Lecture 49 : Problem solving using Euler-Lagrange equations

Hello and welcome to another lecture of classical motion of a single particle. We have discussed various properties of Euler Lagrange equations, we have discussed some application of that. In the last lecture, we also discussed very interesting problems regarding the Lagrangians of simple harmonic oscillator and some other type of oscillators. We have said that although it is sometimes useful to check the symmetry of the Lagrangian, some properties of the Lagrangian and to conclude about the property of the system from that directly. For example, some constants of motion etcetera, there is no substitute of solving the Euler Lagrange equation. Because sometimes the point is that for several Lagrangians which have the a little bit unusual form can be confusing. You have some apparent contradiction between two conclusions which are very hastily taken. Another point can be possible that sometimes some conserved quantities cannot be revealed just by the symmetry of the Lagrangian. Just as we discussed in the very last example of the last lecture that if we take an equivalent Lagrangian of a simple harmonic oscillator, but now it has an explicit time dependence. Although we know that the total mechanical energy of the oscillator actually is conserved in time, the Lagrangian's symmetry cannot reveal that or it is actually revealed, but it is I mean you have to then exactly calculate the form of that quantity that is L minus del L del q i dot times q i dot summation over i. And from that again you have to do some algebraic manipulations to find the proper conserved quantity. This is possible, but what I am saying that without doing any algebra if you directly try to conclude from the Lagrangian sometimes it may be ok confusing or misleading. So, with all these things finally, in this lecture I will attempt to solve two specific problems and they are actually very typical typical problems ok which are being used in different books or in different materials on Lagrangian formalism ok. So, what is the meaning of Lagrangian formalism? Just since I use this phrase this means that we assume that for a system we do not know the equations of motion, but someone has kindly provided us the Lagrangian the form of the Lagrangian and with that we actually dare to solve the Euler Lagrange equation to find the equations of motion ok. Now so the first point is the pendulum a simple pendulum with variable length ok. So, you see that I am just drawing this system. So, a simple pendulum with some constant length L is hanging from this ceiling ok. Now, I am putting some mechanism over here ok which is invisible to any observer by which now the length is no longer constant ok. Now, I can write the law for the variation of this length is L 0 1 plus sin omega t for example, so that at t is equal to 0 the L is always L 0, but when t changes ok L actually varies ok. So, L 0 is the constant length when it was I mean when this type of invisible mechanism was not present ok. Now, for that system of course, again I have to construct first the Lagrangian. So, I now know at least for such a system this type of system I assume that this recipe t minus v also works ok. Now, of course, in this point you can ask this philosophical question why I should assume that is because if you just consider that the friction in this point is negligibly small then just by changing the length of the string of a pendulum cannot cause any dissipation that should be concluded from your common sense of physics ok. So, if you have this you can still use L is equal to t minus v ok, but now you see that v can have a little bit complicated form I am I will come to that and also t will have a little bit complications which you may overlook sometimes and that is why I have chosen this problem. So, first we are trying to say that we by our choice we have already constrain this motion in x z plane. So, if this is the x direction the vertically upward direction will be our z direction as always and we define this length to be the radial length from this point to this bob the bob is actually carrying the whole mass m the string is considering I mean the string is considered to have negligible mass ok. Now let us say the bob is here and it is making with the vertically downward direction some angle theta. So, I know that in at this position the kinetic energy or t is nothing, but half m x dot square plus z dot square that anyone can tell me ok. And if I know for such a system since instantaneously you can think of an arc of a circle at least in this polar system you can think that the instantaneous x is nothing, but L sin theta and the instantaneous z is nothing, but the minus L cos theta I already discussed why we are taking this minus sign over here for z ok. Now this L is no longer a constant here. So, I have to now write this is actually L 0 1 plus sin omega t sin theta and this is L 0 with a minus sign 1 plus sin omega t ok times cos theta ok. Then I have to calculate t. So, x dot will be equal to now L 0 1 plus sin omega t cos theta theta dot plus L 0. Now it will act on this term which will be simply omega cos omega t sin theta ok. I can also have z dot similarly. So, z dot will be given by minus L 0 1 plus sin omega t with a minus sign sin theta theta dot plus L 0. So, there will be a minus. So, that will be ok. So, that will be actually a minus sign over here minus L 0 yes omega cos omega t cos theta ok. So, that much we can write if I do the x dot square plus z dot square what we will get. So, I just put the barrier a little bit longer yeah. So, what will you get? We will simply get that L 0 square 1 plus sin omega t square ok. So, sin square theta plus cos square theta will go away. So, you will have actually theta dot square ok. Then you will have. So, I am just doing this square plus this square ok. So, yes you will have again L 0 square omega square again you will have cos square omega t ok. Again sin square theta plus cos square theta will give you 1 ok. There will be some cross term. So, the first cross term will be 2 L 0 1 plus sin omega t L 0 omega cos omega t cos theta sin theta with a positive sign times theta dot and then we will have. So, there is a minus minus plus. So, I can actually say yeah. So, this is actually a plus sign. So, it should be plus 2 let me just check whether yes ok. So, actually that there will be a minus sign here. So, I just rewrite otherwise yeah. So, minus L 0. So, it will be actually sorry plus L 0 yeah 1 plus sin omega t sin theta theta dot minus L 0 omega cos omega t cos theta yeah ok. So, now, if you have the cross term from this this one you will actually have 2 times L 0 1 plus sin omega t ok. There will be a minus now ok I have L 0 omega I have cos omega t I have cos theta I have sin theta and theta dot. So, we can get rid of this 2 ok. So, I will have simply. So, t will be half m. So, this thing L 0 square 1 plus sin omega t whole square theta dot square plus L 0 square omega square cos square omega t what will be v ok. So, again if you take. So, it depends on you how you choose your reference point. So, if you choose your reference point here at this point and. So, v will simply be m g z and. So, I mean. So, if this is for example, I am just taking z to be negative in this case. So, I mean for me v should be equal to simply m g z because that will take care of the negative sign ok. So, if I now say that v will be minus m g L cos theta yeah v will be minus m g z which is sorry plus m g z actually ok because if I just yes I can say m g z yes right and that z is negative ok. So, it will give me minus m g L cos theta. I have finally, the Lagrangian which is half m L 0 square 1 plus sin omega t whole square theta dot square plus L 0 square omega square cos square omega t and v will be sorry v minus m g L is L 0 1 plus sin omega t cos theta there is a minus sign already there is a minus sign already here and that is why you have a plus sign over here ok. So, that is the L you have to use for solving the Euler Lagrange equation. Now, what is the generalized coordinate in this whole business? It is only theta because other than that only t is appearing and we have said that t is just if you know L as an explicit function of t it is actually not it is just actually knowing the law ok. So, that means, that it is not giving you any extra degrees of freedom ok. So, that is why the only generalized coordinate is generalized coordinate is theta and you have to now calculate theta dot minus del L del theta is equal to 0 ok. What will be the equations of motion? I let you to solve ok. I will give this as an question in the assignment ok. So, you have to somehow solve this in order to get the marks of course, you have to solve this correctly ok. I am going to another problem that is the problem of spherical pendulum ok. So, you see I have just showed you how to construct the Lagrangian. Now, once you know the Lagrangian then of course, this is up to you that and if you know the generalized coordinate then the business is actually done. Then you can just have you have to just do some partial derivatives and total derivatives with respect to t ok, but with caution. Spherical pendulum is a generalization of a usual pendulum ok, which is supposed to oscillate in a 2D plane ok like this. Now, how does a spherical pendulum look like? So, again you think that the total mass is concentrated at this point and this is the mass less. So, spherical pendulum is actually can move in all the 3D space available ok and one point of the pendulum is fixed at a point ok, which is the center and the pendulum can move this. So, it is actually not at all different from the case of a particle, which is moving on the surface of the sphere. Now, if the length of the string is constant then thus it is actually equivalent to move on a sphere of constant radius. If this is not a constant then it is simply again. So, if the length of the pendulum is changing. So, then it is equivalent to move on the surface of a sphere with variable radius, but what is the difference? Whenever we talk of a pendulum ok. So, we actually think that the pendulum is actually moving under the effect of gravity, which is always vertically downward and due to some initial conditions well chosen or smartly chosen initial conditions the pendulum is actually moving in all the possible directions ok. So, with that I will try to tell you that how we should construct the Lagrangian. So, how to construct the Lagrangian of a spherical pendulum? So, spherical pendulum is actually nothing but this yeah nothing but a pendulum, but which can actually now move to here I mean sorry here here here in this direction, but always being on the surface of the sphere of a constant radius. If we think that this is a spherical pendulum of constant string length ok yeah. Then of course, the Lagrangian will be equal to half. So, this is actually now covering three dimensions. So, Lagrangians will be plus z dot square times the mass. This is the kinetic energy minus the potential energy and you just see that if you now think that the reference point for calculating the potential energy is the center of the sphere, then at any point and if you are thinking that the gravitational acceleration is vertically downward ok. So, at any point you can actually think that the g is acting like this and z is directed again sorry z is directed again vertically upward direction ok. Then when the particle let us say the particle is p is at this position, what will be the potential energy? The potential energy will be then simply m g z this ok. It will have a positive potential energy ok. May be at this point for gaining this position it should have some enough initial condition that means someone has give a push. So, that this particle is now here ok. So, this will be simply m g z. Now, from the symmetry of the system since we are saying that the length is fixed, so it is actually equivalent to moving on the surface of a sphere of constant radius. What we will do? We will simply say that we choose spherical polar coordinates and now you know the number of generalized coordinates are reduced to 2 which are theta and phi. So, once again for you to recapitulate a little bit. So, what is theta? Theta is the angle ok between the position vector of the particle and the z axis ok this angle and this is theta it is called the co latitude and the azimuthal angle is the angle if you remember a little bit ok. This is the called azimuthal angle and this is the angle it makes with some reference direction here which we can call actually it is like some reference direction and if you make a projection of this point on this x y plane, then from that reference direction if the projection is making some angle phi on this plane which is perpendicular to this plane of the paper and this plane actually carries this azimuthal angle phi. So, with this we can actually have two generalized coordinates and we know that the standard transformation from a Cartesian system to spherical polar is as simple as just writing x is equal to. So, I am writing for this system actually r sin theta cos phi y is equal to r sin theta sin phi and z is equal to r cos theta ok. Here again all the r's are constants for our case ok. Since we have this I can actually write you can actually check that by explicit calculations that if you do I can actually say that I can do one calculation and then you can do the rest. So, x dot will be then r cos theta cos phi theta dot ok plus r sin theta minus sin phi phi dot, y dot will be equal to r cos theta sin phi theta dot yes plus r sin theta cos phi phi dot ok have I done it correctly. So, first thing is r yes cos theta yes exactly and then sin theta cos phi right yes. Finally, z dot is equal to minus r sin theta theta dot. If you now do x square x dot square plus y dot square plus z dot square remember x square plus y square plus z square is r square. So, here you will have r square actually r dot square plus r square theta dot square plus r square sin square theta phi dot square ok. Now, you know that this r is constant. So, this term will go away this is 0. So, you will actually have your Lagrangian to be half m r square theta dot square plus r square sin square theta phi dot square ok minus v v is m g z. So, it will be m g r cos theta ok. So, now, directly from this Lagrangian of course, you can have some conclusions and I mean the again first thing according to my suggestion this is always the best way to solve the Euler Lagrange equation for this Lagrangian ok that you will do I will also do for that problem. But before even solving the for the Euler Lagrange equation what to conclude about any symmetry of the Lagrangian. So, the first thing is that this has one cyclic coordinate can you say which one. So, this is phi because this Lagrangian is actually containing both theta dot and phi dot, but it contains only theta, but here as well, but no phi ok. So, phi is a cyclic coordinate and that is why the first conserved quantity will be del L del phi dot which is nothing, but half m r square sin square theta phi dot ok. So, this is a constant for that system. Can you see any similarity from your knowledge of central force can it also be thought as the angular momentum you think ok. Then I can say just a minute yeah I can say that this L is also not containing time explicitly minus del L del phi dot phi dot plus del L del theta dot theta dot is conserved. I also suggest you to check this conservation and to conclude that whether this conservation is equivalent is it equivalent to the conservation of the total mechanical energy ok check. Finally I will somehow do some calculations. So, I will just do the calculation with respect to theta. So, you see the only meaningful generalized coordinate is theta because this is not cyclic one. So, I will just do. So, for us the non-trivial Euler Lagrange equation will be actually d d t of del L del theta dot minus del L del theta is equal to 0. You see this is the marvel of this whole formalism although these two systems this pendulum in a plane with variable length and this spherical pendulum they are quite different. At one point you see that it has also only the generalized coordinate is theta angle theta it is only consisting of only one meaningful generalized coordinate which is theta and finally, the main I mean here you have the only one Euler Lagrange equation you have here two, but one just gives you the conservation ok trivially. So, the only one non-trivial Euler Lagrange equation is this one ok. So, you see that ah from the point of view of from the operational point of view both of them are actually very similar ok. So, we have to now calculate. So, if we do del L del theta dot that is simply equal to m r square theta dot. So, d d t of that ok del L del theta dot will be simply given by m r square theta double dot you will have del L del theta and that is a little bit tricky here you have to be a little bit careful. So, then you will have actually two terms here and here. So, from the first one you will have. So, yeah m r square sin theta cos theta yeah phi dot square and you have plus m g ah r sin theta yeah ok I just extend the calculation over here in this page. So, here you will see that if I do this you will have m r square theta double dot and then that will be equal to del L del theta. So, you will have actually m r square sin theta cos theta phi dot square plus m g r sin theta ok. So, this is the equation of motion for a spherical pendulum equation of dynamics for a spherical pendulum. Now, here the caution is that you actually should use the quantity or the constraint you get from this is not a constraint, but the constant ok you get from this relation ok. So, every time you see a phi dot you can actually get rid of this phi dot by just saying that this constant is some let us say m. So, phi dot will be m by half m r square sin square theta ok. So, basically you will get rid of this phi dot square as well ok. So, you will finally, have some relation including theta double dot theta and theta like this ok under which condition this goes to or reduces to the case of a simple pendulum. If we say phi is constant ok then this phi dot is actually 0 and then you have finally, theta double dot will be equal to ok g by r sin theta ok Then of course, you can say that for small theta it will have g by r theta ok and then you can say that this is the an equation which is very similar to that of a simple pendulum, but here the only problem is in general we see there is a minus sign and here we do not get that minus sign that is only because that in case of a pendulum ok in a plane the traditionally the angle we define this angle is from this instantaneous position vector of the particle to the vertical which is vertical downward vertical. So, this is an angle which is measured in the clockwise direction here the angle is measured ok from the position vector to the positive z direction which is anticlockwise ok. So, there is actually a minus difference between this two and you see that that causes this problem whenever you. So, finally, you will see that if you take into account those minus angles you will actually have this minus sign be taken care of and you will actually have the equation of a simple harmonic oscillator. How to do that again this is a small exercise for you, but for that you just do not write at the final step because you see if you just say that your theta is minus theta then here it will be also minus theta. So, from both sides minus theta will go away right from the beginning you have to change those theta. So, you have to take care of the fact that the relation between co latitude type of angle and the angle which we are talking here they are actually related by this. So, an angle like this. So, one is that this angle is 180 minus this angle. So, for example, if I still say that this angle let us say I am saying theta 2 and this co latitude. So, for when the when the position vector is like here then the co latitude is let us say theta 1. So, the first thing is that theta 1 plus theta 2 is equal to 180 degree magnitude wise I agree, but then of course, one of them is calculated in this direction the other one in this direction one of them is positive another one is negative. How to take into account of that you should try it and finally, come up with an equation where you actually find the final equation should be consistent with theta double dot is equal to g by r theta. With that I am wrapping up this lecture from the next week we will start discussing the physics of phase space where you will see that how the mechanics of a particle can be studied in a space where the coordinates or the directions are made by both position and velocity and eventually time. Thank you very much.