Solved
Internal Forces Within Rotating Rigid Body
Posted on 2010-08-24
When a rod undergoes rotational acceleration, I cannot understand how the tangential internal forces get distributed the way they do.
A uniform density thin rod is pinned at its center. If a perpendicular force, F, is applied to a rod a distance d from the center then
Torque = d × F = I_cm * a_rot
(BTW, I_cm = (1/12) M * L² )
The ith atom in the rod has a tangential acceleration, a_i = a_rot × r_i, where r_i is the distance vector from the rod's center for the ith atom.
If the atoms in the rod have mass, µ, then since the ith atom has a tangential acceleration of a_i, then there must be tangential force on that ith atom:
F_i = µ * a_i
(I know there is also a centripetal acceleration and a corresponding centripetal force.)
So, relating the applied force to the resulting tangential force on individual atoms, I get:
F_i = µ * a_i = µ * a_rot × r_i = µ * (d × F / I_cm) × r_i
F_i = (µ/I_cm) * (d × F) × r_i
What this tells me is that if F is constant, then the tangential force on the individual atoms is proportional to d. I've derived it from basic equations which I saw derivations of. Yet, although the derivations of these equations made sense at the time, I do not understand how the application of a force at d got distributed in such a way so that this last observation is true. I push on a point at a distance d from the CM; and that force get propagated down towards the CM in such a way that as I get closer to the CM, this tangential force gets smaller.
And on the other side of the CM, moving further away from the application point of the force, then those atoms start experiencing a greater tangential force as r_i gets larger.
I suppose one answer is that the internal forces align themselves to satisfy Newton's Laws or Conservation Laws; but I was hoping to get more of an intuitive feeling for how this works internally. In HS, I think I learned how to work out mechanical advantage of tools, yet now I am not sure why I feel less resistance opening a door near the doorknob than pushing the door open near the hinge. (Again, high level concepts like conservation of energy probably simplify the answer, and that is great when solving problems - the high level equations simplify getting an answer to a complex system; but somewhere along the line, I have lost any intuitive notions.)
Granted, even if I apply a force horizontally to a block on a table, internally, each atom must feel the same translational force as every other atom - not clear as to how the laws of nature make that work (although it must because the equations say so), but at least each force in this scenario is the same on each atom (assuming each atom has equal mass, µ).
I'm hoping I'll get an answer that I can understand; I'll up the points as I get more insight.