https://youtu.be/5cAPE8YimKQ
PHYS 1101: Lecture Twenty-One, Part Two
Now let’s get started with our new material. What we’re up to now is building a connection between those angle quantities. Angle in radians, angular velocity, omega, radians per second, alpha in units of radians per second squared. They’re angle quantities because we’re not describing a distance in meters, but rather an angle in radians. So hence, you see the presence of the angle unit for all three. We now want to connect those, as I said above, to the distance quantities, the motion along the circular line or path that this point of interest is following.
We’re back to quantities that are measured in real length and distance. We’re going to have variables that represent a position in meters along this path. This variable, “s,” that we’re going to use is the same thing as the x0 and the x from chapter two. S is just commonly used to describe an arc length, or a distance, a coordinate along an arc.
We’re back to the familiar “v”s to describe the velocity in meters per second at any instant along this path. Meters per second. We’re going to call that the tangential velocity. It just means the velocity vector that is aligned at a given instant, the velocity vector that’s tangent to this motion. In fact, if this point is constrained to follow a circular path, at any instant its velocity is always tangent to that path.
At this instant, its velocity is pointing in this direction. The velocity always points in the direction that that object is headed at that instant. If it’s following a circular path, that means that direction is changing all the time, and at each instant it’s tangent. That’s why we call it the tangential velocity. Then we’re going to have to acceleration, again, that’s along the circular path. That also translates to at each instant I can talk about the tangential acceleration vector, which is also tangent to the circle.
That’s the acceleration that would tell me how my tangential velocity is changing. Units of meters per second per second, or meters per second squared. So in a problem, I’m going to have, again, a time duration of this motion that’s of interest. There’ll be an initial snapshot in time, and a final snapshot in time. At each position, at each snapshot, I can talk about the time, the position, and the velocity at this instant.
At the end of the problem, the end scope, I can talk about the time there, the position there, and the velocity. So the absence of the subzero here, those represent the variables at the end snapshot. This is my start over here. And then in between these two snapshots, I have the acceleration, which is constant the duration of the problem. So I’ll label that “aT” and draw an arrow just indicating the scope, that the magnitude of aT has to be the same from start to end.
So if it helps, think of unrolling this motion along this circle into a straight line. Then it’s perfectly analogous to chapter two. The only difference is we’re labeling the position s instead of x, and I’m just calling these, I’m putting aT here to remind myself it’s tangential. The name of the game is going to be of course reading the problem, and being able to map the wording and what’s going on physically into these variables.
I must be able to identify the values for these variables that represent these different physical quantities. And I have to identify what variable I’m going to go after. Here are a couple of quiz questions to give you some practice in starting to do that. I have a wheel with a certain radius that rotates freely about a fixed axle.
There’s a rope wound around the wheel. Starting from rest at t=0, the rope is pulled such that it has a constant acceleration of 4 m/s2. This problem would be to determine the number of revolutions that the wheel has made after 10 seconds.
Let me draw you a quick sketch. I highly recommend always starting with a sketch. Here’s my object, my wheel, and I do have this rope that’s wound around this wheel, and the rope is being pulled down with an acceleration of 4 m/s2. That’s acceleration straight along a line for this rope, but the bulk of the problem is asking about the rotational motion of this wheel. So how is this acceleration related to the motion of the wheel?
Well, the acceleration of the rope, if you think about it, it has to be the same tangential acceleration at the edge of the wheel. In other words, if you were to follow a scratch mark on the edge of that wheel… The rope isn’t slipping on this wheel. If the wheel goes around it with a certain acceleration, that’s going to be the same acceleration that this rope is going to peel off from this wheel with. That’s going to be the same acceleration that the rope is pulling down with.
Okay, question three. For this problem, what variable would you assign to the quantity 4 that’s there in the problem? What variable would you assign to the quantity that you’re trying to determine? That’s question four. Let me go through each of these new quantities in a little more detail with a few more quiz questions to solidify their meaning and their physical connection for you.
Tangential velocity. The tangential velocity literally has to be the velocity vector that points in the direction of motion. If something is restricted to follow a circular path, at any instant, it’s always headed tangent to that circle. The next instant, it’s changed, it’s rolled over to a new part of the motion, and it’s headed now in a different direction. But still tangent to the circle, but now at that point. Always points in the direction of motion.
So here’s a question for you. I’m showing you a wheel rotating about this axis. It’s rotating with an angular velocity of plus omega, and I’ve got two spots on the wheel, a blue one and a green one. When this green spot passes this horizontal spot shown, it’s going around. When it gets to a point just at this instant, what direction is its velocity vector pointing? You’ve got many choices here. Is it one of the specific or approximate vector directions?
Or does it depend on how fast the wheel is rotating? Or does it depend on whether the wheel is winding up or winding down? vT is what we’re calling this tangential speed. Here is a very important connection for you. On the left, notice, is this new variable that we’re talking about, the tangential speed. Tangential speed of course would be the magnitude of this vector, this velocity vector that you’re identifying in question five. This is the magnitude.
The magnitude of the tangential velocity isn’t arbitrary. It depends on the radius of the circle, and the angular velocity. It’s the simple product. So here is an important connection between the angle variable we learned about last lecture and the tangential variable that describes the motion along the circle. This left side has units of meters per second, as I’ve pointed out.
The right side is going to have units of meters per second. Because here I have radians per second. Radians isn’t really a unit. It really drops out. It’s just a place-holder there. The distance, the radius of the circle, is in units of meters, so I have meters per second.
Here’s my next question for you. Given this point, that the tangential speed depends on the radius, answer this question. vT depends on how fast it’s rotating, omega, and it depends on the distance from the center. Which of those two spots, the blue spot or the green spot has the largest speed? These are both attached to a rigid, rotating object. They both have to have the same omega. They both sweep out the same angle as time goes on. So their angular information is always the same. All of their angle variables will be the same. Those don’t depend on r. But the tangential speed does. So which has the larger vT?
Now let’s talk about acceleration. What’s the acceleration vector in real meters per second squared in the context of an object that’s forced to follow a circular path? We’ve looked at it in the context of uniform circular motion, but now we want to include the possibility of something speeding up along this circular path, or slowing down. So we need to look at the acceleration vector for more general types of motion.
The tangential velocity, let me point out, always is along the circular path. It’s always tangent to the circle. It’s always going to point, in the examples, as I show below. The acceleration vector, remember, tells us then how the velocity changes. It’s going to be important to think about the contribution to a because the magnitude of the tangential velocity is changing. It could be that this object is speeding up as it goes along this circle, and the consequences on the direction of this change because it’s forced to follow this circular path.
Let’s see what that is. Let’s go back and first start with the scenario we’ve looked at before, which is uniform circular motion. Notice I’ve depicted that properly here by showing you snapshots out of a motion diagram where the velocity vector, this would be my total velocity vector, which is my total tangential velocity because it’s tangent to the circular path. That velocity vector has the same magnitude between all snapshots. Just the direction is changing.
Remember in this context, when we went through and determined what the acceleration was, we arrived at what we called the centripetal acceleration. But let me remind you how we need to do that. To figure out the direction for acceleration for any scenario, I can compare two adjacent velocity vectors. And what I want to do is draw the same velocity vector at the end of the first one. So I’m going to repeat this vector starting here. I need to repeat the length and the direction.
This is what the velocity would continue to do if I didn’t have any forces, I didn’t have acceleration. But we know that it doesn’t do that. The next velocity vector is pulled over to the left. So to this initial velocity, I have to add the vector Δv, which we’ve come to appreciate. I can think of or look at that vector as being the same direction as the acceleration vector. For uniform circular motion, what that exercise told us, that the total acceleration vector always points to the center, for uniform circular motion, this tail to tip analysis verifies that.
That has to be the real Δv, and therefore a, to represent this kind of motion. We learned that the magnitude of this acceleration was the speed squared divided by r. In the context of this new material, I just want to emphasize that that speed would be the tangential speed squared. The magnitude of the tangential velocity.
Just so you don’t get confused with notations, let me jot that to the side. Note vT in chapter eight is the same v in chapter five. So now though, we need to consider the more general case where something may not be just going around at a steady pace, but it can be following the circular path, and at the same time speeding up or slowing down. Let’s consider a scenario of winding up. Notice with my snapshots here for my motion diagram, that the length of my vectors properly represent winding up.
This object is going faster. It travels farther around the circle in the second snapshot than in the first. So it’s speeding up. Given these two velocity vectors, do the tail to tip analysis that I walked you through here, do it carefully so that you can have a careful assessment of the orientation of this Δv vector, and therefore the orientation of the acceleration. The question is what’s the direction of acceleration? If something is winding up, it’s not going to pointing straight to the center. Let me give you that hint.
It’s different now because this velocity vector at the next instant, not only has changed direction, but it’s also larger. So question seven is do the analysis and then estimate a through e, your arrow choices here, what best represents your acceleration now when something is winding up?
Okay, let me walk you through one way to think about this analysis. Once you go to this more general case where something can be speeding up along this circular path or slowing down, my acceleration vector can now point in some arbitrary direction determined by these velocity vectors before and after. It’s not going to be pointing just toward the center anymore. It’s only going to point to the center if I have no tangential acceleration, if the thing is going around at a steady pace.
So let me just draw an acceleration vector off at some angle. Here’s a useful way to think of how to break this up. Sketch yourself a line to represent an axis that’s tangent at this instant, when the object’s at that point. This line is tangent to the circular path at this point. Now I want to draw the line perpendicular to that. Think of these of these as two axes. This is a vector that’s off at some angle to these two axes.
I want to consider the component of this vector that’s aligned with this axis, and the component that’s aligned with this axis. Well, how did I do that? In order to find the components, I take my vector, and at one end, I dash a line that’s parallel to one axis. In this case, I happened to choose this one. And then at the other end of the vector, I dash a line parallel to the other axis. That actually is along this axis as I’ve drawn it.
The two components now are this vector and this vector. Let me dash my lines here along the other end, just so you can see that. This component is the part of the total acceleration that’s in the tangential direction. We label this the vector aT. This is the tangential part. The other component of the acceleration vector is the part that points towards the center. It’s the centripetal component of the acceleration. I’m going to label that ac.
So an equivalent description of our resultant acceleration, our overall a, is these two components along these two axes instead of the resultant. I can replace my resultant with the aT component and the aC component instead of the resultant.
What’s useful about that is one can then think, or consider, that the centripetal part is the part that is responsible for keeping the object following a circular path, and then you can think about the tangential component as being the acceleration that describes the speeding up or the slowing down along the path. Of course the magnitude, the hypotenuse, the resultant acceleration vector a, is the vector sum of these two components, and the magnitude of a is equal, by Pythagorean, the square root of the two components squared.
So these are the two sides of my right triangle and a is my hypotenuse. So I’ve added here to the sketch hypotenuse to emphasize to you that our total acceleration, the genuine direction of a when I calculate my Δv from a real motion diagram, that total acceleration vector has these two components. The total acceleration is the hypotenuse, and I’ve got a tangential component, which speeds me up along the circle, and the centripetal component, which is responsible for keeping me following the circle.
I’m going to highlight in yellow an equation you may have need for. It’s an important relationship to remember. The connection between the tangential component, the centripetal component, and the hypotenuse, if you will. So, the same comments we’ve made in the past about acceleration and the connection to forces is true. In other words, for a tangential acceleration component to exist, for there to be any speeding up along the path, I have to have a tangential force component. The direction of the force always has to match or agree with a direction of acceleration.
So let me highlight that connection for you. If you’re told there’s some tangential force in some problem, right away from , you can calculate what my acceleration has to be along the circular path. Here, then, is an important equation that connects also this tangential acceleration to our angular acceleration, alpha. The left side has units of meters per second squared. The right side does as well because r is in units of meters, and alpha is radians per meters squared.
Radians, again, just drops out. So here’s our equation connects tangential to angle. It connects this lecture to the last lecture. The next point is about the centripetal component to the acceleration, that that has a magnitude, which we learned about in chapter five. The centripetal acceleration always has to agree with the tangential speed at that instant squared, divided by r. Let me highlight that for you. I’m going to highlight in blue again this other important equation, which again says nothing new.
We saw this in chapter five. But if I have acceleration in some direction, I have to have a net force component in that direction. Your book, again, wants to call this the centripetal force, and I just remind you that that’s somewhat of an artificial label. There’s no new forces here, just like above, there has to be a real, physical mechanism that’s exerting contact, and exerting a force in the tangential direction to give me an at.
There has to be real pull or a real push in the centripetal or the center-oriented direction to give me a centripetal acceleration. For example, tension would be a common cause, or the actual centripetal force would be a tension force. Okay, I’m going to show you some animations now so we can picture how this acceleration vector behaves.
And you can see the consequences of what happens when something is winding up or winding down. This is an animation of circular motion where something is going to be winding. I’m going to play the end part here for you. After it winds up, it continues to go around at then a steady pace. So this is uniform circular motion.
This is the object. The arrow that’s drawn there is the acceleration vector. Notice that that vector, they drew it a bit excessively long. Nevertheless, you can appreciate at any instant, it’s pointing directly to the center. This object is going around at a uniform speed. Every little delta v you would imagine there would show you that the acceleration at every spot has to point to the center. So if you play this, and you watch it, that line, that vector, always points to the center of that circle.
That’s not the case when the object is winding up. We’re going to start this object out at rest, and it has some tangential force that supplies a small amount of tangential acceleration. So the arrow that you see here is my total acceleration initially. I have this force that continues to provide this tangential acceleration, but as I start to speed up, we’re going to see that I start to gain a part of my acceleration that’s toward the center.
Let’s play it, and watch it. I always have this tangential component, but as I go faster, I know I have to have an increasing centripetal component. And that’s what causes this vector to get larger and spiral in toward the center of that circle. Let me show you here in some snapshots what I mean. Here was the final scenario where I was up to speed, and this was the picture for uniform circular motion.
So here’s the center of my circle, and what you see in orange here, I’m going to draw it in orange, is my acceleration vector that points to the center of that circle. That was always true. It’s going around uniform. Here are three snapshots though before that where I’m winding up. Here I’m at the start. This is in the middle. And then here, I’m almost up to speed. Not uniform circular motion yet, but I’m almost there.
Let’s do a force analysis, or a component analysis of each of these. So I start out from rest, and I’m going to draw it in black so you can see it well. My acceleration vector, drawn in black here, only has a tangential component. Let me write at the top here. Remember that aC is equal to vT2 over r, and at is related to r times alpha, if we need to.
So, my total acceleration vector is all in the tangential direction here. I don’t have any directed towards the center. That makes sense because at this snapshot, I don’t have any tangential speed yet. So zero on the left equals zero on the right equals zero on the left. Zero centripetal acceleration. Let me go to this spot now, and find the components at this instant. Here’s my tangential direction. Here’s my center-oriented direction. Now, let me sketch in here for you, this is my aC component.
This would be my aT component. Hopefully you can see that this vector is one component of the hypotenuse. Here’s my total acceleration off at that angle. And here’s my tangential component. The tangential component stays the same, but because I’ve picked up some speed now, I have some tangential velocity. So I do have a centripetal acceleration.
So let me write here for you aC larger now with, now that we have some tangential velocity. As we went to a next snapshot. Let’s do the quick analysis over here where two components…now my centripetal component is even larger. My tangential component is still the same. It’s still small, but it’s there nonetheless.
So I’m still winding up. So aC even larger, and this is because vT is larger. So when you compare how these components are changing, you appreciate that my total acceleration vector then has to have to have this interesting behavior that it starts out tangent, but then the centripetal part grows, which looks to your eye like the acceleration is getting larger and getting pulled toward the center of the circle.
Let’s watch it again real quick. You’ll see that behavior. The magnitude is getting larger, and it gets pulled toward the center because it’s the centripetal component that gets larger. As soon as I reach a constant speed, all I have is the centripetal acceleration v2 over r. So when we do some of these problems, we’re going to be faced with having to think about both components of the acceleration vector.
Then we’re going to see that in an example. Let me first point out. If we think about just the tangential components. So my velocity is always just tangential, but if it’s getting larger or smaller, I will have a tangential acceleration that captures that effect, that slowing down or speeding up along the circular path. We can treat this just like chapter two. I’ll have an initial snapshot in time. I’ll have time-positioned velocity.
At the end snapshot of my problem, time, position, velocity. As long as I have constant acceleration in between that’s causing vT to change, then I know I can use these three equations to work out the kinematic relationship between these quantities. I can figure out what a final distance along the circular path is or what the initial velocity had to have been with so much tangential acceleration and a final vT, for example. Let’s do an example.