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Trying to understand rotational movement

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Hello, I have a "point" A, which lets say, connected by a "wire", to another "point" B which is immovable. If i apply a force on pA, unless the force is enough to move point B or break the wire, it will be transformed into rotational movement. So i will have a rotational movement, which circle radius is the size of the wire. My doubts is, If i have a force (gravity) which is constantly attracting everything in a certain vector direction. When pA is rotating, what will be is vector direction in that moment, so i can reduce gravity from it? I mean if i pause (pause, not stop) for a second, what will be pA vector direction in that moment, it will be the linear momentum? And if when pA is rotating, i remove pB, what direction will it go? It will be the vector point-pivot? PS: The wire is not elastic! [Edited by - GPxz on October 31, 2007 12:03:40 PM]

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The mentioned force is not the only one that effects the mass of pA. The wire, if straightened, also applies a force. Also friction may be applied.

The mentioned force can be separated into a part parallel to the "wire" line, and a part orthogonal to it. If the parallel part is directed from pB to pA (i.e. "against" the wire) then the wire applies a opposite force annuling the parallel part. If the parallel part is directed from pA to pB (i.e. it is anti-parallel) then the wire has no effect (assuming it is absolutely flexible), i.e. it doesn't applies another force. Now, add the parallel force, the orthogonal force, and the "wire force" to get the total force applied.

From this you can see that the mass pA doesn't generally move on a circular lane if the original force is constant (e.g. gravity). It does so only if the motion guarantees that the wire is straightened.

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The direction of movement of pA is always perpendicular to the radial vector (A-B). Remember that the torque at any point in time is the sum of all forces acting on the point (including that caused by gravity), crossed with the radial vector. The point can't move in anything but the θ-hat direction. This is assuming the wire is a rigid constraint of course.

If the point pB is suddenly removed, then you no longer have a centripetal force keeping the point in radial motion. It will only be under the influence of gravity, and as such behave like a falling particle with some initial velocity, depending on when pB was removed.

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Original post by Zipster
The direction of movement of pA is always perpendicular to the radial vector (A-B). Remember that the torque at any point in time is the sum of all forces acting on the point (including that caused by gravity), crossed with the radial vector. The point can't move in anything but the θ-hat direction. This is assuming the wire is a rigid constraint of course.


I think i understood

But whats the influence of torque/angular momentum on it, is it a force happening too?

Cause in the Wikipedia article http://en.wikipedia.org/wiki/Angular_momentum

It shows angular momentum as the cross product of the radial vector and linear momentum as you said, and in the picture its a direction vector up, this would explain why the gyroscope remains upright, cause it would anulate gravity, since its opposite to it, is it?

Quote:
Original post by Zipster
If the point pB is suddenly removed, then you no longer have a centripetal force keeping the point in radial motion. It will only be under the influence of gravity, and as such behave like a falling particle with some initial velocity, depending on when pB was removed.


You mean if i remove pB, pA speed will be completely null?

This is not what happen in real life, it seems it follow the radial vector

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Original post by GPxz
I think i understood

But whats the influence of torque/angular momentum on it, is it a force happening too?

Cause in the Wikipedia article http://en.wikipedia.org/wiki/Angular_momentum

It shows angular momentum as the cross product of the radial vector and linear momentum as you said, and in the picture its a direction vector up, this would explain why the gyroscope remains upright, cause it would anulate gravity, since its opposite to it, is it?

Angular momentum (a vector) is always perpendicular to the direction of motion, since it's defined as r x p (where p = μv). Torque is the time derivative of angular momentum, r x F. Since you've established a hypothetical situation in which all the forces and velocities/momentums exist in a single 2D plane, torque and angular momentum will always either be parallel or anti-parallel. This means that angular momentum never "wobbles", and the motion always exists in this same 2D plane. It's the same logic with the gyroscope, since the angular momentum vector is rather constant over a short period of time, the plane of motion doesn't "wobble" and it maintains a balanced spin.

Quote:

You mean if i remove pB, pA speed will be completely null?

This is not what happen in real life, it seems it follow the radial vector

No, like I said it will be at some velocity depending on when pB was removed. In the absence of frictional forces, the motion of pA will either behave like a pendulum, or if you apply enough initial force, it will keep spinning around pB. In the latter case, it will always have a non-zero velocity, so no matter when pB is removed you'll have some initial velocity. In the case of a pendulum, it will only have zero velocity at maximum height.

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