Does a rod rotate under gravity?

[albertrichardf]

Hello.
Say you have a uniform rod in free-fall. Would it rotate?
Suppose you compute the rotation around the CM. I computed that the lever arm is the same on either side, regardless of the angle at which the rod falls, so there can't be any rotation. So if I just dropped a 30º to the horizontal rod, it would fall and stay at 30º to the horizontal. Is that correct?
Also, suppose I compute the rotation around one end of the rod. The torque is non-zero, but the rod is not rotating around that point. How is that possible? Furthermore, since the rod is falling, the end is falling as well, so my reference point is moving. If the reference point was stationary (the end of the rod is falling, it does not necessarily coincide with the point), does that change anything?

Thank you for answering

 

[Bandersnatch]

Suppose you compute the rotation around the CM. I computed that the lever arm is the same on either side, regardless of the angle at which the rod falls, so there can't be any rotation. So if I just dropped a 30º to the horizontal rod, it would fall and stay at 30º to the horizontal. Is that correct?

For simplicity, assume the rod consists of two point masses at each end, connected by a rigid massless string. Compute the magnitude of the gravitational force on each end. Are they equal?

 

[albertrichardf]

The gravitational force should be equal at either end. So I guess there would be no net torque either

 

[Bandersnatch]

The gravitational force should be equal at either end.

It isn't.

The gravitational force is given by
$$F=GMm/R^2$$
Where $R$ is the distance from the source of gravity to the CoM of the rod. If the rod of length 2x is inclined at any angle $\alpha$ other than horizontal, the closer end will experience
$$F=GMm/(R-xsin\alpha)^2$$
the farther end will have
$$F=GMm/(R+xsin\alpha)^2$$
This difference is what is called tidal force. It produces torque on the rod and acts to align it vertically (parallel to the direction to the source of gravity). This is used to stabilise satellites (cf: https://en.wikipedia.org/wiki/Gravity-gradient_stabilization) and a part of tidal locking mechanism of natural satellites.
Under an additional assumption that the source of gravity is a point-mass (or equivalently - spherically symmetric), with careful analysis you should notice that the torque is not only due to the difference in magnitude of the force, but also due to the difference in direction of the force on each end, as from either end's perspective the direction to the source of gravity is different.

 

[albertrichardf]

When the rod's length compared to the distance is non-negligible, then the torque isn't. But I was talking about a rod much smaller than Earth, so that $F≈mg$. Then based on the your analysis, the force should be equal on both ends and there is no net torque.
Thank you about the tidal effect link.

 

[Dale]

So if I just dropped a 30º to the horizontal rod, it would fall and stay at 30º to the horizontal. Is that correct

Yes, under the assumptions that the gravitational field is uniform and that it is initially not rotating.

Also, suppose I compute the rotation around one end of the rod. The torque is non-zero, but the rod is not rotating around that point. How is that possible? Furthermore, since the rod is falling, the end is falling as well, so my reference point is moving

This type of body centered computation is possible, but you need to carefully account for the reference frame. In this case (assuming Newtonian mechanics) the reference frame is non inertial. There is a fictitious force pointing upwards, exactly equal to gravity. So the torque about the end is zero in that frame, which is consistent with the observed motion.

 

[albertrichardf]

Alright. But what if I did not choose an end of the rod as a computation point, but I chose a point that was completely removed from the rod altogether? The point would not be accelerating so there could not be a fictitious force.

 

[Dale]

Alright. But what if I did not choose an end of the rod as a computation point, but I chose a point that was completely removed from the rod altogether? The point would not be accelerating so there could not be a fictitious force.

If there is no fictitious force then you are talking about an inertial frame. For such a frame there is only the force of gravity, and depending on where you choose your origin gravity can exert a torque. If it does then the angular momentum about the origin will increase as the rod falls.

To see how this works, it is best to actually work the problem. An object does not need to spin about its own axis to have angular momentum about an external point.

 

[jbriggs444]

Alright. But what if I did not choose an end of the rod as a computation point, but I chose a point that was completely removed from the rod altogether? The point would not be accelerating so there could not be a fictitious force.

The angular momentum of a non-rotating, falling brick as computed about a point some horizontal distance away from the brick is not constant.

 

[albertrichardf]

If the angular momentum changes, does that necessarily imply that the centre of mass of the rod is rotating around that point?

 

[Dale]

If the angular momentum changes, does that necessarily imply that the centre of mass of the rod is rotating around that point?

In the sense that it has a non zero angular momentum and non zero angular velocity, yes. In the sense that its path is curved, no.

 

[albertrichardf]

That's what I meant. Thank you for answering

 

[Dale]

You are welcome, you write a lot of good questions.

 

[albertrichardf]

Thanks. And you write a lot of good answers