answer

Last reply by: Professor Dan Fullerton

Fri Apr 8, 2016 6:14 AM

Post by Zhe Tian on April 7, 2016

On example 12, why couldn't we have tilted the plane and made it so that FN goes straight up and mg becomes tilted?

1 answer

Last reply by: Professor Dan Fullerton

Fri Apr 3, 2015 8:50 AM

Post by BRAD POOLE on April 3, 2015

Hey,

On example 5 what direction is the centripetal force? It is opposite the normal force right?

3 answers

Last reply by: Professor Dan Fullerton

Thu Jan 1, 2015 8:01 AM

Post by Micheal Bingham on December 30, 2014

Hello Professor, in example 10, when spinning the bucket of water horizontally, is tension what is causing the centripetal force?

1 answer

Last reply by: Professor Dan Fullerton

Mon Oct 13, 2014 4:31 PM

Post by Foaad Zaid on October 13, 2014

Hello, so for the Example 1: Acceleration: if the object is moving at constant speed throughout the circle, its accelerating because of the change in direction. However, in the last one with an object in a circle, I'm not understanding what we can infer about its actual SPEED. Is it possible for the car to be accelerating in a circle, without constant speed? Thank you.

1 answer

Last reply by: Professor Dan Fullerton

Wed Oct 8, 2014 6:28 AM

Post by Jamal Tischler on October 8, 2014

Why arent you prooving the acceleration formuna ? It would be important.

1 answer

Last reply by: Professor Dan Fullerton

Sun Aug 10, 2014 6:52 AM

Post by Jungle Jones on August 9, 2014

Why did you choose to not angle the centripetal force to be parallel to the banked curve?

1 answer

Last reply by: Professor Dan Fullerton

Sun Aug 10, 2014 6:52 AM

Post by Jungle Jones on August 9, 2014

For the demon drop, why is there a large normal force if they are spinning quickly?

1 answer

Last reply by: Professor Dan Fullerton

Fri Jul 11, 2014 8:27 AM

Post by Jamal Tischler on July 11, 2014

But the centrifugal acceleration ? Can you explain it ?

1 answer

Last reply by: Professor Dan Fullerton

Tue Dec 17, 2013 7:54 AM

Post by Burhan Akram on December 16, 2013

Hey Prof,

Is tan(theta)= v^2/r*g equation applys to any banked curve? I mean can we blindly use it for any banked curve? I know it's not a good idea but i meant by blindly is that, no matter how the question is put forward for us, as long it's a banked curve question, we are always going to end up with this default equation? u know, on exam time matter, so we wont waste time deriving the equation which is going to have the same variable at the end.