Types of Acceleration in Rotation

Concept: Types of Acceleration in Rotation

14m
Video Transcript

Hey guys! Now we're going to talk about the four different types of accelerations you'll see in rotation problems. It can get a little overwhelming because it's a lot of letters, a lot of variables, a lot of equations but I'm going to try to simplify this as much as possible. Let's check it out. There are four, as I said, types of acceleration rotation problems. Most of them have two names. In total if you look you have seven different names and you have to know how the pairs, which pair of names, go together. Centripetal acceleration, ac, also referred to as radial acceleration a rad, those are the same thing. Tangential acceleration is at, sometimes this is referred to as linear acceleration. Let me actually add that there. Sometimes you might see this called linear acceleration. The other one is total acceleration, which sometimes is referred to either as total acceleration or just acceleration. Sometimes it's referred to as the total linear acceleration. This is just a. It's the total acceleration. Then the last one is rotational or angular acceleration. YouÕre used this one. This is _. I want to point out that these guys are measured in m/s^2 and this is measured in rad/s^2. All these three here, these three are linear accelerations, so they're represented with sort of a straight line as opposed to _ which is an angular acceleration so it's measured this way. If it's an a, this is ac, a rad, at, a Ð if it's an a, it's meters per second, so it's an English letter so it's m/s. If it's an _, itÕs Greek letter, it's in rad/s. That's one way to think about it. There's these four types but depending on what kind of situation you have, you're not going to have all four types. What that means is that some of them will just be zero. First, you always have VT and ac. If you're going around a circle, got a little object that's going around the circle let's say this way. At this point this object has a tangential velocity and it has a centripetal acceleration pointing towards the middle and it's spinning this way with w. Remember that at this point, VT = rw, remember also that ac = V^2 / r, this is from a while ago. What I can do is I can actually rewrite V as rw. Let's do that real quick. It's going to look like this. (rw)^2 / r. The square goes on r and the w, so there's actually two r's and two w's but what one r is going to cancel with the r at the bottom. It's going to be like this. r without the square w^2. This is old school old news but this is a new version. This is a brand new version of writing ac. You can write it in terms of r & w instead of the V. These always exist. You always have ac because ac, if you remember from F = ma, you have ac because you have Fc, a centripetal force. There's a force that pulls you towards the middle therefore you have a centripetal acceleration. ac is responsible for maintaining the circular path. ac maintains circular path. Another way to think about this is that it keeps the object spinning. If ac doesn't exist, you can think of it as there's no longer a force pulling you to the middle, so this obviously is just going to go straight up in this direction. ac it's there and it maintains a circular path. As long as you move in a circle, you have to have ac and you have to have VT. These things always exist and you have a w because obviously you're spinning. If you're accelerating, you are also going to have at and _. The converse of that is that if you're not accelerating, these guys are zero. Otherwise, let me put that here, meaning if you're not accelerating then at = _ = 0. What does at look like? The idea is that you're not just spinning in place, but at this point the object is actually getting faster. If this thing is rotating like this, you can think of it that to get faster you have to like something has to push the object this way, some force. If you push the object this way, it's going to spin faster so you have an acceleration here which is a tangential acceleration. If the object is going faster this way, it's also rotating faster so there's also an alpha this way. You got an omega and you got an _ in that same direction. You've got these two guys here. Remember, you always have your ac so this sets up an interesting situation where you have an at this way and an ac this way. They make an angle of 90 degrees with each other right there. What happens is you have two arrows this way so you can combine the two of them using vector compositions. I can make a little triangle here and this combines using vector composition. This is what we call the total acceleration, sometimes referred as the total linear acceleration or sometimes referred simply as acceleration. If you see just acceleration without the word centripetal, radial, tangential, linear, rotational, angular, if you don't see any of these words you can assume it's just regular acceleration, the total acceleration, which is a combination of these two. It's a by itself. This is just vector addition. a is the square root, it's the Pythagorean of the two sides, at^2 + ac^2. That's what the total acceleration. Whereas ac maintains your circular path, these guys here are responsible for changing angular speed. These guys at changes angular speed or velocity. Technically it changes angular velocity so let's write that there. Changes means it could be going faster or slower. Two more points then we're going to do an example. This equation at = r_, which is one that I gave you earlier, is a good way to remember that at and _ are connected. TheyÕre linked up together, which is what we see here. You have this and this or they're both 0. These guys are both either nonzero together or they're both 0 together. Then the last point I want to make is that if your at is in fact 0, which by the way this would mean that _ is 0. Just look at this equation right here. Then your total a which is usually the square root of at^2 + ac^2 simplifies. If your at is zero, look what happens. You get simply 0 + ac^2 which is the square root of ac^2, which is simply ac. I'm just pointing that out in case you see that and you don't think it's weird but there's two a's inside of the square root, but one goes away and then your a becomes just your ac. That's totally doable. In fact that's what happens here. Here your a is only ac, so it's like you have these two arrows stacked on top of each other. Let's not draw that there so it doesn't make this more confusing. Four accelerations and I gave you two new equations, one new way to rewrite ac and an equation for the total a. Let's do a problem here. It's got five parts. It's kind of annoying but hopefully you see that it's not that bad. I have a carousel 10 meters in radius, so big circle radius = 10 meters. I know I don't have a lot of space here but you actually don't need that much room, but let's write small. Completes one cycle every 45 seconds, so that's the period. T = 45 seconds. I want to know what is the tangential velocity. There's a boy that stands at the edge. If the boy is at the edge, the boy is at a distance r = 10. This thing is 10 meters long in radius. If you are sitting at the edge you sit at the 10 meter distance from the center. I want to know his tangential velocity. Tangential velocity is VT, and remember the tangential velocity of a point, a person, an object, whatever on a circle on any kind of spinning object is rw where w is the w of the boy which is the same as the w of the disc. The boy is at a distance 10, but I don't know w. However, remember I can get w from t because w, f, t and RPM are all interconnected. w is 2¹ / T, so 2¹ / 45 and if you do that you get 0.14 and that's what goes here, 0.14 therefore the answer is 1.4 m/s as the velocity that the boy experiences. For Part B, I want to know what is the angular acceleration. Angular acceleration is _. Remember you only have alpha if you're actually spinning faster. Once it says that it completes one cycle every 45 seconds, first 45 seconds is one cycle, second 45 seconds is another cycle. It implies that this is a constant movement at a constant rate at a constant velocity, so _ is actually 0. Radial acceleration is a rad which is the same as ac, which is v^2/r or rw, whichever you prefer. I'm going to use rw just so I donÕt have to square this number but it's the same exact thing. r is 10 and w isÉ Sorry it's rw^2. I'm going to have to square something either way, 0.14^2. But anyway if we do this, we get that the answer is 0.196 m/s. I almost used the wrong equation there. Either one of these works. For Part D, the tangential acceleration is at. Remember at only exists if you're actually pushing this thing to spin faster. Another way to think of this is that at is r_ and _ is 0, therefore at = 0. Part of the reason why this question doesn't require that much space is because some of the answer is just 0. The total linear acceleration is a which is the square root of at^2 + ac^2, and at is 0 so you're left with the square root of ac which is just ac, which is the same thing as a rad which is 0.196 m/s^2. Here's the five answers. This, 0, this, 0 and the total a right here is that. Just a bunch of equations and knowing how to link everything together. It's good to do some practice to make sure you know how to do this. Pretty straightforward, but it's kind of annoying. Hopefully this makes sense. Let me know if you guys have any questions.

Example: Accelerations of boy on carousel

5m
Video Transcript

In the second example, we have a carousel now 16 meters in radius. Let's draw a circle here and the radius is 16. It accelerates from rest with this. w initial is 0 and _ is 0.05. A boy stands at the edge again, so the little r the distance between the center and the boy is just the same, 16 meters. We want to know after 10 seconds, _t = 10 seconds what is his tangential velocity. I'm going to put it over here to the side because tangential velocity VT is not one of his 5 motion equations. I'm just organizing this a little bit. I want to know what is its tangential velocity. Remember, tangential velocity or linear velocity of an object at a point on a disk is just given by rw, where r is the distance from the center and omega is the angular speed. I know r, itÕs 16, but I don't have W. What we're going to have to do here is we're going to have to find w final, which we're going to do using motion equations. The variable thatÕs missing here is __, so I'm going to put a little sad face. This tells me that the equation I should use is the first one, because itÕs the only equation that doesn't have __. w final = w initial + _t, and we're looking for w final. This is 0. _t is 0.05*10, so w is just 0.5 radians per second. I can plug this in here. This is 16*0.5, which means the velocity will be 8 meters per second. For Part B, I want to know the tangential acceleration. at is r_, and I have both numbers so we can just plug it in. 16 times _ is 0.05, so this is going to be 0.8 m/s^2. ThatÕs it. Then C, I'm looking for the radial acceleration, a rad, which is the same thing as a centripetal, ac, and I can use V^2 /r or I can use on rw^2. IÕm just going to use this one. I already have V, so it's going to be 8^2, r is 16, so the answer will be 8^2/16 is 4, so 4 m/s^2. For Part D, I want to know the angular acceleration. Angular acceleration is _. We actually already have the angular acceleration, so it's just going to be 0.05. This one is the freebie. Just make sure you know what you're doing, rad/s^2. The last one is the total linear acceleration, which is a and a is the square root of the other two linear accelerations. ThatÕs the only way to think about this. There's three linear. One of them combines the other two. It's at^2 + ac^2, and if you plug this in, itÕs 0.8^2 + 4^2. You combine all of this and you get 4.1 m/s^2. These are the final answers. That's it for this stuff. Let me know if you have any questions.

Problem: A large disc of radius 10 m initially at rest takes 200 full revolutions to reach 30 RPM. Calculate the total linear acceleration of a point at half way between the disc’s center and its edge, once the disc reaches 30 RPM. (You may assume it continues accelerating past that point)

8m

Problem: An object of negligible size moves in a circular path of radius 20 m with 90 RPM. Find its radial acceleration.

2m