Hi, I'm Kate. I'm a first year mathematician at Trinity College. Hi, I'm Fareed, I'm a first year mathematician at Trinity College. Hi, I'm in. I'm a teacher in applied maths there at Trinity College. OK, so we've got the dynamic sheet and the PD sheet. So that's year and that's you. Five, Which was some good work. So we're going to spend most of the time on dynamics because I think it's quite interesting.
Yeah, let's look at some of these questions went right in in the last one was not quite right. Some of you don't. So this is all about it's constrained motion and stuff. So you've got this particle going around to a cone. Mm-Hmm. And what was the important thing about the normal reaction force? Person, that is normal. Yeah. It's normal to the surface. So, yeah, I think if I said things which are not quite right here, sir, do you mean? I think it's easy to see to zero. Yeah.
So you've got this thing, which is a cone and then the reaction forces that way you define it this way. And is that way and that way, that's got to be two vectors. The end is perpendicular to and one of them is easy to reach into the page there. And then what do you use? Just the other one? Or what's what else is it perpendicular to? The Republican. Yeah, so it comes out of that direction, which is useful to think of it as like the tangent thing to the current of the.
It's a plane, it's a plane. These are easy to direction and that direction. And but what you're often use for these questions is that is that the direction of the velocity of gold is perpendicular to end because you know that it's moving on the cone, then you know, the velocity must be perpendicular to the normal. So you use that statement a lot, and that's where you really get the fact that this force doesn't do any work.
It's the fact that it's perpendicular to the velocity is why it doesn't do any work. And. So and you had to use that when you were showing that the energy was concerned. So the first thing that you always do in these questions is to write down this Newton's second law. So when you get stuck with these questions, that's always the starting point. So in this case, you've got gravity.
And then you've got this normal reaction, so you would always start with that, and then you should usually be trying to write down what ah is. And you want to decide what coordinates it continues and in this case, in your own cylindrical coordinates, so you wouldn't do this. R e r Trump said E Z. And then you have to differentiate that and.
So the first part was fine change that when you got this feature, you get zero because this doesn't have any competitive features, actual list of known directly and that tells you the angular momentum is conserved. So this. I mean, this is so this is the same as this. So this is not true here. That makes sense. Yes, and it is not zero.
And we get this this one. Whenever you say so, you talked here about saying I was going to equate components in the different directions, and that is effectively what you're doing, but you should think of it more as taking a product that's taking a product is taking the component in the future direction. And I think that's the way that you want to try and think about these things that you always have to differentiate this.
And then finally, expressions and I tend to just think it's a good idea to write this down straight away. Because you need to use them. Also, the question so this one is double minus three seconds squared. And then you get one overall independently of R-Squared feature dot. I just remember that you can watch it like that. The feature component, because it's helpful for them to write it like that because of that fact that it often means that this quantity angular momentum is conserved.
So, so that bit was fine, then it should explain why the total energy is conserved from the site, as you said, because because the thing is smooth, therefore it doesn't do any work and that's really because it's perpendicular to the velocity. And I think that you both might. You just argue that that was the case. And then you said, I can differentiate us and show zero. I tend to think that there's a there's two different ways you can think about all of these problems because you can either say,
I'm going to start by saying energy is conserved. And then I divide my equation from that, or you can say, I'm going to start with Newton's second law and then I can derive that the energy is conserved. And they're kind of equivalent, and it's almost a matter of taste as to which one you say is where you're going to say is your starting point. Most of these questions, I kind of referred to say, Well, I've I've started writing down in a second, or I should just use that as my.
That's the thing I know is true. And if I'm going to derive anything else, it should come from there. And there's always an easy way to get from the second row to energy. Do you know what that is? How can I get to seeing what the energy is from that equation? Well, this is equal to force, which is equal to minus Steve by. D, r d. Well, it's a conservative force than we could, right, as the gradient of a potential.
Mm hmm. You've done that. OK. So that's like gravity is can be thought of as the gradient of the potential energy. But do you know how I get to that from this equation? There's a there's a kind of a trick where it's not really a trick, but it was multiplied by the velocity and becomes a vector equation. Multiply totting. So if you don't the equation with an adult, then you go and double dot dot dot. There's a lot of dots in a different many different things. Minus M D is dot dot dot dot dot.
And there is this this thing here? Is the derivative of a half an hour dot squared? That's always the case, does that make sense because there are dots squared is all dot dot dot. And if you differentiate that you get to dot out of all the law. And so if you remember that you can always write that is that that's quite useful. This, of course, is a kinetic energy. Yeah. So this is really the rate of change of the kinetic energy is all block times acceleration.
And then this term here. So that it's just got so you can think of that as DVD of and GZ. So that's the major change in potential energy. And then this time we argue for zero. Yeah. So that's that's kind of like doing the dot of this with the velocity is a good thing to do because of the fact that we know. And not the velocity is zero. And often you don't often you really care what end is kind of what you are limiting, because that's the whole point.
It's just a constraint that too that we're stuck on the cone. But that you can always derive the energy. This just tells us that divide of half an hour dot squared plus and GZ zero energy is conserved, which is exactly what you did just backwards. Then you can see. This also would tell you if if you had a friction force and if not pointed in the normal direction, then this would tell you what the rate of change of energy is due to that friction force.
So that should tell you how, how quickly your energy. I don't think we have actually have to deal with friction, but that would that would come in so, so often with these things, you you can kind of argue that energy is conserved, but often you don't actually need to do that. You can just derive whether it is or not from the equations. And that's kind of irrelevant to Question three, because that one, you have to do it this way.
So anyway, so you derived that equation and that was OK. And then there was a showing that the partnership between the two. So you managed to show that you got an expression for Z Dot squared? And then I wasn't quite convinced about this. I mean, there was on you. You did. It's easily manipulated, I guess. Yes, it is. I think something went wrong there. Yeah. But what were you trying to argue?
I was trying to get it. I know that that that's quite an inspiration, zero, so then I was going to got quadratic with that. Yeah, that would be. So you can't say it can't be quadratic, you know, cubic. Yeah, but I'm going to get three heights. It's got to be a cubic. And so b three heights at which what happens? At which? That is right. That is after the three free routes of that, Quebec are places where there is zero.
And then it didn't work out quite right, but it turns out that one of those heights is negative and therefore doesn't really make any sense because that's on the bottom of the cone. So there are two places where does that go to zero? And then so you then just said it must be the case that it's between those two heights that wasn't kind of completely obvious to me.
Why if I find places representative zero, how do I know that I haven't just gone to that height and then kept kept going a bit higher? I could have, just like a stationary point doesn't necessarily mean it's a maximum or minimum. Mm-Hmm. So we need to check. So you don't need to check that because you. I guess you could check that you can find out what that antibodies. You don't really want to do that because you have to differentiate again.
But it is kind of there anyway because you got this inequality. So you're saying a cubic. Function has to be greater than or equal to zero. And so I think the easiest way to do that is to draw the graph because I think this things going to about something like this. As a function of Z. And it has it has one negative ruin, has two positive reviews. And this this thing here is, I think it's actually z squared, the dot squared because you have to multiply that z squared.
Somewhere along the way, the one that's when. But this thing got to be positive. And so you must have to be in between those two places to try to zero because you can't you can't possibly go through to zero because then that got spread would be negative, which doesn't make any sense. So I think if you if you draw a little graph and you know, that's clear enough to say that they've got it, you've got to be between those two points.
Does that make sense? Yeah. So this thing? So actually, what is one of these things must be right? I guess it's. This one. Yeah. So it was. A. So you noticed I think that verdict was as a solution of this. Yeah. And sometimes you can forget that the fact is that there is a solution is clear from the initial condition because you set it off going horizontally. And that's quite useful. But if it's a big of otherwise, it might be quite difficult to find all the routes.
So what was the rest of it? It was. Remind us. And then. There's a minus Z Times, two G and then plus B squared one minus h spread on z squared. So that's two g. That squared a minus, put the whole thing on top of that squared. Plus, B squared that squared minus h squared. This is. Two gentlemen subsequent. And so that means that to that square that squared is. A-minus is average practise out of everything. Times two g.
That's Fred. And then we factored a minus that we got minus v squared and plus the Z, I think, is the same as what you got faced. Yeah, looks like a. And so this thing is definitely positive, so this thing has always got to be positive. And this is a negative Chebet, so that's what I've drawn there, and it's got three solutions because it equals a. And the other two, which is there's a sequel to the same nice.
Yeah, it's just a funny expression. Yeah, that's the plus or minus Mr. President B squared plus h. That's just the quadratic formula for the solution to those. And then one of those that's bigger than v squared. So that's the minus one is definitely negative. So the minus one is definitely that route. And then the other one is the positive route. So that equals items that equals that one with the plus sign of those two, which is not obvious which one's which.
Yeah, that depends on if you make v big enough, this one will be big enough. You might be small enough. This one will be close to zero. So these two reach. And so that means that when you when you take this thing round the circle, it's not dependent on how fast you flip it either. It's going to start going up and go to a higher height and then back down to again and it's going to do this kind of thing.
Or if you flip it slowly enough, it's a it's going to be its highest height and it's going to go down. Hmm. And then come back up again. It's going to do this kind of thing. That make sense. Yeah, makes sense. And I guess if they're the same when V is equal to. The greatest guy. Then there's two things at the same. So these collide when you scratch and you. What would happen in that case, what what would keep the same height?
Yeah, we just go around so it would just go around in a circle, so you'd have chosen exactly the right speed so that it can just do second. It doesn't have to come down to. OK, good. So that's the kind of question comes up a lot in the exam since it's quite fun. Well, it is not fun. OK, so then the next question was about this. Thing, so, yeah, that's fun. That's really what this thing that's going to roll off the sphere.
And you just have to come up with some reasonable explanation as to why it leaves on the gates of [INAUDIBLE]. Kind of obvious. Yeah. Yeah, it's going to be so difficult putting into words that it's one of those things. It's really difficult to explain. I mean, it's possible that what they want to do to get out here is about conservation of angular momentum. Again, in the sense that if it if it starts rolling off down the sphere, it never has any angular momentum and angry momentum is conserved.
So if it starts off with zero angular momentum, then it can never get any because there is no forces going around. But it was also fine just to argue that whichever way it goes, it's going to be confined to that plane. So if you if you do find your X coordinate to be the way that it starts rolling, then you've only ever got any forces in the X directly on the Z direction.
So why is it ever going to move anywhere else? And if it's confined to a plane and a sphere, then a great circle, then it's a great circle. The the definition. So I bet, anyway, both of your answers that I thought it sounded fine to explain that. And then you had to find these equations. So this was good.
So here, Kate, I thought. It was better to try and use vectors rather than you read everything out in Cartesian components, which kind of works, but usually with these kinds of questions, it's better to try and write things in terms of the vectors and in this case, are any feature very useful. And part of the reason for that is that the end end is pointing perpendicular to the sphere. So when your thing has fallen to here, you know that the end is going to be going that way.
And if I do find it all to be going that way. And I think it said it said it when it ceases to be that hangover. And so we are. Actually, this one's not totally standard, which way the coordinates are so er, in this case, what's going to be fine, Peter? There are constant. And then, Peter, if it comes this way. It's going to be costly to a minus sign contracts for particular.
But I think it's kind of better to work in terms of those, and then you and then you can say that the position of that is all is equal to actually because, you know, it's on the on the sphere. So it's just a PR. And then it's kind of funny, nice and then and then you can say, well, I've got a constant, so that's just a lot. And if you do er, you get. Featured not to. A you're going to sign, right, because it's not complete, yeah, this is not the normal.
Normally you have air going that way and easy to go in this way, but also features different. So if if you, regardless of which we put, would you always get the same derivatives? So like, we all don't go too easy to yeah, but do that if I might see to go that way in this case, I would have got air dotted minus seatbelt, etc. So to see I could always have defined easy to to be going that way.
I mean, it wouldn't really make sense to do that because it would not be in the direction of increasing feature. But you might do that and it wouldn't wouldn't go wrong. We just have to we just have to keep trying to get it. Yeah, I mean, that's one of the reasons why is that stupid ideas sometimes to write these down? Yeah. Just because you can check so I can I can do it in my head.
If I differentiate that with respect to time, I'm going to get costly to feature that I get minus ninety two them. So I know that I get $50 a feature. But if you if you're not, if you're using normal circuit court news, then you should just remember that that's the case. I mean, it's quite straightforward to derive their. And then and then you are double dot in this case. And the derivative of that, so you get a seat in a double dot theatre and then you get seated the citadel,
which is always going to then be minus $50 E.R. So they give you minus a feature dot squared off, which is what you've got here. Yeah, it's just it's written that in more competition that there isn't part of the reason why you do that is then your niche and secondary is going to be more double dot is minus energy. Let me it this time. OK, right that way. Plus end. But you know, the end is in the air direction, so you can immediately write it and scale it.
And you are. Because then when you take components, this is where I say you're taking components, you're ready to drop product and you can do products even if you have not written it all out in the coordinate system. So I've written all doubled up in terms of feature in air and gravity is written in terms of. So you might say, well, I need you rewrite K in terms of any feature to be consistent, but you don't really need to do that. You can just say, Well, I'm just going to take the dot product.
So I come and say, let's take the product with the feature and I'm going to get a of double. Is equal to minus energy. And then I got Katie Keita to feature resign feature, so this is just Angie Sign Theatre and I had plus and he hardly featured of decision-makers. That's nothing tequila. So that gives you that aggression. And then you take the air component, the other one.
But you had to do sort of cross multiply and at the end that if you if you've got your coordinates and directions which are suitable for doing that, you mentioned so this this gives us give us minus a feature dot squared is minus energy k ah, plus NW. And that is. Minus energy cost and. So that gives you a featured object, is Jehovah a science teacher? You know, that's the equation of the pendulum. Yeah, and. If you don't depend on them, and then I just. It's like this upside down.
Yeah, that's a good thing. Minus a heated up squared. So if you only cared about her future changes and you wouldn't worry about this situation because all this equation touches were you. But if you want to know when's it going to leave the surface, then obviously you live in that equation. Right. And then for the next fight you, I think you both said this use energy conservation. Yeah. So this is again, I kind of think it's a bit.
There's so much that you involved in too many things, if you now say I'm going to have energy conservation because you've already really got that here. So I would prefer to at this point say, Well, I've got these two requirements. I just need to solve them because this one really gives you energy conservation if you multiply that one by feature, just kind of the same idea. If I take this and multiply by future thought.
And it will give me conservation of energy. Procedures, lots, effectively a velocity again. So this gives me the vitality of the whole future dot squared is equal to Dubai Duty of. So this is the same treaty, because this is actually the same thing, and that will always give you energy. Well, in fact, the reason why it is sometimes it's helpful to do is because sometimes you might not be sure whether energy is conserved or not.
And this should tell you in a very quick check. Well, it might not be that quick. So let's hope that this equation is more complicated. Let's take a look back to make it work. In this case, it was finally to say I know a news concert because of the Phoenix move, but it seems to me it feels a little like you're invoking more ingredients than you need because you've already really got it here, too.
So this doesn't stop the half featured dot squared plus g over a cost feature is a constant, and we know that when we start, we start at the top. The stage doesn't zero and cost each one. So that's a bet. You just have to take that and stick to that feature dot squared. There's two g of a one minus CONSTATER. And then there was that of a. When you plug that into there and it tells you what. And. And.
It doesn't tell you this doesn't tell you how long it's taken. So if you wanted to know how long did it take for it to get to that, then you would need to actually solve this equation, which we have not quite done here. We found a relationship between Peter, Dutton said, but we've not found what Peter is as a function of time yet. Right, so and then it falls off when end is. Zero, when apparently goes negative. What happens after that? For really?
Yeah. So what what kind of path will follow? Falls off. Did you do this? What did you do to him? And two or three or something? I did. I do not remember. So we found that it happens at two thirds of Z, apparently, so two thirds of A. And I think we're measured from here. So. So. Two thirds of age is not here. So it of. And then it will fall. And this so which forces are acting on it at that point? Gravity, just grab just gravity. Yeah. So you just have A. Minus energy.
And so it will follow a parabola. Because of the X component of its velocity will stay constant. And the Y component of the Z component will be increasing in time. So at least in principle, you could kind of work out where it will hit the ground. By working out what its velocity is, what effect we know what it's velocity is when it leaves that, but we can work that out from feet apart. It tells us what it's lost in in the feature direction. And then you could carry on.
What I would have to figure out really upset about me, but in principle, it's quite straightforward. Once it's not on the circle, then it would be a bit crazy to use second coordinates. So if you actually wanted to do that, then you're best to go back to Cartesian coordinates again and just work in terms of action X and Z. But if you're writing everything in terms of age and things, that's that's not so complicated. You can just switch between them if you want to. And.
If the motion was on like an ellipse, Lloyd, would you also use cylindrical coordinates if it was something ellipse or a similar round object? That's an interesting question. I might do. Yeah, I guess it's different. It's an irony feature, I'm not going to be so useful if it's deployed, because because the point of those here is the air is perpendicular to the surface and the feature is tangent.
And if the thing was elliptical, then it's sort of equivalent of E and E feature are not going to be they're not going to be perpendicular to the surface. So if you imagine that, it's kind of like that. And then the perpendicular direction. Coming from some origin. So I think that that's not going to be very useful. So in that case, would we just imagine that it might be better to go in Cartesian?
Yeah, what you might want to do is write a lot of stuff in terms of the normal tangent at each point. And you might parameterised those in terms of some feature. Remember, you can write the equation of the lips as like X is equal to a cost to Y is equal to being signed into parameters the surface of the thing in terms of feature.
And she might be useful to use some kind of feature, but then the normal in the tangent direction she would want to write in terms of what you have to wear, what they are and then you've use those to say, Well, I'm going to write down, I'm just going to do this because this is true regardless of what coordinate system we then write terms of.
And then we say, let's take a let's take a component of that in the normal direction, and that will tell us what enters and competitive in the general election. But the normal and attendant directions are just not going to be given by these four minutes. I can think about something else. And they're going to change in a different way with position. So I think, yeah, I think it would be helpful if you're constrained to something you want to find,
what's a nice way of parameters being on that surface. So if you know that you're on the lips, then using some kind of feature is going to be good because there's a single variable that tells you where you are on it in the same way that seek to here tells us where we are in this great circle. I've never seen a question asked, but I mean, in principle, but it's the same theoretically, if the we tried to do this before with my globe.
What's happened to my ball? The ball, this big model, it would fall off a bit and suddenly that the where do you think we've got these people filming us for a Range Rover and look at the slow motion? To me, it doesn't look like it falls off the field after two sets of the height. Try and catch it. And also, how about here? I figured for I was quite surprised that it was too layoffs.
But the one thing that I did think is that if you if you account for friction and I think you can argue that it must be a bit further down here, can you see why I can say that? Yes, it's going to lose it some of its energy, so it will have less energy in the horizontal direction. So what do you mean by energy in the horizontal direction as they will have less velocity in the horizontal direction? So it's going to stay on for longer? I think that's sort of true, but what does that mean?
It turns down for a longer. You mean, because because it's also going to have less energy and it's going to have less Palestinian actions. You're thinking because it has a feels that way, its acceleration due to gravity in the vertical direction. It's like decelerating only. Yeah. I don't care. Hmm. I think that's true. I always think it's a way right away with the friction force would come into anything that we wrote down here. In the. Yeah, so I have another force, which is.
In a theatre director, yeah, obviously frictional force. And so what difference would it make to this? The energy conservation. What could we say about the kinetic energy? And we know that was a friction force. How would it compare with this expression? I would feel it would be less if if if there were some friction than the energy at some later time must be less than what we started with because we had to do some work against the friction.
So it would mean that featured dot squared is less than what we get from that expression. And so that would mean. So when you come back and stick that into here, it would mean that. And it's going to be. Bigger than what we got from here. For any given theatre. And so end would be bigger than what it is in the friction this case awaits. And so presumably that means we have to go for longer before and resistance to zero zero.
So I think you I think you could you can see that from from the fact that you're having to subtract off the kind of energy in that. So maybe that's partly why it doesn't look like it leaves quite so quickly. OK. Yeah, the main thing that was that I think it might be trying to use irony feature as much as you can, even if you got a bit inventive in making up what is the best government. OK, but then the third one is with this thing on the wire.
And so you made a you made a classical mistake here. Everyone makes. Which is to say the end has to be perpendicular to our dot. And in this particular question, that's not true of the topics of the rotation. Yeah. So we've got these beads sitting on this parabolic wire and the whole thing's rotating. So which direction is don't? This thing, that permission. It's time to put a. So, yeah, well, well, end is moving around as well.
Yeah, but the component in this plane is tangent to the parabola, but the whole thing is is rotating. So are doctors actually pointing some? We've got some component along the tender. It's also got some component into the page. And and that's and the normal in this case is all we know is that it's normal to the wire. We don't know that it's normal to a plane like the one you're on a cone and you know that this normal forces tangent to the plane.
In this case, it's a wire. So all we know is that the force is perpendicular to the wire, but we don't. It could be at a point in any direction around the wire and it could. In fact, it has to have some component into the page as well. And so the DOT product with our DOT in this case is not zero. Which then means that energy is not conserved, because that's what we saw on this stuff. And because of the work associated with then Dot got to zero.
And that's the essence why you ended up with a minus sign. It didn't work. So your expression here is conservation of energy. And where's the expression you're aiming for? It's actually not conservation of energy. It's something that looks almost like it, except that it's not quite the same. And that's because in this case, we're forcing the things that we're forcing this way to go round and round around. And so we're actually we're doing work on the particle in this case.
So you avoided this because you made a new tangent, correct? Yeah. So the issue here was that so end, it's quite hard to draw this in this if you're like me, you're going to be drawn to the site and is going in some direction, which is great. It got a component into the page. So what do you know about RN? In this case, the only thing you know is the end is perpendicular to the wire. So I would write that is and T is equal to zero, where T? T is the tangent. To the wire.
So you don't know if it was they or in this case, either. In fact, it is not. There is a component in that way. And so you have you have to work out something to do with attendant victim for the wire. Which, you know, is what was the formula for this? That equals. Two today. So do you know how to work out a tangent to that jury to get to the tension will be that is proportional to one and PR, which is all of a. In terms of arms, so then I want to write that is the R plus R of a z.
Because we're going to dock with this. We're not we don't really need to make a unit normal and unattended. We just need something which is in the right direction. So how about air? I always think it's useful to draw these diagrams and label everything. As much as you can. Confused about the angles. So this is exactly the right coordinate system to use, and then this stuff is all fine so that you not a movie that fun.
And to this point, I think you need to say I'm going to start with a tangent and these questions is always just a case of working out what you want to start with. Get this to work. And so if you do that, so let's just do that because I want to check that we got the right equation. So with the end double dot dotted with that tangent is minus energy. Aged. Plus, NN Typekit in attendance and AZT is just minus energy of a. And up to zero. That's very rare.
And this thing is that's the are dotted with the all component of that. So it's going to be M Dot minus omega squared from that thing. And then it said the Z component, so it's going to be plus over a set of about. And then you had to do a little bit of work to work out where all the that were related to each other, which I think we could decide which one you're going to get it in terms of. But I think if we do, all of that then gives us that thing. And it's an it's a bit weird.
The only difference is that sign. But there's that sort of makes sense, especially because when you start with the. So it's minus and plus, and if I actually did my energy kind of argument here and said, let's do it with our dollars. Then we would get half an hour dot squared differentiated. That's what I get from Dot Double Dot. Is equal to and then this time, which still be minus Z, so we'd still get minus d by D, T and GZ. But then I would have plus and dot dot there.
Which is no longer there. And that climate is not there, and in fact, you can kind of work out what it is if you relate to this equation and. So this one is. Not zero in this case. OK, so you then did what you should do, which is said, let's just ignore the fact that the wrong equation and then go with the equation.
Oh yeah, you just you were using dots when I just think it just gets a bit confusing if you start putting dots in there and there's so many dots in this, so they just write numbers next to each other that media notification that sort of. Or write letters next to each other. So then we had this thing about stability. So first of all, do you know where this equation comes from, the Taylor expansion, the Taylor expansion of the previous regime?
This one? Can you show me how that works? It comes from. Retailer expanding the equation for all. Right. Yeah. The reason I'm asking is because it's not totally that like, it's not the standard for you, see these things and you're probably used to seeing something like. But we've got this issue of, Ah. Yeah. And then so what would be the definition of an equilibrium point in that case? When? That is zero one zero.
And always a concern. So if we call it a an equilibrium point is just a constant solution of the equation, the R equals a. It's just any point that satisfies if there is zero and then if you want to analyse the stability, what do you say? We've seen that be a plus a small perturbation, and for some reason we always call a small patch, pessimistic sign, even though your hate writing something is very difficult to write.
I think it's easy. You just write a script of it, but you assume that size small that the whole point of. That's why I called linearity and you plug that in. And so then you get because it's a constant, you get excitable f a value plus sign and then you take the expand that and that gets you f of a which is zero by definition of a. Plus, a prime time site and then you ignore the higher order terms. The terms and so you just end that risk side, everybody is primed of a sigh.
And then that's an equation that has. Either trigonometric or exponential of this, and I made a comment here that I think you need to spend a little bit more about why the sign of this time then determines the stability of. So it's unstable for the exponential solution, because, yeah, why, what occurs exponentially? Yeah, I mean, the question is just if I make a small perturbation like this, does oxide grow or not? And so it's unstable if the solutions are exponentially growing.
And for this particular equation, you always get solutions which either trigonometric or exponential. If they're exponential, one of the exponential might decay the other one. If it's this equation you factor in, one of them will decay. But one of them will grow. And if there's any component of the solution which grows, then it means it's unstable. So I think you should always right when you get that kind of thing. So you had side double dot.
Is that a squared minus J? Or do we know we lost the H omega squared minus G of XIV? What you should write is just the omega squared is bigger than grey means that term is positive. That means my solutions are trigonometric. That means that exponential that I would say exponential solutions. And then that means it's unstable. But I would write that rather than just saying that means it's unstable, that that gives me no explanation as to why.
If you just say this is bigger than this means it's unstable, I don't understand why. And. Whereas if I'm a square, there's less than a I'd get tricked solutions. So Clyde is also right. So that means you're right, perturbed by a small amount, it will just stay close by.
And so that means it's it's stable. OK. But I just drove back to this, so we didn't really quite had this, we had something that is actually it looks a little bit more complicated because we've got all of this kind of funny stuff on the left hand side is equal to this. So there isn't it's not really obvious what is. In fact, this is not an equation of that form you because you can rearrange it, but you still have double but as a function of all and dots squared.
So it's more complicated, but still you're doing the same, you're basically doing the same thing. And so in this case, we said R equals zero. Is an equilibrium. And then so we say that I. There's going to be small. There's a small perturbation from zero, and then that means that you can ignore anything which is more than linear in sight. So you're putting our equals sign. You almost could just leave it at all, but you can just then ignore anything which is. Squared, Ohio squared or.
And in fact, there isn't anything squared, but with two cubic times you can all squared off double dot and you've got R dot squared. And you're always when you say when you take science, where you always assuming it's derivatives are small as well. So then so you then you say a linear rise, meaning not just keep. Turns up to what extent, and then so you just end up with a squared side of adult age and scratch them with the scrap minus sign, which is.
Why can we assume that it's derivatives also as well? So it's an assumption that you would then you then check. OK, so you were as long as PSI itself had sufficiently small? Then the solutions of this at for small time will be small and their derivatives will also be small. So you do this, find the solution and then you can check that it's consistent to say derivatives of that solution were small and therefore you were OK to assume that they were small.
The first was like a consistent. It's a bit like they often do that when we're solving differential equations where we don't, we don't deduce the solution, we say, let me guess the solution and then I find that it works. And then I rely upon someone else who's proven that there is any competition that allows me to do that. And. For discussing the stability, we also talk about the case where they are equal. I think you don't need to talk about that.
I don't think you ever expected to damage that because because it because you then need to look at the terms. So that then becomes a non-linear calculation for you. We're not at that level. So you just do the two cases. I decide and don't worry about the intermediate one. Well, what we say, what happens in this case, if it's unstable? So that means it's rotating fast enough and it is big enough that that means the thing's rotating fast enough for what would happen.
Beats me up. Yeah, we keep going up for a while. Yeah, yeah, it's because there's actually only that's the only equilibrium solution here. If it's unstable, then it's just going to keep going out and then we get faster and faster and get higher and higher. And that's kind of showing you that the energy can't be conserved because because actually the limit of this is that the thing would go off to infinity. And that's because you're pumping energy into the system all the time.
And that's that's associated with the fact that this energy got this component in the direction of things moving. And so you definitely we're not going to get that here with the fact that energy concert means it can't get it, can't get that high. OK, so we've done all of that, so I just spend a few minutes on the PDF, so. The first question I was driving away you, Cleveland, and I think I just wanted to say something about that because I mean,
it was about this dimension, so this bit was not quite like it. You you. I think you just forgot to get rid of that when you differentiated at the time. So because this this equation would certainly alarmed me because you've got all these components of generation and you've got an isolated one in the other. So you were fine to do that, but it would be an issue that shouldn't, shouldn't be there. But yeah, this non dimensional ization that you had to do was lost. The. Senator Richard.
These alpha and beta that you had to come up with, which the non dimensional equivalence of gravity and areas and it says what, what the conditions under which there are negligible. And so what do you want to use to say there was an alpha and beta would need to be small, then you could ignore them. So did you have to go into detail about the physical? Implications of alpha and beta things won't fly.
No, no, it just wants to just be careful because you then sort of try to convert that back into saying something else. And this is precisely why you do it on demonstration because you've now written something. It doesn't make sense because this is an acceleration and this is a speed and this has got some other funny units.
So, yeah, Gamera, but it doesn't make sense to and that celebration is less than the speed because I mean, it's like saying it's like it's like saying I weigh less than your height. It doesn't make any sense. So so what you actually mean there is to say gravity is small, and this alpha is kind of telling you what that what it would mean to say that gravity is small because you're comparing gravity to something else. So in this case, you compare it to some other force, which is due the tension.
So how that happens could be in a number of different ways, but no alpha versus alpha is. G L squared, h. C. C squared, but that c c where this is T on the road, rogue l spread over h t is just sort of the dimension this measure of how much gravity so you could you could have gravity not being important because the string is very short, having will be very small. Or you could have it not be important because tension is really large, set up already a big long string.
It might weigh quite a lot. But if I if I make the tension large enough, I pull it really taught, then gravity becomes irrelevant, even though it might wake the rogue element be quite big. That's the way to the string, but the potential is big enough. That doesn't matter for the way that the equation works. So this is kind of telling you how to how to say that gravity's. Not important.
I've just completely take it all from when you lost the gamma, but then I and how that could work because I don't think any government that. But actually of this, this doesn't work. So karma is whatever makes the work. How did you work out the dimensions of government? So they said that. This is force per unit length. I worked out the dimensions of that and then we know what the dimension of that is. So then we must have that for the dimension of coming back to you.
This has got to be a force that can work out what needs to be. Does that make sense for you? Yeah. And yeah, I mean, even if it not said that, you can look at this equation and say, Well, I know I'm putting this 10 together at this time. So these terms must have the same dimensions, so at least have the same dimensions as argue about changed. So it's just rogue. So if you don't know the units of something, that would be kind of how you can work them out.
He's been making everything work. Then I think, yes, are you plotting the wrong this is dropped, so if you are told that if it is zero for Model X? Bigger than L. It doesn't matter what F is that for. Between Al and minus L. It's definitely zero. Then what can you say about motive of X minus C T? You know, this was all originally just translated it, and then I think parents convinced otherwise.
I think it was because of this thing. They say, you know that f of X, Y and Z, that just means put in its mind to just translate into F. So you know that this is going to be zero for X minus c t bigger than that. So this is the argument to the functions of it goes in there. And so when you plot this as a function of X, then it's the region between C T, C T plus L and C T minus L is the region where it might not be zero.
And it's got to be zero everywhere else. Yeah. And so that question you had this thing, which was this funny shape. That fits in that. Yeah, I can see you up to that. Have you changed your mind? But yeah, sometimes these things, you just have to be almost like a computer. You've got to say, well, f is a function which takes this argument. So if I put it in a different argument, I've just got to replace that argument everywhere in my expression and to be really systematic about.
Right. I think that. You may be out of time. And the last person was, OK, I made some comments about food and part of it. And we calculated the criticism. Not quite right. Yeah, but I'm not too fussed about going to each of those wrong and. Oh, yeah, there was just this last part, so let's say it said it, so can you break down this function into two functions? Can you tell me what these functions are and you found what they were in terms of some very attractive issues?
Can you tell me what those functions are? What what was the definition of the end? In terms of a. Very long question. It was a long questionnaire. Did you remember where the three of coefficients came from? And the thing? OK. So H and XS iPhone X, I think it's almost almost effortless, whereas f of X defined. For X within L minus of or something. This your lessons, just as much as we we're only ever given anything on this fixed domain between zero and L.
And so. At this point when you've tried to apply the initial conditions to the problem, you said, I need to try and find some and such that F of X is the sum of the aims sign. And do you thinking, is that going to say, well, I know because we were taught this at the start of term.
I know how to choose and such that that is the case. Those aliens are going to be the free sign coefficients of F. And then you do the same thing with this one, you say, Well, I need the derivative to be G and the derivative is this.
So I need to choose these coefficients to be the free assign coefficients of the G. And the function that you're then creating when you do the some of the and scientists and an extension of the F. Because you made it periodic, which extension is if you make, it seems there is. I thought extension, which is the old extension of that, that's a you're on minus Al Jazeera, you flipped it over.
So these functions that you ended up with at the end is HMX and you have there for h of X is the odd extension of F and U events is. The odd extension of G. They're related to the initial conditions, and then that actually kind of goes back to the question to that because the way that you can think of these things is that the initial shape splits in two and then moves sideways.
And in this case, we have boundaries, but the boundaries are kind of reflecting and they're allowing the same wave to come back in again upside down. And so this is saying, actually, you can think of the solution that you end up with by this method is kind of in the same framework as the solution that you had here. Even though it looks very different. OK, and that's enough. So next week, we might concentrate a bit more on this stuff and not the dynamic. OK.