[MUSIC] Okay, circular motion. This course is called particles to plants and we'll need circular motion for planets and satellites. So, we need some quite important results that we'll derive here. So, concentrate, have your pencil and paper ready because you don't really understand until. All right, here we go. The marked section is travelling in a circle. If the rotation rate is constant, we call it uniform circular motion and we limit ourselves to that case today. Here's a really important point. Even when its speed is constant, an object in circular motion is accelerating. Acceleration is the rate of change in velocity. Velocity is a vector, so a change in direction of velocity is an acceleration. The velocity of this point is up, across, down, back, its velocity is changing, so it is accelerating. We can make this clear using Newton's second law, which we'll need next week, force equals mass times acceleration. Suppose a ball hits a wall and rebounds with the same speed. Is there an acceleration and is there a force? Let's see if I can feel the force. Ouch, because of the time is short the acceleration is large and it's a large force. A ball is travelling in a roughly horizontal circle. The forces on the ball are its weight and the tension in the string. Look at the direction of the string which is the direction of the tension force. This force has an upwards component which balances the weight and the large horizontal component. This horizontal component provides the acceleration which we'll soon see is towards the center. But to see that, we'll need to analyze uniform circular motion. Let's define some terms in symbols, r is the radius and capital T is the period for 1 complete revolution, omega is the angular speed. The change in angle divided by the change in time, delta theta over delta t. Omega is another Greek letter. Make sure that you distinguish omega from W in your handwriting. Make omega the round one, and W the pointy one. Now, we go one full circle in a period of capital T. So omega = 2 pi r/T. Now, the speed v is just one circumference in the time for one revolution. 2 pi r/T. So that give us v = r omega Now for the direction of the acceleration. The velocity is the tangent to the circle. Let's say we have velocity v1 at some time and v2 short time later. The change in velocity is delta v equals v2 minus v1. Well, we saw how to do vector subtraction in week 1. This diagram shows how v1 plus delta v equals v2. It's uniform circular motion, constant speed, so the velocity vectors have equal length. Look at the direction of delta v. If we consider a very short time interval, v1 and v2 would be nearly parallel so the change in velocity would be at right angles to the velocity. Velocity is tangential so the acceleration is towards the center. We call it centripetal acceleration. So the direction is indeed toward the center as we argued originally. If there were no string to provide centripetal acceleration, the ball would stop travelling in a circle and become a projectile travelling in a vertical plane. Uniform circular motion, the speed is constant but the acceleration is at right angles to the velocity so it accelerates sideways, always changing it's direction so that it follows a circle. These diagrams can give us the magnitude of the acceleration. Call the change in angle delta theta. From this small triangle and the definition of angle, we have magnitude of the change in velocity, delta v = v delta theta. Then write the definition of acceleration. Delta v over delta t for small change. We substitute using omega and we have a = v omega. Previously, we also had v = i omega, so that gives us the magnitude of the centripetal acceleration. A equals rw square and if we substitute for omega, we have a equals v square over r. That was a bit of work but those equations are quite important and we need them for circular motion. Hey, have you checked the units on those equations yet. Now it should be automatic by now. Omega is in second to the -1, so do you agree? Good. Now let's go to an important question? Aristotle and many others thought that the Earth was stationary because we couldn't feel it moving. So, over to you. What is your acceleration due to the earth's rotation on its axis? And due to its orbit around the sun? Here is another problem. A curve that has a hill, whose crest has a radius of curvature of 30 Metres. How fast does it have to travel further so that it becomes airborne at the very highest point. Let's make that the first question in the quiz. And, after that, there is a test because we're finished with projectiles in circular motion for now. Next week, we'll meet Newton's Laws of Motion and we'll start explaining the motions that we've been observing and quantifying so far. Good luck for the test, and see you next week. [MUSIC]
