OK, should we make a start? Welcome back to dynamics and for those watching online?
Welcome. So last week we started to study constrained dynamics and in particular we were looking at the motion of a particle.
This act is done by the force of gravity, where the particles then further constrain to move along a particular curve or surface in three.
We looked in detail at the example of a simple pendulum, and there the pendulum constrains the particle to move on a circle in a vertical plane.
So that's an example of constrained motion on a curve, namely a circle.
In today's lecture, instead, I want to look at constrained motion on a surface.
So you might imagine this is modelling a small bead that's free to move on a nice, smooth friction the surface.
And for simplicity, we're going to take that surface to be a surface of revolution.
So we start the lecture today by reminding ourselves what a surface of revolution is from the geometry course last term.
All right.
So I'm going to start on the right hand board state because I'm going to draw a picture and then refer back to the picture throughout the lecture.
All right. So we study. A particle.
Moving under gravity. OK.
On the smooth. Inside of a surface of revolution.
So on the line, that's i.e. a surface defined by.
Z is Capital H of R, where are Peter and said that Triple R cylindrical polar coordinates?
OK, so remind you, cylindrical polar coordinates Z here, that's the usual height coordinates measured vertically upwards from some origin.
And then ah, and thetr they denote polar coordinates in the horizontal x y plane, and I'm going to draw a picture of that.
So if you're going to draw this picture, then try and draw it big because there's quite a lot going on the picture.
So the Z-axis runs vertically up the boards, that's the upwards direction, and then I'm going to draw the y axis horizontally this way.
That's why and then the x axis. Runs this way.
And the axes intersect at the origin, oh, so this is the X y plane down here at Z equals zero.
Remember that they are coordinating polar coordinates. That's the horizontal distance of points from the origin o in the X y plane.
So that means if I draw this straight line here in the X Y plane from the origin to another point, the length of that line is all.
And then the thetr coordinates in polar coordinates is measured this way.
So theatric with zero is the positive x axis and we measure thetr in the anti-clockwise direction and the x y plane as viewed from above.
I was in the picture here. OK, so next I'm going to draw my surface of revolution on this picture and I'm going to draw it in blue.
All right. It's like that.
OK, so that's supposed to be a surface and the height of points on this surface, so the height z is given by Capital H of R.
So this function capital H, that's going to determine the shape of my surface of revolution.
Now if you look at the set of points in the X Y plane that are at a constant distance are from the origin.
That's a circle of Radius R in the X y plane and the height of points on my surface.
They depend just on that radius R of the circle. So that means that all the points on the circle of radius are they all mapped to the same height.
That also means then that if I cut my surface with a horizontal plane,
so at some fixed height, then the cross section of the surface is always a circle.
It's a circle of radius R, where the height z and the radius of the circle of related by this equation.
And that's also what makes this a surface of revolution.
So it's invariant into rotating the surface about the z axis and those rotations about the z axis.
They precisely shift the theatre coordinate here by a constant and that rotates the circles that are the horizontal cross-sections.
So let me write that as a remark underneath. So you all k remark the surface is invariant.
Under rotation about the Z-axis. OK, next, I'm going to draw on my point particle.
So that's the up there, that's the point particle that's going to move on the surface and then I drop a perpendicular down here.
So that's a right angle from the line between the particle down to the X y plane.
Then next, I'm going to introduce an author normal basis for these cylindrical polar coordinates.
So is it is a unit vector pointing vertically upwards along the Z-axis?
In previous lectures, I've called that vector. But at least for today's lecture is it will be a unit vector pointing vertically upwards.
That's E-Z.
And then if my base is vectors in the horizontal plane, I'm going to use polar coordinate basis that we introduced in the lecture last Thursday.
So you might remember that that is air, which is a unit vector pointing radially outwards.
So again, if I continue that line here onto the z axis, that's a right angle.
And moreover, the length of that line, there is precisely this distance off.
And then finally, if theatre is a unit vector, that's perpendicular to that in the direction of increasing theatre.
So is it on a theatre unit, vectors and then mutually perpendicular?
So they form and also normal basis? And I can then use that to rate the position vector of my particle.
So the position vector that's a vector are of the particle is scalar r e r plus z z.
So scalar air that's the position of the particle in the horizontal plane. And Z is that gives its height.
And finally, the last thing to draw on is the forces that are acting.
So we're told here it's moving under gravity. So that means they will have a wait and.
Acting downwards. So M is the mass of the particle little g acceleration due to gravity as usual.
And then there must be some kind of contact force between the surface and the particle.
And I'm going to draw that contact force just going this way.
And that's going to be Victor, and that's the push of the surface on the particle, the contact force.
All right. Any questions on the set up so far is that is that clear, you want to ask anything?
OK, so we can now write down Newton's second law for the particle.
And so and two is Newton's second law, the mass of the particle little m times acceleration to our double dot is the total force acting.
That's Vector F. And there are two forces, there's gravity.
So that's minus MJG times E Z and then plus the contact force.
And the next step is the right acceleration here in terms of our cylindrical polar coordinate basis.
And we can do that using the results from section 4.1. So that's earlier in the course.
So that allows us to right the left hand side as mass. And then in brackets.
So it's scalar are double dot minus our theatre dot squared off plus one over r d, d t of our squared fita dot he theta plus z double dot.
He said. So that's the left hand side here of the equation, in particular in the square brackets, that's just acceleration in cylindrical policy.
So to compute that when I take a double dot here, well, I need to take two time derivatives of this first term r e r.
And that's precisely the first two terms that I've written here,
and we derive this expression for acceleration in polar coordinates in Thursday's lecture last week and then finally,
taking two time derivatives of Z, he said, is that it's just a unit vector pointing upwards.
So that said, double dots. He said, OK, so that's the left hand side.
And on the right hand side, we have the total force acting minus energy Z Plus and the contact force.
All right. So that's a vector differential equation. It's not immediately obvious what to do with it.
So you might decide to write it out in Cartesian components.
But just as an example, the simple pendulum that we looked at last week,
we should be a bit smarter about what it is that we're actually trying to do with this equation.
What do we want from it? In particular, if we're not interested in working out what's this contact force and is,
we can immediately eliminate it from the equation by taking a dot product of this equation with a tangent vector to the surface.
That's exactly what we did for the simple pendulum last week, and this is a surface of revolution and there's an obvious tangent vector.
So remark our MK e theta is tangent to the surface.
And we can see that from the picture here, if, Peter,
that's this direction that's always tangent to a circle of constant radius in a horizontal plane.
And so it's also tangent to the surface. So that's the first remark.
And the second remark is that we're told that it's a smooth surface in the preamble at the top there.
So it's a smooth surface. That means physically that there's no frictional force acting, so if and only if, no friction.
And last week we said that that means that the contact force end is perpendicular.
Or equivalently normal means the same thing.
Quiver. Mentally normal and is for normal to the surface.
So as we were discussing last Friday, if I had any component of my contact force, that's tangent to motion along the surface,
then physically that would have to be some kind of frictional force or resistive force of the surface of the particle.
So when we say it's moving on a smooth surface or friction, the surface geometrically,
that means that my contact force is perpendicular to the surface or in other words, there's no component tangent to the surface.
And so I immediately deduce that PN tilted into a theatre is zero.
So it is tangent to the surface and is normal. So that dot product to zero in there orthogonal.
OK, so that's the key point. If I now take Newton's second law above and I dot it with theatre.
Well, Easter is orthogonal to air and it's orthogonal to E-Z.
And so we'll say to this term. And it's also worth talking all day and as well.
So the only term that survives when you take the DOT product is this one proportionality feature.
So we immediately deduce that one over R D by d t of R squared theatre dots is zero.
So obviously we can multiply through by all here. And then this is just saying that our squared theta dot is independent of time.
It's a constant. So all square at Theatre Dot. And I'm going to define that quantity to be a little h.
And we've just deduced that it's constant. We met this quantity little h briefly earlier, and there it was related to angular momentum.
And that's also true here. And we can see that by computing the angular momentum of the particle.
So let's compute the angular momentum. Of the political.
About the origin of. So the angular momentum that's a vector L.
It's defined as, oh, that's the position of the particle relative to the origin.
And then you take a cross product with the linear momentum of the particle, and that's just mass times its velocity dot.
That's the definition, and then I can substitute in here for me. These quantities in terms of my cylindrical polar coordinates.
So that's Mars. And then in brackets scalar r e r plus z E-Z, that's position in cylindrical.
And then for the velocity in single payloads that scalar r dot e r.
Plus, Theta Dot Theta Plus Z does is that.
Say the second term here. That's velocity and cylindrical policy, in particular, the first two terms.
That's just components of velocity in the horizontal plane.
And again, we derive this formula in polar coordinates last week and finally the component of velocity in the z direction that's just said Dot.
He said. OK, and if you now multiply out those cross products and simplify, you get the following.
So there's an overall factor of the mass and then our squared theta dot e z plus z or
dot minus z dot e fita minus z are theta dot he r and in deriving that line I've used.
So using e r crossed with e thetr is e z and I right plus the cyclic permutations of that equation.
All right, so what did I do to get this line here? So there's a cross product of two vectors, and I've just expanded that out,
so I have er here crossed with the R zero because the cross product of the vector with itself is zero.
But then I've also got e crossed with the Fita and I've used that that's equal to E-Z.
I can. We've seen that equation, that particular cross product before in previous lectures.
So that gives you this term R-Squared feature Dot E-Z and for the remaining terms, you can use the cyclic permutations of this equation.
So also, if thetr crossed with each set, is it all?
And is it crossed with the R? Is it theatre?
And those just follow by writing out the original definitions of these vectors in terms of Cartesian coordinate basis.
I, J and K. All right.
Any questions on on that so far? All right. All right.
Very good. So we see if you look at that, you see that if you take L, the angular momentum about the origin and you dealt with E-Z,
that just picks out the first term and that's m r squared theta dot.
And that's exactly m little h. Well, I introduce little h is R-Squared Theta Dot above.
And we've just seen that's constant. So i.e. we have shown.
The components. Of angular momentum l along the axis of symmetry.
That's the E-Z direction is conserved.
So that is it's constant when evaluated on a solution to Newton's second law.
So, by the way, the other two terms here in the expression for angular momentum, they're not constant in general.
It's nothing particularly special about those components. All right.
So we've extracted one component of Newton's second law so far,
and it's good to take stock at this point and again, think about what are we trying to do in the problem?
So, all right, this is a remark. So note that Z is capital h of are not just defines the points on my surface of revolution.
So given R of T, we know Z of T.
OK, so, Z, if T is just H of R of T and geometrically here the scale R of T, it's just this distance here.
It's the horizontal distance of my Potts Point particle from the Z-axis since the length of that dotted line.
OK, but also we've just shown that defeated by D T is h over R-Squared of T.
So that's this line here.
If you just write Theta Dot here and take the other two terms to the right hand side, get started on this little h divided by all squared.
So again, so given R of T. We just integrate that equation to get Theatre of tea.
And just by integrating taking the ditty to the right hand side and integrating it, we get theatre as a function of tea.
So in other words, if I know this scale of function of tea, then I know ZF tea and I know theatre of tea.
So I know exactly where the particle is. And I've solved for the dynamics of the particle.
So this implies that we want an equation of motion.
For this function, the scale of tea and on general grounds, we expect that to be a second order ordinary differential equation.
As for the simple pendulum example, last week.
We may extract this. So the equation of motion we're looking for from Newton's second law and to dot it into Vector,
it'll see where this vector little T is another tangent vector to the surface.
OK, so we've already dotted Newton's second law with one tangent vector to the surface, but it's a surface, so there are two tangent vectors.
In general, the position vector of my particle remember so the position vector is scalar.
Ah, he ah, plus capital h of ah, he z.
That's the general position of my particle. If I come over here and again, look at my picture if I fix some constant value of theta.
So let's say thetr equals five or two. So I'm along the y axis here.
Then as a very scalar ah, in that last expression for position, I precisely move along.
Some curve up here like this on my surface.
OK. So if I then want a tangent vector along that curve on the surface, I just have to differentiate with respect to this parameter.
Are you doing that in the geometry course last term?
So this implies the tangent vector T. That I'm looking for is, would you just take the derivative d by d r of the position vector R above?
So here is just a constant vector. And if I differentiate R with respect to R, I get one and then plus h dashed over.
He said. And so here at Daskal tonight, derivative with respect to the arguments are.
So that's another tangent vector to the surface. And so I can take Newton's second law again and dot it with this tangent.
Vector T. R. Newton's second law is disappeared off the top.
Oh, it's a long way off the top. I won't try and find it again, so air and he said to both perpendicular, to each fighter.
And they're also perpendicular to the contact force.
And so quite a few terms drop out when you take the dot product and you get.
And then in brackets, double dots minus 50 dots squared, plus an h dash of R Z double dot.
Is minus m g h dashed.
OK, so that's what you get by dotting Newton's second law with that tangent vector.
OK. And then following our remark above, we can substitute.
Fifita Dots Theatre Dot is little h over R-Squared, so I can substitute that in for theatre dot squared in the line above.
And moreover, remember that Z is capital H of R and so then by the chain rule that implies that Z Dot is h dashte of r o dot.
And similarly, I can compute said double dots by taking another time derivative,
so that's h double dashed of our second derivative of H with respect r o dot squared.
Plus each dashed at a double dose.
So that's the Chambal that we've used them. OK, so we can see 52 dots into the line above and also for that double dots into the line above.
And if you do that, you get the following.
So in brackets one plus h dashed of our all squared close brackets are double dots plus h dash of our h double
dash double all adults squared minus h squared over cubed is equal to minus little g h dash of R and as promised.
So remark that's a second order. Odie Second Order 0d ordinary differential equation for R of T.
It's not a very nice looking equation, the equation, of course, depends on the choice of surface of revolution,
and the choice of surface depends on this capital of our function.
But even for quite simple choices of that function,
this is a pretty nasty differential equation and we're not going to be able to solve it in closed form in general.
But given a differential equation, there are always certain things you can do with it.
So you might be able to deduce some general properties of solutions of this equation.
You might better find approximate solutions,
or you can just stick it onto a computer and solve it numerically and see what the particle actually does.
However, rather than do that, I instead wanted to take a different approach.
So also last week we were discussing conservation of energy, and in particular, we showed that for a particle moving on to gravity,
moving along a smooth curve or a smooth surface, there's always a conserved energy.
It's the first point.
And the second point is we expect that to give a first order differential equation rather than a second order differential equation.
So let's say that that is the case. So there's a better approach.
Then just trying to tackle this second delivery equation directly, and that's to use conservation of energy.
And that was in section five point one from section five point one.
So that was quite a general analysis, and we deduced that E the total energy of the particle,
so that's defined as T plus v, where T is kinetic energy, if the particle and V is the potential energy.
So that's the definition of E. And so the kinetic energy as a half and and then modulus of ah dot squared.
So that's the speed squared of the particle. And the potential energy here is the gravitational potential energy.
So that's m times g times Z with this at the height.
So that term there is the gravitational potential energy.
And we know that's constant. That was a result that we derive last week.
And so we can compute more explicitly what this quantity is.
So I want to write out this speed squared in terms of my cylindrical polar coordinates as a first step.
So modulus of our dot squared that is defined as R dot dot it into our dot and that scalar r dot e r plus R42 dot,
he fita + said dot e said it's a modulus squared.
And so when I take the dot product of that vector with itself, remember the e r e thetr and E's edge.
They're all perpendicular to each other and they're all unit vectors as well.
So actually, this is just the sum of squares of the coefficients. When you make this out.
So it's very simple. It's just scalar dot squared plus are squared, theatre dot squared plus z dot squared.
Some of the squares of the coefficients. OK.
And then just as before, we can substitute. Fifita Dot sets up the eyes of a R-Squared and also said to Dot is H Dash to R R Dot.
So those equations already above. So if you substitute all of that into this conservation of energy equation,
here we get that E is equal to a half times the mass and then we get scalar R dot squared plus and then for our squared theta dot squared,
I can substitute for Theta Dot here. So that's little h squared over our squared.
And then finally, z dot squared from here. That's H dashte of r squared dot squared.
OK, so that's the kinetic energy term plus energy and then Z is closely capital HRR, and that's constant.
And as promised, that is a first order ODI for our team.
So will mark a first order ODI ordinary differential equation for our of two.
OK, so if you take debility of this equation, using the E!
Is constant, I will get our double dot turns on the right hand side.
And you can check if you want that the equation that you get is precisely the equation of motion with odds double dancing.
That's up here from Newton's second law. Another way to say it is we've effectively integrated Newton's second law once to obtain a first order
equation rather than a second order equation where e here on the left hand side is an integration constant.
OK, so the first order equation here is some function of R and R dot is E, which is a constant.
It's a first order equation, though in principle we can integrate this.
But again, we won't be able to do the integral explicitly and find the solution in closed form.
However, we've now seen many times in the lectures that using conservation of energy, we can deduce some qualitative features of the motion.
And also some quantitative features of the Motion two. And so that's what I wanted to demonstrate to you in an example.
So let's actually look at a particular example of this. So it's motion on a parabola weight.
All right, so I'm going to choose a particular surface now, so remember that said is Capitol Hill,
and I'm going to take that to be R-Squared over for a with a some positive constant.
So Z is all squared over 4A. That's a parabola.
And the surface of revelation that you get looks pretty much exactly what, like the one that I've drawn.
So that's why I drew that particular surface that looks pretty much like what that surface looks like.
So the particle is initially at a height z equals a height.
He said he was a and is projected horizontally.
With Speed Little V, and our problem is to show.
The particle moves between two heights, say the particle.
Moves between to heights. In the subsequent motion of.
All right, so this is an initial value problem, we're given the height of the particle initially and also its velocity initially.
So initially. Z equals a.
On the other hand, we also know that R-Squared is for a Z. OK, just by inverting R-Squared here in terms of said.
And so since that is a initially, then all must be to initially remember that R is positive.
So when I take the square root here, I take the positive square root to get R is to a and then we're also told it's projected horizontally.
At speed, we. That tells us there are dots.
It's very easy to initially. So again, if you look at the picture at some height, Zedek was a say this height,
and if it's moving horizontally, that means it's precisely along the east direction.
So tangent to a circle of constant radius, a feature is a unit vector.
So the coefficient here V is precisely the speed.
Moreover, so but in general, we've got the all dots, vector dot is scalar aardvarks e-r plus theatre dots in theatre,
so that tells us that initially scale are dot is zero.
And Theatre Dot is V by just comparing the general expression for velocity here with the initial condition that our Dot is V in theatre.
So scalar dot is zero means there's no component of velocity in the outwards radial direction.
OK, then the problem tells us to show the particle moves between two heights in the subsequent motion.
And so I'm going to eliminate or in terms of Z. In this case, because Z is the height and it's that that I'm interested in in the problem.
So R is to square root of a z in general.
So solving for R in terms of Z and so are dots is equal to the square root of over Z.
Said Dot, using the chain rule. And we can substitute all of that into conservation of energy.
So conservation of energy implies E!
Is a half times the nice and then for.
So I've got the expression, Oh, it's off the top there.
So the first time is all dot squared, but I can substitute that is a to Z, said dot squared plus h squared over for a z plus z dot squared.
So that's the kinetic energy term now express there in terms of Z Dot and Z Plus Energy, Z is the potential energy term.
And we know that that's constant. So we can evaluate this at any time that we like and it's always the same constant.
In particular, I can evaluate it at the initial time when I know the initial data.
So the initial time scale are dotted zero set that above.
So we'll say Z Dot is zero. So this term and this term is zero initially.
And so this is a half m little h squared over for a squared putting in Z is a initially plus M.G. A.
So this second line, I've just evaluated it initial time. And finally, we can also work out this constant little h.
Remember that little h is all squared theta dot. That's the definition I can write that is all times theatre dot.
And this is a constant so I can work out the constant by evaluating it at the initial time.
And initially R is to a and our feature is V, so that's too heavy.
So if you substitute that into the equation for energy, so substitute.
And then rearrange the conservation of energy equation, so I'm reading this here is this line here,
which is go to Z Dot squared and it equals the right hand side, which is a constant.
And I solve this for z dot squared. So all the terms with a z squared.
Put those on one side and everything else on to the other side and you get this z dot squared is.
And a little bit of algebra that tells you the right hand side is minus two g over A-plus, said time z minus v squared over two g z minus a.
OK, so that is the conservation of energy equation when you rearrange it for Z Dot squared.
Now notice the following. So this term here is positive.
It'll get positive, A is positive and Z is bigger than or equal to zero members.
Z is a is a parabola. So it's all squared over 4a. So Z is bigger than an equal to zero.
So this term is positive. But then there's a minus sign here.
On the other hand, on the left hand side, z dot squared that's bigger than or equal to zero.
So since said dot squared, that's manifestly bigger than zero.
And this first term out here with the minus sign is less than zero.
That means that these last two terms must be less than or equal to zero.
So we did use this is less than or equal to zero. And this is a quadratic in Z.
So it's a quadratic and Z looks like this if you sketch it and if it's less than or equal to zero,
then you must live between the two roots of the quadratic and that Z equals A and Z equals V squared over two g.
So you can immediately say that the particle. Stays between heights.
Z equals A and Z equals V squared over two g, which are the two routes of that quadratic.
So this shows that the particle always moves in some strip around my surface of evolution.
So if not, so for the dynamics. But it does lie always between these two heights.
Z equals as the initial height, right, as the height that I projected to at if v squared over two g is bigger than a,
so that's for large velocities, then it always stays above the initial height.
OK, because it's got a lie between this and that and that one's bigger.
On the other hand, if V squared over two g is smaller than A, that's the smaller velocities,
then it always stays below the original height that I projected it out. And that does make some intuitive sense.
Any questions on that? Before we change topic.
All right, so there's quite a bit. Oh yeah, there is one, yeah. First of all, thank you very much.
I just want to get back to the first question.
Give us an intuitive reason as to why conservation momentum around the world, the Z-axis is constant.
And then also, you can't be manoeuvred your way without having to consider the normal force itself.
Yes. So basically, we then increase the normal course of various points to the conservation momentum.
So this constellation of simple things are both really great questions.
So first question about conservation, of angular momentum, about the z axis.
Yeah. So I was going to make a comment on that and then I decided not to.
But now you've asked, I will say so.
There's a general theorem in dynamics called an artist's theorem, and it says whenever you've got a symmetry of your dynamics problem.
So in our case, that's the rotational symmetry of our surface of revolution.
Whenever you've got a symmetry like that, there is always a conserved quantity.
And for us, that was the component of angular momentum along the Z-axis. But it's a very general theorem, and it's constructive as well.
So it's given the symmetry it tells you how to find the conserved quantity,
said MINURSO is a female mathematician working in the first third of the 20th century.
So she mainly works in algebra. If you go in and take algebra courses in the second and third year, you'll see nurse's name all over them.
But then she also published this result in dynamics in 1918. So exactly a hundred years ago?
Pretty much. And it's a very general theorem, not just in mathematics, but in physics, too.
And it's played a very prominent role in the development of theoretical physics in the 20th century.
It's I can't possibly give a proof of this because it's in the third year course on classical mechanics.
So if you're interested in a sort of deeper answer to why you there's a conserved, angular momentum, then you should take that course.
I think that's the easiest answer to that. Then the second point you're asking about the push of the surface.
So we could also determine that force.
If you take a dot product of Newton's second law with a normal vector, you'll find an equation for that contact force.
And it's just determined, just like it was for the simple pendulum. There's a formula for it.
If you want to give an extra push like someone is banging the surface or something is the particle goes round,
then there won't be a conserved energy because someone is putting in some extra energy,
and it's the energy we wrote down would be conserved, that's for sure, right?
So you require conserve energy and segmented.
We needed in order to prevent that you mentioned, we need to know that so they're not directly relate.
So in this problem, there are both, but they're not directly related to each other. Thanks.
Thanks for great questions. Any other questions?
All right, so I was going to start on the next. Just briefly on the next topic, but I'm definitely out of time.
So tomorrow we're going to change topic and instead we're going to look at the universal law of gravitation.
So we've had many problems now in this course where we're looking at particles acted on by the force of gravity.
But as gravity near to the Earth's surface, there's a much more general universal law of gravitation due to Newton.
We'll start that in tomorrow's lecture,
and one of the main results of that section will be to show that planets move on elliptical orbits about the Sun.
Under that universal law of gravity. So see you then.