Period and Frequency in Uniform Circular Motion - Online Tutor, Practice Problems & Exam Prep

In uniform circular motion, key variables include circumference, period, and frequency. The circumference (C) is calculated as 2π⁢r, where r is the radius. The period (T) is the time for one complete rotation, while frequency (f) is the number of rotations per second, expressed as f = 1T . Tangential velocity (v) can be derived from these variables, with acceleration (a) given by a = v2r .

Circumference, Period, and Frequency in UCM

Guys, now that we've been introduced to uniform circular motion, in this video, we're going to 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. We have this 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. 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 of this is really just the distance from the edge of the path to the center of the circle. That's r. 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 letter c here. And, really, c is just equal to 2πr. So if you know one of those variables, you can always figure out the other one. Alright?

Let's talk about the other two; they're very closely related to each other and they have to do with time. The first one we'll 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 or per the number of cycles. And so the unit that we'll 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 divide it by the number of cycles. So quick example here. If it took us 2 seconds to complete one lap, then our period is just 2 over 1, and that's just 2 seconds. Alright?

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'll use for this is not going to be a second. It's actually going to be a hertz, and a hertz is essentially just an inverse second. It's 1/s. And so because we're basically just flipping these two things, what we can do is we can take this equation here, seconds per cycle, and we just flip it upside down. So the way you 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 are the equations to do that. If you want the period, you're just gonna take the inverse of the frequency. And if you want the frequency, you're just gonna take the inverse of the period. You'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 4 rotations in 2 seconds. Imagine you were walking in a circle and you did 1, 2, 3, 4 rotations. Right? So this is basically 4 cycles, and it took us 2 seconds to complete these 4 cycles. So how do we calculate the period? Remember, this 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 going to use 2 seconds divided by 4 cycles, and you'll get 1 half. So if it takes you 2 seconds to complete 4 cycles, then basically each one of these is 1 half of a second and so on and so forth. Alright? So that's the period.

The frequency is going to be the opposite, the flipped fraction of that. We're going to do the number of cycles divided by the number of seconds. So we're just really going to do 4 cycles divided by 2 seconds, and then you'll get 2 hertz. So one way you could also have done this is you could just take these fractions. Right? You could have taken this 1/2 of a second and flip the fraction that becomes 2 over 1, and that's really just 2 hertz. Alright? So let's take a look at the next one. Now we're doing 0.5 rotations in 3 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 3 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 the number of cycles. So it's 3 seconds divided by 0.5 cycles, and then you get 6 seconds. Basically, if it takes you 3 seconds to complete half of a circle and it's going to take you another 3 seconds to complete the other half, then your total period will be 6 seconds. Alright? And so the frequency is just going to be 0.5 cycles divided by 3 seconds. Basically, you just flip the fraction, and then you'll get, as you expected, 1/6 of a hertz. Alright? So that was, again, how these two fractions are basically just flipped versions of each other. Right. So that's it for this one, guys. Let me know if you have any questions.

Windmill RPMs

Guys, let's check out this problem here. We're given some information about windmill blades. We're told that they spin at a rate of 20 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 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 number of seconds per number of cycles or rotations, whatever that 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 could just take one over the frequency. Remember that these things are basically just inverses of each other. All you have to do is flip the fraction. So what happens is we want to calculate the period and then we're going to have to calculate the frequency. So the frequency, remember, is just the opposite, the inverse fraction of that, which is really just the number of cycles divided by the number of seconds.

So if we take a look at our problem here, do we have cycles per second? And we actually don't. We actually have 20 revolutions per minute. So basically what happens is that 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 the frequency, you're just going to divide that RPM by 60. The revolutions per minutes, if you want it in revolutions per second, all you have to do is just divide by 60 seconds. Right? Because that's how many are in minutes.

So, really, what happens here is to get the frequency, we're going to do 20 RPMs, so this is RPM, divided by 60, and you're going to get 1 over 3. So that's basically 1 third, and remember the units for that are hertz. Now remember the reason we calculated 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 1 third, that's our hertz. Once you flip the fraction, it really just becomes 3 over 1, and that's just 3 seconds. And so, basically, it just takes 3 seconds for these blades to complete a full rotation. Alright? So that's really all there is to it. Let me know if you guys have any further questions.

Equations for Velocity and Acceleration in UCM

So now that we're familiar with the basics of uniform circular motion and we've also talked about variables like circumference, period, and frequency, we're going to combine these three variables to come up with some more equations for velocity and acceleration that are going to help you solve some problems. So I'm going to give you these equations, we're going to go ahead and solve some examples. Let's check this out. We're going to start off with the tangential velocity. 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 or one full rotation, and that was called the circumference. And the time 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 t.

Often in problems, you won't be given the circumference of a circle, but instead, you will be given the radius. So we can rewrite this equation as 2 π R t . We have this equation 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 inverses. T = 1 f ; f = 1 T . So we can rewrite this equation as 2 π r f .

Let's talk about the acceleration now. We know that the acceleration is v 2 ; r . 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 2 π r t into v tangential, what you end up getting is you end up getting 4 π 2 r 2 ; r t 2 . And so one of the r's is going to cancel from the top and bottom, and then this is what you're going to get. Now if you plug 2 π r f into v tangential squared, you just get another equation which is 4 π 2 r f 2 . 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 2 ; r and 2 π r t , 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 some problems.

So here we have a ball that's moving in a radius of 10 meters, right, in a circle. So we know that r equals 10. We want to figure out the speed if it takes 60 seconds to complete 100 rotations. So we're going to be using our v equation because that's going to be speed. So, really, it just comes down to two choices. We have either 2 π r T or we have 2 π r f . 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 easy for us to find the period. So we're going to use 2 π r T . So this is going to be 2 π times the radius of 10 divided by the period which we can find by just using the number of seconds per number of cycles, which we know this is just going to be 60 divided by 100. So 60 divided by 100 is 0.6, and now we just plug that in for our equation. And so you get a v tangential of 104.7 meters 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 the period, and you would have gotten the exact same answer. So now let's move on to the second one. We're going to be figuring out this centripetal acceleration. So basically we're not figuring out v, we're figuring out ac. And we're given one rotation every 3 minutes. So now we've got a couple of options. Do we use v 2 ; r or do we use one of these equations involving period and frequency? Well, the thing is is that we actually don't know what the velocity is, so we're not going to use v 2 ; r , but we are given some information about how many rotations it completes per some amount of time. So what we can do here is actually use this equation 4 π 2 r f 2 . So we have ac equals this is going to be 4π² 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 complete 1 rotation and the number of seconds, well, we're given minutes. Right? We're given 3 minutes. We can always convert by just multiplying by 60. So this is going to be 3 times 60, and we're going to get a frequency of 1 divided by 180 Hz. So now we just plug that into our formula. So this is going to be frequency 1 over 180, except we have to square that. Alright? So don't forget to square that. And if you work this out in your calculators, you're going to get an acceleration of 0.012 meters per second squared. Alright? So that's it for this one. Again, you could have actually figured this out using this equation over here. All you would have had to do is instead solve for the period instead of frequency. So either one of these equations works, a lot of times there are multiple ways to get these answers. So hopefully that makes sense and I'll see you guys in the next one.

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

• A. 177 m/s²
• B. 118.4 m/s²
• C. 3160 m/s²
• D. 56.5 m/s²

Spinning Space Station

One of the biggest problems facing astronauts in space is the lack of gravity or the lack of apparent gravity. But luckily, we have a solution for this because one of the ideas that we have is we can build a massive space station with a circle of ring around it that rotates. So there's the central part here, and there's, like, a giant arm that goes across the space station, and this whole entire thing will rotate. And, basically, what you'll end up with is kind of like the effect that happens when you turn in a car. Right? As you turn in a car, the wall sort of pushes you toward the center. That's exactly what would happen here. If you were standing on the ring, you would be pushed toward the center. We want to do this to simulate gravity. So, basically, what we want is we want to simulate centripetal acceleration, which is \( g \), which is 9.8. What we have to do in this problem is figure out how fast the ring has to spin in terms of revolutions per minute or RPM so that this happens. Right? So let's go ahead and take a look here. We're told that the diameter of the spaceship is 500 meters, meaning, all the way across here, the diameter is 500. But we actually never use diameter in these types of problems. So instead, what I'm going to do is I'm going to use the radius. The radius of this is going to be exactly half of that, which is 250 meters. That's the number that I'm going to use. Alright? So how do we figure out the frequency or the RPMs? How do we actually get to this value here, which is RPMs? Remember that RPMs is sort of just like a modified version of frequency? It appears in this equation over here. So \( f \) is equal to \( \frac{1}{T} \). We don't know what the period is, but it's related to the RPM divided by 60. And this is really what we're going to sort of, this is what we're aiming for. Right? So in other words, I can rewrite this equation and see that RPMs is actually equal to the frequency times 60. Right? That's just directly from this relationship. So in other words, if I can get the frequency in this problem, I can figure RPMs because all I'm going to do is I'm just going to take whatever number that is and then multiply it by 60, and that's going to be my final answer. Alright? So then the question is, how do I actually get this frequency? That's really what I'm trying to solve for in this problem. So I'm going to go over here and do that. How do we actually solve for frequency? Remember, there's a lot of equations here for circular motion. A lot of them, unfortunately, include frequency. So let's try to see if we can sort of process an elimination of this. Right? We know we can't just directly use one over period because we don't have the orbital or the period of that rotation. Right? So we can't use \( f = \frac{1}{T} \). And we also know that frequency and period are related to the tangential velocity using these two equations over here. The problem is I don't have anything about the tangential velocity. I don't know how fast the points along the ring are moving around the circle, around that rotation. So, unfortunately, I can't use any of my tangential equations, my \( v_{tangential} \). However, remember that the tangential velocity when you divide it by \( r \) and square it, or square divided by \( r \), is equal to the centripetal acceleration. And we actually know something about this centripetal acceleration. We know that we want that to be 9.8, so we have a number for that. So instead of using the \( v^2/r \) relationship, remember that because we can use centripetal acceleration and actually sort of shortcut it to include period and frequency, we're actually going to use this relationship over here. You can actually go directly from acceleration to frequency. So we know that acentripetal is related to \( 4\pi^2r \times \text{frequency}^2 \). Alright? So notice how all of these numbers here I actually know. I have the centripetal acceleration. I want that to be 9.8, and I have what the radius is, which is 250, and the rest are just numbers. Right? So I can actually rearrange for this, and I could just divide by \( 4\pi^2r \). That cancels. This is going to be \( 4\pi^2r \). And then all I have to do is notice that this frequency is squared here, so I have to take the square roots. Alright? So I'm going to take the square roots of both sides. That's going to be \( \sqrt{\frac{a_c}{4\pi^2r}} \). Lots of variables going on here, so just make sure that you're plugging this in correctly. That's going to be frequency. And just go ahead and attach some numbers to this. This is going to be the square root of 9.8, and then I'm going to divide by \( 4\pi^2 \), and then the radius is at 500 or is that 250? Be very careful here. You're going to use the 250, not the 500. And so when you plug this into your calculator, do 9.8 divided by all of this stuff in parentheses and then square root that number. What you should get is you should get a frequency of 0.032 if you round it to the 2nd decimal place. Alright? So is this the answer? Remember, this is not the answer. This is frequency. This is that number that I'm going to plug into this formula over here and multiply by 60 so that I can get my RPMs. Alright? This is how many cycles it does per one second. So I'm going to have to take this number and multiply it by 60 to get how many revolutions we get per minute. And if you actually go ahead and do this, what you're going to see is that this is equal to this is equal to 1.92. So 1.92 is our final answer. So in other words, this space station has to rotate, you could also just round this down to like 1.9 if you wanted to. So, in other words, this space station has to rotate almost twice per minute in order to simulate gravity at 9.8 meters per second squared. Alright. So that's it for this one, folks. Let me know if you have any questions, and I'll see you in the next one.

Here’s what students ask on this topic:

What is the relationship between period and frequency in uniform circular motion?

The period (T) and frequency (f) in uniform circular motion are inversely related. The period is the time it takes to complete one full rotation, while the frequency is the number of rotations per second. Mathematically, this relationship is expressed as:

$f=\frac{1}{T}$

and

$T=\frac{1}{f}$

This means that if you know the period, you can find the frequency by taking the reciprocal, and vice versa.

How do you calculate the tangential velocity in uniform circular motion?

The tangential velocity (v) in uniform circular motion can be calculated using the circumference (C) of the circular path and the period (T). The formula is:

$v=\frac{C}{T}$

Since the circumference is given by $C=2\pi r$, the equation can be rewritten as:

$v=\frac{2\pi r}{T}$

Alternatively, using the frequency (f), the formula becomes:

$v=2\pi rf$

What is the formula for centripetal acceleration in uniform circular motion?

Centripetal acceleration (a_c) in uniform circular motion is given by the formula:

$a{c}_{}=\frac{v^2}{r}$

where v is the tangential velocity and r is the radius of the circular path. Using the relationship between tangential velocity, period, and frequency, the formula can also be expressed as:

$a{c}_{}=\frac{4{\pi }^2{r}^2}{T^2}$

or

$a{c}_{}=4{\pi }^{2}r{f}^2$

How do you calculate the period of an object in uniform circular motion?

The period (T) of an object in uniform circular motion is the time it takes to complete one full rotation. It can be calculated by dividing the total time elapsed by the number of rotations (cycles) completed. The formula is:

$T=\frac{t}{n}$

where t is the total time and n is the number of rotations. For example, if an object completes 4 rotations in 2 seconds, the period is:

$T=\frac{2}{4}=0.5 s$

What is the significance of the circumference in uniform circular motion?

The circumference (C) in uniform circular motion is the distance an object travels in one complete rotation around the circular path. It is directly related to the radius (r) of the circle and is given by the formula:

$C=2\pi r$

The circumference is important because it helps in calculating the tangential velocity (v) and understanding the distance covered per rotation. Knowing the circumference allows you to determine how far an object travels in a given period (T) or frequency (f).