8. Centripetal Forces & Gravitation

# Period and Frequency in Uniform Circular Motion

1

concept

## Circumference, Period, and Frequency in UCM

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Hey, guys. So now that we've been introduced to uniform circular motion in this video, we're gonna talk about three more important variables. You'll need to know circumference, period and frequency. So let's go ahead and check it out. Basically, they are all related to when objects complete a full rotation. So we have these circular path, and once you complete a full lap around this circle, you've completed a full rotation. Some books will call this a revolution, and some will call it a cycle. So let's talk about the first one, which is basically the distance that you travel around the circular path that's called the Circumference. The circumference is given by the letter C, and it's really just related to the radius. We know that the radius this is really just the distance from the edge of the path to the center of the circle that are the circumference is basically going to be the distance all the way around the circle once you've completed that path. So this is our letters see here and really see is just equal to two pi times the radius. So if you know one of those variables, you can always figure out the other one, all right, so let's talk about the other two, and they're very closely related to each other, and they have to do with time. The first one we talk about is called period, and the symbol we use is this capital T. Basically, the period is just how long it takes for you to complete one cycle or one lap around that circle. And so one way you can think about this is that it's the number of seconds divided by Earth per the number of cycles. And so the unit that will use for this because it's seconds per cycle is just the second. And so one really quick way to figure out the period is just if you take the number of seconds elapsed and you divided by the number of cycles. So quick example here. If it took us two seconds to complete one lap, then our period is just to over one, and that's just two seconds. All right, so the other related variable is called the frequency, and the frequency is basically just the opposite of the period. Instead of it being seconds per cycle. It's the number of cycles that you complete per second. So notice how these are basically just opposites of each other. So the unit that we use for this is not going to be a second. It's actually going to be a hurts. And our hurts is essentially just an inverse second. It's one divided by seconds. And so because we're basically just flipping these two things we can do is we can take this equation here, seconds per cycle, and we just flip it upside down. So the way to calculate the frequency is just by doing the number of cycles divided by the number of seconds. So if you take a look at these two variables here, notice how they're basically just opposites of each other, you just flip the fractions. And so in general, you can always figure out the period from the frequency and then vice versa. And here the equations to do that if you want the period, you're just going to take the inverse of the frequency. And if you want the frequency, you're just going to take the inverse of the period. We're basically just gonna flip the fractions. Let me show you how this works, and we'll do a couple of examples here. We're going to calculate the period and frequency of our motion. If we complete four rotations in two seconds, let me show you what this means. Imagine you were walking in a circle and you did. 1234 rotations. Right, So this is basically four cycles, and it took us two seconds to complete these four cycles. So how do we calculate the period? Remember, the period is really just going to be the number of seconds divided by the number of cycles or rotations or whatever word you're using. And so we're just gonna use two seconds divided by four cycles and you'll get one half. So if it takes you two seconds to complete four cycles, then basically each one of these is one half of a second and so on and so forth. All right, so that's the period. The frequency is going to be the opposite. The flipped fraction of that. We're gonna do the number of cycles divided by the number of seconds. So we're just really going to do four cycles divided by two seconds, and then you'll get to hurts. So one way you could also have done this is you could just take these fractions, right? You could have taken this one half of a second and flip the fraction that becomes too over one. And that's really just to hurts. All right, so let's take a look at the next one. Now we're doing 10.5 rotations in three seconds. So what this means is that we got half a rotation, right, but the other half is kind of missing, so we have 0.5 cycles and then we have three seconds to complete, so we'll do the same exact idea here, right? The period is really just going to be the number of seconds divided by a number of cycles. So it's three seconds divided by 0.5 cycles, and then you get six seconds. Basically, if it takes you three seconds to complete half of a circle and it's going to take you another three seconds to complete the other half, then your total period is gonna be six seconds, all right? And so the frequency is just going to be 0.5 cycles divided by three seconds. Basically, you just flip the fraction and then you'll get as you expected. 1/6 of a hurts. All right, So those again, how these two fractions are basically just flip the versions of each other. All right, so that's it for this one. Guys, let me know if you have any questions.

2

example

## Windmill RPMs

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Hey, guys, let's check out this problem here. We're giving some information about windmill blades were told that they spin at a rate of revolutions per minute. And what we want to do is, at this rate, how long? Which is really just a time. Does it take for a windmill blade to complete one full rotation? So let's check this out. If you're ever unsure of what variable that is, just go ahead and write it down. How long a time is really has a unit of a second, so we're calculating the number of seconds to complete a full rotation. So that's the number of seconds per number of cycles or rotations, whatever they have you, and remember that seconds per cycle is actually the period. So that's what we're trying to find here. What is the period if we have these blades spinning at 20 revolutions per minute? So if you take a look at our equations, we know that there's a shortcut to calculate the period. You can just take one over the frequency. Remember that these things are basically just in verses of each other. All you have to do is flip the fraction So what happens is so we want to calculate the period and then so we're gonna have to calculate the frequency. So the frequency, remember, it's just the opposite, the inverse fraction of that which is really just the number of cycles divided by the number of seconds. So if you take a look at our problem here, do we have cycles per second? And we actually don't. We have actually 20 revolutions per minutes. So basically what happens is that whenever whenever you're given revolutions per minute, which is an rpm, it's a pretty common unit that you'll see in these problems. All you have to do is just to get to frequency. You're just going to divide that rpm by 60 the revolutions per minutes if you want. In revolutions per second, all you have to do is just divide by 60 seconds because that's how many are in a minute. So really, what happens here is to get the frequency we're gonna do 20 rpm. So this is our PM divided by 60 and you're going to get 1/3. So that's basically one third and one of the units for that or hurts. Now, remember The reason we calculate this frequency is because once we get this as a fraction or whatever number it is, all we have to do is just take one over that number. So now we can figure out the period. The period is really just Well, if we're taking one third, that's our hurts. Once you flip the fraction, it really just becomes 3/1 and that's just three seconds. And so basically, it just takes three seconds for these blades to complete a full rotation. Alright, so that's really all there is to it. So let me know if you guys have any questions.

3

concept

## Equations for Velocity and Acceleration in UCM

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Hey guys. So now they were familiar with the basics of uniform circular motion, and we've also talked about variables like circumference, period and frequency. We're gonna combine these three variables to come up with some more equations for velocity and acceleration. They're gonna help you solve some problems. So I'm going to give you these equations, and we're gonna go ahead and solve some examples. So let's check this out. We're gonna start off with the tangential velocity. Now, remember that tangential velocity or the velocity for any moving object is always just related to the distance over time. So in circular motion, we've already talked about the distance that you travel when you complete one perfect one full rotation. And that was called the circumference and the time that it takes for you to complete one of those rotations. We've also talked about that as well. That's just called the period. So you can rewrite this as C over tea. Now, a lot of times and problems, you won't be given the circumference of a circle. But instead you will be given the radius. So we can do is we can rewrite this equation as two pi big are divided by t. So we have this equation over here. But there's also another way we could write this because now we know that period and frequency T and F are related to each other. By inverse is t equals one over f and F equals one over t. So we can rewrite this equation as two pi r times the frequency. Either one of these equations will work in your problems. It really just depends on which one of these variables T or F, is just more easily found. So let's talk about the acceleration. Now we know that the acceleration is v tangential square divided by our So now that we've gotten these two equations for our velocity, we can basically just plug them in both for this v tangential squared and get two more equations for the acceleration. So if you plug in two pi r over tea into v tangential, what you end up getting is you end up getting four pi squared r squared over r t squared, and so one of the ours is gonna cancel from the top and bottom. And then this is what you're gonna get now if you plug in two pi R F into v tangential squared. You just get another equation, which is four pi squared R F squared So it might seem like there's a lot of equations to memorize here, but there's actually really not as long as you memorize V squared over r and you memorize two pi r over tea. You can always get back to any one of these equations. And really, these are just useful because sometimes you don't know what the V tangential squared is going to be in a problem. But you do know what the period or frequency is, so you can figure out the acceleration using these equations. Alright, so here are four equations. Let's go ahead and check them out in a in some problems. So here we have a ball that's moving in a radius of 10 m, right in a circle. So we know that our equals 10. We want to figure out the speed it takes 60 seconds to complete 100 rotations. So we know we're gonna be using our vis a vis equation because that's gonna be speed. So really, it just comes down to two choices. We have either two pi r over tea or we have two pi r times the frequency. It all depends on which one of those two is more easily found for us in the problem. So if we take a look at this, what we're given is that it takes 60 seconds to complete 100 rotations or cycles. So really, what this means is that they're giving us the number of seconds for the number of cycles. And if they're presenting the information this way, then it's actually easier for us to find the period. So we're going to use to pie are divided by the period. So this is gonna be two pi times, the radius of 10 divided by the period which we can find by just using the number of seconds per a number of cycles which we know this is just going to be 60 divided by 100 so 60 divided by 100 0. and then we just plug that in for our equation. And so you get a V tangential of 104.7 m per second. So that's our velocity. Now, could you have found this using the frequency? Absolutely all you would have to do is basically just flip this fraction here to find the frequency instead of period. And you would have gotten the exact same answer. So now let's move on to the second one. We're gonna be figuring out this centripetal acceleration, so basically, we're not figuring out V. We're figuring out a C and were given one rotation every three minutes. So now we've got a couple of options. Do we use V squared over R, or do we use one of these equations involving period and frequency? Well, the thing is that we actually don't know what the velocity is, So we're not going to use V squared over r what we are given some information about how many rotations it completes personal amount of time. So what happens here is they're giving us the cycles per number of minutes or seconds. So we can do here is we can actually use this equation. Four pi squared our frequency square so we have a C equals. This is going to be four pi squared times 10. Now we just have to figure out the frequency and remember that the frequency is the number of cycles that you complete in the number of seconds. So we're told we're we We we complete one rotation, and then the number of seconds were given minutes, right? Were given three minutes. We can always convert by just multiplying by 60. So there's gonna be three times 60 and we're gonna get a frequency of one divided by 180 hertz. So now we just plug that into our formulas. This is gonna be frequency 1/1 80 except we have to square that. All right, So don't forget the square. That and if you work this out in your calculators, you're gonna get an acceleration of 0.12 m per second square. All right, so that's it for this one again. You could have actually figure this out. Using this equation over here, all you would have had to do is instead software period instead of frequency. So I had one of these equations works a lot of times. There's multiple ways to get these answers, so hopefully that makes sense. And I'll see you guys the next one

4

Problem

A 3kg rock spins horizontally at the end of a 2m string at 90 RPM. Calculate its centripetal acceleration.

A

177 m/s

^{2}B

118.4 m/s

^{2}C

3160 m/s

^{2}D

56.5 m/s

^{2}5

example

## Spinning Space Station

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Hey, guys, let's check out this problem here. So a big problem for astronauts in space is gonna be the lack of gravity. So one way we can fix this for future space travel is we can build a spinning space station. So the idea here is that you have this big ring like this that's attached to the main part of the hull of the ship or whatever, and you have this arm that basically attach is it to the central part of the ship, and this entire ring here is going to rotate, and as it rotates, it actually produces some centripetal acceleration for the people on the inside, so they start to accelerate towards the center. So we want to figure out basically how fast we have to spin this thing so that the acceleration inwards equals G, which is 9.8. It's kind of like the same effect as when you're going in a car and you're taking a corner. You'll feel like that kind of push from the walls or the door of your car. It's basically kind of like the same effect they're going to be pushed inwards like this from the walls of the space station, and we want the acceleration to be 9.8. So we have to do here is let's take a look at our problem here. Right? We've got this diameter, which equals 500 which is the diameter of the length of the entire thing. So remember that we use the radius in these problems, and that's just gonna be the diameter divided by two. So we're going to actually use 250 in our problems. Now we're trying to find here is how fast in RPMs we have to actually spend this spaceship here. Okay, so how do we figure RPMs? Well, let's take a look at our problems Are equations we have RPMs is equal to, um sorry. Frequency is equal to RPMs over 60. So, you know, you can find the RPMs if we have the frequency. So this is frequency equals RPMs over 60. So we can do here. Is that if I want to solve for r. P M. S, which is really just like a modified version of frequency, all we have to do is right, is get the frequency and then multiplied by 60. So to find my RPMs. I just have to find the frequency like this and then multiply it by 60. So remember, I don't have the frequency in this problem. So how do I go ahead and find that right? So I'm gonna have to go find the frequency. Now, let's take a look at our problems or equations. We know that frequency is related to one over the period one over T. But I don't have tea in this problem either. So this is not gonna be a good equation for us to use Now, we can also look at our velocity equation because our velocity equation has period and also frequency. Now, unfortunately, what happens is I don't really know what the the tangential velocity of any one of the points of the ring are. So this is not going to be a good equation for us to use either. Instead of what we can do is we know that a C is related to V squared over R. And we know that this centripetal acceleration also has a shortcut equation in which we can use this. This guy over here, we actually do know something about the centripetal acceleration. So that's what we're going to use. So we have here is we have a C equals and then we have four pi squared r times the frequency squared. So if you take a look, I know what the centripetal acceleration is gonna be. It's gonna be 90.8. I don't know are so I can figure out the frequency squared and then I can bring it basically back into this equation to solve for r p. M s. All right, so that's what we're gonna do. So I'm just gonna go ahead and solve for the frequency squared and this is gonna be a C divided by four pi squared times are now All I do is I just take the square roots and I'm just gonna start plugging stuff in so you know, this is gonna be 9.8. That's the centripetal acceleration. And then when you plug in all this stuff on the bottom here, you're just gonna basically plug this in as a as a parentheses. Actually, let me go ahead and write this four pi squared times 250 you're gonna have to, like, surround this whole entire thing, a parentheses otherwise, you get the wrong answer and you're gonna get 0.0 99 hertz. So now we just plug this guy right back into here. So we have 0.0, 99 times 60 and we get an rpm equal to 94. So basically, you have to spin this thing at almost six revolutions per minute, even though the space station's almost half a kilometer long or it is half a kilometer long in order to simulate gravity that's similar to that of Earth. All right, so that's it for this one. Guys, let me know if you have any questions.

Additional resources for Period and Frequency in Uniform Circular Motion

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