What is the distribution of force in a rotating object without a fixed axis?

In summary: And then, once the first end starts moving, the end with the force applied will be moving downwards at the same speed as the end without the force applied, so it will also be moving downwards at the same speed as the perpendicular force. So it seems like the impulse would be the same as if the force was just applied at the other end. But if the force is applied at the other end and the rod is already moving, that seems like it would add impulse to the moving end...In summary, the problem is that a rigid object is in free-fall, and a force is applied so that it would be a torque if the object had a fixed axis
  • #1
Bartholomew
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The problem is this: a rigid object is in free-fall, and a force is applied so that it would be a torque if the object had a fixed axis through its center of mass. But the object doesn't have such an axis. How much of that force turns into linear motion, and how much of it goes into rotation?

I'm really only working with the case where the object is a bar with all of the mass divided evenly between the ends, but a general solution would be very welcome.

The basic case for the bar is where the force is applied at one of the ends. That end will start to move a little bit away from the force, thus changing the center of mass (and there's no reason why the other end would move up to compensate)--so there must be at least _some_ linear motion. But then the tension between the ends will slow the first end down and start the second end moving towards the first. What is that tension?

If I can find that tension, the problem's solved, because then I can just plug it into the formula for centripetal force with the radius as half of the bar length and the mass as half of the mass of the bar, and solve for speed, and figure that all the rest of the energy must go into linear motion. But how do I find that tension?

This is for a simulation of bars bouncing against one another, so there's the additional complication that it's not a fixed _force_, it's a fixed mass and velocity. But unless you happen to know what happens there off the top of your head, forget it--I think I have an intuitive idea of how it should work (at the middle the bar would act like it had its full mass, and as you moved towards an end it would linearly decrease to acting [for the purposes of figuring collision force] like it had half of its mass).
 
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  • #2
There's always an axis. Draw the line from the center of mass of the object to the point at which the force is applied. The axis of rotation is perpendicular to that line and to the force vector.
 
  • #3
That's true, however...

There is always an axis, but it's not always a fixed axis. If there were a _fixed_ axis, then 100% of the impulse and energy would go into rotation. That's not the case.
 
  • #4
A few of my "solutiions" (not really)

There are three ideas that I've had about this so far, all saying that it should be 1/2 the impulse, but I don't think any of them is particulary trustworthy.

The first idea is a hoop rolling down a hill. If you work it out, half of the gravitational energy goes into rotation and half goes into motion. The only difference between the hoop sliding and the hoop rolling is that when the hoop rolls, there is a force applied at its base. So that force must be responsible for canceling out half the gravitational energy and adding that same amount of energy to rotation.

The second idea is when the rod is already turning and an impulse is applied perpendicular to it at the center. Assume the impulse applied is equal to its rotational momentum, and break its rotational momentum up into two vectors: one on each end, one going backwards and one forwards. So it seems like half the impulse applied should go to each end. When it goes to the end that's rotating backwards, it cancels that out. When it goes to the other end, it doubles the momentum. So the rod now has just one vector, on that one end. But obviously the result has to be that the rod starts moving forward with the same linear speed as the ends were rotating in the beginning--so somehow that one vector on the end has to be divided up into that result. So, half would go to motion and half to rotation? By energy, that works, but by impulse it doesn't seem to. I've always had kind of a problem reconciling momentum with energy, and how splitting the energy in half leaves you with more momentum than you started with.

The third idea starts when the rod is stationary and a perpendicular force is applied at one end (say the rod is horizontal and the force goes "down"). There are no forces that are going to move the other end up; it will only be pulled towards the first end once the first end starts moving. And for just the first instant, it (the end without the force applied) isn't going to move down, either, because the tension between the ends is not going to start pulling down until the first end is already below the second. So in the very first instant, the rod must rotate at enough speed to exactly cancel out its vertical linear motion. That works out to an even split between rotation and linear motion.

But there's a problem with that last one--I don't think it depends on the moment of inertia. The mass could be distributed any way you choose along the bar and that idea would still say the bar rotates half as fast as it moves. But the first idea, with the rolling hoop, does say that the moment of inertia matters. So I think they contradict one another--which is right?

Ah, I think I just thought my way out of that one. The third idea says it rotates and moves at the same _speed_-so if it had a lower moment of inertia then presumably it would both move and rotate faster for the same energy.

But still, I'm far from sure, and I don't know what happens in more complicated cases.
 
  • #5
Its a pain but you have to combine the equations for linear and rotational motion and solve for linear and rotational (angular) acceleration. And just as linear acceleration depends on mass (inertia) so to does angular acceleration depend on moment of inertia.

A thought experiment: Take a graphite stick (light and strong) and put a single weight in the middle. apply a force tangential to one end of the stick. What happens? How much rotation and now much forward motion?

Now take two weights and place them on the ends of the stick instead of the center. Apply the same force. Now what happens? How much rotational and forward motion?
 
  • #6
For the first one, it would all be rotation, ignoring the graphite's mass. For the second one--well, that's the same question I asked before. It seems that half the energy goes into rotation and half into linear motion, but the only justifications I have for that are somewhat dubious. I don't know what thought process you expected.
 

1. What is torque without fixed axis?

Torque without fixed axis, also known as free torque, is a type of torque that is not constrained to a fixed point or axis. It can be applied to any point or axis in a system, allowing for more flexibility in rotational motion.

2. How is torque without fixed axis different from torque with fixed axis?

The main difference between torque without fixed axis and torque with fixed axis is the point of application. Torque with fixed axis is applied to a specific point or axis, while torque without fixed axis can be applied to any point or axis in the system.

3. What are the applications of torque without fixed axis?

Torque without fixed axis has various applications in physics and engineering, such as in rotational motion, gyroscopic motion, and oscillations. It is also used in machinery, vehicles, and other mechanical systems to control and manipulate rotational motion.

4. How is torque without fixed axis calculated?

The formula for calculating torque without fixed axis is the cross product of the force vector and the position vector. This can be represented as T = r x F, where T is the torque, r is the position vector, and F is the force vector. The unit of torque without fixed axis is Newton-meters (N·m) in the SI system.

5. Can torque without fixed axis be negative?

Yes, torque without fixed axis can be both positive and negative. The direction of the torque depends on the direction of the force and the position vector. If the force and position vector are in the same direction, the torque will be positive. If they are in opposite directions, the torque will be negative.

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