Skip to main content
Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Equations for Velocity and Acceleration in UCM

Patrick Ford
Was this helpful?
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