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I'm trying to wrap my head around a conceptual problem involving a simple pendulum with a rocket attached to its mass. Imagine the rocket expels gas to provide a tangential thrust force. However, the thrust is calibrated such that it's not sufficient to swing the pendulum upward; instead, it only counteracts the gravitational pull, bringing the pendulum to a point of equilibrium (let's say, 45°) where the downward gravitational force equals the upward thrust.

Under ideal conditions, where we negate the effects of friction, air resistance, and other non-conservative forces, I can't seem to understand the conservation of energy in this system. The rocket's fuel conversion into kinetic energy of the expelled gases generates thrust that maintains this equilibrium. Yet, with the pendulum statically held in this equilibrium state:

Where does the energy imparted by the thrust go, considering the pendulum itself does not seem to gain kinetic or potential energy?

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    $\begingroup$ Voting to reopen. A perfectly clear question with a straightforward answer. $\endgroup$
    – gandalf61
    Mar 6 at 15:25
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    $\begingroup$ I wonder if you have a similar worry about the momentum of the system? The exhaust is moving, so the rocket must lose momentum, but the rest of the system (rocket + pendulum) is stationary at all times. Is that problematic? $\endgroup$ Mar 7 at 11:52
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    $\begingroup$ As one answer points out, the important thing to recognize first is that the pendulum is irrelevant. Look at a helicopter (or hummingbird) hovering in place $\endgroup$ Mar 7 at 19:18
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    $\begingroup$ The expelled fuel has mass that will reduce the mass of the pendulum over time. Have you, and are we expected to consider this in the system model? $\endgroup$ Mar 7 at 20:24

3 Answers 3

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If the pendulum is in equilibrium then the rocket motor does no work on the pendulum. It exerts a force on the pendulum, but because the pendulum is not moving, this force does no work on the pendulum. It is exactly as if the pendulum was held by a length of rope - the rope exerts force on the pendulum but does no work on it.

The rocket motor, of course, does work by expelling its exhaust, but the energy that goes into the exhaust is initially seen as kinetic energy of the exhaust, and is eventually dissipated into the environment as sound and heat.

Note that during the initial phase of the motion - as the rocket motor moves the pendulum from vertical to its new equilibrium position - the velocity of the exhaust is lower than in the equilibrium position. This is because the exhaust has a fixed velocity relative to the rocket motor, which is now moving. So in this initial phase the rocket motor does less work on the exhaust and does some work on the pendulum instead - and this energy is stored as the potential energy of the pendulum in its new equilibrium position.

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    $\begingroup$ Yes, I think the explanation is correct: "This is because the exhaust has a fixed velocity relative to the rocket motor, which is now moving." This explains where the energy goes: from a external observer, not the rocket, the propellent has more kinetic energy when the rocket is static than when it's moving. For example, if the system is in equilibrium and you remove the pendulum, the rocket will start accelerating (and the new exhaust will have a smaller speed relative to the ground) $\endgroup$ Mar 6 at 15:06
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    $\begingroup$ That's true for mechanical energy, but not "energy" generally. There it's conserved because you are liberating energy from chemical bonds and turning it into heat and noise. $\endgroup$
    – fectin
    Mar 8 at 0:20
  • $\begingroup$ @EduardoGonzález: Yup. Rockets (and jet engines) are more or less constant thrust, and useful power (work done on the rocket / plane) = force x speed. So they can give huge amounts of power at high speeds. Unlike propeller planes, where the propeller can move enough air to extract the full power of the engine over a wide range of airspeeds. But even then the work done increasing the plane's speed from zero is very inefficient, and with the brakes on you're just pushing the air around without doing any useful work on the plane. $\endgroup$ Mar 9 at 1:39
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All of the energy of the rocket motor in your case goes into moving air/used propellant downward. None of it goes into adding velocity to the rocket or pendulum. Eventually the velocity of that air/propellant starts to get diffused as eddies form or the plume interacts with the ground, and that kinetic energy in the air dissipates, converting into heat energy.

The situation you have could be simplified. The image of a rocket on a pendulum is rather gratifying, but if it is at equilibrium with the forces of gravity, the force diagrams are comparable to a static fire of a rocket engine. In that case, instead of working in a delicate equilibrium, we simply clamp the rocket down and light it. I find the visuals of a static fire do a good job of demonstrating where the energy goes.

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  • $\begingroup$ I agree, both systems are equivalent. However a clamped rocket has other factors (e.g. deformation the clamping structure) that make it more complex. A rocket in a gravitational field, statically pushing against gravity would be maybe better. The thing that bugs me is: does the energy of the propelled gas depend on the movement of the rocket? It seems counterintuitive to me. $\endgroup$ Mar 6 at 14:02
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    $\begingroup$ The energy of the gas does not depend on the movement of the rocket, as the static fire shows. Now momentum also has to be conserved in a closed system, and that starts to get a little nuanced. In theory, your little rocket is "tethered" to the planet by gravity, and when you propel the gas downward, you pull the Earth upwards at an astonishingly small rate. But in the most general sense, you can have force without energy (simply ensure the distance over which it is applied is 0) and you can have kinetic energy without force (moving at a constant velocity with no friction or opposing force) $\endgroup$
    – Cort Ammon
    Mar 6 at 14:29
  • $\begingroup$ And the pattern of the rocket using air to pull on the earth, and that air then colliding with the ground feels a lot like blowing your own sail, but now we're getting into really complex interacting systems that are more complicated than your thoguht experiment. $\endgroup$
    – Cort Ammon
    Mar 6 at 14:34
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    $\begingroup$ See comment below: the kinetic energy of the gas, from the reference frame of the ground, does depend on the movement of the rocket!, since the gas is expelled at a constant velocity from the reference frame of the rocket. So if the rocket is not moving, the gas goes faster (from ground's reference frame). $\endgroup$ Mar 6 at 15:04
  • $\begingroup$ @EduardoGonzález Certainly. Your question seemed to be focused on the equilibrium conditions, so I focused on that. $\endgroup$
    – Cort Ammon
    Mar 7 at 3:08
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Hovering

Your scenario is equivalent to a rocket suspended above a suitable gravitational body, firing its engines so that it may remain in place. Of course, the gravitational field is accelerating the rocket towards the surface, so the engine must exactly counteract that acceleration. Thus, the downward momentum of the exhaust must perfectly balance the momentum of the rocket in free-fall. The COM of the rocket + exhaust system is in free-fall, conserving the energy expended. The pendulum here, is a mere distraction.

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