I Rigid body with center of rotation other than CoM

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1. Dec 10, 2016

Delta²

Suppose we have a rigid body that rotates around a point P other than its center of mass, with point P being a point of the rigid body. This implies that there are external forces to the rigid body. If the external forces cease to exist(like we have an axis passing through P and we suddenly remove the axis) the center of mass will move in a straight line. Will the CoM become the new center of rotation of the rigid body? if yes what happens to the angular momentum, will it remain the same though we ll have a new center of rotation?

Last edited: Dec 10, 2016
2. Dec 11, 2016

A.T.

The angular momentum around which point?

3. Dec 11, 2016

Delta²

Well, say the angular momentum around point P. But first thing first, will the C.O.M become the new center of rotation after the external forces stop to exist? if yes how exactly is the physical process that changes the centre of rotation?

4. Dec 11, 2016

PeroK

You can work this out for yourself with a uniform rod fixed at the origin (let's say at one end) and rotating about the origin. When it becomes aligned with the positive y-axis, say, it comes loose and moves off.

The speed of each point on the rod is proportional to its distance from the origin at this time, all in the positive x-direction, so it's initial state as it moves off will have the COM moving at a certain speed, the top of the rod moving at twice this speed and the bottom of the rod stationary. The rod, therefore, already in a state of rotating about its COM! There is no physical process required.

5. Dec 11, 2016

Delta²

Well, this example is fine but you trying to tell me that regardless of the shape of the rigid body, regardless of what point P is and the centre of mass, and regardless of the moment we stop the external influence, the rigid body is already in a state of rotating about the C.O.M? This seems nice but counterintuitive at first glance.

6. Dec 11, 2016

PeroK

The point is that only rotation about the COM is "stable". If you consider rotation about any other point, then there is a net angular momentum, so that point cannot move in a straight line. In the above example if you consider the bottom end of the rod, say, then it cannot continue to rotate about that point, because if the end of the rod moves in a straight line, then there is net (changing) angular momentum about that point

As a next step, if you think about a non-uniform rod, then as it moves off, only the COM can move in a straight line. This logic extends to any 2D or 3D body.

7. Dec 11, 2016

A.T.

You can decompose the rotation around P into:

1) circular translation of the COM around P
2) rotation around COM

When the centripetal force towards P disappears, 1) becomes translation along a line, while 2) remains unchanged.

8. Dec 11, 2016

Delta²

Can you point me to any document or link on the web with the math for such a decomposition?

9. Dec 11, 2016

PeroK

You don't need any maths, as such. If a body is rotating about a point P, then every point in the body is rotating in a circle about P. If you pick any point Q and look at the motion relative to that point, then the body is rotating about Q. So, you can decompose a rotation about P into:

1) The circular motion of Q about P.

If you remove the centripetal force towards P, then this decomposition is only "stable" when Q is the COM. Otherwise. rotation about Q would require a force.

10. Dec 11, 2016

Delta²

I need the math badly :) but you are saying something that I was afraid it might be true: If the decomposition holds for any point Q, then (I might say silly things from now on but that's what i understand of it) the universe automatically selects as point Q the CoM because its the only stable selection?(doesn't require external force to happen). There isn't any physical process (some transient process that's happening involving internal forces in the rigid body) that establishes CoM as the new centre of rotation?

11. Dec 11, 2016

PeroK

The internal forces are balanced, so you have no net torque and no net translating force on the body as a whole. This fact compels the COM of the body to move in a straight line and the body to rotate only about its COM. Any other motion requires external forces.

If you consider the internal forces and how they change when a body is released from a forced rotation, then that is more complicated. A rigid body behaves differently from a system of points attached by springs or by elastic strings. In those systems, you would see an interplay of internal forces after release, which you do not observe in a perfectly rigid body. But, in all cases, the COM moves in a straight line and the body rotates about the COM.

You could view this as a selection process (in the same way that at each instant the universe decides that a point moving in straight line continues in a straight line). You could analyse the internal forces on a rigid body, but that makes it more complicated than applying the laws of conservation of linear and angular momentum.

12. Dec 11, 2016

A.T.

Depending on how the centripetal force towards P was being applied, the internal forces might change after it disappears.

13. Dec 11, 2016

Capn'Tim

Hello. Perhaps your question is simpler than I suppose it to be.
What is the setting for your question? Is this a body being acted on by gravity or a body in free space with no significant gravity?
The center of rotation is normally the center of mass as previously stated unless other wise constrained to a pivot point elsewhere in the body like an axle or a fulcrum point. If as you state the center of rotation is at a point displaced from COM, and you abruptly remove that pivot point restriction, the rotation point will instantly change to the COM (for all intents and purposes). The momentum achieved during the previous rotation will cause an acceleration as the rotation moves to the center of mass proportional to the shift in COR as a proportion of the offset mass change. The best example I can use which might not perfectly fit the example is when an ice skater doing a spin suddenly pulls their arms in tight and rotation rapidly accelerates. This is caused by the mass distribution being closer to the center of rotation. In your example, because the angular radius of momentum is different with an offset COR than would be once rotating around the COM, there will be a temporary wobble induced in the rotation that will slowly normalize around the COM.
The reason I asked about whether the body is being acted apon by gravity, is that different forces would be in play providing different results as a result of the gravity.

14. Dec 13, 2016

PeroK

Let's assume that the body is rotating about a point - we may as well take that point to be the origin. Pick any other point, $A$. We can take this to lie initially along the x-axis, so the motion of $A$ is given by:

$(r \cos(\omega t), r \sin(\omega t))$

If we pick any other point $B$, then its motion is:

$(R \cos(\theta + \omega t), R \sin(\theta + \omega t))$

And, it's motion relative to $A$ is:

$(R \cos(\theta + \omega t) - r \cos(\omega t), R \sin(\theta + \omega t) - r \sin(\omega t)) \$ (1)

As it is a rigid body, the distance from $A$ to $B$ is fixed and, from the law of cosines or by using the above, it is:

$d = R^2 + r^2 - 2Rr \cos \theta$

And, relative to $A$, $B$ initially has a polar angle $\phi$ given by:

$\cos(\phi) = \frac{R \cos \theta \ - r}{d}, \$ or $\ \sin(\phi) = \frac{R \sin \theta}{d}$

Again, as the body is rigid, this angle between the position vectors of $A$ and $B$ is fixed. So, an alternative description of the locus of $B$ is:

$(r \cos(\omega t), r \sin(\omega t)) + (d \cos(\phi + \omega t), d \sin(\phi + \omega t)) \$ (2)

You can check that equations (1) and (2) describe the same locus for B using some trig if you want to.

Now, if we choose a reference frame defined by translation along the path of $A$, but no rotation of the axes. I.e. we put $A$ at the origin. How does the rest of the body move in this reference frame?

From (2), point $B$ moves according to:

$(d \cos(\phi + \omega t), d \sin(\phi + \omega t)) \$

I.e. it rotates around $A$ with angular velocity $\omega$, which shouldn't be a surprise.

This shows that for any point ($A$) the motion can be decomposed into the translation of that point and a rotation about that point.

Note: you don't, of course, need equation (1) here. You can simply derive equation (2) directly, but it seemed worth confirming that equation (2) is valid by comparing it to (1).

15. Dec 13, 2016

Delta²

Ok , thanks for the math of decomposition, I have to check it later, at first glance seems a bit confusing to me because you work in Cartesian coordinates by employing polar variables but seems correct. Perhaps can you make a figure showing the details about $\phi$, because I am not sure about the cosine and sine of $\phi$ but their values seem to be such as that there is an equivalence between (1) and (2) (for sure (1) is correct).

16. Dec 13, 2016

PeroK

It's a good exercise for you to work through this. $\phi$ is the angle from the x-axis to the line joining $A$ to $B$ initially.