How to calculate torque on a body falling freely through the sky?

[pointdexter16]

TL;DR Summary

if a body is pivoted at a point it is quite easy to calculate the torque as we know the axis of rotation. but, if a body is under free fall and we apply a force of F newtons at a distance of x from its centre of mass , then how do we determine the axis of rotation?

if a body is pivoted at a point it is quite easy to calculate the torque as we know the axis of rotation. but, if a body is under free fall and we apply a force of F Newtons at a distance of x from its centre of mass , then how do we determine the axis of rotation?

 

[Baluncore]

Welcome to PF.

The body will rotate about the centre of mass. As the body falls it also has an aerodynamic centre of drag. The distance between the centre of mass and the centre of drag, provides the torque arm that may cause the object to rotate.

 

[nasu]

The torque depends on the point you choose to calculate it. It does not matter what the body does. The torque will have different values for different reference points (or axes). It's up to you to pick up a point that is convenient or useful for what you want to do with the values of the torque. But you can calculate it around any axis you want.

 

[kuruman]

The angular momentum of a body can be decomposed into two vectors: (a) angular momentum about the center of mass, more descriptively the spin angular momentum and (b) angular momentum of the center of mass, more descriptively the orbital angular momentum. Both vectors require a torque to change. A torque $\vec \tau=\vec r \times \vec F$ will have different effects depending on $\vec r$ which is the position vector from the origin to the point of application of the force.

Consider a fixed force $\vec F$ applied on a body. The question is, "what is the position vector $\vec r~$? The answer is, "it depends on where you choose the origin." If you choose it at the center of mass of the object, its spin angular momentum will change in a direction perpendicular to the plane defined by the force and the position vector; its orbital angular momentum will not change. This is the situation described by @Baluncore in post #2. If you choose the origin about any point other than the center of mass, then both spin and orbital angular momentum will change in a way that depends on your choice.

Specifically, for a mass falling from rest, if you choose the origin directly above the center of mass, the angular momentum will not change about that origin which means, of course, that the torque will be zero. However, if you choose the origin at perpendicular distance $R$ from the straight line trajectory, the magnitude of the torque will be $\tau = mgR$ and the orbital angular momentum about the origin will vary in time according to $L=mgtR$. If you tack on the mechanism suggested by @Baluncore, there will also be a change in the spin angular momentum.

 

[Lnewqban]

Welcome!
Respect to the ground, anybody in free fall, if rotating, will naturally tend to rotate around its center of mass.
During the time an external force other than weight and drag is applied, the body is not free falling.