# Rotation of a Bar by a Moment!

1. May 16, 2012

### Sarin

There is a bar AB. A is fixed and a force F is applied at B perpendicular to the bar.
The result of this can be computed by taking equal and opposite forces acting at A perpendicular to the bar (therefore they cancel out and the net force at A is zero, hence this is equivalent to the initial condition). The opposing forces at A & B form a couple, producing rotation, and the other force left at A causes a reaction. So this is the net effect of the force at B.

My question is :

1. What is happening actually to cause this effect, i.e if we were not to apply equal and opposite forces at A, then how else would we arrive at this conclusion/ result? Is it at the moleculer or lattice level, due to the bonding etc. Can somebody explain the real mechanism?

2. If at A the "fixed" constraint, then what would happen to the bar and at A?

Thanks.

2. May 16, 2012

### haruspex

Not sure I agree with your analysis.
Is this a massive uniform bar, no gravity, free to rotate about A? If so, the reaction at A will not equal F. Were that the case, the centre of the bar would not move. I believe the reaction at A will only be F/2.
Try, instead, reducing it to a couple about the centre of the bar and a force at that point.

Q1. Yes, you can interpret a given force + couple in many ways, and the equivalence can be seen by breaking the object down into rigid elements.

Q2. I'm guessing you meant to say "if the fixed constraint were removed".
In that case the centre of the bar would move in the direction of F, plus the bar would rotate. Question for you: where would the instantaneous centre of rotation be?

3. May 17, 2012

### Sarin

Haha i myself dunno what the second ques is, now that am reading it, i guess i ate some of the words in the middle, however, i f i am understanding you correctly, you are asking if say while rotating the bar would break off from the constraint at A, what would happen.

well in that casethere wont be anything to give reaction to A, so it will start moving forward (translation) and rotating at the same time, then the bar would realize that its unstable, cuz the majority of the mass is right now only on one side, so i n order to balance out (centrifugal force is doing the job here), the spinning axis eventually comes to the centre of mass, obviously al this happens very quickly, and thats that, ofcourse i could be wrong.

I really somoetimes want to do such experiments, these are so so simple to perform and yet give you such an insight into physics fundamental stuff.

4. May 17, 2012

### Sarin

"Not sure I agree with your analysis.
Is this a massive uniform bar, no gravity, free to rotate about A? If so, the reaction at A will not equal F. Were that the case, the centre of the bar would not move. I believe the reaction at A will only be F/2.
Try, instead, reducing it to a couple about the centre of the bar and a force at that point."

I understand what you're saying, just assume the mass has no gravity acting on it and the end A is tied to some rigid but roatatable joint like a pin joint. in this case there is just one force thath is actually acting, its F no gravitational force no nothing.

you can visualize a pin jointed metallic bar and there is a sort of magnetic roof in this setup which attracts it with a force equal to gravitational; force, i bet i ve seen these kinda toys.

now imagine what happens when you push it a lil at the other (free) end.

If Gravity is acting, then depending on the direction of gravity wrt to F, the moent would be F x d + mg x d/2
OR F x d - mg x d/2 ofcourse i havent considered the case when the 2 are perpendicular.

5. May 17, 2012

### haruspex

Not quite. I thought you might be asking what happens to a bar free floating in space if you hit it at right angles at one end.
But ok, let's think about the case you now mention: a bar is rotating around a hinge, then set free from the hinge.
Suppose at the instant of release the bar, length 2L, is on the x-axis and rotating in the xy plane. The hinge is at the origin.
Before release it is rotating with angular velocity w. Its mass centre is therefore moving in the Y direction with speed wL.
Having been released, by conservation of linear momentum, the mass centre will continue at speed wL in the y direction; by conservation of angular momentum it will continue to rotate at speed w. Where is the "spinning axis"? That is, at a given instant, what part of the bar is stationary?
In general, it may be that no part of the bar is stationary, but we can extend the bar by embedding it in an infinite massless disc in the xy plane, rotating with the bar. Now we can ask which part of the disc is stationary at each instant. I'll let you have a go at answering that before continuing.
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