Decoupling Per-Cycle Momemtum Yields From RPM
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OK here's a poser:
• using MoI variations to raise velocity in successive steps causes J/L to square with velocity, because CF force does
• what if, instead, these MoI variations occurred in parallel, ie. all at the same initial speed, and thus J/L, but resulting in the same net rise in velocity?
For instance consider the apparent solution to the 'quarters' riddle:
• suppose a system's net angular momentum were divided into four equal angular momenta, which individually but in concert trade 1 kg-m² of MoI for 1 rad/s of velocity..
• ..would it thus be possible to end with some kind of inelastic collision / braking phase that caused those four increased 'velocity' components to stack up, as if the four discrete accelerations had followed sequentially..?
..IOW paying a linearly-summing cost for a squaring KE value, by applying simultaneous MoI variations that have the same net effect as a consecutive sequence.. ie. achieving the same net velocity but at ie. a quarter of the cost..?
• using MoI variations to raise velocity in successive steps causes J/L to square with velocity, because CF force does
• what if, instead, these MoI variations occurred in parallel, ie. all at the same initial speed, and thus J/L, but resulting in the same net rise in velocity?
For instance consider the apparent solution to the 'quarters' riddle:
• suppose a system's net angular momentum were divided into four equal angular momenta, which individually but in concert trade 1 kg-m² of MoI for 1 rad/s of velocity..
• ..would it thus be possible to end with some kind of inelastic collision / braking phase that caused those four increased 'velocity' components to stack up, as if the four discrete accelerations had followed sequentially..?
..IOW paying a linearly-summing cost for a squaring KE value, by applying simultaneous MoI variations that have the same net effect as a consecutive sequence.. ie. achieving the same net velocity but at ie. a quarter of the cost..?
re: Decoupling Per-Cycle Momemtum Yields From RPM
So you have 4 masses traveling at 2m/s. You brake them and transfer momentum in the hope of ending up with 1 mass traveling at 8m/s.
I unsuccessfully tried something like that using an elastic collision on a lever. If you can do that you truly are a magician.
I unsuccessfully tried something like that using an elastic collision on a lever. If you can do that you truly are a magician.
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Precisely - well, something like this, achieving that net effect.
Obviously, described like that (as just an inertial interaction) it's pure fantasy.. any system just involving interacting masses has to be considered comprehensively closed.
If we want to raise net momentum, our only option is to open the system to gravity & time.
But getting our ducks in a row: if the hypothetical objective is not so much to gain momentum, as to cause the momentum we already have to be spontaneously split into a CoE-violating distribution, then the question still boils down to "can opening the system to G&t help here?"..
IOW, could being in a gravitating FoR somehow assist with trading MoI for velocity, on the cheap, like?
Dunno, just trying to reach outside the box of simply gaining cut-price momentum directly from G&t.. ie. maybe momentum can only be sourced from G&t at unity efficiency (ie. it has to be paid for by work in the ground / KE FoR), and thus the trick (and only option) for a CoE break is somehow plying gravity to cause asymmetric distributions of extant momentum..
After all, actually getting more momentum is trivial - just keep over-balancing - the only hurdle is paying less input energy for it.
For five years + now i've been trying to game the input energy cost of momentum as a function of velocity squared, per the KE equation: at 10 m/s a 1 rad/s acceleration of 1 kg-m² costs 10.5 J, but the same acceleration / momentum rise from a standing start only costs 0.5 J.. so obvioushly we need a system that spoofs a lower-than-actual net velocity, in order to be able to keep buying momentum at its minimum, base rate, net input energy thus summing with velocity whilst net KE squares, right? Why's it taking 5+ years to actually engineer then? If it's so bloody simple?
I've simply been unable to decouple the internal cost of momentum from its GPE / KE values in the ground frame; using gravity to produce asymmetric inertial interactions is trivial, but the KE 'gain' is always equal to the GPE lost. No FoR separation / divergence; input and output workloads remaining doggedly in the same, terrestrial, FoR.
Hence the efforts to try think outside that box, for any alternative hypothetical mechs i may have overlooked.
So, one way to sidestep the whole issue of buying momentum in a moving frame as if we were in a static one (ie. spoofing a pseudo-static FoR), would be to book multiple accelerations together whilst in a slower frame, so that they're all delivered at once; purchasing them in the same FoR simply sums their net cost, whereas buying 'em sequentially, the costs would square with rising net velocity each cycle.
Obviously, CoE has to be satisfied in both input and output velocity frames at all times.. the trick would be spontaneously switching / flipping between those FoR velocities without having to accelerate between them..
Dunno, prolly so much half-thought out word salad, just trying to take a step back for a moment..
Obviously, described like that (as just an inertial interaction) it's pure fantasy.. any system just involving interacting masses has to be considered comprehensively closed.
If we want to raise net momentum, our only option is to open the system to gravity & time.
But getting our ducks in a row: if the hypothetical objective is not so much to gain momentum, as to cause the momentum we already have to be spontaneously split into a CoE-violating distribution, then the question still boils down to "can opening the system to G&t help here?"..
IOW, could being in a gravitating FoR somehow assist with trading MoI for velocity, on the cheap, like?
Dunno, just trying to reach outside the box of simply gaining cut-price momentum directly from G&t.. ie. maybe momentum can only be sourced from G&t at unity efficiency (ie. it has to be paid for by work in the ground / KE FoR), and thus the trick (and only option) for a CoE break is somehow plying gravity to cause asymmetric distributions of extant momentum..
After all, actually getting more momentum is trivial - just keep over-balancing - the only hurdle is paying less input energy for it.
For five years + now i've been trying to game the input energy cost of momentum as a function of velocity squared, per the KE equation: at 10 m/s a 1 rad/s acceleration of 1 kg-m² costs 10.5 J, but the same acceleration / momentum rise from a standing start only costs 0.5 J.. so obvioushly we need a system that spoofs a lower-than-actual net velocity, in order to be able to keep buying momentum at its minimum, base rate, net input energy thus summing with velocity whilst net KE squares, right? Why's it taking 5+ years to actually engineer then? If it's so bloody simple?
I've simply been unable to decouple the internal cost of momentum from its GPE / KE values in the ground frame; using gravity to produce asymmetric inertial interactions is trivial, but the KE 'gain' is always equal to the GPE lost. No FoR separation / divergence; input and output workloads remaining doggedly in the same, terrestrial, FoR.
Hence the efforts to try think outside that box, for any alternative hypothetical mechs i may have overlooked.
So, one way to sidestep the whole issue of buying momentum in a moving frame as if we were in a static one (ie. spoofing a pseudo-static FoR), would be to book multiple accelerations together whilst in a slower frame, so that they're all delivered at once; purchasing them in the same FoR simply sums their net cost, whereas buying 'em sequentially, the costs would square with rising net velocity each cycle.
Obviously, CoE has to be satisfied in both input and output velocity frames at all times.. the trick would be spontaneously switching / flipping between those FoR velocities without having to accelerate between them..
Dunno, prolly so much half-thought out word salad, just trying to take a step back for a moment..
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Just for a mental image of the kinds of systems that might be considered WRT such an objective, perhaps something like the rigs i was playing with a couple of year ago:
..ie. a system with both 'axial and orbital' MoI's, both in-principle variable, and interconnected via clutches.
So, you'd presumably use OB torque to induce angular momentum to the 'orbiting' MoI's, maybe something like this:
..but then trading up 'MoI' and 'velocity' components when these momenta are dumped, all at once, into the central system axis, somehow..?
Again, one of the intrigues of practical mechanics is that it's actually perfectly feasible for the central, main system MoI to be lower than that of the orbiting MoI's - so they could be 4 kg-m² each, say, yet only applying an MoI of 1 kg-m² on the central axis.. quite unintuitive as one might expect that the central MoI would always have to be greater, yet, because MoI=mr², bring the orbiting MoI's radii in close enough to the main axis and the central MoI can be arbitrarily-lower..
Like i say, dunno, just kicking it out there, more looking for a concept than actually having one yet..
..ie. a system with both 'axial and orbital' MoI's, both in-principle variable, and interconnected via clutches.
So, you'd presumably use OB torque to induce angular momentum to the 'orbiting' MoI's, maybe something like this:
..but then trading up 'MoI' and 'velocity' components when these momenta are dumped, all at once, into the central system axis, somehow..?
Again, one of the intrigues of practical mechanics is that it's actually perfectly feasible for the central, main system MoI to be lower than that of the orbiting MoI's - so they could be 4 kg-m² each, say, yet only applying an MoI of 1 kg-m² on the central axis.. quite unintuitive as one might expect that the central MoI would always have to be greater, yet, because MoI=mr², bring the orbiting MoI's radii in close enough to the main axis and the central MoI can be arbitrarily-lower..
Like i say, dunno, just kicking it out there, more looking for a concept than actually having one yet..
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..for instance, what if the orbiting MoI's were reduced - applying positive inertial torque - prior to dropping the clutch? Could we thus spin up & accumulate some 'relative angular momentum' (again, a BS concept unless it successfully generates a divergent inertial FoR).. with input CF workloads thus in a lower velocity frame than that of the rise in net system rotKE post-braking?
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re: Decoupling Per-Cycle Momemtum Yields From RPM
MrVibrating wrote:
What you don't see is that a Hamstercage is a rotating frame.
For everything what is in this wheel the wheel is the ground on which other objects have to move.
Put a stick on the Hamstercage, turn the Hamstercage some degrees and the stick will fall. you have a moving ground and you produce an impact with the stick on the downgoing side.
Depending on the stick length you will get different impact points on the rim of the Hamstercage.
I think you can agree on the explanation so far.
What is different or in addition is the indirect impact.
On the stick you fastened a pendulum.
The stick will make the impact against the Hamstercage, the pendulum is still on the move and is rised with a flash against gravity up.
So you lift up the mass and have created torque for the Hamsterwheel.
Study this movement. It is a tilt oscillation.
t
All what you said is correct.Ja, er hat eine rotierende FoR erstellt, aber sie rotierte immer noch relativ zum Boden FoR, und es ist in diesem, externen FoR, dass die KE-Verstärkung existiert und geerntet wird. Der Energiegewinn ist die Differenz zwischen den Input-Arbeitskosten im rotierenden FoR und ihrem KE-Wert im Boden FoR. Sie müssen in grundlegenderen Begriffen denken - ein Aufprall impliziert einen vorhergehenden Geschwindigkeitsunterschied vor der Kollision (dh. etwas war schneller unterwegs), gefolgt von einer Umverteilung des internen Momentums nach der Kollision (so wurde die leichtere Masse oder MoI verlangsamt, und die schwerere beschleunigt).
Dies sind jedoch nur die Begriffe grundlegender Trägheitsinteraktionen, bei denen die Nettodynamik strikt konstant /netto Null ist. Momentum kann nur aus Schwerkraft und Zeit gewonnen werden - speziell aus einer steigenden vs fallenden G-Zeit-Asymmetrie (denn wenn 'up' & 'down' G-zeiten symmetrisch sind, haben Sie nur ein Pendel).
Daher ist einerseits die mathematische Lösung "OU = die Effizienz der Akkumulierung reaktionsloser Impulse steigt", aber in der Praxis ist es ein Fall von "OU = Fixierung der Einheitsenergiekosten des Momentums aus Schwerkraft und Zeit" (d. h. invariant von steigender Drehzahl).
Begriffe wie "parametrische Schwingung" und sogar "Kollisionen" an sich sind also bestenfalls Nebenthemen;
das grundlegende Problem muss unter seinen ausdrücklichen Begriffen gerahmt und angegangen werden, indem man den Stier prägnant an den Hörnern packt.
What you don't see is that a Hamstercage is a rotating frame.
For everything what is in this wheel the wheel is the ground on which other objects have to move.
Put a stick on the Hamstercage, turn the Hamstercage some degrees and the stick will fall. you have a moving ground and you produce an impact with the stick on the downgoing side.
Depending on the stick length you will get different impact points on the rim of the Hamstercage.
I think you can agree on the explanation so far.
What is different or in addition is the indirect impact.
On the stick you fastened a pendulum.
The stick will make the impact against the Hamstercage, the pendulum is still on the move and is rised with a flash against gravity up.
So you lift up the mass and have created torque for the Hamsterwheel.
Study this movement. It is a tilt oscillation.
t
Best regards
Georg
Georg
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Georg, because of your persistence, i'm willing to measure whatever you suggest.
I still don't quite understand the interaction you're describing though...? Help me understand, and i'll simulate an accurate measurement in WM2D..
An interaction between a weight and wheel, that lifts the weight whilst applying positive torque, is obviously on the right track.
As ever though, what really matters is how its efficiency scales with speed, ie. PE:KE symmetry as RPM's increase..
I still don't quite understand the interaction you're describing though...? Help me understand, and i'll simulate an accurate measurement in WM2D..
An interaction between a weight and wheel, that lifts the weight whilst applying positive torque, is obviously on the right track.
As ever though, what really matters is how its efficiency scales with speed, ie. PE:KE symmetry as RPM's increase..
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Once again, here's a spoof of what OU looks like:
• the actuator is performing ½ J of work on each stroke; so 1 J / cycle
• 5 cycles thus cost 5 J
A hidden second actuator is absorbing counter-momentum, but only when the masses are moving apart, not when they're moving back together; obviously, in reality the full system is at unity.
The key point however is that, as far as the internal actuator is concerned, it's only performing 1 J of work per cycle, hence its net input energy is 1 J * elapsed cycles.
Output KE however is squaring with net system velocity - each 1 J spent internally giving a larger KE profit as net velocity rises in the external FoR.
Note however that to make the 'gain' real, we don't need the external stacking force to be entirely free like this; rather, we only need its cost to also sum linearly..!
In fact, the per-cycle cost of the external stacking force can be arbitrarily greater - so eg. 100 J per cycle would be no problem; provided it's summing linearly then 5 cycles cost 500 J etc., but because KE in the absolute FoR is squaring with velocity, there's always a threshold speed / no. of elapsed cycles beyond which KE>PE.
IOW the FoR of the stacking force cannot be bound to some other, external, inertia. Thus the gravitating / free-fall FoR would present an ideal candidate frame..
IOW, just imagine that the counter-momentum is instead being absorbed by gravity * time..
One way or another, this is what Bessler's wheels were doing, albeit in a rotating FoR..
• the actuator is performing ½ J of work on each stroke; so 1 J / cycle
• 5 cycles thus cost 5 J
A hidden second actuator is absorbing counter-momentum, but only when the masses are moving apart, not when they're moving back together; obviously, in reality the full system is at unity.
The key point however is that, as far as the internal actuator is concerned, it's only performing 1 J of work per cycle, hence its net input energy is 1 J * elapsed cycles.
Output KE however is squaring with net system velocity - each 1 J spent internally giving a larger KE profit as net velocity rises in the external FoR.
Note however that to make the 'gain' real, we don't need the external stacking force to be entirely free like this; rather, we only need its cost to also sum linearly..!
In fact, the per-cycle cost of the external stacking force can be arbitrarily greater - so eg. 100 J per cycle would be no problem; provided it's summing linearly then 5 cycles cost 500 J etc., but because KE in the absolute FoR is squaring with velocity, there's always a threshold speed / no. of elapsed cycles beyond which KE>PE.
IOW the FoR of the stacking force cannot be bound to some other, external, inertia. Thus the gravitating / free-fall FoR would present an ideal candidate frame..
IOW, just imagine that the counter-momentum is instead being absorbed by gravity * time..
One way or another, this is what Bessler's wheels were doing, albeit in a rotating FoR..
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..also note that in a rotating FoR, if the two interacting inertias share the same axis (ie. in co-axial configuration / radial weight levers), then the 'return' stroke may be redundant - the relative angle of separation can then simply accumulate, as if the two masses above were growing further apart; since two rotors sharing an axis always remain adjacent. In this case we only need an 'accelerate & brake' cycle, so wouldn't need to fully reverse the action each cycle.
If the two interacting MoI didn't share the same axis (such as in the case of diametric weight levers), then you obviously still need an oscillating / reciprocating action / 'reset' stroke, as above..
If the two interacting MoI didn't share the same axis (such as in the case of diametric weight levers), then you obviously still need an oscillating / reciprocating action / 'reset' stroke, as above..
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The take-home here is that we only ever require input and output energies / workloads to be at unity efficiency in their respective FoR's.
So the internal workload would be say 1 J / cycle relative to the rotating / accelerating FoR.
The actual / absolute work being done relative to the ground FoR is obviously the rotational KE. But both are entirely 'correct' in their respective FoR's - no magic / free work.
The only requirement is that the internal workload's FoR is inertially isolated from the external / absolute FoR. That is, without incurring Newton's 3rd law..
So the internal workload would be say 1 J / cycle relative to the rotating / accelerating FoR.
The actual / absolute work being done relative to the ground FoR is obviously the rotational KE. But both are entirely 'correct' in their respective FoR's - no magic / free work.
The only requirement is that the internal workload's FoR is inertially isolated from the external / absolute FoR. That is, without incurring Newton's 3rd law..
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re: Decoupling Per-Cycle Momemtum Yields From RPM
Hi MrVibrating,
thanks of the offer,
I have prepared some pictures so that you can understant the actions, reactions which are happen in the wheel.
Given is a Hamstercage which can rotate and a falling stick.
you will get an impact on the downgoing side,
with this function you will get an direct impact on the rim, a loud noise, but also torque to turn the Hamstercage.
The second picture the same situation, but on the stick we have a pendulum.
Again we will get the impact on the rim with the stick.
In addition we will get an upswing of the pendulum.
This is the function which I was alwasy telling, the indirect impact.
We don*t act directly against the mass.
The function is a tilt oscillation.
And as you can see the stick is not bound to the middle axle, the stick can fall forward in the Hamstercage.
A hard stop against the wheel rim of the stick is producing a flash upjump of the pendulum weight against gravity.
This is the essential function in the Besslerwheel.
As bigger the diameter of the Hamstercage as faster the stick will fall against the rim.
Remember Besslers words, the force has to be applied in the correct angle.
The stick will make the impact in the upper half of the Hamstercage, left or right depending on the turning direction of the Hamstercage.
This is one indirect impact, but we need 8 per turn for the complete run.
If you need more explanations let me know.
thanks of the offer,
I have prepared some pictures so that you can understant the actions, reactions which are happen in the wheel.
Given is a Hamstercage which can rotate and a falling stick.
you will get an impact on the downgoing side,
with this function you will get an direct impact on the rim, a loud noise, but also torque to turn the Hamstercage.
The second picture the same situation, but on the stick we have a pendulum.
Again we will get the impact on the rim with the stick.
In addition we will get an upswing of the pendulum.
This is the function which I was alwasy telling, the indirect impact.
We don*t act directly against the mass.
The function is a tilt oscillation.
And as you can see the stick is not bound to the middle axle, the stick can fall forward in the Hamstercage.
A hard stop against the wheel rim of the stick is producing a flash upjump of the pendulum weight against gravity.
This is the essential function in the Besslerwheel.
As bigger the diameter of the Hamstercage as faster the stick will fall against the rim.
Remember Besslers words, the force has to be applied in the correct angle.
The stick will make the impact in the upper half of the Hamstercage, left or right depending on the turning direction of the Hamstercage.
This is one indirect impact, but we need 8 per turn for the complete run.
If you need more explanations let me know.
Best regards
Georg
Georg
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I realised that what i was considering - shrinking the MoI's of the over-balancing weights on the above 2019 rig design, after giving them some OB momentum, then braking 'em back against the main axis - is basically exactly the same thing i was doing just the other week on another rig; essentially trying to accumulate relative angular momentum..
Obviously, retrying the same bust ideas is the very definition of insanity, so i thought of something else instead; i doubt it's anything Bessler would've tried, but it's totally me (surprised i ain't tried it already):
• overbalancing whilst overbalancing!
Right, eh? Ingenious even if i say so meself. So i could take the above style of over-balancing rig, and instead of making the OB weights little vMoI's, i could just make them run a little OB loop as well.. basically, doubling-up the same futile concept has to cancel out the 'stupid', right? Exactly.
So an OB system in which the OB weights themselves are little orbiting OB systems.. thus they spin up, and can be braked against the main axis, which itself is also over-balancing, so it's spinning up, and the weights are also spinning up, and it goes around and around and there's all crazy spikes on the meters and arcane formulas and everything's coming out exactly 10 J over and, ah, you know the drill..
Will prolly make a start in the next few days with a scratch-build (to keep it as simple & focused as possible)..
Obviously, retrying the same bust ideas is the very definition of insanity, so i thought of something else instead; i doubt it's anything Bessler would've tried, but it's totally me (surprised i ain't tried it already):
• overbalancing whilst overbalancing!
Right, eh? Ingenious even if i say so meself. So i could take the above style of over-balancing rig, and instead of making the OB weights little vMoI's, i could just make them run a little OB loop as well.. basically, doubling-up the same futile concept has to cancel out the 'stupid', right? Exactly.
So an OB system in which the OB weights themselves are little orbiting OB systems.. thus they spin up, and can be braked against the main axis, which itself is also over-balancing, so it's spinning up, and the weights are also spinning up, and it goes around and around and there's all crazy spikes on the meters and arcane formulas and everything's coming out exactly 10 J over and, ah, you know the drill..
Will prolly make a start in the next few days with a scratch-build (to keep it as simple & focused as possible)..
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM
Georg Künstler wrote:Hi MrVibrating,
thanks of the offer,
I have prepared some pictures so that you can understant the actions, reactions which are happen in the wheel.
Given is a Hamstercage which can rotate and a falling stick.
you will get an impact on the downgoing side,
with this function you will get an direct impact on the rim, a loud noise, but also torque to turn the Hamstercage.
The second picture the same situation, but on the stick we have a pendulum.
Again we will get the impact on the rim with the stick.
In addition we will get an upswing of the pendulum.
This is the function which I was alwasy telling, the indirect impact.
We don*t act directly against the mass.
The function is a tilt oscillation.
And as you can see the stick is not bound to the middle axle, the stick can fall forward in the Hamstercage.
A hard stop against the wheel rim of the stick is producing a flash upjump of the pendulum weight against gravity.
This is the essential function in the Besslerwheel.
As bigger the diameter of the Hamstercage as faster the stick will fall against the rim.
Remember Besslers words, the force has to be applied in the correct angle.
The stick will make the impact in the upper half of the Hamstercage, left or right depending on the turning direction of the Hamstercage.
This is one indirect impact, but we need 8 per turn for the complete run.
If you need more explanations let me know.
Wheel = 1 kg * 1 m radius = 1 kg-m² MoI
Stick =0.5 kg, 1.65 m x 0.05 m
Orange Ring = 0.05 kg, 0.05 m radius
Black Weight = 1 kg
Elasticity = 0
Sim ends in a stable keeling swing.
Is there some specific aspect of the interaction (some kind of hoped-for advantage) you'd like to see measured?
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re: Decoupling Per-Cycle Momemtum Yields From RPM
Hi MrVibrating,
thanks for the simulation.
Some corrections are necessary.
The length of the stick is not correct in the moment.
It must be adjusted in that way the the stick will make the impact at about 10:30
As you can see in your simulation the pendulum weight is rised up against gravity. The upswing of the pendulum will drive the wheel in the down direction.
The stick is not fastened at the rim of the Hamstercage.
At the beginning the stick is standing vertical in the wheel.
See the picture how the stick is standing in the wheel.
The stick is making an impact on one dowel.
The dowel is like a rod on which the pendulum swings up.
So you get torque to turn the hamstercage.
The swing up is done in an offset position of the center.
Hopefully you see that we have different accelerations, different to g.
thanks for the simulation.
Some corrections are necessary.
The length of the stick is not correct in the moment.
It must be adjusted in that way the the stick will make the impact at about 10:30
As you can see in your simulation the pendulum weight is rised up against gravity. The upswing of the pendulum will drive the wheel in the down direction.
The stick is not fastened at the rim of the Hamstercage.
At the beginning the stick is standing vertical in the wheel.
See the picture how the stick is standing in the wheel.
The stick is making an impact on one dowel.
The dowel is like a rod on which the pendulum swings up.
So you get torque to turn the hamstercage.
The swing up is done in an offset position of the center.
Hopefully you see that we have different accelerations, different to g.
Best regards
Georg
Georg