In this video David shows how to relate the angular displacement to the arc length, angular velocity to the speed, and angular acceleration to the tangential acceleration. Created by David SantoPietro.
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- At2:26, arc length can also be found using a degree.
so why do we use radians if we can find arc length by using
(Theta/360)*2pi*r? I mean why do we have to use radians?(13 votes)
- radians have no units. This makes them much more convenient than degrees.(34 votes)
- How is arc length a distance and not a displacement?(11 votes)
- Displacement is the shortest distance from the initial position to the final position, while Distance is the actual path covered by a moving object. In this case, the tennis ball covers the arc as the actual path, therefore it is the distance. Displacement, in this case, is the line joining the initial position to the tennis ball. Vote up if it helped(41 votes)
- If tangential acceleration and centripetal acceleration are both components of the total acceleration, why tangential acceleration is not a vector whereas centripetal acceleration is a vector?(10 votes)
- You are correct, tangential acceleration is just as much a vector as centripetal acceleration. That's why at11:03in the video, he draws an arrow to represent it. Then, he goes on to say that you would add the tangential acceleration and the centripetal acceleration as vectors to get the total acceleration, just like we do with all vectors and their components. Did he say something somewhere that seemed to imply that tangential acceleration was not a vector?(11 votes)
- if v=r(omega) , then omega has units rad/s and r has units meters then what do we express v in?, (meter*rad)/s ?(4 votes)
- I am a bit confused about the usage of radians for angular measurements. Is pi(π) a unit of radians, or is it not? When converting different units, such as revolutions or degrees, to radians, should I solve for radians in terms of pi or terms of an integer?
[Question]: Convert 3 revolutions to radians.
To solve, one would have to do (3 rev.)(2π/1rev.)
So would the answer be 6π radians or 18.849 radians?(4 votes)
- Radians are technically unitless since it is ratio of radius which is a distance to the circumference which is also a distance. This gives you meter/meter which cancel out leaving no units. π is not a unit it is just a constant of proportion.
How you report the answer depends on who you are reporting it to but in general an answer like 6π is more understandable that the decimal amount since using π you get a better idea of the proportion of the circle it is.(5 votes)
- why cant we just call angular displacement 'angular distance' or angular velocity as 'angular speed'?(4 votes)
- Angular displacement is the displacement of an object from the starting point to a final point. If you are talking about the distance then it's all of distance that it took us to get to a final point from an initial point. This about this in this way; if you're in a circular road driving a car and you complete one lap i.e one revolution then your displacement would be 0 because you are at the same location from where you started but you distance traveled would be the circumference of that circular road track i.e the total distance that it took you to complete one lap.
For Angular velocity, it is the change in your angular displacement per time or in other words, how fast our angle is changing for every change in time(second), and since it is velocity, it must have a direction which are counter-clockwise or clockwise but the Angular speed is simply the magnitude of the angular velocity, not specifying the direction of the motion.(4 votes)
- why R(radius) was involved in every conversion, please?(3 votes)
- Is tangential acceleration equal to the angular acceleration?(3 votes)
- Tangential acceleration is not (numerically) equal to angular acceleration. If the motion is circular (radius is a constant) then they are related by the constant radius. alpha (angular acceleration) times the radius is equal to the tangential acceleration.(4 votes)
- If we're given numerical problem with degrees instead of radian, how do we convert it to radian?(2 votes)
- Great question-
To convert from degrees to radians, multiply the number of degrees by π/180. This will give you the measurement in radians. If you have an angle that's 90 degrees, and you want to know what it is in radians, you multiply 90 by π/180. This gives you π/2.(1 vote)
- how do we relate 2d curviliner motion with angular variables(4 votes)
- Good question. What I understand there are three things: Centripetal Acceleration, Tangential Acceleration and Angular Acceleration. One can calculate tangential acceleration from angular acceleration. Then using tangential acceleration and Centripetal acceleration one can calculate Total acceleration....(2 votes)
- [Instructor] So in the previous video, we defined all the new angular motion variables and we made an argument that those are more useful in many cases to use than the regular motion variables for things that are rotating in a circle. Since every point on the string in tennis ball, let's say this is a tennis ball you tied a string to and you're whirling around in a circle. Every point on the string including the tennis ball will have the same angular displacement, angular velocity and angular acceleration. But even though using angular motion variables is more convenient for these rotational motion problems, it's also really important to know how to translate those angular motion variables back into the regular motion variables. So that's what I wanna show you in this video how to translate angular motion variables back into regular motion variables. So let's do this. The simplest possible angular motion variable was the angular displacement because that just represented how much angle an object has rotated through. So let's say it rotated through this much. We represented the angular displacement with a delta theta and we call it the angular displacement. In Physics, we typically choose to measure this in radians for a reason and I'll show you in just a second. Now, how would we convert this into a regular motion variable? What regular motion variable would that be? If I were to come at this for the first time, I'd be like all right, this is the angular displacement. Let's figure out how to relate it to the regular displacement but that would be weird. Because just think about it, the regular displacement for the ball that started over here and made it over here would be from this point to that point, that would be the regular displacement of the ball, the regular linear displacement of the ball. That's a little weird. I don't wanna show you how to find that for one, you have to use the law of cosines. That's a little more in depth than I'd wanna get to in this video. For two, the better reason this isn't all that useful in turns out. There's a much more useful quantity that would tell you how far the ball went. That's the arc length of the ball. So the ball traced out of path through space around the circle. We call this the arc length that it turns out this is much more useful in a variety of problems. Good news is it's much easier to find than that regular displacement. So this is the arc length. People vary on what letter to use here. I've seen an l but most math books use s so we'll just use s as well. You might think this is hard to find but it's not. In fact, if we use radians and this is why we use radians, it's extremely easy to find. If we wanted to find the arc length of this tennis ball, we're just gonna take the radius of the circular path that tennis ball is tracing out. So in this case it'd be the length of the string. We take that radius. If we're in radians, we just multiply by the angular displacement. This is why the radians are so convenient. We just take that measurement in radians multiplied by the radius of the circular path the object is tracing out. You get the arc length which is the number of meters along this path that the object has traveled. If that seems miraculous, it really isn't. I mean the reason why this works so well because this is how the radian was defined. One radian is defined to be the angle through which you have to travel so that the arc length is equal to the radius of that circle. So this isn't a surprise. This was selected and defined strategically so that we can use this unit and we get a really easy way to convert between the angular displacement, how many radians something has rotated through, and how many meters it has actually traveled through its arc. So this arc length is gonna have units of meters as long as we measure the radius in meters. All right, so that's one relationship between angular displacement, how much angles something has rotated through, and how many meters it has actually traveled. The next relationship I wanna talk about relates the angular velocity to the regular velocity. So remember in the previous video, we defined the angular velocity to be the angular displacement per times. So this is the rate at which something is rotating through a certain amount of angle and the letter we use to represent velocity was the Greek letter omega. So this angular velocity represented the rate at which something in a circle. So it's rotating slowly. It's gonna have a small angular velocity. If it's rotating quickly, it's gonna have a large angular velocity. Obviously the speed and the angular velocity are gonna be related because the higher the angular velocity, the higher the speed. But, what is that relationship? How would we get from angular velocity to the regular velocity? Well it's not actually that hard at all because all we need to do is turn this number of radians per second into meters per second and I can do that. If multiply both sides of this equation down here by R, I'll get R times omega is gonna equal R times delta theta and I still have to divide by delta t. So you just multiply both sides of this equation by R. But look what I get. R times delta theta is just the arc length. So this whole side over here is just how many meters that object has traveled around the edge of the circle divided by the time that it took. But that is just the speed. This arc length is just the distance the object has traveled and the time is the time that it took and distance per time is just speed. So this is the speed of the object. I'm gonna write that as v even though it's not velocity. This is not a vector and it's not velocity because think about it, this arc length isn't displacement. This was the distance the object traveled. Distance per time is the speed. Displacement per time is the velocity. We didn't use displacement. Displacement was this weird one. We didn't wanna deal with that. So since we're choosing to deal with arc length which is distance, what we're gonna do is relate the angular velocity into the speed. Now we have that relationship. Look at this. This is R the radius times the angular velocity equals the speed of the object. So this is the relationship between the angular velocity and the speed. The speed of the object is gonna equal the radius of the circular path the object is traveling in times the angular velocity. I should box these. These are important. This arc length formula was how you relate the number of radians and object has rotated through to how much arc length it's traveled, i.e. how much distance it's gone through. In this formula down here relates the angular velocity omega, the number of radians per second something has rotating with to how many meters per second it's traveling. In other words, how many meters per second it's tracing out along this arc length. So this is good. Now we know how to relate the angular displacement to the distance the object has traveled and we know how to relate the angular velocity to the speed of the object so you probably know what's coming next. We have to relate the angular acceleration to the regular acceleration. So we're called that the angular acceleration which we represented with a Greek letter alpha was defined to be the change in the angular velocity per times. It's the rate at which your angular velocity was changing. So there's moving at the constant rate. You've got no angular acceleration because there's no change in omega. But if omega starts off slow and then it gets faster and faster, you do have angular acceleration. It's probably not a surprise that if you have angular acceleration, this ball is gonna have regular acceleration too because it's speeding up in its angular rotation. It's gonna be changing its velocity as well. So how do we this? How do we relate the angular acceleration to the regular acceleration? Well the simplest thing to try is we go work down here. We multiply both sides of our equation by the radius and we found the relationship the related speed to angular velocity. So let's try it again. Let's multiply both sides of this equation by radius and see what we get. On the left hand side, we can get the radius times the angular acceleration. That's gonna equal the radius times the change in angular velocity over the change in time. So all I've done here is multiply both sides of this equation. This definition of angular acceleration by the radius. So let's see what we get on the right hand side. We got R times delta omega. So this is really R times the change in omega. Well that's just omega final minus omega initial and then divide it by delta t so I can distribute this R and get that. This would equal R times omega final minus R times omega initial divided by the time that it took. But now look what happens. We've got R times omega final and R times omega initial. We know what R times omega is. It's the speed, not the velocity, but the speed. So I could rewrite this. I could say that this is really the final speed minus the initial speed over the time that it took to change by that much speed. So this is, if I just rewrite the left hand side, this is what R times alpha is equal to. Now if I were you, I'd be tempted to just be like, oh look, we did it. That's the acceleration which change in speed over time. But you gotta be careful. Acceleration, the true acceleration vector is the change in velocity per time, but these are not velocity vectors. These were speeds. So this isn't the true acceleration vector. This is something different. This is the change in speed per time. So that's still an acceleration but it's not necessarily the entire acceleration because there's two ways to accelerate. You can change your speed or change your direction. Basically this acceleration we just found doesn't take into account any acceleration that's coming from changing your direction. This is only the acceleration that's gonna be changing your speed. If I were you, I'd probably be confused at this point. So let me try to show you what this means. If this ball is rotating in a circle just by the mere fact that the ball is rotating in a circle, it has to be accelerating even if the ball isn't speeding up or slowing down. There's got to be an acceleration because this ball is changing the direction of its velocity. These are gonna be a force. It's centripetal force, in this case it would be the tension. There's gotta be a centripetal acceleration that's changing the direction of the velocity. That is not this acceleration over here. This is a different acceleration. We know the centripetal acceleration is directed inward. We already know how to find this centripetal acceleration. Remember the formula for centripetal acceleration is the speed squared divided by the radius. This component, this centripetal acceleration is the component of the acceleration that changes the direction of the velocity. So I'm gonna say that again because this is important. The centripetal acceleration which you can find with v squared over R is the component of the acceleration that changes the direction of the velocity. If something is going in a circle, it must have centripetal acceleration. But this acceleration component that we found down here is different. This is what's changing the speed. You don't have to have this if you're going in a circle. You could imagine something going in a circle at a constant rate. If that's happening, it's got centripetal acceleration but it doesn't have this thing down here because this thing we found R times alpha is the change in the speed of the object per time. How would I draw that up here if I wanted to represent this a that we found down here visually up here? I'd draw tangential to the direction of motion, i.e. tangential to the circle because components of acceleration that are directed perpendicular to the velocity change the direction of the velocity but components of acceleration that are directed parallel to the direction of the velocity change the magnitude of the velocity, i.e. the speed, to change the magnitude of the velocity, in other words to speed something up or to slow it down. You need a component of that acceleration that's either in the directional motion or opposite of the directional motion. If it was opposite the directional motion, the acceleration would be slowing the object down. If the component of acceleration is in the direction of motion, then it's speeding the object up. That's what we found down here. That's what this component of acceleration is, R alpha which is why it's often called the tangential acceleration. So I'm gonna write that up here. The tangential acceleration which is equal to R times alpha, the radius times the angular acceleration is the component of the acceleration that's changing the magnitude of the velocity, i.e. it's changing the speed. In order to do that, it's gotta be directed tangential to the direction of motion. That's what this R times alpha represents. So let me box that. That's important. This is the formula to find the tangential acceleration. It doesn't give you the total acceleration. We know that there's always a component of acceleration that's acting centripetally if an object is going in a circle and you could that with v squared over R. But if the object going in a circle is also speeding up not only going in a circle but changing its speed, it's gonna also have this component of acceleration which is the tangential acceleration. So at this point you might be confused like wait, we've got tangential acceleration, we've got centripetal. Which one is the acceleration? Well, they're both just components of the total acceleration which you could find if you really wanted to. You could say that the total acceleration squared. You could use the Pythagorean theorem because these are the two perpendicular components of the total acceleration. We said the total acceleration squared would equal the tangential acceleration squared plus the centripetal acceleration squared. What we would be finding is the total acceleration squared which if you wanted a direction, the direction of that total acceleration would point this way somewhere. If you had centripetal acceleration inward and let's say the object was speeding up. So let's say it wasn't slowing down. So you didn't have this component. You've got this forwards component and inwards component. The total acceleration would be directed here somewhere. Since you could form a right triangle out of this with these two sides, you can imagine moving this centripetal acceleration over to this side and you could find the hypothenuse which would be the total acceleration by just taking the tangential acceleration squared plus the centripetal acceleration squared and then taking a square root would give you the magnitude of the total acceleration. So recapping, there's two components of acceleration, the tangential acceleration which is R times alpha, either speeds an object up or slows it down, the centripetal acceleration works to change the direction of the motion of the object. You can relate the speed of an object to the angular velocity by multiplying by R. You could relate the arc length, i.e. the distance the object traveled around this edge of the circle to the angular displacement by also multiplying by R. So these three equations or how you relate the angular motion variable to its linear counterpart.