1
00:00:11,920 --> 00:00:19,180
OK, should we make a start? Welcome back to dynamics and for those watching online?
2
00:00:19,180 --> 00:00:30,940
Welcome. So last week we started to study constrained dynamics and in particular we were looking at the motion of a particle.
3
00:00:30,940 --> 00:00:40,450
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.
4
00:00:40,450 --> 00:00:50,170
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.
5
00:00:50,170 --> 00:00:55,330
So that's an example of constrained motion on a curve, namely a circle.
6
00:00:55,330 --> 00:01:00,610
In today's lecture, instead, I want to look at constrained motion on a surface.
7
00:01:00,610 --> 00:01:09,910
So you might imagine this is modelling a small bead that's free to move on a nice, smooth friction the surface.
8
00:01:09,910 --> 00:01:13,810
And for simplicity, we're going to take that surface to be a surface of revolution.
9
00:01:13,810 --> 00:01:21,740
So we start the lecture today by reminding ourselves what a surface of revolution is from the geometry course last term.
10
00:01:21,740 --> 00:01:21,970
All right.
11
00:01:21,970 --> 00:01:29,470
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.
12
00:01:29,470 --> 00:01:38,680
All right. So we study. A particle.
13
00:01:38,680 --> 00:01:47,660
Moving under gravity. OK.
14
00:01:47,660 --> 00:02:10,430
On the smooth. Inside of a surface of revolution.
15
00:02:10,430 --> 00:02:23,420
So on the line, that's i.e. a surface defined by.
16
00:02:23,420 --> 00:02:53,990
Z is Capital H of R, where are Peter and said that Triple R cylindrical polar coordinates?
17
00:02:53,990 --> 00:03:03,020
OK, so remind you, cylindrical polar coordinates Z here, that's the usual height coordinates measured vertically upwards from some origin.
18
00:03:03,020 --> 00:03:12,320
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.
19
00:03:12,320 --> 00:03:20,270
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.
20
00:03:20,270 --> 00:03:31,880
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.
21
00:03:31,880 --> 00:03:40,160
That's why and then the x axis. Runs this way.
22
00:03:40,160 --> 00:03:53,160
And the axes intersect at the origin, oh, so this is the X y plane down here at Z equals zero.
23
00:03:53,160 --> 00:04:01,680
Remember that they are coordinating polar coordinates. That's the horizontal distance of points from the origin o in the X y plane.
24
00:04:01,680 --> 00:04:10,980
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.
25
00:04:10,980 --> 00:04:16,210
And then the thetr coordinates in polar coordinates is measured this way.
26
00:04:16,210 --> 00:04:25,440
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.
27
00:04:25,440 --> 00:04:42,170
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.
28
00:04:42,170 --> 00:04:53,450
All right. It's like that.
29
00:04:53,450 --> 00:05:11,910
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.
30
00:05:11,910 --> 00:05:18,360
So this function capital H, that's going to determine the shape of my surface of revolution.
31
00:05:18,360 --> 00:05:24,360
Now if you look at the set of points in the X Y plane that are at a constant distance are from the origin.
32
00:05:24,360 --> 00:05:31,350
That's a circle of Radius R in the X y plane and the height of points on my surface.
33
00:05:31,350 --> 00:05:41,160
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.
34
00:05:41,160 --> 00:05:45,420
That also means then that if I cut my surface with a horizontal plane,
35
00:05:45,420 --> 00:05:50,640
so at some fixed height, then the cross section of the surface is always a circle.
36
00:05:50,640 --> 00:05:57,690
It's a circle of radius R, where the height z and the radius of the circle of related by this equation.
37
00:05:57,690 --> 00:06:00,270
And that's also what makes this a surface of revolution.
38
00:06:00,270 --> 00:06:07,110
So it's invariant into rotating the surface about the z axis and those rotations about the z axis.
39
00:06:07,110 --> 00:06:16,440
They precisely shift the theatre coordinate here by a constant and that rotates the circles that are the horizontal cross-sections.
40
00:06:16,440 --> 00:06:31,650
So let me write that as a remark underneath. So you all k remark the surface is invariant.
41
00:06:31,650 --> 00:06:52,670
Under rotation about the Z-axis. OK, next, I'm going to draw on my point particle.
42
00:06:52,670 --> 00:07:03,440
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.
43
00:07:03,440 --> 00:07:13,760
So that's a right angle from the line between the particle down to the X y plane.
44
00:07:13,760 --> 00:07:20,340
Then next, I'm going to introduce an author normal basis for these cylindrical polar coordinates.
45
00:07:20,340 --> 00:07:28,770
So is it is a unit vector pointing vertically upwards along the Z-axis?
46
00:07:28,770 --> 00:07:36,570
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.
47
00:07:36,570 --> 00:07:37,530
That's E-Z.
48
00:07:37,530 --> 00:07:45,510
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.
49
00:07:45,510 --> 00:07:56,010
So you might remember that that is air, which is a unit vector pointing radially outwards.
50
00:07:56,010 --> 00:08:02,190
So again, if I continue that line here onto the z axis, that's a right angle.
51
00:08:02,190 --> 00:08:09,570
And moreover, the length of that line, there is precisely this distance off.
52
00:08:09,570 --> 00:08:22,050
And then finally, if theatre is a unit vector, that's perpendicular to that in the direction of increasing theatre.
53
00:08:22,050 --> 00:08:26,490
So is it on a theatre unit, vectors and then mutually perpendicular?
54
00:08:26,490 --> 00:08:34,170
So they form and also normal basis? And I can then use that to rate the position vector of my particle.
55
00:08:34,170 --> 00:08:45,630
So the position vector that's a vector are of the particle is scalar r e r plus z z.
56
00:08:45,630 --> 00:08:54,990
So scalar air that's the position of the particle in the horizontal plane. And Z is that gives its height.
57
00:08:54,990 --> 00:08:59,040
And finally, the last thing to draw on is the forces that are acting.
58
00:08:59,040 --> 00:09:09,390
So we're told here it's moving under gravity. So that means they will have a wait and.
59
00:09:09,390 --> 00:09:16,050
Acting downwards. So M is the mass of the particle little g acceleration due to gravity as usual.
60
00:09:16,050 --> 00:09:21,180
And then there must be some kind of contact force between the surface and the particle.
61
00:09:21,180 --> 00:09:29,310
And I'm going to draw that contact force just going this way.
62
00:09:29,310 --> 00:09:41,630
And that's going to be Victor, and that's the push of the surface on the particle, the contact force.
63
00:09:41,630 --> 00:09:50,910
All right. Any questions on the set up so far is that is that clear, you want to ask anything?
64
00:09:50,910 --> 00:09:58,440
OK, so we can now write down Newton's second law for the particle.
65
00:09:58,440 --> 00:10:09,450
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.
66
00:10:09,450 --> 00:10:15,640
That's Vector F. And there are two forces, there's gravity.
67
00:10:15,640 --> 00:10:21,450
So that's minus MJG times E Z and then plus the contact force.
68
00:10:21,450 --> 00:10:32,460
And the next step is the right acceleration here in terms of our cylindrical polar coordinate basis.
69
00:10:32,460 --> 00:10:39,900
And we can do that using the results from section 4.1. So that's earlier in the course.
70
00:10:39,900 --> 00:10:45,840
So that allows us to right the left hand side as mass. And then in brackets.
71
00:10:45,840 --> 00:11:06,930
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.
72
00:11:06,930 --> 00:11:19,270
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.
73
00:11:19,270 --> 00:11:27,220
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.
74
00:11:27,220 --> 00:11:30,490
And that's precisely the first two terms that I've written here,
75
00:11:30,490 --> 00:11:38,470
and we derive this expression for acceleration in polar coordinates in Thursday's lecture last week and then finally,
76
00:11:38,470 --> 00:11:43,630
taking two time derivatives of Z, he said, is that it's just a unit vector pointing upwards.
77
00:11:43,630 --> 00:11:49,390
So that said, double dots. He said, OK, so that's the left hand side.
78
00:11:49,390 --> 00:12:01,390
And on the right hand side, we have the total force acting minus energy Z Plus and the contact force.
79
00:12:01,390 --> 00:12:06,850
All right. So that's a vector differential equation. It's not immediately obvious what to do with it.
80
00:12:06,850 --> 00:12:11,710
So you might decide to write it out in Cartesian components.
81
00:12:11,710 --> 00:12:16,510
But just as an example, the simple pendulum that we looked at last week,
82
00:12:16,510 --> 00:12:20,200
we should be a bit smarter about what it is that we're actually trying to do with this equation.
83
00:12:20,200 --> 00:12:27,310
What do we want from it? In particular, if we're not interested in working out what's this contact force and is,
84
00:12:27,310 --> 00:12:35,920
we can immediately eliminate it from the equation by taking a dot product of this equation with a tangent vector to the surface.
85
00:12:35,920 --> 00:12:45,400
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.
86
00:12:45,400 --> 00:12:58,510
So remark our MK e theta is tangent to the surface.
87
00:12:58,510 --> 00:13:02,890
And we can see that from the picture here, if, Peter,
88
00:13:02,890 --> 00:13:10,060
that's this direction that's always tangent to a circle of constant radius in a horizontal plane.
89
00:13:10,060 --> 00:13:15,430
And so it's also tangent to the surface. So that's the first remark.
90
00:13:15,430 --> 00:13:23,230
And the second remark is that we're told that it's a smooth surface in the preamble at the top there.
91
00:13:23,230 --> 00:13:38,300
So it's a smooth surface. That means physically that there's no frictional force acting, so if and only if, no friction.
92
00:13:38,300 --> 00:13:49,890
And last week we said that that means that the contact force end is perpendicular.
93
00:13:49,890 --> 00:13:54,620
Or equivalently normal means the same thing.
94
00:13:54,620 --> 00:14:11,830
Quiver. Mentally normal and is for normal to the surface.
95
00:14:11,830 --> 00:14:20,410
So as we were discussing last Friday, if I had any component of my contact force, that's tangent to motion along the surface,
96
00:14:20,410 --> 00:14:25,570
then physically that would have to be some kind of frictional force or resistive force of the surface of the particle.
97
00:14:25,570 --> 00:14:30,400
So when we say it's moving on a smooth surface or friction, the surface geometrically,
98
00:14:30,400 --> 00:14:38,950
that means that my contact force is perpendicular to the surface or in other words, there's no component tangent to the surface.
99
00:14:38,950 --> 00:14:45,640
And so I immediately deduce that PN tilted into a theatre is zero.
100
00:14:45,640 --> 00:14:52,450
So it is tangent to the surface and is normal. So that dot product to zero in there orthogonal.
101
00:14:52,450 --> 00:15:02,080
OK, so that's the key point. If I now take Newton's second law above and I dot it with theatre.
102
00:15:02,080 --> 00:15:07,180
Well, Easter is orthogonal to air and it's orthogonal to E-Z.
103
00:15:07,180 --> 00:15:11,080
And so we'll say to this term. And it's also worth talking all day and as well.
104
00:15:11,080 --> 00:15:19,540
So the only term that survives when you take the DOT product is this one proportionality feature.
105
00:15:19,540 --> 00:15:33,490
So we immediately deduce that one over R D by d t of R squared theatre dots is zero.
106
00:15:33,490 --> 00:15:40,390
So obviously we can multiply through by all here. And then this is just saying that our squared theta dot is independent of time.
107
00:15:40,390 --> 00:15:48,100
It's a constant. So all square at Theatre Dot. And I'm going to define that quantity to be a little h.
108
00:15:48,100 --> 00:16:03,330
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.
109
00:16:03,330 --> 00:16:11,010
And that's also true here. And we can see that by computing the angular momentum of the particle.
110
00:16:11,010 --> 00:16:23,910
So let's compute the angular momentum. Of the political.
111
00:16:23,910 --> 00:16:36,980
About the origin of. So the angular momentum that's a vector L.
112
00:16:36,980 --> 00:16:44,930
It's defined as, oh, that's the position of the particle relative to the origin.
113
00:16:44,930 --> 00:16:54,200
And then you take a cross product with the linear momentum of the particle, and that's just mass times its velocity dot.
114
00:16:54,200 --> 00:17:02,160
That's the definition, and then I can substitute in here for me. These quantities in terms of my cylindrical polar coordinates.
115
00:17:02,160 --> 00:17:12,860
So that's Mars. And then in brackets scalar r e r plus z E-Z, that's position in cylindrical.
116
00:17:12,860 --> 00:17:18,860
And then for the velocity in single payloads that scalar r dot e r.
117
00:17:18,860 --> 00:17:30,230
Plus, Theta Dot Theta Plus Z does is that.
118
00:17:30,230 --> 00:17:36,500
Say the second term here. That's velocity and cylindrical policy, in particular, the first two terms.
119
00:17:36,500 --> 00:17:40,100
That's just components of velocity in the horizontal plane.
120
00:17:40,100 --> 00:17:48,230
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.
121
00:17:48,230 --> 00:18:00,480
He said. OK, and if you now multiply out those cross products and simplify, you get the following.
122
00:18:00,480 --> 00:18:08,870
So there's an overall factor of the mass and then our squared theta dot e z plus z or
123
00:18:08,870 --> 00:18:25,850
dot minus z dot e fita minus z are theta dot he r and in deriving that line I've used.
124
00:18:25,850 --> 00:18:45,880
So using e r crossed with e thetr is e z and I right plus the cyclic permutations of that equation.
125
00:18:45,880 --> 00:18:55,220
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,
126
00:18:55,220 --> 00:19:01,700
so I have er here crossed with the R zero because the cross product of the vector with itself is zero.
127
00:19:01,700 --> 00:19:09,080
But then I've also got e crossed with the Fita and I've used that that's equal to E-Z.
128
00:19:09,080 --> 00:19:14,450
I can. We've seen that equation, that particular cross product before in previous lectures.
129
00:19:14,450 --> 00:19:22,880
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.
130
00:19:22,880 --> 00:19:27,530
So also, if thetr crossed with each set, is it all?
131
00:19:27,530 --> 00:19:31,940
And is it crossed with the R? Is it theatre?
132
00:19:31,940 --> 00:19:38,420
And those just follow by writing out the original definitions of these vectors in terms of Cartesian coordinate basis.
133
00:19:38,420 --> 00:19:45,680
I, J and K. All right.
134
00:19:45,680 --> 00:19:50,770
Any questions on on that so far? All right. All right.
135
00:19:50,770 --> 00:20:01,190
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,
136
00:20:01,190 --> 00:20:09,250
that just picks out the first term and that's m r squared theta dot.
137
00:20:09,250 --> 00:20:15,910
And that's exactly m little h. Well, I introduce little h is R-Squared Theta Dot above.
138
00:20:15,910 --> 00:20:25,380
And we've just seen that's constant. So i.e. we have shown.
139
00:20:25,380 --> 00:20:45,240
The components. Of angular momentum l along the axis of symmetry.
140
00:20:45,240 --> 00:20:52,460
That's the E-Z direction is conserved.
141
00:20:52,460 --> 00:21:07,210
So that is it's constant when evaluated on a solution to Newton's second law.
142
00:21:07,210 --> 00:21:13,930
So, by the way, the other two terms here in the expression for angular momentum, they're not constant in general.
143
00:21:13,930 --> 00:21:19,090
It's nothing particularly special about those components. All right.
144
00:21:19,090 --> 00:21:23,680
So we've extracted one component of Newton's second law so far,
145
00:21:23,680 --> 00:21:30,880
and it's good to take stock at this point and again, think about what are we trying to do in the problem?
146
00:21:30,880 --> 00:21:46,450
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.
147
00:21:46,450 --> 00:21:57,700
So given R of T, we know Z of T.
148
00:21:57,700 --> 00:22:06,250
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.
149
00:22:06,250 --> 00:22:15,330
It's the horizontal distance of my Potts Point particle from the Z-axis since the length of that dotted line.
150
00:22:15,330 --> 00:22:27,250
OK, but also we've just shown that defeated by D T is h over R-Squared of T.
151
00:22:27,250 --> 00:22:29,380
So that's this line here.
152
00:22:29,380 --> 00:22:38,890
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.
153
00:22:38,890 --> 00:22:52,720
So again, so given R of T. We just integrate that equation to get Theatre of tea.
154
00:22:52,720 --> 00:22:59,830
And just by integrating taking the ditty to the right hand side and integrating it, we get theatre as a function of tea.
155
00:22:59,830 --> 00:23:05,980
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
00:23:05,980 --> 00:23:12,220
So I know exactly where the particle is. And I've solved for the dynamics of the particle.
157
00:23:12,220 --> 00:23:24,010
So this implies that we want an equation of motion.
158
00:23:24,010 --> 00:23:36,430
For this function, the scale of tea and on general grounds, we expect that to be a second order ordinary differential equation.
159
00:23:36,430 --> 00:23:44,990
As for the simple pendulum example, last week.
160
00:23:44,990 --> 00:24:04,870
We may extract this. So the equation of motion we're looking for from Newton's second law and to dot it into Vector,
161
00:24:04,870 --> 00:24:28,210
it'll see where this vector little T is another tangent vector to the surface.
162
00:24:28,210 --> 00:24:38,710
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.
163
00:24:38,710 --> 00:24:45,070
In general, the position vector of my particle remember so the position vector is scalar.
164
00:24:45,070 --> 00:24:54,430
Ah, he ah, plus capital h of ah, he z.
165
00:24:54,430 --> 00:25:04,150
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.
166
00:25:04,150 --> 00:25:09,640
So let's say thetr equals five or two. So I'm along the y axis here.
167
00:25:09,640 --> 00:25:15,820
Then as a very scalar ah, in that last expression for position, I precisely move along.
168
00:25:15,820 --> 00:25:20,710
Some curve up here like this on my surface.
169
00:25:20,710 --> 00:25:29,620
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.
170
00:25:29,620 --> 00:25:34,240
Are you doing that in the geometry course last term?
171
00:25:34,240 --> 00:25:47,560
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?
172
00:25:47,560 --> 00:26:01,680
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.
173
00:26:01,680 --> 00:26:10,260
He said. And so here at Daskal tonight, derivative with respect to the arguments are.
174
00:26:10,260 --> 00:26:18,780
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.
175
00:26:18,780 --> 00:26:28,740
Vector T. R. Newton's second law is disappeared off the top.
176
00:26:28,740 --> 00:26:38,640
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.
177
00:26:38,640 --> 00:26:42,690
And they're also perpendicular to the contact force.
178
00:26:42,690 --> 00:26:51,180
And so quite a few terms drop out when you take the dot product and you get.
179
00:26:51,180 --> 00:27:04,340
And then in brackets, double dots minus 50 dots squared, plus an h dash of R Z double dot.
180
00:27:04,340 --> 00:27:13,760
Is minus m g h dashed.
181
00:27:13,760 --> 00:27:18,530
OK, so that's what you get by dotting Newton's second law with that tangent vector.
182
00:27:18,530 --> 00:27:25,630
OK. And then following our remark above, we can substitute.
183
00:27:25,630 --> 00:27:38,230
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
00:27:38,230 --> 00:27:57,570
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
00:27:57,570 --> 00:28:03,840
And similarly, I can compute said double dots by taking another time derivative,
186
00:28:03,840 --> 00:28:12,720
so that's h double dashed of our second derivative of H with respect r o dot squared.
187
00:28:12,720 --> 00:28:20,050
Plus each dashed at a double dose.
188
00:28:20,050 --> 00:28:37,420
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.
189
00:28:37,420 --> 00:28:40,060
And if you do that, you get the following.
190
00:28:40,060 --> 00:28:52,660
So in brackets one plus h dashed of our all squared close brackets are double dots plus h dash of our h double
191
00:28:52,660 --> 00:29:09,640
dash double all adults squared minus h squared over cubed is equal to minus little g h dash of R and as promised.
192
00:29:09,640 --> 00:29:27,730
So remark that's a second order. Odie Second Order 0d ordinary differential equation for R of T.
193
00:29:27,730 --> 00:29:35,380
It's not a very nice looking equation, the equation, of course, depends on the choice of surface of revolution,
194
00:29:35,380 --> 00:29:39,760
and the choice of surface depends on this capital of our function.
195
00:29:39,760 --> 00:29:42,970
But even for quite simple choices of that function,
196
00:29:42,970 --> 00:29:49,870
this is a pretty nasty differential equation and we're not going to be able to solve it in closed form in general.
197
00:29:49,870 --> 00:29:54,340
But given a differential equation, there are always certain things you can do with it.
198
00:29:54,340 --> 00:29:59,080
So you might be able to deduce some general properties of solutions of this equation.
199
00:29:59,080 --> 00:30:01,300
You might better find approximate solutions,
200
00:30:01,300 --> 00:30:10,120
or you can just stick it onto a computer and solve it numerically and see what the particle actually does.
201
00:30:10,120 --> 00:30:14,050
However, rather than do that, I instead wanted to take a different approach.
202
00:30:14,050 --> 00:30:21,850
So also last week we were discussing conservation of energy, and in particular, we showed that for a particle moving on to gravity,
203
00:30:21,850 --> 00:30:27,730
moving along a smooth curve or a smooth surface, there's always a conserved energy.
204
00:30:27,730 --> 00:30:29,080
It's the first point.
205
00:30:29,080 --> 00:30:35,980
And the second point is we expect that to give a first order differential equation rather than a second order differential equation.
206
00:30:35,980 --> 00:30:45,850
So let's say that that is the case. So there's a better approach.
207
00:30:45,850 --> 00:31:00,650
Then just trying to tackle this second delivery equation directly, and that's to use conservation of energy.
208
00:31:00,650 --> 00:31:12,920
And that was in section five point one from section five point one.
209
00:31:12,920 --> 00:31:19,310
So that was quite a general analysis, and we deduced that E the total energy of the particle,
210
00:31:19,310 --> 00:31:28,970
so that's defined as T plus v, where T is kinetic energy, if the particle and V is the potential energy.
211
00:31:28,970 --> 00:31:39,200
So that's the definition of E. And so the kinetic energy as a half and and then modulus of ah dot squared.
212
00:31:39,200 --> 00:31:46,850
So that's the speed squared of the particle. And the potential energy here is the gravitational potential energy.
213
00:31:46,850 --> 00:31:51,500
So that's m times g times Z with this at the height.
214
00:31:51,500 --> 00:32:05,310
So that term there is the gravitational potential energy.
215
00:32:05,310 --> 00:32:14,000
And we know that's constant. That was a result that we derive last week.
216
00:32:14,000 --> 00:32:20,100
And so we can compute more explicitly what this quantity is.
217
00:32:20,100 --> 00:32:28,470
So I want to write out this speed squared in terms of my cylindrical polar coordinates as a first step.
218
00:32:28,470 --> 00:32:44,160
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
00:32:44,160 --> 00:32:53,340
he fita + said dot e said it's a modulus squared.
220
00:32:53,340 --> 00:33:00,330
And so when I take the dot product of that vector with itself, remember the e r e thetr and E's edge.
221
00:33:00,330 --> 00:33:04,530
They're all perpendicular to each other and they're all unit vectors as well.
222
00:33:04,530 --> 00:33:10,080
So actually, this is just the sum of squares of the coefficients. When you make this out.
223
00:33:10,080 --> 00:33:19,290
So it's very simple. It's just scalar dot squared plus are squared, theatre dot squared plus z dot squared.
224
00:33:19,290 --> 00:33:26,910
Some of the squares of the coefficients. OK.
225
00:33:26,910 --> 00:33:44,620
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.
226
00:33:44,620 --> 00:33:52,080
So those equations already above. So if you substitute all of that into this conservation of energy equation,
227
00:33:52,080 --> 00:34:08,940
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
00:34:08,940 --> 00:34:16,950
I can substitute for Theta Dot here. So that's little h squared over our squared.
229
00:34:16,950 --> 00:34:27,500
And then finally, z dot squared from here. That's H dashte of r squared dot squared.
230
00:34:27,500 --> 00:34:42,190
OK, so that's the kinetic energy term plus energy and then Z is closely capital HRR, and that's constant.
231
00:34:42,190 --> 00:34:45,970
And as promised, that is a first order ODI for our team.
232
00:34:45,970 --> 00:35:02,420
So will mark a first order ODI ordinary differential equation for our of two.
233
00:35:02,420 --> 00:35:06,890
OK, so if you take debility of this equation, using the E!
234
00:35:06,890 --> 00:35:12,620
Is constant, I will get our double dot turns on the right hand side.
235
00:35:12,620 --> 00:35:19,190
And you can check if you want that the equation that you get is precisely the equation of motion with odds double dancing.
236
00:35:19,190 --> 00:35:28,820
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
237
00:35:28,820 --> 00:35:36,290
equation rather than a second order equation where e here on the left hand side is an integration constant.
238
00:35:36,290 --> 00:35:43,490
OK, so the first order equation here is some function of R and R dot is E, which is a constant.
239
00:35:43,490 --> 00:35:47,300
It's a first order equation, though in principle we can integrate this.
240
00:35:47,300 --> 00:35:53,240
But again, we won't be able to do the integral explicitly and find the solution in closed form.
241
00:35:53,240 --> 00:36:01,070
However, we've now seen many times in the lectures that using conservation of energy, we can deduce some qualitative features of the motion.
242
00:36:01,070 --> 00:36:12,200
And also some quantitative features of the Motion two. And so that's what I wanted to demonstrate to you in an example.
243
00:36:12,200 --> 00:36:29,540
So let's actually look at a particular example of this. So it's motion on a parabola weight.
244
00:36:29,540 --> 00:36:36,380
All right, so I'm going to choose a particular surface now, so remember that said is Capitol Hill,
245
00:36:36,380 --> 00:36:46,600
and I'm going to take that to be R-Squared over for a with a some positive constant.
246
00:36:46,600 --> 00:36:49,760
So Z is all squared over 4A. That's a parabola.
247
00:36:49,760 --> 00:36:55,550
And the surface of revelation that you get looks pretty much exactly what, like the one that I've drawn.
248
00:36:55,550 --> 00:37:05,570
So that's why I drew that particular surface that looks pretty much like what that surface looks like.
249
00:37:05,570 --> 00:37:15,440
So the particle is initially at a height z equals a height.
250
00:37:15,440 --> 00:37:28,570
He said he was a and is projected horizontally.
251
00:37:28,570 --> 00:37:39,160
With Speed Little V, and our problem is to show.
252
00:37:39,160 --> 00:37:47,330
The particle moves between two heights, say the particle.
253
00:37:47,330 --> 00:38:15,320
Moves between to heights. In the subsequent motion of.
254
00:38:15,320 --> 00:38:27,460
All right, so this is an initial value problem, we're given the height of the particle initially and also its velocity initially.
255
00:38:27,460 --> 00:38:38,500
So initially. Z equals a.
256
00:38:38,500 --> 00:38:48,370
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.
257
00:38:48,370 --> 00:38:56,410
And so since that is a initially, then all must be to initially remember that R is positive.
258
00:38:56,410 --> 00:39:10,710
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.
259
00:39:10,710 --> 00:39:18,400
At speed, we. That tells us there are dots.
260
00:39:18,400 --> 00:39:28,180
It's very easy to initially. So again, if you look at the picture at some height, Zedek was a say this height,
261
00:39:28,180 --> 00:39:32,720
and if it's moving horizontally, that means it's precisely along the east direction.
262
00:39:32,720 --> 00:39:38,920
So tangent to a circle of constant radius, a feature is a unit vector.
263
00:39:38,920 --> 00:39:46,460
So the coefficient here V is precisely the speed.
264
00:39:46,460 --> 00:39:58,220
Moreover, so but in general, we've got the all dots, vector dot is scalar aardvarks e-r plus theatre dots in theatre,
265
00:39:58,220 --> 00:40:05,900
so that tells us that initially scale are dot is zero.
266
00:40:05,900 --> 00:40:22,700
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.
267
00:40:22,700 --> 00:40:30,810
So scalar dot is zero means there's no component of velocity in the outwards radial direction.
268
00:40:30,810 --> 00:40:37,070
OK, then the problem tells us to show the particle moves between two heights in the subsequent motion.
269
00:40:37,070 --> 00:40:45,950
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.
270
00:40:45,950 --> 00:40:53,450
So R is to square root of a z in general.
271
00:40:53,450 --> 00:41:04,440
So solving for R in terms of Z and so are dots is equal to the square root of over Z.
272
00:41:04,440 --> 00:41:21,380
Said Dot, using the chain rule. And we can substitute all of that into conservation of energy.
273
00:41:21,380 --> 00:41:28,210
So conservation of energy implies E!
274
00:41:28,210 --> 00:41:33,110
Is a half times the nice and then for.
275
00:41:33,110 --> 00:41:37,360
So I've got the expression, Oh, it's off the top there.
276
00:41:37,360 --> 00:41:52,810
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.
277
00:41:52,810 --> 00:42:03,370
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.
278
00:42:03,370 --> 00:42:11,650
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.
279
00:42:11,650 --> 00:42:16,930
In particular, I can evaluate it at the initial time when I know the initial data.
280
00:42:16,930 --> 00:42:20,890
So the initial time scale are dotted zero set that above.
281
00:42:20,890 --> 00:42:26,560
So we'll say Z Dot is zero. So this term and this term is zero initially.
282
00:42:26,560 --> 00:42:40,690
And so this is a half m little h squared over for a squared putting in Z is a initially plus M.G. A.
283
00:42:40,690 --> 00:43:01,460
So this second line, I've just evaluated it initial time. And finally, we can also work out this constant little h.
284
00:43:01,460 --> 00:43:12,410
Remember that little h is all squared theta dot. That's the definition I can write that is all times theatre dot.
285
00:43:12,410 --> 00:43:17,240
And this is a constant so I can work out the constant by evaluating it at the initial time.
286
00:43:17,240 --> 00:43:27,680
And initially R is to a and our feature is V, so that's too heavy.
287
00:43:27,680 --> 00:43:34,820
So if you substitute that into the equation for energy, so substitute.
288
00:43:34,820 --> 00:43:45,890
And then rearrange the conservation of energy equation, so I'm reading this here is this line here,
289
00:43:45,890 --> 00:43:50,900
which is go to Z Dot squared and it equals the right hand side, which is a constant.
290
00:43:50,900 --> 00:43:55,250
And I solve this for z dot squared. So all the terms with a z squared.
291
00:43:55,250 --> 00:44:04,910
Put those on one side and everything else on to the other side and you get this z dot squared is.
292
00:44:04,910 --> 00:44:25,300
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.
293
00:44:25,300 --> 00:44:30,520
OK, so that is the conservation of energy equation when you rearrange it for Z Dot squared.
294
00:44:30,520 --> 00:44:36,950
Now notice the following. So this term here is positive.
295
00:44:36,950 --> 00:44:43,460
It'll get positive, A is positive and Z is bigger than or equal to zero members.
296
00:44:43,460 --> 00:44:48,750
Z is a is a parabola. So it's all squared over 4a. So Z is bigger than an equal to zero.
297
00:44:48,750 --> 00:44:53,180
So this term is positive. But then there's a minus sign here.
298
00:44:53,180 --> 00:44:58,880
On the other hand, on the left hand side, z dot squared that's bigger than or equal to zero.
299
00:44:58,880 --> 00:45:05,270
So since said dot squared, that's manifestly bigger than zero.
300
00:45:05,270 --> 00:45:09,980
And this first term out here with the minus sign is less than zero.
301
00:45:09,980 --> 00:45:28,120
That means that these last two terms must be less than or equal to zero.
302
00:45:28,120 --> 00:45:33,190
So we did use this is less than or equal to zero. And this is a quadratic in Z.
303
00:45:33,190 --> 00:45:38,530
So it's a quadratic and Z looks like this if you sketch it and if it's less than or equal to zero,
304
00:45:38,530 --> 00:45:47,290
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
00:45:47,290 --> 00:46:01,680
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
00:46:13,920 --> 00:46:20,430
So this shows that the particle always moves in some strip around my surface of evolution.
308
00:46:20,430 --> 00:46:25,710
So if not, so for the dynamics. But it does lie always between these two heights.
309
00:46:25,710 --> 00:46:35,310
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,
310
00:46:35,310 --> 00:46:40,560
so that's for large velocities, then it always stays above the initial height.
311
00:46:40,560 --> 00:46:44,070
OK, because it's got a lie between this and that and that one's bigger.
312
00:46:44,070 --> 00:46:49,440
On the other hand, if V squared over two g is smaller than A, that's the smaller velocities,
313
00:46:49,440 --> 00:46:58,110
then it always stays below the original height that I projected it out. And that does make some intuitive sense.
314
00:46:58,110 --> 00:47:04,130
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.
317
00:47:13,090 --> 00:47:20,930
Give us an intuitive reason as to why conservation momentum around the world, the Z-axis is constant.
318
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.
319
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.
325
00:48:00,890 --> 00:48:05,900
So in our case, that's the rotational symmetry of our surface of revolution.
326
00:48:05,900 --> 00:48:10,490
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
00:50:15,220 --> 00:50:20,500
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,
352
00:50:29,710 --> 00:50:35,500
and one of the main results of that section will be to show that planets move on elliptical orbits about the Sun.
353
00:50:35,500 --> 00:50:58,973
Under that universal law of gravity. So see you then.