@andrerenault
When the graph came up, I figured "that's assuming it's continuous" and even though I couldn't have given a proof, I had a feeling there was a discontinuous point on the graph. Cool solution!
@helenross3037
last time I was this early pendulums obeyed the laws of gravity
@PapaFlammy69
I love that problem! :) The lagrangian approach is great! :3
@rafaelmramblas
Always when i have some unknown problem with math, the first try is "let's look if the function is continuous", it's always the pitfall when your dealing with controlled systems or signal processing.
@tonyhakston536
6:22 bold assertion.
@MrDowntownjbrown
This doesn't prove that we can't always keep the pendulum in the air, it only shows that it can't be proved using the intermediate value theorem. So is there an example of some motion where we can show it's impossible to keep the pendulum from falling?
@brandonn6099
The correct way to prove this is through induction. First take any simple element of movement. It could be a pause in movement, a constant speed, or an acceleration. Now, it is trivial to keep the pendulum up through that movement. And the final angle as a function of the initial angle IS continuous through such a simple movement. This is plain to imagine. Now, add any next movement to the object. Because the first movement enabled any output, we can give the second movement any initial angle. And I've already proven that the function is continuous given any simple movement. Any complicated movement need merely be broken down into simpler movements till the conditions are met. QED
@robstamm60
Mathematician: There must be a solution for every f(x) Mathematician2: Oh no for some f(x) you might hit a discontinouity Engineer: Ah guys a inverted pendulum is inherently unstable if you drive slowly it will fall down anyway!
@shk439
Zach took a course in Real Analysis, expect more continuity videos
@iwersonsch5131
A really nice question! In this case, though, we don't just have a free rolling marble on a flat semicircle with sticky ends. We have gravity applying a force downwards and movement applying a force left or right, and that changes the whole situation. The closer the pendulum gets to 180° or 0°, the less it will be affected by the cart movement as opposed to gravity. Spinning this further, we do actually get to a situation where given any cart movement with finite acceleration, there will be an angle at which the pendulum will not fall down on one side, but won't either swing over to the other side until the acceleration changes signs. The key to getting this answer of "yes" lies in understanding that the restriction not only applies at 0° and 180°, but also affects how the curves can behave near those values. The areas near those edge values apply a pull, which, divided by the force applied by the cart, tends towards infinity and as such can counteract any cart movement at some in-between value.
@sighko
"The function is continuous" Unit step function: "Allow me to introduce myself"
@nodrogj1
Pretty sure it is continuous though. If you actually try constructing a cart path that always leads to 0 or 180 you end up with a problem: namely that there's no way to actually collapse the possibilities. Changing the cart's acceleration can move the unstable balancing point but it never removes it. But we can think of time progressing as a sort of 'zooming in' on points near that instability, since over time points always fall away from that unstable point. But no matter how you move that unstable balancing point, or how long you zoom in on it, you'll always have more points since the starting angle was continuous. It makes it somewhat more complicated when you consider position and velocity of the pendulum, but the same argument holds albeit in the 2D state space with a quasistable line instead of a point. So in effect your picture that breaks continuity doesn't exist in the actual problem space because coming back from near 0 or 180 is actually really hard. Those 'paths' that just barely skim one or the other but don't collapse must spend a lot of time near the balancing point, which means nearby states include all possible future paths of the pendulum, including ones where it doesn't ever fall to 0 or 180
@thedoublehelix5661
I thought I was being an idiot when I tried to prove that the final angle was a continuous function of the initial one, but I guess I was right to be careful this time lmao
@RC32Smiths01
Really interesting problem, something to really think about and change your perspective!
@jamesahibbard
"It seems like a physics problem, but we're going to treat it like a maths problem." There's a difference?
@kingacrisius
I love how when he was giving the first explanation I literally said to myself, "But how do we know it's continuous?"
@Smithers888
I still believe the function is continuous. I'm not sure I can prove that right now, but I can at least counter your argument against it. Let's look at the graph at 10:00 and actually consider what motion of the cart must be. Around t=3 to t=4 the (unrestricted) dark blue pendulum is just below the horizontal on the left side and is accelerating upwards. This means the cart must be accelerating to the right to pull it back up. But now look at the cyan pendulum at the same time. It's above the horizontal and also accelerating upwards, which means the cart is accelerating left to push it up. The cart cannot cause both of these effects with the same motion. In fact: the white curve cannot exist at all. At the point where it touches 180° with zero angular momentum, how can it accelerate upwards? The forces acting on it are gravity pulling it down and the force from the cart which can only be parallel to the direction of the pendulum rod (i.e. horizontally), so the resultant force must be downwards. And that's basically the crux of my belief that the function is continuous: as the pendulum gets closer to the horizontal on either side, it gets arbitrarily hard to pull it back away from that, so no matter how hard one portion of the cart's motion tries to yank the pendulum over to one side, it must be possible to have been close enough to the other side that it will never go all the way.
@danielchoi2345
Interesting approach to the problem. A similar question to this was asked in my classical mechanics test to be solved with lagrangian, which solves it analytically (assuming the integral can be solved).
@BrownBoy02
This was one of my most memorable control systems labs at uni.
@zachstar