I am wondering if rotating flyweights would slow down the descent of a object that is free falling down to the Earth's surface.

Please reference the conceptual drawing below:

enter image description here

This drawing is showing a cross-sectional view of a sealed hollow cylinder. Inside this hollow cylinder would be two DC electric motors with each having two rods attached to their rotor shaft and would have two thrust bearings sandwiching each rod which would allow each rod to freely pivot up or down as the motor shaft rotates. At the end of each rod will be a flyweight.

There would be a battery attached to the bottom of the cylinder to power the DC motors and to also act as a dead-weight so the cylinder remains in this upward position as it free falls down to Earth's surface. The drawing is also showing a metal plate with a hole in it located at the top of the cylinder to enable this to be dropped from a balloon or helicopter. The DC motors would be counter-rotating to each other in order to keep the cylinder from spinning around.

A balloon or helicopter would take this cylinder up to a certain height and then release it. Just before releasing it, the two DC motors will be turned on and the flyweights will be rotating at a 90 degree angle to the motors' rotor shafts.

Once the cylinder is released and starts to accelerate downward, the rotating flyweights should pivot upwards to a certain degree (as shown in the drawing) and should act as a drag on each of the motors' rotor shafts and thus a drag on the falling cylinder, thus slowing down its descent. I believe the faster the flyweights rotate, the more drag they will produce and the slower the cylinder will descend.

Would rotating flyweights slow down the descent of an object that is free falling down to the Earth's surface?


I have created a revised design that I believe should slow down the descent of the falling object.

enter image description here

The revised drawing above is showing that a round disc with an O-Ring around it which been inserted into the cylinder. This disc has two metal posts attached to the bottom of it. Each post has a thrust bearing and a shaft bearing on top of it. Attached to the shaft bearings are steel cables that are attached to the rods that are attached to the motors' rotor shafts. The inner lining of the cylinder would have some kind of lubricant on it to allow the disc to slide up or down.

The concept is that the air pressure inside the cylinder, below the O-Ring, could be set at ground level pressure (14.7 psia) while the air pressure above the O-Ring will be lower due to the lower ambient air pressure outside of the cylinder. There will be holes in the top of the cylinder to allow air to enter in or exit out of the top section of the cylinder.

In order to create a decelerating force, the DC motors would need to be turned on and off in a continuous loop. When the motors are turned off, the disc will move towards the top of the cylinder and when the motors are turned on, the disc will be pulled back down. A temporary decelerating force should be created during the time periods when the disc is being pulled down.

  • $\begingroup$ Are you shielding one side or the rotating mass? Otherwise the drag effect will cancel. $\endgroup$
    – Solar Mike
    Commented Aug 18, 2020 at 4:34
  • $\begingroup$ @Solar Mike, that’s interesting, but I wouldn’t know how that could be done. $\endgroup$
    – user26803
    Commented Aug 18, 2020 at 12:16
  • $\begingroup$ I spent most of 1970-71 unable to make this scheme work as a freshman physics student. The first elementary dynamics class I took taught me why. $\endgroup$ Commented Aug 18, 2020 at 15:57
  • 1
    $\begingroup$ If it was at rest in outer space far from anything else, would you expect it to start moving because of rotation of its internal parts? $\endgroup$
    – J...
    Commented Aug 18, 2020 at 18:50
  • 2
    $\begingroup$ This looks like a 'Novel Idea' question. As demostrated by the latest edit, such questions tend to become moving targets and lead to discussions, neither of which are a good fit for our format. See if you can edit your question to make it specific and answerable. $\endgroup$
    – Wasabi
    Commented Aug 19, 2020 at 15:01

5 Answers 5


Once the cylinder is released and starts to accelerate downward, the rotating flyweights should pivot upwards to a certain degree (as shown in the drawing) and should act as a drag on each of the motors' rotor shafts and thus a drag on the falling cylinder

No, there is no reason for them to pivot upwards. The gravity pulls the flyweights with the same acceleration as it does the box.

Another way to think about it: being inside a falling box is equivalent to a box in zero gravity. In zero gravity the flyweights would remain in in the 90 degree orientation.

  • $\begingroup$ that's a good point you make about the flyweights not pivoting upwards. I can see that now. $\endgroup$
    – user26803
    Commented Aug 18, 2020 at 14:18
  • 1
    $\begingroup$ But before the box is released, don't the flyweights hang slightly below the 90 degree mark? If the rods are perfectly horizontal, there's no force holding the flyweights up against gravity. I agree that when the box is released, the flyweights are indeed perfectly horizontal, since you don't need any force to support the weight, as the effective weight goes to 0 in a freefall. $\endgroup$ Commented Aug 18, 2020 at 19:10
  • $\begingroup$ @NuclearWang Yes, if they can rotate in that direction. But in the situation in the question, they start at 90 degrees: "the flyweights will be rotating at a 90 degree angle to the motors' rotor shafts". What you are asking about is probably what the second part of kamran's answer deals with, though I did not understand it fully. In any case, it would be similar to having a spring and weight inside the box: the center of mass of the system would fall at usual acceleration, while the relative positions of box and weight would change relative to center of mass. $\endgroup$
    – jpa
    Commented Aug 18, 2020 at 19:29
  • $\begingroup$ I am making this the answer for this question based on what it said concerning the original design -- before the Edit section was added and the new drawing was added. $\endgroup$
    – user26803
    Commented Aug 19, 2020 at 23:28

jpa made a good answer on why the weights wouldn't flex upwards. I'll add another point. Newton's third law says that for every action, there is an equal and opposite reaction. For the box to be slowed by this mechanism, it must exert a force on something else, in the same way gravity pulls up on earth and aerodynamic drag moves the air. However, this mechanism does not interact with anything outside of the box, and so it can't cause an acceleration.

  • $\begingroup$ I agree with what you said. I'm pretty sure that if the motors and the flyweights are rotating out in the open air, like a helicopter's rotor and blades, then the interaction of the air pushing against the rotating flyweights as the mechanism descends will generate air resistance which should slow the descent of the mechanism. It would be similar to a man parachuting down to the ground with his arms extended out, his arms will cause some air resistance and drag which will slow down his descent a little bit. $\endgroup$
    – user26803
    Commented Aug 18, 2020 at 17:16
  • $\begingroup$ Drag would work the same, although you describe the device as a sealed hollow cylinder, and so you wouldn't get any drag from the internals. $\endgroup$ Commented Aug 18, 2020 at 19:09

This will make the box hit the ground slightly faster.

Before you release the box, the flyweights are not quite horizontal (unless they are constrained to be - the question makes incompatible statements that the rods can "freely pivot up or down" but also that the flyweights "will be rotating at 90 degrees"). There must be some force supporting their weight, which comes from tension in the rods. If the rods are completely horizontal, there is no force counteracting gravity, so it must be the case that the flyweights hang somewhere below 90 degrees, no matter how fast they spin. Either the rods do not hang at 90 degrees, or they do not move freely up and down - this answer assumes the former.

When the box is released, the only force acting on the flyweights is the tension in the rods, so they will naturally orient to a horizontal position. We can see that the distribution of mass inside the box has moved upwards - but because there was no external force involved, the center of mass hasn't moved. We can see that as the rods pull the flyweights upward, they also pull the body of the box downwards, keeping the center of mass in the same position.

Overall, the box's center of mass will fall at the same rate, but due to the redistribution of mass inside the box, the bottom edge of the box gets pushed downward slightly and will hit the ground sooner than an empty box. It's similar to how one could make landfall in a canoe slightly faster by walking from the front of the boat to the back, exerting no force on the center of mass but redistributing that mass to push the canoe forward while you move backward. Another example presented in a comment by @jpa is to replace the flywheel apparatus with a mass on a spring - the spring is normally compressed under the force of gravity, but when the box starts to fall, there is no effective weight on the spring, so it expands, pushing the mass up and the box down, although the center of mass remains stationary. The center of mass follows the exact same path regardless of what's in the box, but the bottom edge of the box may not.


I don't quite understand what your rotating things are doing exactly, but the only thing a closed system like this can do is change the center of gravity of the box. If your rotating things moved the center of gravity down you could theoretically make the bottom of the box hit the ground a fraction of a second slower.

This is the same concept as a high-jumper whose center of gravity actually passes below the bar.

  • $\begingroup$ I agree with you on that, yet please see my revised design which I believe should slow down the descent of the falling object. $\endgroup$
    – user26803
    Commented Aug 19, 2020 at 14:23
  • $\begingroup$ you cannot slow down the descent of an object by moving things inside of it. It defies Newton's laws of motion. Unless you can apply a force to the object you cannot slow it down. An object cannot apply a force to itself. $\endgroup$
    – Tiger Guy
    Commented Aug 25, 2020 at 6:15

Let's say you rotate the wheels with the speed $\omega $

So our centripetal force $F= mr\omega^2$

The direction of this force is orthogonal to the force of gravity pulling the flywheel down so it would not impede the fall.

However, if we imagine the same m as the mass of the flywheel is two balls attached to a thin massless rigid string (total of 4 balls) the force of gravity will do some work to change the angle of the strings to an angle

$ \theta= arctan(mg/mr\omega^2)$

But even this will not slow down the descent. The work done to deflect the wires is maximum $W_{max}=mg*r$

The gravity will do a total of work during this transition period as

$W=W_{work to deflect}+W{potential}= mg*r +mg*r =2mgr$

This means the gravity has treated the work done to bend the wires as inertial work, in other words, the box mass by all intents and purposes had increased during this transition.


After some comments about why the balls won't go up.

The Balls initially while rotating are being pulled down by gravity but are supported by the wires. Once we release the box the balls will start to move up because the force of gravity holding them down is removed.

Edit I think an example would help. Lets annotate the total strain in the box de. And the combined stiffness of all the plates and walls and levers and wires K.

Jist before the dropping of the box it has strain energy of E= de^1/2*k.

This energy will work instantly after the drop and try to pull the balls up. And the system will start a harmonic motion as it free falls. The first half period is covered in my answer.

  • $\begingroup$ *intents and purposes 🙃 $\endgroup$ Commented Aug 18, 2020 at 7:26
  • $\begingroup$ thanks @JonathanRSwift. $\endgroup$
    – kamran
    Commented Aug 18, 2020 at 7:35