High School Physics 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) / speed0446

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) = (8²) / 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 m²/s²) / 150 m = 400/150 = 2.67 m/s/s .0674

m/s² 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