# Rigid body with center of rotation other than CoM

• Delta2
In summary, the conversation discusses a rigid body rotating around a point P that is not its center of mass and the implications of external forces on its motion. It is explained that when the external forces stop, the center of mass will move in a straight line and become the new center of rotation. The concept of decomposing rotation into circular motion and rotation about the center of mass is also discussed. The conversation ends with a request for a resource with the mathematical details of this decomposition.
Delta2
Gold Member
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?

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The angular momentum around which point?

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?

Delta² said:
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?

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.

PeroK said:
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.
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.

Delta² said:
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.

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.

Delta² said:
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?
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.

Delta2
A.T. said:
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.
Can you point me to any document or link on the web with the math for such a decomposition?

Delta² said:
Can you point me to any document or link on the web with the math for such a decomposition?

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.
2) Rotation about Q.

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.

PeroK said:
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.
2) Rotation about Q.

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.
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?

Delta² said:
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?

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.

Delta² said:
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?
Depending on how the centripetal force towards P was being applied, the internal forces might change after it disappears.

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.

Delta² said:
I need the math badly

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).

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).

Delta² said:
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).

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.

## 1. What is a rigid body with a center of rotation other than the center of mass?

A rigid body is an object that does not change its shape or orientation when subjected to external forces. The center of mass (CoM) is the point where the entire mass of the object can be considered to be concentrated. When a rigid body rotates around an axis that is not passing through its CoM, it is said to have a center of rotation other than the CoM. This means that different points on the object will have different linear and angular velocities.

## 2. How does the location of the center of rotation affect the motion of a rigid body?

The location of the center of rotation plays a crucial role in determining the motion of a rigid body. When the center of rotation is at the CoM, the object will only experience rotational motion without any translation. However, when the center of rotation is at a different point, the object will also experience linear motion in addition to rotational motion.

## 3. Can the center of rotation coincide with a fixed point on a rigid body?

Yes, it is possible for the center of rotation to coincide with a fixed point on a rigid body. In this case, the object will only experience rotational motion around that fixed point without any translation. This is commonly seen in objects like wheels or gears, where the axis of rotation is fixed and coincides with a fixed point on the object.

## 4. How is the motion of a rigid body with a center of rotation other than the CoM calculated?

The motion of a rigid body with a center of rotation other than the CoM can be calculated using the principles of kinematics and dynamics. The linear and angular velocities of different points on the object can be determined using the relative distances and angles from the center of rotation. The equations of motion, such as Newton's laws and the conservation of angular momentum, can then be used to analyze the motion of the object.

## 5. What are some real-life examples of objects with a center of rotation other than the CoM?

There are several examples of objects with a center of rotation other than the CoM in our daily lives. A common example is a seesaw, where the center of rotation is at the pivot point, which is not at the CoM of the plank. Other examples include a swinging pendulum, a spinning top, and a spinning gyroscope. In all of these cases, the center of rotation is not at the CoM, resulting in interesting and complex motion patterns.

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