If a shuttlecock is dropped(not giving it any angular momentum wrt it's COM) head up(i.e. the heavier part up) without tilting it, it always falls head down(i.e. the heavier part down). This fact it's not consistent with conservation of angular momentum. Why is this so? Drop it this way

Ends up this way

  • $\begingroup$ How do you drop the shuttlecock with the heavy end up, with absolutely no tilting ? Not even at the sub-atomic level? $\endgroup$
    – DJohnM
    Jul 9, 2017 at 5:10
  • $\begingroup$ What do you think would happen if dropped in a vacuum and constant gravitational field? $\endgroup$
    – Bill N
    Jul 9, 2017 at 18:52

1 Answer 1


Short answer: Air resistance pushes it to its stable configuration. If you dropped it in a vacuum, it wouldn't turn over.

Long answer:

As the shuttlecock falls, the air exerts an upward force on all parts of the shuttlecock. The net effect of all of these upward forces is a torque on the shuttlecock that depends on its orientation. There are two orientations in which there is no torque on the shuttlecock at all. The first is with the head facing perfectly straight up, and the second is with the head perfectly straight down.

The difference between these two equilibria is their stability under small changes in orientation. The orientation with the head facing up is unstable under small changes in orientation; any small rotation to one side increases the directness of the force on that side, meaning that the small rotation is amplified. In contrast, the orientation with its head facing down is stable. Any small rotation to one side deflects that side back toward the head-down configuration. So, when you drop it head-up, if you don't do it exactly perfectly (or if there is some miniscule air current on the way down, or if literally anything interferes with its descent in any way), it'll revert to its stable equilibrium, the head-down configuration.


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