Dynamics, Lectures 9 & 10: Oxford Mathematics 1st Year Student Lecture

[00:00:07][music] >> So, I'm going to give you another challenge problem. This is kind of a It's actually a really good problem.

[00:00:16]This is actually a problem posed by Charles Dodgson or Lewis Carroll of Alice in Wonderland, who was, if you didn't know, a math tutor at Christ Church once upon a time.

[00:00:28]So, the idea is I have a rope going around a pulley, frictionless pulley, and on one end of the rope I've got a mass, whatever, say it's 100 100 kg mass pulled down by gravity.

[00:00:43]And on the other end of the rope, this is going to stretch my artistic abilities, but there's a monkey. And the monkey is 100 kg.

[00:00:50]Oh.

[00:00:54]Okay.

[00:00:57]That's sort of Okay.

[00:01:08]So, this is a constrained system.

[00:01:09]There's a tension in the rope, which is uh connecting these two objects that are the exact same mass.

[00:01:18]All right. So, then what uh what Dodgson suggested is that the monkey starts to climb the rope. And then the question is, what happens?

[00:01:53]Okay.

[00:01:58]I will let you I'll let that just sit there, let you think about that.

[00:02:02]It is a non-trivial uh question, and it kind of depends on how exactly what we mean by, you know, how exactly the climbing is happening and so on, but um yeah, let's leave it there.

[00:02:15]This is what we were doing last time.

[00:02:18]Pendulum, we were talking about constrained systems. Here's a pendulum on the end of a massless rod.

[00:02:24]Uh we worked out the Newton's second law has the radial form, which is a second-order equation for theta, the angle with the vertical, and oops, that's not a zero. That should be a T.

[00:02:38]The e theta version of the equation um looked like this. T is the tension in the rod, right? The the object is pulling on the rod, and the rod is pulling back with a tension T, which must be positive, otherwise it's not pulling.

[00:02:53]And what we said is that point is I don't know T. So, really you can view it as I solved the first equation for the motion theta, and once I know theta, then I can work out what the tension is at any given moment.

[00:03:05]Okay?

[00:03:06]Um so, there's actually a lot more kind of nice calculations one can do with this system. So, one one calculation is we can we can look at equilibrium and stability, and this is kind of uh I guess obvious and intuitive, there are two equilibria. So, theta e equals zero and pi are both equilibrium points because the the sign goes away, so I get no forces, theta dot dot equals zero.

[00:03:45]>> [clears throat] >> And >> [clears throat] >> I won't put all the details on the board cuz it's effectively very standard exercise with the tools we developed about a week ago. If you do linear stability So theta equals theta e plus some perturbed variable psi what you find is that zero is stable with a frequency frequency of oscillation which is 2 pi Sorry. Sorry. Omega equals square root of g over l which is which is quite almost immediate because I'm doing linear stability. That sine theta linearizes to a theta and that's simple harmonic motion with with a root g over l.

[00:04:58]That means that the period of oscillation which you might call t is just 2 pi over omega.

[00:05:11]So that's 2 pi root l on g.

[00:05:16]And that's the bottom. So that means if I give it a small oscillation from being happily sitting at the bottom, that's the amount of time it should take to do one cycle.

[00:05:26]Theta e Theta equals pi is also an equilibrium that's sitting at the very top and it won't surprise you that that one is unstable and if you do the calculation, you confirm that. Theta e equals pi is unstable.

[00:05:45]If I give it a little nudge from sitting right at the top, it doesn't come back, right?

[00:05:51]Okay. Um Let's talk about energy in this context.

[00:05:56]Find a better pen.

[00:06:09]So, E 1/2 M We've argued last time that a constraint like that does not uh do any work, so it doesn't impact our energy, so we do have the kinetic energy plus potential energy is going to be a constant, and here V, the only potential energy is gravitational.

[00:06:37]So, V which is like the Z coordinate uh which in that coordinate system mgz which is minus mgL cosine theta.

[00:06:54]The way we've oriented that and R R dot, as we've worked out before, since the radius stays fixed is L theta dot E theta.

[00:07:09]So, therefore conservation of energy looks like uh M over 2 L theta dot squared plus um Sorry, it is minus mgL L cosine theta equals E constant.

[00:07:49]Since that's not changing, I can work out what the constant is by taking any time point. Most naturally, I set E from T = 0.

[00:08:01]So, E whatever initial angle and theta dot I give at T = 0, E is going to be M L squared theta dot of 0 squared over 2 - M G L cosine theta 0.

[00:08:25]Sorry, theta of 0.

[00:08:36]So, one kind of thing we could do with that expression is ask Think about the following thought experiment. So, if I start at some angle at theta naught and let's say what I want to do I'm at a I'm at the playground and I want to swing the swing all the way around. So, how hard do I have to push?

[00:09:02]What theta dot of 0 do I need to give?

[00:09:05]Just different pushing to get this thing to swing all the way to the top.

[00:09:16]So, question.

[00:09:19]Uh >> Well, so if I'm going to reach that point, then of course this is the point corresponding to theta equals pi.

[00:09:59]And what do I need to be true? I need to have reached that point uh and still have theta dot non-zero.

[00:10:09]Right?

[00:10:13]So, the pendulum stops when theta dot equals zero.

[00:10:34]And what we need is >> [clears throat] >> the energy to be high enough so that I get all the way to theta equals pi when for that when that happens.

[00:10:46]Right?

[00:10:49]So, if I put in theta equals pi, that's like the statement that minus m g l l okay, theta stops. Sorry. The pendulum stops when theta dot is zero, which is the which is when mgl cosine theta equals e.

[00:11:07]So, we require I put in pi here. I need e uh to be at least mgl.

[00:11:17]Right? So, if e is greater than m g l, then our pendulum makes it all the way to the top and still has a little bit of energy.

[00:11:32]Yeah.

[00:11:38]In fact, if E is greater than MGL, I don't just make it to the top, I swing a little bit over the top.

[00:11:51]What happens next?

[00:11:53]I've given this thing a big push, it's made it all the way to the top.

[00:11:57]Then what?

[00:12:04]Okay, you want to think it just keeps going, right? Given enough energy. A way to see that is uh think about like an energy landscape.

[00:12:14]So, >> [clears throat] >> if I if I plot V of theta, V is just a minus cosine, right?

[00:12:28]So, V is looking like that, and this maximum point is MGL.

[00:12:51]And I guess what I'm doing in the experiment is starting I'm starting here, and if I just let go, this is our usual picture, ball rolling in a potential field, it's going to do that, but I'm starting here and giving it some velocity so that it gets past here, and we see that if my total energy is greater than MGL, then every time it gets up to here, it still has some theta dot zero, which means it's just going to continue flipping over and over, cuz there's no energy being lost, right?

[00:13:27]So, my total energy is sitting somewhere above there, then this is the experiment where I started with some velocity is the same as the experiment where I'm up here with a little bit of velocity.

[00:13:36]And it's just going to keep keep doing this loop.

[00:13:45]Mhm.

[00:13:47]Um When we do that calculation up there, the linear stability, we get a period, right? From the simple harmonic motion.

[00:14:00]But, that's linear stability, which means that's an approximation. I've I've linearized the sign.

[00:14:07]What if I want to know the actual period? There's sort of the difference between an actual pendulum pendulum, which has the sine theta, and the sort of pendulum we think about that you've done, say, in A-level physics, which which we've already linearized uh to simple harmonic motion. If I want to work out the actual period, I can.

[00:14:38]In other words, if I don't linearize, I keep the full theta dot dot plus the g over l sine theta, what period, how long does it take?

[00:14:47]I'm imagining I'm imagining I start somewhere without any velocity, so the the pendulum is going back and forth in my potential field. How long does it take to get one cycle if I have the sine theta there?

[00:15:03]So, we can let theta be theta zero.

[00:15:09]Let's take theta dot zero to be zero.

[00:15:16]Uh all right, theta of zero.

[00:15:21]Theta zero, theta dot of zero, So, that means that our total energy in this system is >> [clears throat] >> just the minus mgl.

[00:15:40]And the key >> All the way down.

[00:15:48]>> Okay, I don't know how to solve that equation and it's second order. I'm a bit stuck looking at that equation.

[00:15:55]But, uh this equation has time living in the d theta dt and I can we we did this before generically. I can extract time by thinking about integrating that equation implicitly with theta and t.

[00:16:14]Right? So, theta dot squared if I plug in that value of E and solve this energy equation for theta dot, I get theta dot squared is 2 E >> [clears throat] >> over ml squared and plus a 2 g over L cosine theta.

[00:16:43]Right? That's d theta dt.

[00:16:46]So, if I take a square root and divide this is independent of t, so I can divide this whole thing below the d theta and end up with a dt sitting on its own. So, that's a time and I end up with an ugly integral.

[00:17:01]So, think of that as d theta dt squared.

[00:17:08]Put the take a square root, put the dt on the right-hand side and we get an expression t equals And there's a plus or minus, which is really the time irreversibility of the system.

[00:17:20]Plus or minus the integral Let's view it as a um indefinite [clears throat] integral for the moment of this rather ugly square root in the denominator 2e over m L squared plus 2g Okay, that's not very nice. If I could just integrate that, then I would have an expression that tells me the time as a function of theta and if I can invert that, I would get the total path theta of t.

[00:18:07]Right?

[00:18:08]Um If I'm just interested in computing the period then I can integrate over a specific range of theta, which is theta Well, if I'm doing the period um >> [clears throat] >> Go here.

[00:18:38]Starting here, it's going to there.

[00:18:42]It's going to come back.

[00:18:44]Notice that the period is There's a symmetry there in the form of that ener- energy potential.

[00:18:50]So, theta minus theta naught's going to go to theta naught and so on.

[00:18:54]So, I could write the period as the integral from minus theta naught to theta naught or from theta naught to minus theta naught.

[00:19:04]Uh Oh, sorry. That's That's halfway, right?

[00:19:08]So, theta naught to minus theta naught and then it goes back to theta naught.

[00:19:14]So, I can write the period as two times the integral from theta naught to minus theta naught of that monster. I don't want to write it again.

[00:19:27]Or I can use the symmetry even further, which is to say that because this thing is perfectly symmetric about theta equals zero, however long it takes me to get to zero, it's going to take me the same amount of time to get to theta naught. And you can argue on the symmetry of V and what theta dot is doing there.

[00:19:46]So, that's the same as four times the integral of theta naught to zero or zero to theta naught of this ugly integral.

[00:20:05]And and d theta there.

[00:20:17]>> [snorts] >> Yeah, so let's just take that one step further so I can My E has an has a an MGL cosine theta naught, which is going to live here.

[00:20:33]Um Cancel out the M and then you have an L over L squared, which means both of these terms have a G over L if I sitting inside a square root. So, I can pull out the root L over G, I flip it. And actually that's not surprising.

[00:20:55]Cuz that's the same term there, right?

[00:20:57]My linear stability, the scaling of my linear stability I expect to work the same. What I expect to be different is that 2 pi came from the approximation.

[00:21:07]and want to get the sort of proper coefficient there.

[00:21:11]So, if I pull out that root L over G, then I get four root L over G times the integral zero theta naught of square root of two cosine theta naught Sorry, theta. I think it works better to flip that.

[00:21:39]one half.

[00:21:45]Okay.

[00:21:46]There's probably a nice trick to integrate that. I don't know. You can put it in a computer and get that number. Depends on the theta naught.

[00:21:53]Um I'm going to leave the calculation here. The observation though is that the units, this is dimensionless, right?

[00:21:59]This is just an angle. So, the units of T are the same as the units of square root of L over G.

[00:22:13]Which is as I said, what we expect to be consistent with the linear stability. And that is a time. If you work out a length over a length per time squared square root is a time.

[00:22:25]And effectively that four times the integral sets the sets the constant.

[00:22:42]Okay.

[00:22:45]>> [clears throat] >> Right. So, I think we'll leave the pendulum problem there. So, that's kind of uh in a way [clears throat] a nice simple constraint. I'm always fixing my radius and just letting the angle vary.

[00:23:09]As I said briefly the other day, and you'll you'll play with this in a problem sheet, we have to be a little bit careful in this motion that T is staying the right sign, right?

[00:23:17]What happens if T flips sign?

[00:23:19]So, I said T, the way I've set it up, T should be positive. That's the uh the the the rod is pulling the object back towards the origin.

[00:23:30]If I solve this system and I find that T becomes negative, what would that indicate?

[00:23:40]It's pushing instead of pulling.

[00:23:42]And you can think if I'm if I'm on a swing at the playground, this is like um because it's not rigid, if it's no longer pulling me, it's pushing me, that's like the the I'm pushing it and it's going the other way, so that's like the the chain has become slack, which means that the model is probably not valid anymore.

[00:24:01]Right? Same as we said for the the snowboarder, a flipping sign of the reaction force is the condition that you leave the surface.

[00:24:11]Right, speaking about that, let's do an example for motion on a surface.

[00:24:17]Actually, I want not an example, but I want to do a sort of setup the generic system.

[00:24:27]And >> [clears throat] >> I guess we could do this for more generic force, but gravity is the natural one to think about.

[00:24:40]So, I have a I have some surface.

[00:24:48]I have my mass constrained to be sitting on that surface. And let's say gravity is pulling it down.

[00:24:57]Call that vector G down.

[00:25:00]Normal reaction force pointing that way.

[00:25:08]And I didn't say it, but I mean a frictionless or smooth surface. So, there's no, again, no force in the plane of the surface. There's no loss of energy, no friction.

[00:25:18]So, actually, let's just write it here.

[00:25:19]So, a smooth surface implies that the constraint force n >> [clears throat] >> is normal to the surface.

[00:25:56]And the one way to express this condition that I need to keep track that n doesn't flip is that it you can see it in the picture. At any given time, n should be pointing the opposite direction from gravity.

[00:26:10]All right, if n ever points in the direction along Remember with a a dot product that's positive with gravity, then things have flipped and the surface is now grabbing instead of pushing.

[00:26:20]Right?

[00:26:21]So, I must have n dot Let's write it as vector G negative.

[00:26:33]>> [clears throat] >> Otherwise, >> [snorts] >> And again, simple way to think of that intuitively is our our particle is on top of the surface relative to gravity.

[00:26:58]Gravity's pulling the particle into the surface and the surface is pushing back the other way.

[00:27:02]Right?

[00:27:05]Right. So, by construction, that's a two-dimensional motion.

[00:27:15]Which is to say that the surface is a two-dimensional space, and as long as my particle is living in that space, then it's 2D, though embedded in 3D, right?

[00:27:27]Okay. So, depending on the surface, this can be quite messy to set up. It's a much more complicated constraint than our pendulum.

[00:27:35]But, there is a class of surface for which the equations look much nicer, and that's a surface of revolution.

[00:27:50]>> [clears throat] >> So, what we're going to do is play with the equations in the case of a surface of revolution, and really the idea is to exploit the axial symmetry.

[00:28:09]>> [clears throat] >> To simplify.

[00:28:19]Okay.

[00:28:21]So, let's draw a little picture. So, I'm going to let Z, the Z axis, be my axis of revolution. So, I'll put an X and a Y like that.

[00:28:34]And I'm going to suppose that we have a curve. I think that's not [clears throat] very good. I want it to be a bit smoother at the bottom.

[00:28:51]Okay, so I take some curve, rotate it around the z-axis, that's going to define my surface.

[00:28:59]>> [clears throat] >> So, um really x and y don't matter. It's just If I'm looking at this from above, there's a radial symmetry.

[00:29:07]And so if I define the surface by rotating the curve Let's write it as z is some function, I'm going to call it capital H of r around the z-axis.

[00:29:43]So, this um Yeah, so take a distance r or for instance, just go out in the y direction.

[00:29:54]And this is >> [clears throat] >> for a given rate any radius h of r.

[00:30:09]>> [clears throat] >> Right, so that symmetry suggests um I should exploit polar coordinates, really cylindrical coordinates for a surface of revolution.

[00:30:26]So, let's let's see what Newton looks like in cylindrical coordinates.

[00:30:44]Mhm.

[00:30:47]>> Uh we haven't done it before, but cylinder cylindrical coordinates is just polar coordinates plus a z, right?

[00:30:54]Which is Cartesian. So, that's our position vector r would be r er plus z ez.

[00:31:06]Where now we remind ourselves that that er looks the same as it did before having a cosine theta sine theta in Cartesian plus an extra z component, which is zero.

[00:31:25]In Cartesian.

[00:31:35]So, all I've done is add the z, but that's that's a fixed vector ez always pointing up, which means when I start taking derivatives, I get the same pattern taking derivatives. Here, er dot becomes e theta with a theta dot.

[00:31:48]E theta dot becomes er with a minus theta dot and so on. And then the z just becomes z dot and z dot dot. So, everything's uh very simple extension from polar coordinates, which means that's I'll just write them both. So, r dot is r dot er plus r theta dot e theta plus z dot ez.

[00:32:18]Acceleration r dot dot >> And I've used our trick where I combine the terms in the e theta to be a DDT. R squared theta dot.

[00:32:55]Right. So, if I write down Newton's second law, >> [clears throat] >> I get m times that equals the forces.

[00:33:04]I'm not going to write that whole expression again. m r dot dot Forces, so I have gravity, which we're orienting I didn't say it, but let's have gravity in the z, so minus mgz ez.

[00:33:24]And I have n.

[00:33:27]n is my normal reaction, which I don't know. All I know is that it should be pointing away from gravity and normal to the surface.

[00:33:44]Right. So, really the question is what do I do with this? Right? That the there are a couple of challenges, right? If you want to kind of list list the problems or the the aspects of this that makes it hard to solve, one is I don't know n.

[00:34:14]And two, uh I mean, I have three variables, r, theta, and z, and they're all kind of tangled up, well, at least the r's and the theta's are tangled up in this quite messy, very non-linear expression.

[00:34:33]So, it's a non-linear equation for three variables r of t theta of t, z of t. But that doesn't feel quite right because I've It's a two-dimensional problem.

[00:34:53]Right? So I shouldn't really have three distinct variables. They They're linked. Right? So part of our challenge is figuring out how uh how do I get to the properly two-dimensional problem here?

[00:35:10]Okay.

[00:35:11]So these are sort of the challenges of the system, but let's start collecting some useful observations. So the keys Well, let's take things we know about n.

[00:35:25]So we know that n >> [clears throat] >> I know n is orthogonal to the surface.

[00:35:36]Let me state that in a different way.

[00:35:38]Um r dot, the the tangent, the velocity vector is always in the surface.

[00:35:47]Right? And whatever the motion is, the tangent vector points into the surface and n is orthogonal to the surface, so I can write that as the statement that n dot r dot is zero.

[00:36:01]We'll think about how we might use that fact.

[00:36:05]Um I can say more about n for a surface of revolution. So this is always true.

[00:36:14]For a surface of revolution, I can make another statement about n which is that it's going to have no components going around the surface.

[00:36:25]Right? In other words, just geometrically, the normal to a surface of revolution does not point in the circumferential direction at all.

[00:36:35]And so, N in this notation, we can say that N dot E theta is zero by the symmetry of the surface.

[00:36:52]Okay, that's going to be quite useful.

[00:36:54]Uh three, I think anytime solving these problems, you want to always have it in your head that there's two kind of distinct approaches you can take.

[00:37:05]I can look at Newton's second equations and see how far I can get in solving them, or I can recast the problem into the energy domain, and maybe in the energy domain some nice tricks pop out like we've seen over here.

[00:37:18]As long as we have conservative system, uh no energy being lost, smooth surface, smooth constraints, I have an energy conservation going to apply. So, energy conservation >> [clears throat] >> applies. In other words, 1/2 M R dot squared plus V of R is constant.

[00:37:50]And since I only have gravity as the only conservative force in the system, V is just MGY Z.

[00:38:05]And one more useful observation to make up front.

[00:38:14]This comes down to the dimensionality.

[00:38:15]So, I say it's not It can't possibly be actually three independent variables R, r, and Z.

[00:38:21]So, where what can I use to reduce that space from three dimensions to two dimensions?

[00:38:35]Let's Any given point, the the mass is on the surface and the surface is defined by revolving this curve, whatever I called it z equals h of r, around the z axis.

[00:38:47]All right, that means z and r are not independent, they're linked by that function h.

[00:38:52]Right?

[00:38:53]So, z we can write it this way.

[00:38:56]z of t equals h of r of t at all times that the particle is on the surface.

[00:39:23]And by the way, that's why it's better to think of revolving a curve h of r around the z axis.

[00:39:29]Sometimes we construct surface of revolution by taking z equals h of y, for instance, and revolving around the axis, but that's arbitrary and it's really h of r. You get the same thing.

[00:39:41]Okay, so I have these four facts at my disposal. Let's play around.

[00:39:47]Uh maybe I'm going to leave this where we can see it and come over here.

[00:39:56]So, one thing I can do is I can take Newton's second law and I can either dot with r dot, which is a little bit messy, or this one is much easier, dot with e theta. And if I dot n two with e theta and use fact two Looking at our nice form for the e theta component, we get the same kind of thing we've had before, which is the DDT of r squared theta dot is zero.

[00:40:41]So, r squared theta dot is a constant. So, let's call that constant little h as we have before.

[00:41:00]When we saw this before, we understood that as a conservation of angular momentum. And indeed we can uh we can we can derive that in a similar way here. So, if I think about the angular momentum about So, angular momentum is r So, angular momentum about the origin, so L0 is r cross m r dot Right?

[00:41:46]And if I want the components, if I want to think about the rotation around the z axis, I take the z component of that vector, dot this thing with ez.

[00:41:58]And let's just do that. So, you get ez dot r er plus z ez cross m times R. That's Okay, a lot of terms in that cross product, but I can immediately discount most of them because I'm taking the EZ.

[00:42:37]So, anything that has an EZ um I'm going to ignore because when I dot with EZ, it's going to be zero.

[00:42:46]The only thing that survives is when I take um the ER cross an E theta cuz that's going to give me an EZ, which dots with an EZ.

[00:43:03]Yeah?

[00:43:04]So, that's uh >> [cough and clears throat] >> R M So, I get M R squared theta dot.

[00:43:18]Or from what we've just worked out, that's M times H.

[00:43:32]In other words we can say generically that if I have motion on a surface of revolution angular momentum about the symmetry axis is conserved.

[00:44:21]>> [clears throat] >> In other words, it's got the symmetry axis. If I set it sort of spinning in that around that axis, there's nothing um that energy of rotation around the axis is is not disappearing anywhere.

[00:44:42]You can understand it that way.

[00:44:46]Okay, so that's going to become quite a useful statement.

[00:44:50]Uh what next to do with this equation?

[00:44:55]So, I've used this.

[00:44:57]I can try to use this.

[00:45:01]>> [clears throat] >> It's a bit Well, yeah. Okay, so I could take this equation and dot with ER and dot with EZ.

[00:45:14]The problem with doing that is I'm going to have these unknown components of N, the ER and the EZ components of N.

[00:45:21]So, if two components that I don't know and these two separate equations, I'm going to be a bit stuck.

[00:45:27]So, what I'd rather use is perhaps that fact.

[00:45:31]N.R.

[00:45:32]So, how do I use N.R. = 0?

[00:45:35]Um well, I can just work out R. So, I mean, I have an expression for R. So, I can take R.

[00:45:42]dot Newton's second law and I know the N bit goes away, but that's quite messy.

[00:45:48]Let me see if I can use this fact in a cleaner way by using the symmetry um of the surface. So, here's one approach.

[00:46:02]is Let's think in the plane.

[00:46:08]That's R and that's Z.

[00:46:12]And this is a curve Z equals H of R.

[00:46:17]I can define Let's call it vector tau as a tangent vector to the curve of revolution.

[00:46:43]So, in my picture here I'm going to define that as tau.

[00:46:57]So, tau just by that geometry there is just a derivative H prime of R. That's in the Z component, right? So, that would be like ER plus H prime of R EZ.

[00:47:27]And actually, we can work out exactly what N is in that picture, too, or at least the direction of N.

[00:47:36]I don't know its magnitude, but I know that N is pointing that way because the other component would be into the board, which is the E theta. I'm rotating this thing.

[00:47:46]So, N uh is Think of this as just a vector 1 H' in E R E Z components. I flip the two and put a minus sign. So, N is going to be some scalar N times -H' E R + E Z.

[00:48:15]That vector dot that one is zero.

[00:48:19]Yeah.

[00:48:20]And I've chosen the sign so that it's pointing up in the E Z, which is what I need so that that should be positive.

[00:48:28]Um Right, I don't necessarily want to use that. What I want to do is get rid of the N from Newton's second law cuz I don't know it and it's it's causing me trouble. So, if I take Newton's second law and dot with this tau, then the N is going to go away.

[00:48:43]So, N dot tau is zero. So, let's dot N two with tau.

[00:48:56]So, that gives me um So, I'm going to take this thing and dot with tau. Tau is in E R and in H' of R E Z. So, I just get this guy R double dot minus R theta dot squared.

[00:49:22]And then that goes away. Z dot dot The E Z gets an H' of R. So, I get plus H' of R Z dot dot.

[00:49:38]Equals That goes away and that just gives me the gravity dotted with that tau, which is a minus M G.

[00:49:48]Uh becomes plus Sorry, minus M G H' of R.

[00:50:05]Okay.

[00:50:13]That's a big messy equation, but I'm getting quite close to something that's at least one variable.

[00:50:20]So, uh theta dot is still living in this equation, but on the other hand, r squared theta dot is a constant.

[00:50:29]So, I can look at this as r squared theta dot squared divided by an r cubed.

[00:50:40]Right? So, this is r squared theta dot squared divided by r cubed, which is h squared over r cubed.

[00:50:54]Which means I've gotten rid of the theta.

[00:50:56]And that h I work out from whatever my initial conditions are.

[00:51:00]I've still got r's and z's, but as we've said, r and z are not really independent. They're connected via the definition of the surface. So, one thing I could do is get rid of the z's. So, z of t here.

[00:51:22]Single ODE for r of t.

[00:51:28]I can use z is h of r of t, which means if I take a z dot I get h prime r r dot.

[00:51:45]And I can go further, zed.dot gives me an h double prime r. squared plus an h prime uh r.dot Ah.

[00:52:17]Right. So, in principle, I plug that into the zed.dot I get rid of the theta in terms of the h, and now I have one second-order ODE for r.

[00:52:30]And you give me whatever initial conditions principle, I solve that and I've got the motion.

[00:52:37]Um Right. Problem with that calculation is it's very, very messy. I think that's a very ugly It's got an r cubed in the denominator. It's got all sorts of nonlinearities, r. squared, and so on.

[00:52:53]Uh so, possible to do it that way. You probably have to do it numerically.

[00:52:59]We will pick up on um Thursday with the the energy approach and see if we can make a better progress. Let's stop there.

[00:53:15]Okay, shall we get going?

[00:53:19]So, we've been looking at constrained motion, and we've kind of worked through some of the equations of a particle on a surface of revolution.

[00:53:28]Let's do an example. So, I want to imagine the surface is zed equals r squared over 2a. So, this is like a bowl, parabolic shaped bowl.

[00:53:43]A some constant.

[00:53:45]And >> [clears throat] >> let's uh a particle is projected horizontally from Z equals at uh with speed V.

[00:54:21]So, I've got a marble in a bowl and at time zero the marble is at height Z and I give it a little horizontal push with speed V. And the question, well, different questions we might ask, but the question I want to answer is does the marble ever make it to the bottom of the bowl?

[00:54:56]Okay.

[00:54:58]Of course, we expect yes if I'm doing this with a real marble, but that's real marbles are subject to drag, right? This marble is living in a vacuum with no air resistance. So, it's not it's not necessarily obvious that it should ever make it to the bottom of the bowl.

[00:55:14]Okay, so uh I'm not going to go through every equation.

[00:55:20]For the most part, we've already worked out the equations last time. We just have a particular form of surface here.

[00:55:26]Uh last time we were doing Newton's equation and we were getting doing some tricks to get rid of the normal reaction and we end up with the second-order equation, which is quite ugly. And I said another way we could go about that is to think about conservation of energy.

[00:55:43]And conservation of energy, which is just going to of course say 1/2 magnitude of r.

[00:55:51]squared plus potential energy is just gravitational, which is mgz.

[00:55:58]is a constant E.

[00:56:02]>> [clears throat] >> We can work out that E from thinking about the initial conditions. So, our initial conditions We have to [clears throat] make sense of this. So, we're told that So, z equals a. So, z of 0 is a.

[00:56:22]Projected horizontally, what does that mean? With speed v, I need to turn that into a an initial condition.

[00:56:30]So, if we think r uh r is r e r plus z e z, right?

[00:56:44]r.

[00:56:47]r. e r plus r theta. e z plus z.

[00:56:57]e z. Sorry, this is an e theta.

[00:57:01]Yeah?

[00:57:03]What does it mean projected horizontally? We're saying something about this vector.

[00:57:08]Right? Which is the horizontal direction.

[00:57:12]theta, yeah?

[00:57:14]So, that statement projected horizontally with speed v means the vector at t equals 0 is only in the theta direction and its magnitude is v.

[00:57:25]Right?

[00:57:27]So, at t equals 0 r dot of zero must be just r uh r of zero theta dot of zero e theta and that must have magnitude of v.

[00:57:48]Right?

[00:57:50]>> [clears throat] >> Okay.

[00:57:57]Well, so my surface is z equals r squared over 2a and z is starting at a.

[00:58:03]So if z is a, r is 2a.

[00:58:08]Right?

[00:58:09]r of zero is 2a and r of zero times theta dot of zero uh is is v.

[00:58:23]Right?

[00:58:26]um Now actually what what we need to use is the fact that we have conservation of angular momentum about the z. Right? This is the statement that r squared theta dot is h. This is what we derived last time from thinking about the theta component of Newton's second law. Question?

[00:58:51]>> Why is r of zero equal to 2a?

[00:58:53]>> Say? Where's this coming from?

[00:58:56]Uh so if I put z equals a and solve for r, r squared over 2a 2a squared over Oh.

[00:59:07]4a. You're right. This was wrong.

[00:59:10]The mistake is up here. I want the surface to be 4a. Thank you.

[00:59:16]That should work.

[00:59:19]Yeah.

[00:59:27]Good now?

[00:59:30]So, we're going to need to use this this H and effectively what I can do is I can work out what the value of H is because I know that at T equals 0 uh R of 0 theta dot of 0 is V and R of 0 is 2A. So, therefore this H is R of 0 which is 2A times R of 0 theta dot of 0 which is V.

[01:00:04]So, H is 2AV.

[01:00:13]Okay.

[01:00:17]So, therefore my conservation of energy uh R dot squared is Sorry, and I'm missing um I'm missing an M in here.

[01:00:42]The M is not going to matter cuz the M is also going to show up in the E and can cancel out of everything, but it should be there.

[01:00:50]Uh Yeah, so I end up with 1/2 R dot squared plus R [clears throat] theta dot squared uh But, sorry. I want to write So, this is this is my take the square of each of those components sum of the squares of those components for for kinetic energy and this R squared theta dot squared I want to write as an h squared divided by r squared.

[01:01:36]And then I get a z dot squared plus Okay, we'll keep the m.

[01:01:44]plus mg z equals e.

[01:01:52]And the e I can just put in the values of these components at z e or sorry, at t equals 0.

[01:02:00]And at t equals 0 r dot is 0, z dot is 0.

[01:02:04]Um and I'm just left with the h squared over r squared plus mgz. So that's m over 2 >> [clears throat] >> times h squared, which is 4 a squared v squared over r squared, which is 4 a squared.

[01:02:27]plus m g a.

[01:02:35]So e our constant total energy is What are we left with? The a squareds, the four go away.

[01:02:45]There's an m times v squared over 2 plus ga.

[01:03:01]Okay.

[01:03:06]I still have r's and z's and as we said before really they're linked to each other by the fact that they're stuck. It's on the surface and the surface has a relationship between r and z. So I can eliminate uh I can eliminate R or Z, so let's eliminate R in this case because I want to think about going value of Z.

[01:03:30]So, to eliminate R, I just need to think about this relationship, which just looks a little weird, so I'm going to write I'm going to solve for R from that Z equals R squared over 4A, so R is 4AZ to the 1/2, which means that R dot take a DDT, >> [clears throat] >> we get a a 1/2 Z to the minus 1/2, so that becomes a 2 root A Z dot over root over two root Z, I think.

[01:04:14]I'm pulling out the root 4A and taking 1/2 Z to the minus 1/2.

[01:04:20]And >> [clears throat] >> Right, so I can plug that in there.

[01:04:31]And now I have a single equation for Z.

[01:04:34]And on the one hand, the benefit of conservation of energy is that this is a first order equation, right? It's just Z dot. Effectively, if you're working with energy, you've already integrated once.

[01:04:45]So, I don't have second derivatives, but it's quite messy. I've got I could try to solve this equation for Z dot.

[01:04:52]It says Z dot squared, and then I could try to separate and integrate. It's possible. It's messy.

[01:04:58]Um Actually, there's a little trick we can use to answer our question, which isn't going to be fully resolving the trajectory, but it'll answer our question.

[01:05:07]So, I take this, I shove it in there, and I'm going to shuffle things around where I solve I I get a coefficient of the z dot.

[01:05:14]And after a little bit of algebra, you can write this as 1/2 1 + a over z z dot squared plus a b squared over 2 z plus g z equals v squared over 2 plus g over a. Sorry, g times a.

[01:05:47]So, all I've done is I've factored out the things that have a z dot squared.

[01:05:50]These are the terms that involve z.

[01:05:52]And then this is the constant e. And I've gotten rid of the m.

[01:05:57]Okay.

[01:06:03]Now, this is still as I said, this is quite hard to solve exactly.

[01:06:08]But my question was does the does the marble reach the bottom of the bowl?

[01:06:12]And this equation looks a bit like t + v equals constant. It's It's of course it's come from kinetic energy plus potential energy equals constant.

[01:06:25]And I've shuffled things around a bit.

[01:06:28]So, if I call this thing I'm going to call it sort of curly t cuz it's not quite kinetic energy. It's kinetic energy where I where I moved one of the terms away.

[01:06:38]And this thing is not quite potential energy, but it's close. Sort of looks like p a v. So, let's call that curly v.

[01:06:45]And then this is like our e.

[01:06:47]In fact, that is just the e.

[01:06:54]And the the key is that so this this has that kind of form where the t has like kinetic energy z dot squared plus this kind of weird factor. But nevertheless, this thing is positive or non-negative.

[01:07:09]And this looks like a potential field.

[01:07:13]And so I can think about the shape of this and make some progress on what what's allowed to happen with Z.

[01:07:20]In fact, you can already see that there's something funny happening if Z really did go to the bottom of the bottom of the bowl, Z goes to zero, there's singularities, right? So, it already suggests that that probably can't happen.

[01:07:33]So, if I plot V of Z, >> [clears throat] >> so that's a function that's diverging at zero and then as I go away in Z, it looks linear. So, it's doing something like this.

[01:07:56]And >> [clears throat] >> if you do some calculus, you can work out that that point right there where V prime is zero has solution Z equals root a V squared over 2g.

[01:08:17]>> [clears throat] >> Now, as I said, this is not exactly kinetic energy plus potential energy equals constant, but it has the same kind of mathematical flavor.

[01:08:29]And we can really argue the same kind of statements about the the trajectory. So, what we're doing in our experiment is we're starting at Z equals A. Let's suppose that Z equals A with no with T equals zero.

[01:08:44]Right? So, at T equals zero, E is V of A and T of A is zero.

[01:08:56]So, it's very much like starting a ball rolling in a potential field that has that shape.

[01:09:02]And in fact, all of the arguments we've used before work the same way. The ball is going to start moving this way.

[01:09:09]It's going to be trading V. V is going to be going down. T is going to be going up.

[01:09:14]Until it comes up to this point.

[01:09:18]And at that point, E is back to V, which means T kinetic energy or our sort of pseudo kinetic energy goes back to zero.

[01:09:27]So, at this point uh V is E, which implies T is zero, which means it stops. And then it goes back down again.

[01:09:38]Okay?

[01:09:42]So, indeed, the particle can't reach zero.

[01:09:47]Partly cuz there's a singularity there, but even if it weren't a singularity, um this graph just isn't going to get to zero.

[01:09:56]And we can even go further. So, you can do more calculus. You can say I can work out what that point is. And that's the other place where Let's call that Z star.

[01:10:08]Z star is such that um all the energy is in the V.

[01:10:17]Right? So, I set V equals E, solve for Z. One of those solutions is Z equals A. That's by construction. And the other solution, if you do a little bit of uh algebra on that, you get Z Z star is V squared over 2g.

[01:10:42]>> [clears throat] >> In my picture, I've drawn Z equals A to the right of this point. That's not doesn't have to be that way.

[01:10:56]But actually, there are two scenarios.

[01:10:58]If if A is less than this value, then that means I started it over here, and that means In fact, it's kind of interesting two experiments. If A is less than uh what do we want to call this guy?

[01:11:13]Let's call this guy Well, I don't know. Let's leave it like that. So, two scenarios. If A is less than that value, root A V squared over 2G.

[01:11:29]And that means if the particle's always rolling to the right of A, which means that Z is going to stay between A and V squared over 2G.

[01:11:46]And if A is greater than that value AV squared over 2G.

[01:11:53]>> [snorts] >> Then Z stays between Oh, yeah. You can see this is the same thing as A uh less than or greater than V squared over 2G.

[01:12:17]So, the interesting thing about the experiment, one, the particle never makes it to the bottom of the bowl. And that's partly because the normal reaction force, as you get lower, you're doing these smaller circles, the normal reaction force is pointing more up, and it pushes you back up the bowl.

[01:12:31]Basically, it's always going down, and then it comes back up, and then it goes back down, and then it comes back up.

[01:12:36]In this scenario, uh in fact, you can really think of this as a statement on how hard I pushed. If I give it enough velocity initially, it never goes down at all. All right, I give it horizontal push, it goes up the bowl, then it comes back down, then it goes up the bowl, then it comes back down.

[01:12:52]It's quite I think an unexpected result.

[01:12:56]Okay.

[01:12:59]Want to move to our next topic unless there any questions on what we've done there.

[01:13:11]So, what we want to do This is part six of the of the course.

[01:13:34]is to connect to Newton's second law to Newton's um law of gravitation or more generally inverse square law.

[01:13:42]So, Newton's law of gravitation I have Let's put an origin here.

[01:14:04]I'm going to put a mass, I'll call it mass one there at position vector R1 and another mass here with mass two at position vector R2.

[01:14:19]The law of gravitation says that the force the gravitational force >> [clears throat] >> of point mass M1 located at position vector R1 due to Sorry, gravitational force on point mass M1 due to point mass M2 at R2 Let's write it this way is F 21 F Let me write that F12, the force on one due to two >> [snorts] >> is constant called GN product of the masses M1 M2 divided by the distance between them squared. It's the inverse square part of the law, so that's magnitude of I take the vector R1 minus R2 magnitude squared.

[01:15:39]That's the distance between them squared.

[01:15:43]And that vector is going to point uh towards R2.

[01:15:51]So, this is the vector R2 minus R1.

[01:16:00]>> [clears throat] >> And I need to make that a unit vector since I've already taken care of the magnitude here.

[01:16:19]Okay?

[01:16:21]So, of course, we can bring that together and this is the same as R2 minus R R1 minus R2, so I can write it as R2 minus one over the magnitude cubed, but it's a bit nicer to write it this way cuz it reminds us that the really the magnitude of the force goes with the square of the distance between them, and this is just needed to make it a unit vector.

[01:16:42]Right, so that G is sort of as we understand it a universal constant.

[01:16:50]So, that's approximately has a very specific value 6.67 * 10 -11 N m ^ 2 kg to the minus two. That is the gravitational constant.

[01:17:17]Same constant will apply for the attraction between Jupiter and the Sun as for any other matter in the universe.

[01:17:29]At least within Newtonian gravity.

[01:17:31]So, a few notes.

[01:17:35]By Newton three, Newton's third law So, now the easiest uh expression of Newton's third law, the force on particle two due to particle one is the exact same thing just flipped around. So, F 21 is minus F 12.

[01:18:01]Two, gravity as we know it is always attractive.

[01:18:14]Particles, the the masses are being attracted to each other, not repelled.

[01:18:19]Uh three, this is more of a an approximation that's very useful.

[01:18:25]Let's consider the case where mass two is much greater than mass one. So, if mass two is the Sun and mass one is the Earth.

[01:18:36]Um the claim or the statement is that particle two is approximately stationary.

[01:18:57]And at first that might seem a bit curious cuz we've just said they have the same force acting on each of them just flipped around.

[01:19:03]But then the key is that if you think about Newton's second law on particle two, that says m2 r2 dot dots equals f 21.

[01:19:17]Because there's m2 over here, cancels out the m2 in the force, which means that the acceleration of particle two is only seeing m1 over here.

[01:19:29]Whereas uh, so if the if I cancel out the m1s and m2 is very big, acceleration of particle one is very much higher than the particle acceleration of particle two.

[01:19:38]So to a very good approximation, though it is an approximation. So actually the sun and the earth, if we had nothing else in our solar system, the sun is not sitting stationary. It is also moving based on the gravity of the earth, but it's much much much smaller than the motion of the earth. So it's mathematically we can effectively treat particle two as being stationary, and then the nice thing to do is take that to be the origin to work out the path of particle one.

[01:20:05]>> [snorts] [clears throat] >> Right, so m2 cancels and uh, r2 dot dot stay close to zero if it just has the m1 there.

[01:20:29]So what we can do in that case is we can set R2 as I say to be the origin.

[01:20:36]And then I can relabel.

[01:20:42]I don't need the subscript now. I just have R1. So, let's just call its position vector R.

[01:20:48]And let's define M capital M to be M2 and I'll use little M for M1.

[01:21:00]And then we can write that uh little M R dot dot is minus K E R over R squared where K is M M G.

[01:21:34]So, particle one, which I'm labeling R, its force is in the radial direction as an inverse square, one over R squared, and the constant as the sum the product of the two masses and the gravitational constant.

[01:21:48]And this is the equation we want to make some sense of, but this fits into what we've been thinking about cuz that's a central force.

[01:21:55]Right? And it just has this very particular form of the one over R squared.

[01:21:59]Um and then make a note, this has potential V of R is the antiderivative of the K over R squared which is a such a minus the antiderivative of that, which is a minus K over R.

[01:22:26]Take the derivative of minus K over R.

[01:22:29]You get a positive K over R squared the minus that gives me the force there.

[01:22:36]>> [clears throat] >> Okay, and all of these statements we've been making apply in the past few lectures. So energy is conserved in the system.

[01:22:47]It's a central force. So that means the motion is planar.

[01:22:54]And what we want to do is kind of work out the possible paths of of particle one.

[01:23:01]Uh but first let's do an aside or a couple of aside. So one aside is to ask the question of uh what we've what we've mentioned at the very beginning of the course is there's kind of these two different laws we might use for gravity. When do we use this F equals minus mg E Z versus F equals minus K E R over R squared.

[01:23:36]E Z was that's what we've been using for gravity. This is what I'm telling you is actually the accurate description of gravity at least within Newtonian mechanics.

[01:23:47]And it's about scale. So if I draw a circle and call that the Earth and then let's give that radius R.

[01:23:56]And here I am sitting up here and I'm going to do a little experiment of dropping something.

[01:24:01]So I'm sitting here and I call that E Z.

[01:24:05]And I I do an experiment where I drop something from height H.

[01:24:12]>> [snorts] >> When I drop the pen in the classroom it's being attracted to the center of the Earth.

[01:24:18]So I the actual gravitational law of pulling the pin down is this one.

[01:24:25]But the R is not the distance from the pin to the floor, it's the distance from the pin to the center of the Earth.

[01:24:32]And that's where the dis- the distinction comes in. So really R little R is big R, the radius of the Earth plus whatever height I have there.

[01:24:42]And then of course if I'm doing an experiment anywhere near the surface of the Earth H over R is a dimensionless number which we might call epsilon which is of course very small.

[01:24:56]Right?

[01:24:58]So then the claim is that I can get that law as an approximation of that one.

[01:25:04]So if you if you if you take M G over little R squared so that's M G over big R squared times 1 plus epsilon squared to the minus 1.

[01:25:24]Right? I'm just factoring out I'm writing that way and factoring out the R so I get a epsilon squared to the minus 1.

[01:25:32]And I can do a Taylor series of that epsilon squared so that's approximately M G over R squared times 1 minus epsilon dot dot dot.

[01:25:50]And actually if you take the mass of the Earth times the gravitational universal gravitational constant divided by the radius of the Earth squared and plug those all in, you get about 9.8 m/s squared which is the little G we use.

[01:26:19]Indeed this little calculation also tells you if you wanted to go a bit further in your approximation, if you were to say you know, above the surface of the Earth by a substantial amount, but not say far enough to be in some sort of orbit, you could you can do this and see what the next term you need is.

[01:26:35]But, that's the connection between those two formulations.

[01:26:41]Another comment, which is sort of an aside, we will primarily be thinking about gravity, but there are other uh laws that follow the same form, and in particular Coulomb's law.

[01:27:03]This is if I have two point charges, this is in electrostatics or electromagnetism, Q1 and Q2, um >> [clears throat] >> I have two point charges Q1 and Q2 at rest, then there's a there's a force between them, which has the exact same sort of form. So, F 1 2, the force on the first particle due to the second, satisfies So, you get Uh, another [clears throat] universal constant sitting out front, which is written as 1 over 4 pi epsilon naught. I'll define that epsilon naught in a moment.

[01:28:23]Product of the charges, Q1 over 2 Q1 * Q2 divided by the distance between them.

[01:28:29]So, again, we need the R1 or the distance squared.

[01:28:37]And careful if I get on the first R1 That epsilon naught is called the permittivity of free space. It's just, again, a universal constant.

[01:29:10]>> [clears throat] >> Permit- Where's the double the double T?

[01:29:23]Um The only difference between this law and the gravitational law is that Q can be positive or negative. So, the charge can be a positive or negative charge.

[01:29:37]Which means that if one if both charges Now, if Q1 Q2 have the same sign In other words both charges are positive or negative then F12 Why am I writing it that way?

[01:30:15]>> I think I have my notes have it written the wrong way. I think so if they have the same sign it should be repulsive, right?

[01:30:24]I'll switch for that.

[01:30:32]Have I written that right? So R1 Have I written that right? So if they have opposite signs that's negative which means that's pointing R2 minus R1 and R2 minus R1 is pointing that way which is attractive. I think that's good.

[01:31:20]As I say, we won't really play with that but the mathematics works the same except that you can have of course this additional feature of attractive versus repulsive.

[01:31:29]What we really want to do so our goal for the next couple of lectures is to use some tools and that we've developed already and build some new tools to understand planetary orbits and in particular what's called the Kepler problem.

[01:31:54]Oops, planetary Okay.

[01:32:05]So, I'm going to imagine a large object here. Maybe it's the sun.

[01:32:16]And we'll put the origin there.

[01:32:18]And then we'll have a single particle, a mass.

[01:32:22]Might be a comet, might be a planet.

[01:32:24]It's the only other thing that's in our picture, sitting here at vector R with little mass little m.

[01:32:35]This is a planet, a comet, etc. Of course, if it's a planet, then I know that it's orbiting the sun and I want to say something about the properties of that orbit.

[01:32:49]If it's a comet, it might be orbiting the sun, but not all comets orbit the sun. Some comets come in and then they shoot off and they never come back again. So, there's a distinction we want to be able to make.

[01:33:03]And so, our goal is to find the path of that object R satisfying satisfying this inverse square law.

[01:33:35]Where K is M M GN.

[01:33:46]Okay. As I said, just to repeat, this is an approximation of the reality in various ways, but in particular, that really there's an orbit that this this object is doing due to the gravity of that one, but it's much much much smaller than the orbit that that one's doing. So, we're going to fix that one as our origin and just focus on the motion of that one.

[01:34:07]Um it's called the Kepler problem. So, in 16 early 17th century, Johann Kepler devised three scientific laws about the properties of planetary orbits.

[01:34:21]Um the fascinating thing was this is before Newton even had done calculus. So, he was really just examining data that other people had collected and processing that data to come up with these laws.

[01:34:34]Our goal is to reproduce these same laws but within the framework of solving these equations.

[01:34:41]Okay?

[01:34:42]And we'll list those laws that Kepler found uh a bit later.

[01:34:47]Right. So, uh if I define this just for notation as say capital F of R, that's our central force.

[01:35:10]So, right away we can conclude based on things we've said in previous lectures that this angular momentum about the object at M, the origin, R cross M R dot uh is constant.

[01:35:33]And that further implies that the motion is planar, as we've seen before.

[01:35:44]Which means that we can work in what we might as well call the XY plane, but we'll really do this in polar coordinates.

[01:36:03]Okay.

[01:36:08]So, let's start with Newton's second law. So, N2 with R = RER.

[01:36:18]We've already done this exercise a few times. So, I get M R dot dot is M R little scalar R dot dot minus R theta dot squared ER + 1 over R DDT R squared theta dot E theta.

[01:36:44]Same formula we've seen a few times.

[01:36:47]= S. Let's Let's write it as F of R.

[01:36:51]>> [clears throat] >> ER.

[01:36:57]And some of our same tricks are going to work. There's no E theta component to the force, which means that this thing is not balanced by anything, which means that that DDT is zero, which means that R squared theta dot is constant, which is what we've been calling H is going to apply as usual.

[01:37:17]So, the E theta component we get R squared theta dot = H is a constant.

[01:37:27]>> [clears throat] >> Let's label that as kind of fact one or equation one.

[01:37:36]And if we look at the ER component, things are a bit less nice. We have M R dot dot minus R theta dot squared which I can write as m r dot dot minus same trick. I don't like the theta dot there, so I'm going to view that as an r theta dot squared.

[01:38:02]But then I get an extra uh I get too many r's, so I have to divide by an r cubed.

[01:38:07]So that's h squared over r cubed equals f of r.

[01:38:16]And that's let me call that equation two.

[01:38:22]>> [snorts] >> Okay.

[01:38:29]So uh in principle, I could try to solve that equation. It's a second order equation for r of t. Right?

[01:38:40]It's not a nice one. It's got a one over r cubed here. It's got a one over r squared over there. This is very non-linear. I don't know any any methods to solve that exactly.

[01:38:52]Uh and indeed once I solve it, I still then need to work out theta of t, right? Which I could get from here. So if I knew r of t, I could stick it in there, and then I could solve this equation for theta of t, and now I've got the path r and theta of t.

[01:39:06]But that's very hard.

[01:39:08]So we need a trick.

[01:39:10]Um Before we get to a trick, let's take the simplest case.

[01:39:21]The simple case is where theta dot is zero. So if h is zero, >> [snorts] >> and when does the simple case hold? The simple case this theta dot is zero at t equals zero.

[01:39:52]So, I have an object far from the sun and I just let it go from stationary position. It's going to have no angular component. It's just going to go straight towards the sun. It's going to be a radial path. Right? So, theta dot is zero and h is zero and then we get a radial path.

[01:40:16]>> [clears throat] >> In other words, uh this equation becomes a lot easier.

[01:40:22]It just says r dot dot is f of r.

[01:40:31]And that's called a radial Kepler trajectory.

[01:40:49]I'll say that's not trivial because the f of r still has the non-linearity, the one over r squared.

[01:40:55]So, you still have to figure out how to solve this, but that is a solvable equation. I think I put that in a problem sheet. So, you work out those paths.

[01:41:03]Okay?

[01:41:05]>> [clears throat] >> But, that only works in the case where the theta dot is zero and the the trajectory is very simple. Right? So, that's um that's the less interesting one. If h is non-zero we need a trick. And here's the trick.

[01:41:30]The trick is I'm going to define u as the variable one over r.

[01:41:36]And what we'll see is in terms of U um So, I'm going to do two things. I'm going to I'm going to define U as the inverse of R.

[01:41:47]And my idea is to eliminate T make T sort of a hidden variable and solve for the path as a curve U of theta.

[01:42:11]So, we'll eliminate the independent variable and see if I can get a equation for U of theta.

[01:42:19]Okay.

[01:42:21]I think we should have just about enough time to see how this trick works. So, it's just being careful about how I convert into the U via chain rule. So, R is 1 over U which means that dR dT is -1 over U squared dU dT.

[01:42:42]And I'm going to write that as -1 over U squared dU d But, I know that R squared theta dot which is R squared d theta dT is H.

[01:43:05]So, therefore dR dT has a d theta dT which is an H over R squared.

[01:43:14]But, R squared 1 over R squared is U squared.

[01:43:19]So, that means that d dR dT Sorry, d theta dT I can write as H U squared.

[01:43:32]And therefore, the U squared and the u squared cancel and I get this nice expression that dr dt is just minus h du d theta.

[01:43:53]With me?

[01:43:54]Now we need to go second derivative.

[01:43:56]Second derivative r dot dot that's d dt of dr dt.

[01:44:06]>> [clears throat] >> And as we've just worked out when I take a d dt um I pick up this d theta dt factor.

[01:44:19]I say so I'm going to write d dt is d d d theta d theta dt.

[01:44:26]And dr dt we've just written as minus h du d theta.

[01:44:38]But this is nice because I'm taking a d d theta h is constant so there's just like a second derivative of u with respect to theta and then I've got this d theta dt which as before is h u squared.

[01:44:53]So therefore r dot dot becomes I have an h there I have an h that I can pull out there so that's minus h squared u squared d squared u d theta squared.

[01:45:09]So therefore let's see what Newton's second equation well not that one but this one look at the er component there two becomes this is where the all the magic works.

[01:45:24]Put Put these things together you get a minus h, so there's an m minus h squared u squared d squared u d theta squared r dot dot minus h squared over r cubed but that's just like a u cubed, so that's minus h squared u cubed equals f, but f is really k minus k over r squared, so that's minus k u squared.

[01:46:07]And that's lovely because I can cancel out a u squared from everything and I'm left with a very manageable equation m d squared u d theta squared plus I'm going to put the h on the other side cancel out a minus everywhere so I get plus u actually let's get rid of that as well plus u equals uh k over h squared.

[01:46:55]Sorry?

[01:46:55]>> You forgot the mass.

[01:46:58]>> Yes, sorry. I was going to cancel the mass out of the k, but I'll do we'll do that in we'll do that next. So the mass there.

[01:47:08]Okay? We'll pick up with this next time, but as you can see this is magic because that's a second order that has solutions of signs and cosines. That thing's just a constant and then we need to put the pieces together to see what the possibilities are.

[01:47:21]Good. Let's stop there. See you next time.