WEBVTT physics/ap-physics-1-2/fullerton
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Hi and welcome back to Educator.com
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I'm Dan Fullerton and I would love to talk to you now about describing circular motion.
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Our objectives are going to be to calculate the speed of an object traveling in a circular path or a portion of a circular path.
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And also to calculate the period and frequency for objects moving in circles at constant speed.
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So uniform circular motion has to do with an object traveling in a circular path at a constant speed.
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First thing we have to know. The distance around the circle is its circumference.
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That is equal to 2π times the radius of the circle.
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Or it is also equal to π times the diameter or the diameter is the distance through the circle.
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The average speed formula we learned from kinematics still applies.
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If average speed is the distance travelled divided by time, for something moving in a circle you have to take into account it is 2π times the radius, the distance around the circle, divided by the time.
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Frequency is a number which describes the number of revolutions or cycles that you can get in in 1 second.
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So if an object travels 3 times in a circle in 1 second, it would have a frequency of 3 cycles per second, or 3 hertz.
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H-E-R-T-Z, abbreviated Hz, where a Hertz is the same as 1 over a second.
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The symbol for frequency is *f*, so frequency is the number of cycles per second, or the number of revolutions per second.
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In similar fashion, period is the time it takes for an object to take one complete trip around the circle, or do one complete revolution or cycle.
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Its symbol is capital T and its units are seconds. The amount of time it takes to go once around the circle.
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Now what is really nice here is that frequency and period of course are closely related.
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Frequency is 1 divided by the period and period is 1 divided by the frequency.
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Once you know one, you automatically know how to figure out the other.
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Let us take an example by looking at a car on a track.
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Miranda drives her car clockwise around a circular track of radius 30 meters.
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She completes 10 laps around the track in 2 minutes.
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Let us find Miranda's total distance travelled and average speed and centripetal acceleration.
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Well to find the distance travelled, her distance is 2π times the radius and she does 10 laps.
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So circumference times 10 laps is going to be 2π times 30 meters, her radius times 10 laps. 300 times 2π is about 1885 meters.
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Average speed then is the distance she travels divided by the time, or 1885 meters divided by 120 seconds, 2 minutes or 15.7 meters per second.
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Now centripetal acceleration. That is going to describe how quickly she is accelerating.
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Since she is moving in a circle, her velocity is changing, there must be an acceleration.
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Centripetal acceleration, we will talk more about it later, is given by the formula v² over r. Square the speed divided by the radius.
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So in this case, the speed is 15.7 meters per second squared, divided by the radius of the circle, 30 meters.
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That is going to be 8.22 meters per second squared.
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And the direction of centripetal acceleration is always going to be toward the center of the circle.
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All right, let us take a look at a race car problem.
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The combined mass of a race car and its driver is 600 kilograms.
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Travelling at constant speed, the car completes one lap around the circular track of radius 160 meters in 36 seconds.
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Calculate the speed of the car.
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All right, well, average speed is just distance travelled divided by the time, completes 1 lap or 1 circumference, 2πr in 36 seconds.
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So that is going to be 2π times 160 meters over 36 seconds or about 27.9 meters per second.
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Take a look at the toy train.
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A half-kilogram toy train completes 10 laps of its circular track in 1 minute and 40 seconds.
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If the diameter of the track is 1 meter, let us find the train's period and frequency.
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Let us start with period. Period is how long it takes for 1 lap.
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It takes it 100 seconds to go 10 laps, so the period must be 10 seconds per lap or revolution, and once we know period, frequency is easy.
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Frequency is 1 divided by the period or 1 divided by 10 seconds which is 0.1, 1 over seconds, which is also equal to 0.1 hertz.
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Named of course after the famous rental car company. I'm kidding.
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Another example here, a roundabout on a playground.
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Allan makes 38 complete revolutions on the playground roundabout in 30 seconds.
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If the radius of the roundabout is 1 meter, let us determine the period of motion, the frequency of the motion and the speed at which Allan revolves.
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Well, here is our diagram. Roundabout, the radius of 1 meter.
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The period of the motion, how long it takes to go once around.
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He makes 38 revolutions in 30 seconds, so that is 30 seconds for 38 revolutions or 0.789 seconds.
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Frequency then 1 over period, or 1 over 0.789 seconds which is 1.27 hertz.
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And the speed at which Allan revolves.
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Speed is distance travelled divided by the time, that is going to be 2π times the radius 1.
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He does 38 laps, 38 times around, and the total time to do all that is 30 seconds.
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So 2π times 1 times 38 divided by 30, I get a speed of about 7.96 meters per second.
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Hopefully that gets you going with describing circular motion.
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We are worried about things like speed, period, frequency and we'll tackle a little bit more about centripetal acceleration coming up quickly.
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Thanks for watching Educator.com. Make it a great day.