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The question is more specifically to explain why the box doesn't weigh more, because if it did then I could revolutionize space travel.

Consider a rigid box suspending two counter oscillating pendulums of mass m/2. Starting against the top, at the bottom of their swing the masses will be moving with a speed of $\sqrt{2gl}$ where $l$ is the length of the pendulum. The sum tension in the lines at this moment should be equal to the weight of the masses plus the centrifugal force: $T = mg + \frac{mv^2}{r}$. Plugging in the speed at the bottom gives a tension of $T = 3mg$. I figure the weight of the full box is the weight of the box material plus the downward component of the tension, as any horizontal components are canceled by the counter-oscillating setup.

So it seems at the bottom of the pendulums swing, the box weighs more than when with stationary pendulums. And, intuitively, the box will weigh less as the tension goes to zero at the apex.

I can envision a box suspending two sets (four total) of counter rotating powered rods that are out of phase such that at any moment there is a non-zero downward component of tension, making this box always heavier when the rods are moving.

Why couldn't I use this "extra weight" to transfer momentum?

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    $\begingroup$ you probably could. What 's wrong with that? How will you make the pendulums swing in space? $\endgroup$
    – user20574
    Commented Mar 1, 2023 at 18:20
  • $\begingroup$ The same centrifugal forces would be generated if the pendulums were motorized rods, this is the case I describe in the second to last paragraph. I'm sure the reaction forces on the box must have something to do with it, but I can't resolve how. $\endgroup$
    – A McKelvy
    Commented Mar 1, 2023 at 18:29
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    $\begingroup$ Why do you need counter-rotating pendulums to generate an oscillating force on the box? Why not just one heavy ball bouncing up and down on a spring? $\endgroup$
    – user20574
    Commented Mar 2, 2023 at 13:00
  • $\begingroup$ That is a very good point. I already understand where my reasoning failed, but this example highlights it well. $\endgroup$
    – A McKelvy
    Commented Mar 2, 2023 at 13:01

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The double pendulum apparatus has an apparent weight equal to the sum of the weights of the box and two pendulums when at rest, and weighs more than that at the bottom of the swing, but weighs less than that at the top of the swing. Also note that the bob moves through the bottom of the swing very quickly, meaning there is a very brief time when the apparent weight increases, but that the bob spends comparatively more time at the extremes of the swing when its velocity is low and apparent weight is less than its true weight.

As such, it is not the case that starting two double pendulum apparatuses will make a box appear heavier overall at all times. The key is that you need to time-average the effects of the pendulums - while one very briefly makes the box heavier, the other is making the box lighter, by a smaller amount but for a longer time. The apparatus' apparent weight oscillates between slightly heavier and slightly lighter at any instant of time as the pendulums swing, but on average, is no different than with the pendulums at rest.

It is possible to transfer momentum out of a system like this, but you're just decreasing the amount of energy in the system, and will eventually run it down. You could transfer some momentum out of the box at the bottom of a pendulum swing, but the pendulum will of course not continue to swing with the same amplitude. You don't get energy "for free" by starting a double pendulum and having the box magically increase in weight in the absence of other external forces.

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  • $\begingroup$ Thanks for the answer. Are you sure the answer averages to zero, and for arbitrarily many pendulums in the box? Also if you consider nothing else, could you please address the last paragraph: could a box of motorized oscillators propel a spacecraft? I know the answer is no, but I don't know why. $\endgroup$
    – A McKelvy
    Commented Mar 1, 2023 at 18:45
  • $\begingroup$ @AMcKelvy Every individual pendulum time-averages to zero, so it doesn't matter how many you have. You could do a one-time momentum transfer from a moving oscillator to another object like a spacecraft, but after that, the spacecraft is moving but the oscillator isn't (or at least is moving slower). You can't repeat the process, since to restart the pendulum you'd need to take momentum back from the ship, putting you back where you started. $\endgroup$ Commented Mar 1, 2023 at 18:54
  • $\begingroup$ Thank you. I understand now. $\endgroup$
    – A McKelvy
    Commented Mar 1, 2023 at 19:00

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