AP Physics 1 & 2 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 = V²/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_netx = MAx.0282

Then we did that in the y direction, F_nety = 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_netc = 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_netc = MAc.0374

But we also just learned that the magnitude of centripetal acceleration is the speed² divided by the radius.0388

Therefore, F_net c = M × V²/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-m² per m/s².0420

One of the meters, we will make a ratio of 1 and I will be left with kg-m/s² which is a Newton (N).0428

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

So F_net c = MV²/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 = V²/r; that is 8.6 m/s² divided by the radius (2.5 m) or about 29.6 m/s².0716

How about the centripetal force acting on the student?0737

To find that -- F_netc = 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 = V²/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 = V²/r, so that is going to be 4 m/s²/0.25 m -- 16/0.25 will be 64 m/s².0916

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/s² -- or about 80 kg.0960

To find the running back's centripetal acceleration -- Ac = v²/r.0984

The Velocity is 8 m/s²/radius (1 m) -- 64 m/s².0991

The centripetal force -- F_netc = MAc.1006

So, the mass of our running back (80 kg), the centripetal acceleration (64 m/s²) -- 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_netc, is the frictional force, which is MV²/r.1069

Our mass is 1200 kg, our speed is 9 m/s² 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_netc = MAc, which is MV²/r.1138

Our mass is 5 kg, our speed is 2 m/s²/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_netc, which is all the forces now pointing toward the center of the circle is going to be t + mg and F_netc is always equal to MV²/r.1192

So if I want to solve for the tension, tension is going to be mv²/r - mg.1209

The mass is 5, speed is 3² 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 MV²/r.1265

Therefore, tension = mg + mv²/r or tension = 5 kg × 10 m/s² + 5 × speed (3²)/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^netc equals...1445

Well, I am going to replace F^netc 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 MV²/r.1463

Let us do my F^nety 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 MV²/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 = V²/r or to solve for the velocity, V² = 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