For a very important reason, you are driving your car around in a circle at constant velocity. The radius of the circle is R, and its circumference is C equals 2 pi R. The time taken for an object to make one complete revolution is referred to as a period T of its motion. You use a capital T to represent period, which is measured in seconds. This is really seconds per complete cycle, but we just say seconds. For example, 2.3 trips around the circle, and 154 seconds has period T equals 154 seconds divided by 2.3 cycles equals 70 seconds. That's really 70 seconds per complete cycle, but we just refer to the period as 70 seconds. The number of times per second that a complete cycle is made is the frequency F equals 1 over the period, and measured in inverse seconds, or hertz, abbreviated HZ. Continuing the example, we add a period of 70 seconds, and this means that the frequency F equals 1 over the period, equals 1 over 70 seconds, equals 0.014 hertz. If an object is moving through one complete circle in period T seconds, then its tangential speed is V equals distance divided by time, equals 1s around the circle in a period T equals 2 pi r over period T, which is also 2 pi r times the frequency F. This is referred to as a tangential velocity, because it lies along a line tangential to the path. The velocity vector and the car's headlights are always tangential to the path. Continuing the example, for a radius r equal 50 meters, the tangential velocity V equals 2 pi r divided by a period T equals 2 pi times 50 meters divided by 70 seconds equals 4.5 meters per second. The angular frequency, omega equals 2 pi times the regular frequency F, and is measured in inverse seconds, or hertz. Omega looks like a W, but it's not, and it gets its feelings heard if you call it a W. Continuing the example, the angular frequency omega equals 2 pi times the frequency F, equals 2 pi times 0.014 hertz, equals 0.088 hertz. Throughout the year-long course, we will use period T, frequency F, angular frequency omega, and tangential velocity V sub T equals 2 pi r divided by the period T. Centripetal acceleration is a sub C equal V squared over r, which will now show. Your speedometer shows the linear or tangential velocity of the car. At moment number 1, here is a velocity vector V1, and the velocity vector V2 at a slightly later moment, say one-tenth of a second later. The direction of the velocity vector V is changing, but the magnitude of the velocity is not changing because we're moving at constant speed. The magnitude of vector velocity V1 equals the magnitude of velocity vector V2 equals the constant velocity V, which is tangential to the path. If we slide the velocity vectors V1 and V2 to form a triangle, delta V points radially inward. Here is the arc length s equal theta r. In this figure, we have arc length delta s equals r delta theta, which is a curving distance along the circumference of the circle. The car moves at velocity V and wants to traverse a straight line distance delta x equals V delta T. For tiny time increments, the curving and straight line distances are about equal, giving delta s equals delta x, or r delta theta equals V delta T, which we arrange as delta theta equals V delta T over r. A similar arc length equation for the velocity vectors is delta theta equals delta V over V. Equating the two incremental angles gives delta theta equals V delta T over r equals delta V over V, or delta V over delta T, which is an acceleration, equals V squared over r equals a sub c, which is this intripetal acceleration and a points radially inward toward the center of the circle. Whenever an object is moving in a circle, we know from geometry that the magnitude of its inwards intripetal acceleration is a sub c equals V squared over r. In circular motion, the velocity is V equals 2 pi times the radius of the circle r, divided by the period T of the motion equals 2 pi r times the frequency F equals r times the angular frequency omega. This intripetal acceleration is a sub c equals V squared over r equals r omega squared equals 4 pi squared r times f squared equals 4 pi squared r over T squared, and it always points radially inward toward the center of the circle. We keep the string of equal signs handy for use in various homework problems. Continuing the example, the centripetal acceleration could be calculated from any of the formulas. We'll use the most common version, a sub c equals V squared over r equals 4.5 meters per second that we have to square, divided by 50 meters equals 0.41 meters per second squared. Anyone inside the car feels thrown slightly outward. In this example of circular motion, the period is 2.9 seconds and the radius is 1.3 meters. Please supply the numbers to show that we have frequency F equals 0.34 hertz tangential velocity V equals 2.8 meters per second. Angler frequency omega equals 2.2 hertz, and centripetal acceleration A sub c equals 6.0 meters per second squared. Put a cylinder like this one into space. People live on the inside surface of this cylinder. The cylinder has a radius r equal 1,000 meters. To picture the size of this cylinder, think of an object that is 1 kilometer away from you. We make so-called artificial gravity by spinning the cylinder about its long axis. At what tangential velocity should it rotate to give an acceleration of 1g? We have V squared over r equals the centripetal acceleration A sub c, which we want to be equal to g, or V equals the square root of g r equals the square root of 9.8 meters per second squared times 1,000 meters equals 100 meters per second, which is 225 miles per hour. The cylinder will be rotating at a large speed. The velocity is also V equals 2 pi r divided by the period T. The period T is the time taken to make one revolution, which is T equals 2 pi r divided by V equals 2 pi times 1,000 meters divided by 100 meters per second equals 63 seconds. How many revolutions per second is this? The cylinder is spinning at a rate of 1 revolution per 63 seconds, or about 1 revolution per minute. People walk around the inner surface and fill exactly the same weight as occurs when walking on the surface of the earth. However, imagine throwing a ball toward the center of the cylinder. When the ball leaves your hand, it will move in a straight line due to its inertia, but the cylinder would rotate below it. If you throw the ball such that it takes 31.5 seconds to travel 2 kilometers, then you would arrive to catch the ball at the end of its flight. From your point of view, the ball would appear to curve behind as it travels toward the center of the cylinder. Like a ball tossed on a merry-go-round, in the 1890s there was a bit of a fashion to play billiards on rotating tables.