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Let's say that we drop a stick down with an angle, meaning that the stick is not parallel to the ground. Assume there is only gravity. Is it correct that the orientation of the stick will not change as it drops down? Since, to a good approximation, each part of the stick is pulled down by gravity with the same force. I assume that the difference in height with respect to the ground does not really matter due to gravity being very weak (and the stick being pretty light).

In practice, we do see that the orientation of the stick changes as it falls down, but I am assuming that this is due to air resistance/wind for example (assuming the stick is dropped from a large height).

Here, I assume a uniform mass distribution of the stick.

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    $\begingroup$ each part of the stick is pulled down by gravity with the same force. Elementary mass of stick part at $x$ is $\pi r^2 \rho (x) dx$, so it depends how density is distributed along stick. If it's ideally uniform, then yes- attracting force per each stick part will be same. Otherwise if mass fluctuations gets out of acceptable error level- it should be taken into consideration while calculating Earth gravitational force over each stick part. Besides as stick is at angle,- distances from ends to Earth center differs, and as such gravity force should differ on them too, as it is $\propto 1/r^2$. $\endgroup$
    – Agnius Vasiliauskas
    Commented May 6 at 6:49
  • $\begingroup$ @AgniusVasiliauskas - regardless of the density distribution the $$\text{(center of gravity)} \equiv \text{(center of mass)}$$ $\endgroup$
    – John Alexiou
    Commented May 6 at 15:35
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    $\begingroup$ @JohnAlexiou It's not about COM, neither about COG, but will gravity induce some torque to the stick. And for that type of question mass distribution along the rod is very cruicial as it is also important in the tidal forces effect, which in case of the moon produces negative (stopping) torque for the Earth. I believe tidal forces and the like should rotate stick especially if say mass increases linearly along the stick. $\endgroup$
    – Agnius Vasiliauskas
    Commented May 6 at 18:40
  • $\begingroup$ Yes indeed, if one end is heavier than the other end then the gravitational forces are not balanced causing a torque which causes the heavier end to point down towards the center of the Earth. Here, I have assumed the mass distribution to be uniform, which I have added to my question now explicitly. $\endgroup$
    – Stallmp
    Commented May 6 at 18:55
  • $\begingroup$ Are you saying that if I drop a hammer pointing horizontally, then while falling it will swivel so the heavy end is pointing downward? I suggest you try the experiment yourself and see. $\endgroup$
    – RC_23
    Commented May 7 at 3:31

2 Answers 2

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Air resistance force is one possible effect that leads to a torque, but I think that the bigger effect comes from the torque which you apply when throwing the stick. Unless you hold it exactly in its center of mass, you will not only give it momentum but also an initial angular momentum, such that it will rotate while falling down.

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  • $\begingroup$ I see, so if the stick is dropped at an angle while not applying an initial torque, it will just drop down without rotating right? So the reason why a stick usually does rotate is 1) one end being closer to the center of the earth (small effect) 2) giving it initial angular momentum 3) air resistance/wind $\endgroup$
    – Stallmp
    Commented May 6 at 9:32
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    $\begingroup$ Yes, I think so! $\endgroup$
    – Photon
    Commented May 6 at 17:40
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The tidal effects are negligible; the difference in gravitational attraction on a 2m stick dropped vertically is less than 1 part in 1 million from one end to the other.
The distribution of mass along the stick is immaterial; the acceleration due to gravity is independent of the mass when considering a stick’s and the earth’s comparative masses.
If the stick has a uniform cross-section, drag due to air resistance will be uniform along it. The air flow over the stick should be uniform.

There is one other consideration; suppose that the ends of the stick are cut flat and perpendicular to its length, then when the stick is falling at an angle the lower end of the stick will have a force applied to it due to it deflecting air away from its face which will cause a torque and rotate the stick about its COM. The stick will gain angular momentum and will continue to turn past horizontal to become tilted in the opposite direction. This will cause the other end of the stick to be exposed to the air flow and the motion will be reversed. Under this scenario the stick will oscillate back and forth as it falls.

It would be interesting to test this experimentally.

EDIT
Here is the pressure distribution of a cylinder in a wind tunnel.
from here
wind tunnel test
Depending on the Reynold's number, the pressure on the back side of the cylinder, may be positive or negative. If the falling stick is in an area where the Reynold's number is such that the pressure on the back of the stick is negative while that on the front is positive, the stick will tend to straighten out, falling end first. As it will not be in equilibrium it will gyrate as it falls.

So my conclusion is that the stick will tend to fall end first and gyrate about its axis.

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  • $\begingroup$ I am confused why the mass distribution is not significant. Why can't you say that there is an imbalance of forces which causes a torque and therefore the stick to rotate? Just like the situation with a lever and a pivot/fulcrum, why is that different from here? I do get that mg=ma and therefore mass cancels out, but it seems to contradict my earlier statement $\endgroup$
    – Stallmp
    Commented May 6 at 22:15
  • $\begingroup$ Different masses fall with the same acceleration. Suppose you had two masses, one twice that of the other connected by a light rod, both masses have the same acceleration, there is no torque. Remember Galileo. $\endgroup$
    – Rich
    Commented May 6 at 22:30
  • $\begingroup$ But then how does a lever work with a fulcrum where one side will rotate due to torque if it's heavier? $\endgroup$
    – Stallmp
    Commented May 6 at 22:55
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    $\begingroup$ That's because it is supported at one point about which it will rotate. That's not true for falling bodies. You have the upward normal force at the fulcrum and downward forces where the masses are. Those opposing forces cause the torque. On falling bodies there are only the downward gravitational forces, which produce the same acceleration independent of the masses involved. $\endgroup$
    – Rich
    Commented May 6 at 23:13
  • $\begingroup$ But the upward normal force is at the fulcrum which is the pivot, so it can't exert a torque right? Or is the pivot exerting a force on the masses which makes it go up/down? $\endgroup$
    – Stallmp
    Commented May 6 at 23:50

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