analytical mechanics of force couple on rigid body

The sum of the moments in the z (out of page) direction (and the parallel axis theorem, if needed) can give you the instantaneous angular acceleration about that point. The sum of the moments about x and y gives nothing. The sum of the forces in all directions gives nothing, unless there is only one force, in which case you will get a net acceleration of the center of mass in that direction.

I have never had to solve a problem where the body wasn't constrained in any way. I suggest doing the sum of moments about three different points, just so see if the result for the net angular acceleration is the same (those three points would be at the application of force 1, at the application of force 2, and at the center of mass). I expect them to be the same number. If so, I would say that it rotates about the center of mass. Why?

Well, sum of the forces in the y direction (along the length of the bar) results in no net force in that direction. In order to spin about a point other than the center of mass, there must be a net acceleration of the center of mass along the y direction, which couldn't be caused by the forces shown in the diagram.

If you want a full solution as a function of time for the translation of the center of mass and the rotation about the center of mass in an intertial coordinate system, you will have to transform the local variables x,y,z to that coordinate system. I have no idea how to do this at the moment (unless you're talking about a computer program that takes explicit finite time steps).

========= A crash course on rigid-body mechanics ===========

About forces and couples:

(A) The effect of a force is entirely determined by its magnitude, direction and line of action, and is independent of what specific point on the line of action is its point of application. One may say: a force is represented by a =sliding= vector, bound to its line of action.

(B) The effect of a couple is entirely determined by its moment vector. It is independent of what specific pair of forces with opposite force vectors constitutes the couple. One may say: a couple is represented by a =free= vector, not bound to any specific point or line in space.

About the action of a force on a rigid body:

one may transfer a force F from its original point of application to its new point of application, the centre of mass of the rigid body, as follows:

Let the original force be determined by its vector F and its line of action "a". Draw the line "b" parallel to "a" thru the centre of mass G of the rigid body. Imagine two equal and opposite forces F and -F applied at G and working along line "b". Their net effect is of course zero, but the force F along "a" is in its effect equivalent to a force F working along line "b" together with the couple consisting of the original force F along line "a" and the force -F along line "b".

The instantaneous acceleration of the rigid body consists of...

(1) a translational instantaneous acceleration of the body as a whole: F/Mass, combined with...

(2) a rotational instantaneous acceleration determined in the well-known (?) Eulerian way from the moment of the couple, its orientation relative to the rigid body, and the body's principal moments of inertia.

All this stuff is elementary and advanced rational mechanics. Excellent books dating back to around 1900 are the treatises by Appell (in French), Routh, Whittaker (both of them in English and translated into German), Hamel (in German). More recent treatises are by Sommerfeld (1930s) and Goldstein (around 1950).

In theory the free-floating bar in J Jensen's problem will indeed start and keep rotating about its centre of mass, which will in turn keep moving on at uniform speed in a constant direction. The practice may of course widely diverge from this...

Johan E. Mebius

J Jensen wrote:

Thanks for your reply, Chris. That is an interesting idea in the paragraph that I quoted above, although I still don't quite see the solution to my problem. For example, by that logic, no point on the bar would have an acceleration in the y direction,but yet it would rotate!

--Jeff

Thanks for your reply, Johan. I am familiar with the theorems that you have stated, although I can't always say that I truly "understand" them in an intuitive way. That is actually what I am trying to do with this problem: to come up with the simplest illustrative example of rigid body motion, and try to understand it from first principles.

Maybe a different tact would be fruitful. We could ask, what do the forces have to be in order for the bar TO rotate about the midpoint of the force couple in my original diagram? I've been thinking about it, but that problem seems hard too!

--Jeff

It will rotate about it's center of mass.

Just do your force and moment balances.

There is no net force in the X direction, therefore the CG will not move left or right.

There is no net force in the Y direction, therefore the CG will not move up or down.

There is a net moment in the plus Z direction, therefore the bar will rotate around its CG. The point of application of the couple doesn't matter...you can prove this by taking sum of moments around various points, they'll all come out the same.

It depends on whether the force direction is fixed relative to space, or relative to the bar. If fixed relative to space, the CG will accelerate in the X-axis due to the force and the bar will rotate around the CG. If it's relative to the bar, the bar will go into a spiral of some kind, spinning on its CG while the CG itself moves in a curve.

Tom.

"J Jensen" wrote

Any time you take force/moment balances for the whole body, you can only look at motion of the whole body relative to the CG.

Just because the CG isn't accelerating doesn't mean that points on the bar can't be accelerating. They just have to sum to zero and not violate the rigid body criteria (i.e. the rod doesn't break apart). If the bar is rotating points will have an accleration in the Y direction, but as long as the CG isn't accelerating in the Y direction you meet your force balance.

When a high jumper arcs their body to clear the bar their CG actually passes below the bar...the CG isn't even inside the body. That allows them to clear a bar higher than they're capable of jump their CG up to.

Tom.

I think the true question being asked is "what causes the rotation?" If there was no bar, just two rockets, then the two rockets would move off in straight lines. The bar is somehow causing the rotation. And that is the real question being asked.

I would recommend replacing the bar with a weight tethered to an inelastic string. A rocket would move in a straight line until the string was pulled tight. At that time, the momentum of the ball would increase and this would affect the momentum of the rocket.

The bar causes the rotation by putting a constraint on the point of action of the forces. Two rockets without the bar would just sail past each other in straight lines. With the bar, the distance between the rockets (and their angle relative to the bar) becomes fixed so the straight-line solution is kinematically impossible.

You're applying a torque to a free-floating bar with no reaction force...how could it not rotate?

Tom.

I think your use of the word "constraint" answers the heart of the original question. Replace the bar with a massless bar and you have constrained the motion to be circular about the center of the bar. Adding mass to the bar, shifting the rocket positions, etc. are all secondary questions.

Thanks for your reply, Tom.

I guess I was so "familiar" with the center of mass that I didn't actually understand it :) Its claim to fame is precisely that it is THE point which can't be accelerated without a net external force on the body.

And for my problem, there are obviously unbalanced torques around this fixed point, so the bar will rotate around it, no matter where we put the couple.

--Jeff