Understanding of Wave swing physics

[Kakainsu]

Homework Statement:

Hi guys,

Currently, we are discussing the physics of wave swing with and without the constant changing of the rotating axis. Former situation gives me problems. Our teacher presented a solution (attached), in which he explained the relationship of the actual tilt between the rotating axis and the chains (beta) and the tilt between rotating axis and the vertical axis (alpha). He makes use of the law of sine ("Sinussatz" in German).

I do understand the explanations with one exception: How can we assume, that beta is the same at the lowest and highest point of the ride? To go one step further, does that mean, that beta is constant at a given alpha for all "positions" on the described circle at that moment? I hope you know what I mean, it's hard to express :)...

Edit: I now included another "commented" pic, maybe it now is easier to understand...

Relevant Equations:

No equations needed, I guess...

Yeah, as I said, i have no idea why we can assume beta to be the same for any position at given alpha.

 

[haruspex]

Please explain more about the physical system being analysed.
What is a "wave swing"? What are the chains doing and what are they attached to? What is this rotating axis?

 

[Delta2]

I also have trouble to understand what is a wave swing. Please post an image of the system itself , images of the forces in the system is good but first we need to understand what the system is.

 

[Kakainsu]

Oh, okay, sorry. A wave swing is an amusement park ride. It's a variation of a simple carousel: The guests are rotating around the axis with constant angular velocity while being held by chains
. However at the same time, the axis itself is periodically changing and tilting from the maximum tilt alpha to maximum tilt minus alpha.

What we were trying to analyse, was, how the constant change in tilt of the axis influenced the tilt of the guests in relation to the axis. Hope it makes sense now.

 

[Kakainsu]

This is another impression.

 

[haruspex]

Ok, now it's clear.

There would not be a unique behaviour. Consider the much simpler case of a pendulum attached at a fixed point but, instead of simply swinging back and forth in a vertical plane, able to move around in x and y. In general, it can perform complicated motions: .

So we need to think what the simplest behaviour might be. An obvious problem with the teacher's solution is that the passengers are experiencing changes in GPE as they rotate about the tilted axis. This will likely lead to greater speed at the lowest point, so a greater angle to the axis.

 

[Delta2]

@haruspex i think the spherical pendulum system is a generalization and more complex system than the carousel with tilting axis. I don't know the full analysis of the two systems, but my intuition tells me that the spherical pendulum can be found anywhere within the hemisphere, while in the tilting carousel i would expect the rotating body to be found near the circumference of the circle.

 

[haruspex]

@haruspex i think the spherical pendulum system is a generalization and more complex system than the carousel with tilting axis. I don't know the full analysis of the two systems, but my intuition tells me that the spherical pendulum can be found anywhere within the hemisphere, while in the tilting carousel i would expect the rotating body to be found near the circumference of the circle.

Ah, the perils of intuition.
The spherical pendulum is a special case of the tilting carousel. Shrink the canopy to a point and turn off the tilt.

Unless... maybe we should assume that each rider is constrained to stay in a vertical plane that rotates with the central axis. I'll have to think about that... it's late here.

 

[Kakainsu]

Ok, now it's clear.

There would not be a unique behaviour. Consider the much simpler case of a pendulum attached at a fixed point but, instead of simply swinging back and forth in a vertical plane, able to move around in x and y. In general, it can perform complicated motions: .

So we need to think what the simplest behaviour might be. An obvious problem with the teacher's solution is that the passengers are experiencing changes in GPE as they rotate about the tilted axis. This will likely lead to greater speed at the lowest point, so a greater angle to the axis.

Alright, so apparently beta is different for any position on the circle at given alpha. Could someone give me an explanation as to why?

 

[Kakainsu]

Alright, so apparently beta is different for any position on the circle at given alpha. Could someone give me an explanation as to why?

Especially considering that the velocity is dictated by the rotating axis and is constant

 

[haruspex]

Especially considering that the velocity is dictated by the rotating axis and is constant

With the simplifying assumption I suggested at the end of post #8, the angular velocity is constant; but the distance from the central axis varies, so both tangential velocity about the central axis and tangential velocity (in the vertical plane) about the point of attachment of the cable can vary.

 

[haruspex]

Alright, so apparently beta is different for any position on the circle at given alpha. Could someone give me an explanation as to why?

So far, I only claim it can vary. There may be many possible behaviours. What happens in practice may be an even more complicated problem, involving consideration of perturbations and stability.
If we can develop the appropriate equation, it may be possible to say whether the angle can remain constant.

 

[Kakainsu]

So far, I only claim it can vary. There may be many possible behaviours. What happens in practice may be an even more complicated problem, involving consideration of perturbations and stability.
If we can develop the appropriate equation, it may be possible to say whether the angle can remain constant.

No, sorry, I don't understand. I do understand This is true for alpha changing constantly, as the two equation I attached suggest that beta is dependent on alpha, thus beta changes with changing alpha. HOWEVER, I considered a 'stopped' moment with alpha being given. Apparently, beta is still different for every position on the trajectory. This is what I do not understand.

 

[Kakainsu]

I think, I made a mistake by saying that the teacher suggested those two equations for the system as a whole. This was a mistake on my side, those equations represent the wave swing in a 'stopped' moment, like freezing time.

 

[haruspex]

No, sorry, I don't understand. I do understand This is true for alpha changing constantly, as the two equation I attached suggest that beta is dependent on alpha, thus beta changes with changing alpha. HOWEVER, I considered a 'stopped' moment with alpha being given. Apparently, beta is still different for every position on the trajectory. This is what I do not understand.

I am assuming alpha changes sufficiently slowly that we can treat it as constant.

I've just thought of a simple demonstration that beta will not be constant.
Imagine a very slow rotation. The seats will just hang vertically. β varies between -α and +α.

 

[Kakainsu]

I am assuming alpha changes sufficiently slowly that we can treat it as constant.

I've just thought of a simple demonstration that beta will not be constant.
Imagine a very slow rotation. The seats will just hang vertically. β varies between -α and +α.

Yes, I do understand that when alpha is changing beta is changing. But again, after asking my teacher it turns out, that every position on the 'stopped' wave swing's trajectory has different beta from another. This is what I can't get my head around.

 

[haruspex]

Yes, I do understand that when alpha is changing beta is changing. But again, after asking my teacher it turns out, that every position on the 'stopped' wave swing's trajectory has different beta from another. This is what I can't get my head around.

As I wrote, I am not changing alpha.

 

[Kakainsu]

As I wrote, I am not changing alpha.

Hmm...I would be very happy if you could explain for the last time why this is...I think I got lost in the split responses...

 

[haruspex]

Hmm...I would be very happy if you could explain for the last time why this is...I think I got lost in the split responses...

If I understand the original problem correctly, you have a central axis leaning at some angle α to the vertical. In reality this can vary, but let's keep it fixed, and in a fixed direction: say it leans to the north always.
Rotating around this axis at a constant rate ω is a disc, the canopy of the ride.
Suspended at intervals around the perimeter of the canopy are cables with masses attached.
Suppose a particular cable makes angle β to the axis when at its northmost position, so makes angle β-α to the vertical. What angle do we expect that cable to be at when in its southmost position? Will it still be at angle β to the axis (so β+α to the vertical) or still at β-α to the vertical, or something else?

In an attempt to analyse it I simplified by shrinking the canopy to a point and requiring that the plane containing the axis and the cable rotates at ω, but I still have not found the equation.
So instead I went back to the problem as I described it above and considered the case of small ω. There is little centrifugal effect, so gravity dominates and the cable should remain nearly vertical.
At the other extreme, large ω, the centrifuge dominates and β will be nearly constant.
In the ride in the video, the cables look to be about 50 degrees to the vertical when α is zero, implying the centrifugal acceleration is more than g, but not much more.

 

[haruspex]

Yet another simplification option:
In reality, the behaviour at any point in the cycle depends on the history. E.g. in going from the high (south) position to the low position the rider will gain speed, so swing out further. But suppose we treat the north and south positions, independently, as snapshots in a motion in which the seat is describing circles around the axis, normal to it.
If the canopy has radius r and the cable has length L then the radius of the circle is $r+L\sin(\beta)$. The centripetal acceleration is $\omega^2(r+L\sin(\beta))$, and the component of this normal to the cable is $\omega^2(r+L\sin(\beta))\cos(\beta)$.
The tension has no component in that direction, so this must equal the component of g in that direction, $g\sin(\alpha+\beta)$:
$\omega^2(r+L\sin(\beta))\cos(\beta)=g\sin(\alpha+\beta)$.
The same calculation for the north position gives $\omega^2(r+L\sin(\beta'))\cos(\beta')=g\sin(-\alpha+\beta')$.
Clearly, β and β' are different unless α=0.

Relating this to the two extremes I described earlier, as ω→0, β+α and β'-α tend to 0 (i.e. hangs vertically) and as ω→∞, β→π/2, β'→π/2.