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OK, should we make a start? Welcome back to dynamics and for those watching online?

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Welcome. So last week we started to study constrained dynamics and in particular we were looking at the motion of a particle.

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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.

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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.

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So that's an example of constrained motion on a curve, namely a circle.

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In today's lecture, instead, I want to look at constrained motion on a surface.

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So you might imagine this is modelling a small bead that's free to move on a nice, smooth friction the surface.

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And for simplicity, we're going to take that surface to be a surface of revolution.

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So we start the lecture today by reminding ourselves what a surface of revolution is from the geometry course last term.

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All right.

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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.

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All right. So we study. A particle.

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Moving under gravity. OK.

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On the smooth. Inside of a surface of revolution.

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So on the line, that's i.e. a surface defined by.

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Z is Capital H of R, where are Peter and said that Triple R cylindrical polar coordinates?

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OK, so remind you, cylindrical polar coordinates Z here, that's the usual height coordinates measured vertically upwards from some origin.

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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.

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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.

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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.

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That's why and then the x axis. Runs this way.

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And the axes intersect at the origin, oh, so this is the X y plane down here at Z equals zero.

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Remember that they are coordinating polar coordinates. That's the horizontal distance of points from the origin o in the X y plane.

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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.

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And then the thetr coordinates in polar coordinates is measured this way.

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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.

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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.

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All right. It's like that.

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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.

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So this function capital H, that's going to determine the shape of my surface of revolution.

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Now if you look at the set of points in the X Y plane that are at a constant distance are from the origin.

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That's a circle of Radius R in the X y plane and the height of points on my surface.

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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.

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That also means then that if I cut my surface with a horizontal plane,

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so at some fixed height, then the cross section of the surface is always a circle.

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It's a circle of radius R, where the height z and the radius of the circle of related by this equation.

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And that's also what makes this a surface of revolution.

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So it's invariant into rotating the surface about the z axis and those rotations about the z axis.

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They precisely shift the theatre coordinate here by a constant and that rotates the circles that are the horizontal cross-sections.

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So let me write that as a remark underneath. So you all k remark the surface is invariant.

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Under rotation about the Z-axis. OK, next, I'm going to draw on my point particle.

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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.

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So that's a right angle from the line between the particle down to the X y plane.

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Then next, I'm going to introduce an author normal basis for these cylindrical polar coordinates.

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So is it is a unit vector pointing vertically upwards along the Z-axis?

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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.

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That's E-Z.

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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.

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So you might remember that that is air, which is a unit vector pointing radially outwards.

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So again, if I continue that line here onto the z axis, that's a right angle.

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And moreover, the length of that line, there is precisely this distance off.

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And then finally, if theatre is a unit vector, that's perpendicular to that in the direction of increasing theatre.

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So is it on a theatre unit, vectors and then mutually perpendicular?

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So they form and also normal basis? And I can then use that to rate the position vector of my particle.

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So the position vector that's a vector are of the particle is scalar r e r plus z z.

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So scalar air that's the position of the particle in the horizontal plane. And Z is that gives its height.

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And finally, the last thing to draw on is the forces that are acting.

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So we're told here it's moving under gravity. So that means they will have a wait and.

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Acting downwards. So M is the mass of the particle little g acceleration due to gravity as usual.

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And then there must be some kind of contact force between the surface and the particle.

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And I'm going to draw that contact force just going this way.

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And that's going to be Victor, and that's the push of the surface on the particle, the contact force.

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All right. Any questions on the set up so far is that is that clear, you want to ask anything?

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OK, so we can now write down Newton's second law for the particle.

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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.

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That's Vector F. And there are two forces, there's gravity.

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So that's minus MJG times E Z and then plus the contact force.

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And the next step is the right acceleration here in terms of our cylindrical polar coordinate basis.

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And we can do that using the results from section 4.1. So that's earlier in the course.

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So that allows us to right the left hand side as mass. And then in brackets.

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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.

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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.

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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.

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And that's precisely the first two terms that I've written here,

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and we derive this expression for acceleration in polar coordinates in Thursday's lecture last week and then finally,

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taking two time derivatives of Z, he said, is that it's just a unit vector pointing upwards.

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So that said, double dots. He said, OK, so that's the left hand side.

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And on the right hand side, we have the total force acting minus energy Z Plus and the contact force.

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All right. So that's a vector differential equation. It's not immediately obvious what to do with it.

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So you might decide to write it out in Cartesian components.

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But just as an example, the simple pendulum that we looked at last week,

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we should be a bit smarter about what it is that we're actually trying to do with this equation.

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What do we want from it? In particular, if we're not interested in working out what's this contact force and is,

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we can immediately eliminate it from the equation by taking a dot product of this equation with a tangent vector to the surface.

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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.

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So remark our MK e theta is tangent to the surface.

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And we can see that from the picture here, if, Peter,

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that's this direction that's always tangent to a circle of constant radius in a horizontal plane.

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And so it's also tangent to the surface. So that's the first remark.

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And the second remark is that we're told that it's a smooth surface in the preamble at the top there.

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So it's a smooth surface. That means physically that there's no frictional force acting, so if and only if, no friction.

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And last week we said that that means that the contact force end is perpendicular.

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Or equivalently normal means the same thing.

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Quiver. Mentally normal and is for normal to the surface.

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So as we were discussing last Friday, if I had any component of my contact force, that's tangent to motion along the surface,

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then physically that would have to be some kind of frictional force or resistive force of the surface of the particle.

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So when we say it's moving on a smooth surface or friction, the surface geometrically,

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that means that my contact force is perpendicular to the surface or in other words, there's no component tangent to the surface.

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And so I immediately deduce that PN tilted into a theatre is zero.

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So it is tangent to the surface and is normal. So that dot product to zero in there orthogonal.

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OK, so that's the key point. If I now take Newton's second law above and I dot it with theatre.

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Well, Easter is orthogonal to air and it's orthogonal to E-Z.

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And so we'll say to this term. And it's also worth talking all day and as well.

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So the only term that survives when you take the DOT product is this one proportionality feature.

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So we immediately deduce that one over R D by d t of R squared theatre dots is zero.

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So obviously we can multiply through by all here. And then this is just saying that our squared theta dot is independent of time.

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It's a constant. So all square at Theatre Dot. And I'm going to define that quantity to be a little h.

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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.

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And that's also true here. And we can see that by computing the angular momentum of the particle.

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So let's compute the angular momentum. Of the political.

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About the origin of. So the angular momentum that's a vector L.

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It's defined as, oh, that's the position of the particle relative to the origin.

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And then you take a cross product with the linear momentum of the particle, and that's just mass times its velocity dot.

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That's the definition, and then I can substitute in here for me. These quantities in terms of my cylindrical polar coordinates.

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So that's Mars. And then in brackets scalar r e r plus z E-Z, that's position in cylindrical.

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And then for the velocity in single payloads that scalar r dot e r.

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Plus, Theta Dot Theta Plus Z does is that.

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Say the second term here. That's velocity and cylindrical policy, in particular, the first two terms.

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That's just components of velocity in the horizontal plane.

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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.

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He said. OK, and if you now multiply out those cross products and simplify, you get the following.

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So there's an overall factor of the mass and then our squared theta dot e z plus z or

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dot minus z dot e fita minus z are theta dot he r and in deriving that line I've used.

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So using e r crossed with e thetr is e z and I right plus the cyclic permutations of that equation.

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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,

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so I have er here crossed with the R zero because the cross product of the vector with itself is zero.

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But then I've also got e crossed with the Fita and I've used that that's equal to E-Z.

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I can. We've seen that equation, that particular cross product before in previous lectures.

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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.

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So also, if thetr crossed with each set, is it all?

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And is it crossed with the R? Is it theatre?

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And those just follow by writing out the original definitions of these vectors in terms of Cartesian coordinate basis.

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I, J and K. All right.

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Any questions on on that so far? All right. All right.

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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,

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that just picks out the first term and that's m r squared theta dot.

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And that's exactly m little h. Well, I introduce little h is R-Squared Theta Dot above.

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And we've just seen that's constant. So i.e. we have shown.

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The components. Of angular momentum l along the axis of symmetry.

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That's the E-Z direction is conserved.

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So that is it's constant when evaluated on a solution to Newton's second law.

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So, by the way, the other two terms here in the expression for angular momentum, they're not constant in general.

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It's nothing particularly special about those components. All right.

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So we've extracted one component of Newton's second law so far,

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and it's good to take stock at this point and again, think about what are we trying to do in the problem?

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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.

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So given R of T, we know Z of T.

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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.

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It's the horizontal distance of my Potts Point particle from the Z-axis since the length of that dotted line.

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OK, but also we've just shown that defeated by D T is h over R-Squared of T.

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So that's this line here.

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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.

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So again, so given R of T. We just integrate that equation to get Theatre of tea.

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And just by integrating taking the ditty to the right hand side and integrating it, we get theatre as a function of tea.

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So in other words, if I know this scale of function of tea, then I know ZF tea and I know theatre of tea.

156
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So I know exactly where the particle is. And I've solved for the dynamics of the particle.

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So this implies that we want an equation of motion.

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For this function, the scale of tea and on general grounds, we expect that to be a second order ordinary differential equation.

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As for the simple pendulum example, last week.

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We may extract this. So the equation of motion we're looking for from Newton's second law and to dot it into Vector,

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it'll see where this vector little T is another tangent vector to the surface.

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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.

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In general, the position vector of my particle remember so the position vector is scalar.

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Ah, he ah, plus capital h of ah, he z.

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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.

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So let's say thetr equals five or two. So I'm along the y axis here.

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Then as a very scalar ah, in that last expression for position, I precisely move along.

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Some curve up here like this on my surface.

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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.

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Are you doing that in the geometry course last term?

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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?

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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.

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He said. And so here at Daskal tonight, derivative with respect to the arguments are.

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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.

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Vector T. R. Newton's second law is disappeared off the top.

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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.

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And they're also perpendicular to the contact force.

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And so quite a few terms drop out when you take the dot product and you get.

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And then in brackets, double dots minus 50 dots squared, plus an h dash of R Z double dot.

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Is minus m g h dashed.

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OK, so that's what you get by dotting Newton's second law with that tangent vector.

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OK. And then following our remark above, we can substitute.

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Fifita Dots Theatre Dot is little h over R-Squared, so I can substitute that in for theatre dot squared in the line above.

184
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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.

185
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And similarly, I can compute said double dots by taking another time derivative,

186
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so that's h double dashed of our second derivative of H with respect r o dot squared.

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Plus each dashed at a double dose.

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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.

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And if you do that, you get the following.

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So in brackets one plus h dashed of our all squared close brackets are double dots plus h dash of our h double

191
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dash double all adults squared minus h squared over cubed is equal to minus little g h dash of R and as promised.

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So remark that's a second order. Odie Second Order 0d ordinary differential equation for R of T.

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It's not a very nice looking equation, the equation, of course, depends on the choice of surface of revolution,

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and the choice of surface depends on this capital of our function.

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But even for quite simple choices of that function,

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this is a pretty nasty differential equation and we're not going to be able to solve it in closed form in general.

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But given a differential equation, there are always certain things you can do with it.

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So you might be able to deduce some general properties of solutions of this equation.

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You might better find approximate solutions,

200
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or you can just stick it onto a computer and solve it numerically and see what the particle actually does.

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However, rather than do that, I instead wanted to take a different approach.

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So also last week we were discussing conservation of energy, and in particular, we showed that for a particle moving on to gravity,

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moving along a smooth curve or a smooth surface, there's always a conserved energy.

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It's the first point.

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And the second point is we expect that to give a first order differential equation rather than a second order differential equation.

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So let's say that that is the case. So there's a better approach.

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Then just trying to tackle this second delivery equation directly, and that's to use conservation of energy.

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And that was in section five point one from section five point one.

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So that was quite a general analysis, and we deduced that E the total energy of the particle,

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so that's defined as T plus v, where T is kinetic energy, if the particle and V is the potential energy.

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So that's the definition of E. And so the kinetic energy as a half and and then modulus of ah dot squared.

212
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So that's the speed squared of the particle. And the potential energy here is the gravitational potential energy.

213
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So that's m times g times Z with this at the height.

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So that term there is the gravitational potential energy.

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And we know that's constant. That was a result that we derive last week.

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And so we can compute more explicitly what this quantity is.

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So I want to write out this speed squared in terms of my cylindrical polar coordinates as a first step.

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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,

219
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he fita + said dot e said it's a modulus squared.

220
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And so when I take the dot product of that vector with itself, remember the e r e thetr and E's edge.

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They're all perpendicular to each other and they're all unit vectors as well.

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So actually, this is just the sum of squares of the coefficients. When you make this out.

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So it's very simple. It's just scalar dot squared plus are squared, theatre dot squared plus z dot squared.

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Some of the squares of the coefficients. OK.

225
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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.

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So those equations already above. So if you substitute all of that into this conservation of energy equation,

227
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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,

228
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I can substitute for Theta Dot here. So that's little h squared over our squared.

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And then finally, z dot squared from here. That's H dashte of r squared dot squared.

230
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OK, so that's the kinetic energy term plus energy and then Z is closely capital HRR, and that's constant.

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And as promised, that is a first order ODI for our team.

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So will mark a first order ODI ordinary differential equation for our of two.

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OK, so if you take debility of this equation, using the E!

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Is constant, I will get our double dot turns on the right hand side.

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And you can check if you want that the equation that you get is precisely the equation of motion with odds double dancing.

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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

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equation rather than a second order equation where e here on the left hand side is an integration constant.

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OK, so the first order equation here is some function of R and R dot is E, which is a constant.

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It's a first order equation, though in principle we can integrate this.

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But again, we won't be able to do the integral explicitly and find the solution in closed form.

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However, we've now seen many times in the lectures that using conservation of energy, we can deduce some qualitative features of the motion.

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And also some quantitative features of the Motion two. And so that's what I wanted to demonstrate to you in an example.

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So let's actually look at a particular example of this. So it's motion on a parabola weight.

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All right, so I'm going to choose a particular surface now, so remember that said is Capitol Hill,

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and I'm going to take that to be R-Squared over for a with a some positive constant.

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So Z is all squared over 4A. That's a parabola.

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And the surface of revelation that you get looks pretty much exactly what, like the one that I've drawn.

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So that's why I drew that particular surface that looks pretty much like what that surface looks like.

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So the particle is initially at a height z equals a height.

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He said he was a and is projected horizontally.

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With Speed Little V, and our problem is to show.

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The particle moves between two heights, say the particle.

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Moves between to heights. In the subsequent motion of.

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All right, so this is an initial value problem, we're given the height of the particle initially and also its velocity initially.

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So initially. Z equals a.

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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.

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And so since that is a initially, then all must be to initially remember that R is positive.

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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.

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At speed, we. That tells us there are dots.

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It's very easy to initially. So again, if you look at the picture at some height, Zedek was a say this height,

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and if it's moving horizontally, that means it's precisely along the east direction.

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So tangent to a circle of constant radius, a feature is a unit vector.

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So the coefficient here V is precisely the speed.

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Moreover, so but in general, we've got the all dots, vector dot is scalar aardvarks e-r plus theatre dots in theatre,

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so that tells us that initially scale are dot is zero.

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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.

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So scalar dot is zero means there's no component of velocity in the outwards radial direction.

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OK, then the problem tells us to show the particle moves between two heights in the subsequent motion.

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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.

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So R is to square root of a z in general.

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So solving for R in terms of Z and so are dots is equal to the square root of over Z.

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Said Dot, using the chain rule. And we can substitute all of that into conservation of energy.

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So conservation of energy implies E!

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Is a half times the nice and then for.

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So I've got the expression, Oh, it's off the top there.

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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.

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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.

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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.

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In particular, I can evaluate it at the initial time when I know the initial data.

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So the initial time scale are dotted zero set that above.

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So we'll say Z Dot is zero. So this term and this term is zero initially.

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And so this is a half m little h squared over for a squared putting in Z is a initially plus M.G. A.

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So this second line, I've just evaluated it initial time. And finally, we can also work out this constant little h.

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Remember that little h is all squared theta dot. That's the definition I can write that is all times theatre dot.

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And this is a constant so I can work out the constant by evaluating it at the initial time.

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And initially R is to a and our feature is V, so that's too heavy.

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So if you substitute that into the equation for energy, so substitute.

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And then rearrange the conservation of energy equation, so I'm reading this here is this line here,

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which is go to Z Dot squared and it equals the right hand side, which is a constant.

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And I solve this for z dot squared. So all the terms with a z squared.

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Put those on one side and everything else on to the other side and you get this z dot squared is.

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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.

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OK, so that is the conservation of energy equation when you rearrange it for Z Dot squared.

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Now notice the following. So this term here is positive.

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It'll get positive, A is positive and Z is bigger than or equal to zero members.

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Z is a is a parabola. So it's all squared over 4a. So Z is bigger than an equal to zero.

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So this term is positive. But then there's a minus sign here.

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On the other hand, on the left hand side, z dot squared that's bigger than or equal to zero.

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So since said dot squared, that's manifestly bigger than zero.

300
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And this first term out here with the minus sign is less than zero.

301
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That means that these last two terms must be less than or equal to zero.

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So we did use this is less than or equal to zero. And this is a quadratic in Z.

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So it's a quadratic and Z looks like this if you sketch it and if it's less than or equal to zero,

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then you must live between the two roots of the quadratic and that Z equals A and Z equals V squared over two g.

305
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So you can immediately say that the particle. Stays between heights.

306
00:46:01,680 --> 00:46:13,920
Z equals A and Z equals V squared over two g, which are the two routes of that quadratic.

307
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So this shows that the particle always moves in some strip around my surface of evolution.

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So if not, so for the dynamics. But it does lie always between these two heights.

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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,

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so that's for large velocities, then it always stays above the initial height.

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OK, because it's got a lie between this and that and that one's bigger.

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On the other hand, if V squared over two g is smaller than A, that's the smaller velocities,

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then it always stays below the original height that I projected it out. And that does make some intuitive sense.

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Any questions on that? Before we change topic.

315
00:47:04,130 --> 00:47:07,940
All right, so there's quite a bit. Oh yeah, there is one, yeah. First of all, thank you very much.

316
00:47:07,940 --> 00:47:13,090
I just want to get back to the first question.

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Give us an intuitive reason as to why conservation momentum around the world, the Z-axis is constant.

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00:47:20,930 --> 00:47:28,250
And then also, you can't be manoeuvred your way without having to consider the normal force itself.

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00:47:28,250 --> 00:47:34,880
Yes. So basically, we then increase the normal course of various points to the conservation momentum.

320
00:47:34,880 --> 00:47:40,400
So this constellation of simple things are both really great questions.

321
00:47:40,400 --> 00:47:45,410
So first question about conservation, of angular momentum, about the z axis.

322
00:47:45,410 --> 00:47:49,250
Yeah. So I was going to make a comment on that and then I decided not to.

323
00:47:49,250 --> 00:47:52,700
But now you've asked, I will say so.

324
00:47:52,700 --> 00:48:00,890
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.

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So in our case, that's the rotational symmetry of our surface of revolution.

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Whenever you've got a symmetry like that, there is always a conserved quantity.

327
00:48:10,490 --> 00:48:18,080
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.

328
00:48:18,080 --> 00:48:23,720
So it's given the symmetry it tells you how to find the conserved quantity,

329
00:48:23,720 --> 00:48:29,000
said MINURSO is a female mathematician working in the first third of the 20th century.

330
00:48:29,000 --> 00:48:35,780
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.

331
00:48:35,780 --> 00:48:40,340
But then she also published this result in dynamics in 1918. So exactly a hundred years ago?

332
00:48:40,340 --> 00:48:46,250
Pretty much. And it's a very general theorem, not just in mathematics, but in physics, too.

333
00:48:46,250 --> 00:48:51,980
And it's played a very prominent role in the development of theoretical physics in the 20th century.

334
00:48:51,980 --> 00:48:57,050
It's I can't possibly give a proof of this because it's in the third year course on classical mechanics.

335
00:48:57,050 --> 00:49:03,830
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.

336
00:49:03,830 --> 00:49:10,760
I think that's the easiest answer to that. Then the second point you're asking about the push of the surface.

337
00:49:10,760 --> 00:49:13,130
So we could also determine that force.

338
00:49:13,130 --> 00:49:21,980
If you take a dot product of Newton's second law with a normal vector, you'll find an equation for that contact force.

339
00:49:21,980 --> 00:49:26,180
And it's just determined, just like it was for the simple pendulum. There's a formula for it.

340
00:49:26,180 --> 00:49:31,550
If you want to give an extra push like someone is banging the surface or something is the particle goes round,

341
00:49:31,550 --> 00:49:35,930
then there won't be a conserved energy because someone is putting in some extra energy,

342
00:49:35,930 --> 00:49:39,320
and it's the energy we wrote down would be conserved, that's for sure, right?

343
00:49:39,320 --> 00:49:44,930
So you require conserve energy and segmented.

344
00:49:44,930 --> 00:49:52,120
We needed in order to prevent that you mentioned, we need to know that so they're not directly relate.

345
00:49:52,120 --> 00:49:57,390
So in this problem, there are both, but they're not directly related to each other. Thanks.

346
00:49:57,390 --> 00:50:03,310
Thanks for great questions. Any other questions?

347
00:50:03,310 --> 00:50:08,140
All right, so I was going to start on the next. Just briefly on the next topic, but I'm definitely out of time.

348
00:50:08,140 --> 00:50:15,220
So tomorrow we're going to change topic and instead we're going to look at the universal law of gravitation.

349
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So we've had many problems now in this course where we're looking at particles acted on by the force of gravity.

350
00:50:20,500 --> 00:50:27,610
But as gravity near to the Earth's surface, there's a much more general universal law of gravitation due to Newton.

351
00:50:27,610 --> 00:50:29,710
We'll start that in tomorrow's lecture,

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and one of the main results of that section will be to show that planets move on elliptical orbits about the Sun.

353
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Under that universal law of gravity. So see you then.