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# 5: Uniform Circular Motion

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Today we will discuss what we call "uniform circular motion." What is uniform circular motion? An object goes around in a circle, has radius r and the object is here.This is the velocity.It's a vector, perpendicular. And later in time when the object is here the velocity has changed, but the speed has not changed. We introduce T, what we call the period-- of course it's in seconds-- which is the time to go around once.

We introduce the frequency, f, which we call the frequency which is the number of rotations per second. And so the units are either seconds minus one or, as most physicists will call it, "hertz" and so frequency is one divided by T. We also introduce angular velocity, omega which we call angular velocity.

Angular velocity means not how many meters per second but how many radians per second. So since there are two pi radians in one circumference-- in one full circle-- and it takes T seconds to go around once it is immediately obvious that omega equals two pi divided by T. This is something that I would like you to remember. Omega equals two pi divided by T-- two pi radians in capital T seconds.

The speed, v, is, of course, the circumference two pi r divided by the time to go around once but since two pi divided by T is omega you can also write for this "omega r." And this is also something that I want you to remember.

These two things you really want to remember. The speed is not changing, but the velocity vector is changing. Therefore there must be an acceleration. That is non-negotiable. You can derive what that acceleration must be in terms of magnitude and in terms of direction. It's about a five, six minutes derivation. You'll find it in your book. I have decided to give you the results so that you read up on the book so that we can more talk about the physics rather than on the derivation.

This acceleration that is necessary to make the change in the velocity vector is always pointing towards the center of the circle. We call it "centripetal acceleration." Centripetal, pointing towards the center. And here, also pointing towards the center. It's a vector.

And the magnitude of the centripetal acceleration equals v squared divided by r, which is this v and therefore it's also omega squared r. And so now we have three equations and those are the only three you really would like to remember. We can have a simple example.

Let's have a vacuum cleaner, which has a rotor inside which scoops the air out or in, whichever way you look at it. And let's assume that the vacuum cleaner these scoops have a radius r of about ten centimeters and that it goes around 600 revolutions per minute, 600 rpm.

600 rpm would translate into a frequency, f, of 10 Hz so it would translate into a period going around in one-tenth of a second. So omega, angular velocity, which is two pi divided by T is then approximately 63 radians per second and the speed, v, equals omega r is then roughly 6.3 meters per second.

The centripetal acceleration--

and that's really my goal--

the centripetal acceleration would be omega squared r or if you prefer, you can take v squared over r. You will get the same answer, of course, and you will find that that is about 400 meters per second squared.

And that is huge.

That is 40 times the acceleration due to gravity. It's a phenomenal acceleration, the simple vacuum cleaner. Notice that the acceleration, the centripetal acceleration is linear in r. Don't think that it is inversely proportional with r. That's a mistake, because v itself is a function of r.

If you were sitting here then your velocity would be lower.

Since omega is the same for the entire motion you really have to look at this equation and you see that the centripetal acceleration is proportional with r.

Therefore, if you were...

if this were a disc which was rotating and you were at the center of the disc the centripetal acceleration would be zero. And as you were to walk out, further out, it would increase. Now, the acceleration must be caused by something.

There is no such thing as a free lunch.

There is something that must be responsible for the change in this velocity and that something I will call either a pull or I will call it a push. In our next lecture, when we deal with Newton's laws we will introduce the word "force." Today we will only deal with the words "pull" and "push." So there must be a pull or a push. Imagine that this is a turntable and you are sitting here on the turntable on a chair. It's going around with angular velocity omega and your distance to the center, let's say, is little r.

You're sitting on this chair and you must experience--

that is non-negotiable--

centripetal acceleration A of c, which is omega squared times r.

Where do you get it from? Well, if your seat is bolted to the turntable then you will feel a push in your back so you're sitting on this thing, you're going around and you will feel that the seat is pushing you in your back and so you feel a push, and that gives the push out.

Yeah, I can give this a red color for now.

So you feel a push in your back.

That push, apparently, is necessary for the acceleration.

Alternatively, suppose you had in front of you a stick.

You're not sitting on a chair.

You don't get a push from your back.

But you hold onto the stick and now you can go around by holding onto the stick.

Now the stick is pulling on you in this same direction.

So now you would say, aha, someone is pulling on you.

Whether it is the pull or whether it is the push one of... either one of the two is necessary for you to go around in that circle on that turntable with that constant speed.

Now, the classic question comes up, which we often ask to people who have no scientific background.

If you were to go around like this and something is either pushing on you or is pulling on you to make this possible suppose you took that push out, all of a sudden.

The pull is gone.

[makes whooshing sound]

What is now the motion of the person who is sitting on the turntable? And many non-scientists say, "Well, it will do like this." That's sort of what your intuition says.

You go around in a circle, and all of a sudden you no longer have the pull or the push and you go around in a spiral and obviously, that is not the case.

What will happen is, if you have, at this moment in time a velocity in this direction and you take the pull or the push out you will start flying off in that direction and depending upon whether there is gravity or no gravity there may be a change, but if this were...

if there were no gravity you would just continue to go along that line and you would not make this crazy spiral motion.

I have here a disc, which we will rotate and at the end...

the edge of the disc here we have a little ball.

And the ball is attached to that disc with string.

So now this is vertical, and so this is going to go around with angular velocity omega.

And we have a string here and the string is attached to this ball and the whole thing is going around and so at one moment in time this has a velocity, like so.

And therefore there must be non-negotiable centripetal acceleration which in magnitude is omega squared r or, if you want to, v squared divided by r.

Now I cut it and that's like taking away the push and the pull. The string that you have here is providing the pull on this ball. This ball is feeling a pull from the string and that provides it with the centripetal acceleration. Cut the string and the pull is gone and the object will take off. And if there were gravity here, as there is in 26.100 it would become a parabola and it would end up here. If, however, I cut the ball exactly when it is here--

not the ball, but I cut the string--

then, of course, it would fly straight up gravity would act on it, it would come to a halt and it would come back. So it really would then go along a straight line. But you would clearly see, then that it's not going to do what many people think-- that it would start to swirl around.

It would just go...

[makes whooshing sound]

and comes back. Let's look at that. We have that here. So here is that ball. The string is behind here; you cannot see the string. I will rotate it, wait for it to pick up a little speed and the knife, that you can't see either, is behind here and when I push the knife in, I do it exactly here.

It cuts the string and it goes up. You ready for this? You sure you're ready? Three, two, one, zero. Wow! That was very high. So you see, it's nothing like this. It simply continued on in the direction that it was going. It wasn't going into a parabola because I was shooting it straight up. The string forms the connection between the rotating disc and the ball and therefore, the pull is responsible for the centripetal acceleration.

Let's now think about planets.

Planets go around the sun.

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