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.