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@amaanverma8248 Says:
I solved this question with physics approach before your solution . With three equation
1) Mglcos theta+1/2mv^2=1/2mv2^2 (it hit the ground so assume the pe at hitting point is zero)
2 t-mgcos thera=mv^2/R
3) then use ldtheta/dt . You never get a real solution for T to go from any angle between 0 and pi . So it does not hit.
@FishSticker Says:
This video is wrong.
After a lot of thought, I conclude that the only way for a path to have no stable starting angle is for the path to wait at one point for an infinite amount of time.
@FuoChEsplode Says:
8:30 hold up, how can the angle become greater than 180° and then start decelerating, and accelerate torwards 0°, when, in the same fragment of time, at 10:00, the angle minor than 180° accelerate torwards 0° too? Really, I don't find that explanation consistent with the problem. I'll add, it should be possible to use the intermediate value theorem of angle and velocity to prove that it's possible to stabilize the pendulum, by first deviding the movement of the cart in segments in which the direction of the acceleration is consistent, because after every segment it's possible to have values with the angle approaching 0 and with a velocity to the right, and an angle approaching 180 with a velocity to the left, and intermediate values. Am I wrong about that? I still belive that the ending angle is continuous, if someone has something to say to convince me otherwise I'm happy to hear
@timstanglow3847 Says:
Sorry, but this Video is wrong, the intermediate value theorem does apply, because it is in fact continuous.
Here's why the "proof" by counterexample from the Video is wrong: At t=3 the first graph of the unrestrained pendulum has a negative second deritive while it's above 180°. In order for this to be the case, the cart has to moving to the right. However, if the cart is moving to the right, the pendulums at angles between 90 and 180° always accelerate towards 180°. So the light blue graph would also hit 180°
@rbbza2749 Says:
Loooool.
45°, A=G no?
😂
@hfs-lk5ip Says:
I havent watched to the end but i assume the path has to take a finite amount of time to complete
@nodrogj1 Says:
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
@koenth2359 Says:
He're my first thought: of course in the given example there is such a trajectory. Just let the cart accelerate fast enough to the left until the pendulum is at 90° upright, maybe, a little bit leaning over to the right. Next, accelerate it to the right in a very short time interval, just to stop the angular velocity of the pendulum, and finally with an acceleration just enough to compensate for gravity alone.
I found the graph (and parts of the explanation) very confusing, because the relation between starting and final angle are treated is as if the final angle is a single function of initial angle. In fact, it is highly dependent on the trajectory, so there is really not such a function, as long as the trajectory is not fixed. When connecting the starting angle with the final angle in the graph even enhances this confusion.
@FishSticker Says:
I just made up a path that couldn’t be completed, have the cart stop in the middle for a second, so that the pendulum must be pointing straight up to not fall over, then accelerate to the end, knocking the 90 degree pendulum over
@Prinrin Says:
I think there's an issue with the argument in this video, too: it assumes that the function describing the "unrestricted path of the pendulum" can be arbitrary, but that isn't the case. The easiest way to see this is to imagine a 3-dimensional plot: The x-axis is the initial angle [0,180], the y-axis is the time axis, and the z-axis is the angle of that starting location at that current time [0,180]. Let the function f(x,t) give the position of the point on the z-axis. Because of the specifics of this problem, f has to be continuous (shown below).
Claim: for any time t, there exists an open interval (a,b) such that the image of (a,b) under f is (0,180).
Proof: We wish to show this by showing there cannot be a point where it stops being the case.
At t=0, the interval is (0,180). Also note that at t=0, the angular velocity at any point in the interval is 0.
For any t, the angular acceleration ranges from some negative number (at a) to some positive number (at b). This is because the acceleration only has two contributing factors: gravity and and change from the motion of the cart. Gravity always contributes negative angular acceleration for the low angles and positive angular acceleration for the high angles. The acceleration from the movement of the cart can take three forms, depending on how the cart is accelerating.
If the cart is accelerating forwards (i.e. in the direction of angle 0), then the contribution to the angular acceleration by movement of the cart for angles near 0 is some small, positive value (to see that this approaches 0, split the force vector into components based on the pendulum shaft; as the angle of the shaft approaches 0, the component perpendicular to the pendulum approaches 0, so the acceleration does as well). Thus, the angular acceleration is a nondecreasing continuous function over the whole interval [0,180] that takes value 0 at a.
We can do similar logic for no acceleration and negative acceleration.
Since gravity and the cart's movement are the only two things affecting the pendulum's angular acceleration, the total acceleration ranges from negative to positive, is continuous over the interval, and is monotonic.
This shows that for all t>0, the angular velocity also ranges from negative to positive over the interval (a,b) and is also monotonic and continuous.
Hopefully this is sufficient to convince you that the function f is continuous, and therefore the pendulum must always have a valid value. It isn't something to assume for free, but it is still true.
@Omlet221 Says:
My idea was to just find a way for the cart to move that would guarantee the pendulum falls. Just imagine the cart goes really far left before coming back. If the pendulum is right of center it will fall on the initial trip left. If it’s left of center it might stay up going left but then it would fall going back right. And if it’s balanced center it would clearly fall immediately. But I’m not sure that this is really a solid proof.
@jffrysith4365 Says:
I knew there was an issue with their being a solution for any x(t) because of this scenario:
Consider if the cart starts by doing nothing for a long time, then it moves to the finish directly.
When the cart is stationary, the pendulum will fall at any angle except 90 degrees. However, if the angle is 90 degrees, it will not stay standing after the cart moves. Therefore, there is no initial angle such that after x(t) the pendulum is still standing.
@Manabender Says:
The answer is clearly NO: There do exist some paths x(t) for which there exists no starting pendulum angle that will end at an angle other than 0 or 180 deg.
Consider an x(t) that remains at point A for several seconds, then moves a small amount towards B, stops again for several seconds, then finishes the journey. During the first rest at point A, any angle other than 90 deg will fall to 0 or 180, whichever is closer. Once it starts moving, if the angle was 90, it will no longer be 90. During the second rest, it will fall back to 0 or 180.
@PhilippeCarphin Says:
I had my doubts the whole time but I couldn't point out exactly what was wrong. It was obvious to me that the assumption of continuity was wrong but not obvious how. Showing the version without constraints and how adding the constraints breaks continuity made it very clear.
Also intuitively from the start from math experience it felt like the problem was too hard to be solved by invoking the intermediate value theorem.
@Graknorke Says:
I know you said it isn't a physics problem, but is there actually any guarantee that your θ(t) was actually physically possible? after all changing the angle wouldn't just shift the curve up or down, the easy example being you can imagine a X(t) where the cart accelerates at a rate that keeps the pendulum balanced at exactly the same angle the whole time, whereas at a different angle it would move. obviously not just a translation of the same θ(t) curve since one's a horizontal line and one isn't.
@Frits34000 Says:
If the cart goes to the middle of the track with acceleration approaching infinity and then stops, then goes back half the track and then to the end I reckon its difficult to find a working starting angle lol
@gershommaes902 Says:
It seems like a matter of working backwards; at some points the movement of the cart will be so severe that the pendulum will have to be pre-emptively counterbalanced such that it is hardly any distance from its terminal values at all. In the world of math, ensuring these extremely sensitive values get achieved at the necessary times seems trivial, as long as "hardly any distance" is anything other than "0 distance".
@misspotato813 Says:
Haven't watched far into the video yet, but isn't the answer yes as this is the talent where you balance plate on a long stick is achieving and move while keeping it from toppling over?
@jacko2131 Says:
Wait so you're just assuming there will be a curve at 4:20? Just straight up discarding the idea that the pendulum falls down at every angle?
@Smithers888 Says:
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.
@aDifferentJT Says:
Actually your proposed discontinuous function isn’t possible due to physics, consider the pendulum at the turning point where it just grazes 180, it has no angular momentum, as it’s a turning point, and, because it’s horizontal, the acceleration of the cart isn’t imparting any vertical forces on the pendulum, so it can’t go back up.
@aDifferentJT Says:
You haven’t actually solved the problem, just because you haven’t proved that the function is continuous doesn’t mean that you’ve proved that it can’t be done.
@brandonn6099 Says:
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
@lukejohnson29 Says:
Your problem is assuming that gravity exists in this problem.
@thomaskrude5599 Says:
The time angle graph makes absolutely no sense. This drawn changes of angels only can happen with accelaration of the cart. But this would affect the pendulum in a different direction depending if it is above or below 180°. So the lighter blue line must have a completely different shape than the blue/brown/green lines. Also it has maybe a great effect on how close it is to this 180°. Maybe the small difference between wight and lightblue covers the whole range of outcomes without a jump.
@mercrazzle Says:
Might be a silly question but here goes.
You phrased the actual problem as this: "given any motion that starts at point A and ends at point B, can we set the pendulum's starting angle such that it always stays in the air", and then spent ages proving that the range of starting angles to finishing angles isn't in fact continuous. my question is why does it matter that it isn't continuous... Just because the curve on the Theta initial vs Theta final graph isn't continuous doesn't mean that it definitely can't be done.
Given a complete step function, you would have no solution, so I understand the point is to say you can't guarantee there is a starting angle for ANY x(t), but I think it would be useful to prove mathematically that you could physically achieve a perfect step function given the rules of the pendulum, with some x(t). Because it seems unfeasible
@Hi_Brien Says:
For every... there exists... so if we negate that, it's false if we can think up q situation where there is no angle that works. If you jitter it and then make it move slowly, the pendulum is gonna fall.
@TheFinagle Says:
My instinct is no, not every x(t) will have some starting angle where the pendulum is in the air at the end. Since stopping is a valid motion you could stop for some amount of time that the pendulum would hit the bottom unless its perfectly balanced, move some then stop again where in that time it would either already be at rest or be jogged out of the perfect balance and end up tipped over. Now to watch the video
@angelatheist Says:
The given correction in this video doesn't work. The problem is that it's impossible for the graph of the pendulum to be tangent at 180° or 0° when coming from within that range. At that point gravity is pulling it down and it has momentum pushing it down too and the movement of the cart has no effect at that point so its impossible for it to change directions there. The graph is continuous but adjusting the angle just pushes the point it crosses to a later time, it won't stop it from crossing unless the crossing point is pushed passed the stopping time.
@lightknightgames Says:
It's easy to create a movement path that cannot be stood up in.
Move the cart at the maximum speed, to make it easier to think about, we can use c the speed of light.
And make AB (c * 1second) + 1 meter.
Now if the pendulum starts at 179.99999999999999... degrees, there is no way for it to avoid ending at 0 degrees, and if it starts at 180, it will end at 180.
I'm sure you can figure out the limit of the distance required if you move at the speed of light, since it is an equation that has a limit, but I'm fairly sure it cannot surpass c+1
@asatht55 Says:
wait but arent pendulums made of string???
@crb233 Says:
I think this argument is flawed and that the output angle is a continuous function of input angle. I think this is because the example he gave isn't physically realizable. Haven't formalized the reasoning but the basic idea is that in order for the example plots that he gave to be valid paths for the pendulum, the pendulum must feel an upward acceleration at points arbitrarily close to 180°. However, this is physically impossible given the setup of the problem because as the angle approaches 180°, the force on the pendulum from the cart's acceleration must approach 0 and so there's a point at which the acceleration due to gravity takes over and pulls the pendulum downward.
This doesn't mean that the function is necessarily continuous, but just that the example he gave doesn't work. Still, I think it must be continuous and but don't have a complete explanation yet.
@wovaka Says:
Well if the pendulum length isn't set all you would need to do is increase the length of it. Such that it is a little bit longer than the required distance. Start at 90° Then make sure that the cart moves forward at the rate the pendulum falls vertically. (Alternatively you could make a track for the pendulum weight and nudge the cart forward) and the cart would reach point B before the pendulum shaft hits the cart
@TomtheMagician21 Says:
Don't try to understand it, feel it
@upsilan_mitstrima Says:
I may not understand the question but what if you move at a very high soeed to the left for an hour and then move back to point b
@MechMK1 Says:
A very simple illustration of one x(t) which clearly shows that the pendulum ends in either 0° or 180° is a movement which is very very very slow. So no matter the starting angle, it would fall down to either side before it reaches the end.
@killhour Says:
Even if the pendulum didn't have any restrictions on angle, it would still be discontinuous.
@JakeHowlHero Says:
I don't like thinking too much but I am positive that if you move it fast enough then it shold move any distance without the pendulum ever touching the ground. I only based my answer on the thumbnail/title
@radiationpony8449 Says:
assuming no maximum acceleration limit, then unless the pengilum starts at 0 or 180 degrees then the cart can travel from A to B no matter the distance with the following method: if the pegilum is in the first quadrant (IE 0
@TheVergile Says:
the function is so obviously NOT continuous that im kinda confused how mathematicians in the 40s even considered that as a valid assumption. And if you assume it is continuous it isnt even worth asking the question anymore because per definition the answer must be yes. If it is continuous then per definition there must be an angle that works.
@danielrhouck Says:
5:00 You’re assuming continuity which seems plausible but is not something you can wave away as trivial (and even if you could you haven’t yet).
@skipper8745 Says:
Am i the only person who immediately thought of a counter example. If you choose the path, where the cart remains motionless for a sufficient amount of time, then most initial angles will fall to 0 or 180 with gravity. The only initial angle that doesnt is 90°. If you then move a small amount forward, the pendulum will no longer be at 90° due to inertia. Repeat the first step to cause it to fall. Ergo it is not possible for every path.
@raphaelnej8387 Says:
the explain is really complicated while i believe it could be explained much more easily
it s a given x problem so you simply have to find a counter example
and the counter example is the car staying still for 5
minutes than going to the point B
the pendulum cannot stay still
the fact that it is not continuous comes from the fact that an inverted pendulum has equilibrium points but no stable ones
@FinetalPies Says:
The answer to the thumbnail is yeah but you gotta go backwards at first. Was kinda hoping to see some math of the physics of acceleration and inverted pendulums but oh well
@GRAYgauss Says:
Didn't watch passed initial problem intrudoction... so I don't know if you're allowed to: Just drive backwards (enough that we are driving into the falling stick) then forwards, since driving will impart a torque counter-wise to the direction of movement...If not allowed to drive backwards....If the pendulum state starts "behind" the perpendicular relative to direction of movement...It will probably fall, but can we brake, then this function space isn't a problem to solve either... Otherwise.. distance, mass, acceleration, drag, pendulum length all will allow for various edge cases which make me think "fall or not" is a red herring... If it starts ahead of the perpendicular, a clever combination of acceleration and no acceleration should be enough, but if we can brake, even easier...
@theoverseer393 Says:
Continuous is only applicable if it’s less chaotic. Makes a lot if sense, only few angles would work for some of them, since they’d skirt the line.
I think you explained it better than I could ever. I didn’t think about the “179 vs 180” issue
@BrokenLifeCycle Says:
Well, I've seen segways...
@Pheonix1328 Says:
I don't really get what the unrestricted pendulum proved. If you pick any angle between 0 and 180 you can always move the cart in such a way that it fails but that doesn't mean it's discontinuous. If anything, the only thing you proved is that no mater what angle you pick you can always reach a fail state, which isn't what the problem is asking.
@staticcactus6029 Says:
Assuming the cart starts perpendicular to the floor, and is perfectly balanced you can actually induce a stable state by first rolling back wards, causing the pendulum to move towards the right, and at the appropriate angle you can then begin accelerating “forward” towards the right so long as you maintain an appropriate acceleration the force of acceleration will be transmitted to the pendulum. This is my answer not watching this video, assuming gravity is perpendicular to the “floor” as well. Obviously this answer may not be accurate to the actual context of the video, but it’s my initial thought. If we assume you cannot start by first reversing then the answer if you can reach the point, is directly related to the distance needed to travel, any acceleration will induce the pendulum to fall, and assuming that greater acceleration causes the pendulums distance to ground to decrease proportionally to the carts distance to destination then this is unlikely. However assuming that it’s a real world sort of problem due to the angle of force applied to the pendulum (assuming it’s the “pull” of the cart at point attached, then this may be feasible)
I learned this intuitively from balancing a broom on my fingertip because a broom irl will act like an inverse pendulum.
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