Why do objects always rotate about their centre of mass?

[vanhees71]

In addition if you take the center of mass as the body-fixed point of reference, there's no spin-orbit coupling.

 

[Dale]

In addition if you take the center of mass as the body-fixed point of reference, there's no spin-orbit coupling.

Yes, which again makes the analysis simpler.

 

[John Mcrain]

Any infinitesimal rigid body motion can be described as an infinitesimal translation of any arbitrary material point on the body and an infinitesimal rotation about that point.

The thing that makes the center of mass special is not the rotation, that happens about every point. The thing that makes the center of mass special is that its translational motion is simple. Rotation occurs about every point, as does translation of that point. But the simple translation of the center of mass makes analysis easier if we choose that as our reference.

If referenece frame is ground, about what arrow rotate?

 

[Dale]

If referenece frame is ground, about what arrow rotate?

It rotates about any point.

 

[John Mcrain]

It rotates about any point.

C.m. of arrow rotate about any point? what gives centripetal force for such rotation?

 

[Dale]

C.m. of arrow rotate about any point? what gives centripetal force for such rotation?

Drag and gravity exert any necessary force. But recall that angular momentum is conserved even in the absence of an external torque. So you have to be careful in demanding a centripetal force.

 

[John Mcrain]

Drag and gravity exert any necessary force. But recall that angular momentum is conserved even in the absence of an external torque. So you have to be careful in demanding a centripetal force.

Dont understand this concept "it can rotate around any point"...
There is huge difference is arrow rotate around cm or rotate around some other point, because in second case we must have centripetal force...
Dont understand this concept where you can invent forces at will..

 

[Ibix]

C.m. of arrow rotate about any point?

Sure. Imagine the arrow is rotating in the vertical plane and there's an ant standing on the arrow. The ant is sometimes going to see the tip of the arrow between it and the sky and sometimes between it and the ground. And this is true wherever the ant is along the arrow - hence the arrow is rotating about any arbitrary point.

The key point about the center of mass is that (edit: in free fall) this point is not moving in circles itself. So an ant standing there, alone among all points, will see no force (or "force", because it's an inertial force and not a real one) pushing it towards the end of the arrow.

 

[John Mcrain]

Sure. Imagine the arrow is rotating in the vertical plane and there's an ant standing on the arrow. The ant is sometimes going to see the tip of the arrow between it and the sky and sometimes between it and the ground. And this is true wherever the ant is along the arrow - hence the arrow is rotating about any arbitrary point.

The key point about the center of mass is that (edit: in free fall) this point is not moving in circles itself. So an ant standing there, alone among all points, will see no force (or "force", because it's an inertial force and not a real one) pushing it towards the end of the arrow.

How can you say here that triangle rotate about any point?
I see only rotate around center of mass, and if cm rotate around some other point then we must have centripetal force, path of cm will be curve not straight line
You cant move cm in curve path without external force.

 

[Ibix]

How can you say here that triangle rotate about any point?

Imagine being an ant sitting somewhere on that triangle. Is the nearest tip of the triangle sometimes between you and the left edge and sometimes between you and the right edge? Yes! So the triangle is rotating about you. If it's not rotating, how is the tip pointing in different directions at different times?

You are implicitly adding another constraint, that to be "a point a body rotates around", that point must move in a straight line. Neither @Dale nor I are including that restriction. Thus we can say that the body rotates around any point and that point moves in a cycloid.

 

[vanhees71]

I forgot that I've already translated the chapter on rigid-body motion of my mechanics manuscript to English. Maybe that helps:

https://itp.uni-frankfurt.de/~hees/pf-faq/spinning-top.pdf

 

[John Mcrain]

Imagine being an ant sitting somewhere on that triangle. Is the nearest tip of the triangle sometimes between you and the left edge and sometimes between you and the right edge? Yes! So the triangle is rotating about you. If it's not rotating, how is the tip pointing in different directions at different times?

You are implicitly adding another constraint, that to be "a point a body rotates around", that point must move in a straight line. Neither @Dale nor I are including that restriction. Thus we can say that the body rotates around any point and that point moves in a cycloid.

If I am ant I will see outside world rotate around me not trinagle about me.
Same like at merry go round.

 

[Ibix]

If I am ant I will see outside world rotate around me not trinagle about me.
Same like at merry go round.

That's true, and it's true at the center of mass too.

If you want, give the ant a compass so it has an axis fixed in the inertial frame to work with. Or better yet, glue a compass to the triangle and have the ant sit on the compass needle. It'll see the triangle rotate under it wherever it is, and if it isn't at the center of mass it'll also "feel" inertial forces.

 

[John Mcrain]

If they did not then the center of mass would not be traveling in a straight line. This would violate Newton’s first law.

Didnt we just say that object rotate about any point?
What is difference in instantaneous center of rotation and center of rotation?

 

[jbriggs444]

Didnt we just say that object rotate about any point?
What is difference in instantaneous center of rotation and center of rotation?

For me, "instantaneous center of rotation" means that one has already chosen a frame of reference. The "instantaneous center of rotation" is the point where the rotating object (or a wire frame extension thereof) is momentarily motionless according to that choice of frame.

So if you pick the road frame, the center of the tire's contact patch is at the "instantaneous center of rotation". If you pick the car frame, the axle is at the "instantaneous center of rotation". If you pick the rest frame of an aircraft passing the car at a 600 mph while the car is moving at 60 mph and the tire has a radius of 12 inches then the "instantaneous center of rotation" will be at a point 10 feet above the axle.

By contrast, I would take "center of rotation" to mean that you have instead first chosen the point and then allowed that choice to dictate your frame of reference.

If you choose a point that is fixed to the wheel at the axle, you get a nice inertial frame of reference moving at 60 mph down the road. If you choose a point that is fixed to the surface of the tire then you get a nasty non-inertial (but non-rotating) reference frame whose origin is tracing a cycloidal path down the highway.

Looking at it another way, I do not think that either "center of rotation" or "instantaneous center of rotation" are physical attributes of an object at all. They are merely choices about how to describe the motion of that object. You are free to choose. Some choices make for easy calculations. Other choices do not.

This last is what @Dale said, much more succinctly, in #60.

 

[John Mcrain]

For me, "instantaneous center of rotation" means that one has already chosen a frame of reference. The "instantaneous center of rotation" is the point where the rotating object (or a wire frame extension thereof) is momentarily motionless according to that choice of frame.

So if you pick the road frame, the center of the tire's contact patch is at the "instantaneous center of rotation". If you pick the car frame, the axle is at the "instantaneous center of rotation". If you pick the rest frame of an aircraft passing the car at a 600 mph while the car is moving at 60 mph and the tire has a radius of 12 inches then the "instantaneous center of rotation" will be at a point 10 feet above the axle.

By contrast, I would take "center of rotation" to mean that you have instead first chosen the point and then allowed that choice to dictate your frame of reference.

If you choose a point that is fixed to the wheel at the axle, you get a nice inertial frame of reference moving at 60 mph down the road. If you choose a point that is fixed to the surface of the tire then you get a nasty non-inertial (but non-rotating) reference frame whose origin is tracing a cycloidal path down the highway.

Looking at it another way, I do not think that either "center of rotation" or "instantaneous center of rotation" are physical attributes of an object at all. They are merely choices about how to describe the motion of that object. You are free to choose. Some choices make for easy calculations. Other choices do not.

This last is what @Dale said, much more succinctly, in #60

Chief ask engineer: about what point will tanker rotate if our wokers push with tugboat 30m from stern?
Engineer answer: tanker will rotate about any point

Chief: What??

Tell that object rotate about any point is useless ,complety unphysical?

 

[jbriggs444]

Chief ask engineer: about what point will tanker rotate if our wokers push with tugboat 30m from stern?
Engineer answer: tanker will rotate about any point

If one is docking a tanker, the frame of reference of the dock would be a good choice for describing the tanker's motion.

Tell that object rotate about any point is useless ,complety unphysical?

Unphysical and useless are not synonymous

For instance, potential energy is unphysical. Only differences in potential energy are physically meaningful. But that does not stop us from setting an arbitrary zero, pretending that the resulting figure for potential energy means something and using it in a calculation that produces a useful result.

 

[John Mcrain]

If one is docking a tanker, the frame of reference of the dock would be a good choice for describing the tanker's motion.

Lets say if tanker has diffrent location of c.m. and center of water lateral resistance (clr).

If you slowly push with only one tugboat at tanker clr, tanker will only translate,no rotation.

But if you push at tanker c.m., it will rotate around some other point that is not c.m. this mean c.m. is moving in curve path.

How is this possible, tugboat push perpedicular to radius of turn, so what produce centripetal force?

 

[Dale]

Any infinitesimal rigid body motion can be described as an infinitesimal translation of any arbitrary material point on the body and an infinitesimal rotation about that point.

Dont understand this concept "it can rotate around any point"...

This is just a mathematical fact of rigid body motion, purely kinematically. It has nothing to do with forces, just the way that rigid body motion behaves mathematically.

Suppose I have a rigid disk which is spinning about its center of mass. At a given instant I can plot the velocity of each point on the disk as follows:

The formula for the velocity field is $\vec v=(y,-x)$. This is rotation around the center, as expected.

However, suppose instead of a disk rotating, we have a wheel rolling. Kinematically these are the same motion in different reference frames. Then at any given instant I can plot the velocity of each point on the wheel as follows:

The formula for the velocity field is $\vec v=(y,-x)+(1,0)$. Notice that this motion is also a pure rotation, but about the bottom of the wheel, the point that it contacts the ground. Again, this is kinematically identical to the disk rotating about the center in a different reference frame.

These two points are not special. In fact, for any point on the wheel you can pick a reference frame where that point is momentarily at rest. When you do so the motion is as follows:

The formula for this velocity field is $\vec v=(y,-x)+(0,0.7)$. Notice again that this motion is momentarily a pure rotation about the chosen point which happens to be $(0.7,0)$.

This is a general feature of rigid body motion. You can always decompose the velocity of the material points in a rigid body into a pure rotation about any point and a rigid translation.

So again, it isn't that the rotation is about the center of mass, but that the translation of the center of mass is particularly simple.

 

[John Mcrain]

This is just a mathematical fact of rigid body motion, purely kinematically. It has nothing to do with forces, just the way that rigid body motion behaves mathematically.

Suppose I have a rigid disk which is spinning about its center of mass. At a given instant I can plot the velocity of each point on the disk as follows:

View attachment 319993
This is rotation around the center, as expected.

However, suppose instead of a disk rotating, we have a wheel rolling. Kinematically these are the same motion in different reference frames. Then at any given instant I can plot the velocity of each point on the wheel as follows:
View attachment 319994
Notice that this motion is also a pure rotation, but about the bottom of the wheel, the point that it contacts the ground. Again, this is kinematically identical to the disk rotating about the center in a different reference frame.

These two points are not special. In fact, for any point on the wheel you can pick a reference frame where that point is momentarily at rest. When you do so the motion is as follows:
View attachment 319996
Notice again that this motion is momentarily a pure rotation about the chosen point.

This is a general feature of rigid body motion. You can always decompose the velocity of the material points in a rigid body into a pure rotation about any point and a rigid translation.

So again, it isn't that the rotation is about the center of mass, but that the translation of the center of mass is particularly simple.

This last diagram is hard to coprehend..

 

[Dale]

This last diagram is hard to coprehend..

There exists a frame where the motion is momentarily pure rotation about the random point. It is just to show that neither the center nor the edge are special points. Any point can be considered the center of rotation.

 

[hmmm27]

Chief ask engineer: about what point will tanker rotate if our wokers push with tugboat 30m from stern?
Engineer answer: tanker will rotate about any point

Chief: What??

Tell that object rotate about any point is useless ,complety unphysical?

Chief didn't say angle of push.

 

[John Mcrain]

Chief didn't say angle of push.

Obviusly perpendicular to tanker, if not, tugboat will slide and scratsh tunker, worker knows that.
Worker must know how tunker is rotate, othervise it will crash into dock.

So tell them tunker rotate about any point doesnt help at all.

So mathematicians wonder why they even ask so stupid question about what point will tanker rotate, chief and worker dont understand mathematicians answer at all!

There is only one position about tanker will rotate for each position of tugboat push.
Tanker will not rotate only when you push at center of lateral resistance,but with constant speed.
During acceleration phase it will slightly rotate becuase c.m. is not at CLR.

Obviusly relevant reference frame for workers is dock, not Jupiter or maybe rotating cranckshaft inside engine of tugboat.

 

[Dale]

So mathematicians wonder why they even ask so stupid question about what point will tanker rotate, chief and worker dont understand mathematicians answer at all!

This is not conducive to a good discussion. You will need more than petulance and stubbornness to learn physics and math.

The topic of the OP is actually more clear than the tugboat, so the tugboat isn't necessarily a good example. The topic of the thread is a free object. In a free object the choice of the center of mass as a convenient reference point is clear. That is the unique point which travels in a straight line at constant speed. So although for a free object you can express the motion as rotation around any point, the translation becomes particularly simple for a single point.

For the tugboat there is no point which travels in a straight line at constant speed. So the motion is less straightforward. You have to deal with acceleration regardless of which point you choose for analyzing the rotation. Correctly judging that motion is not trivial and a simple stubborn "center of mass" doesn't necessarily reflect what is actually going on in the mind of the chief and worker.

 

[vanhees71]

Let's take a closed system of point particles with only pair-particle interactions at a distance, i.e., the paradigmatic Newtonian model. Then you have
$$m_i \ddot{\vec{x}_i}=\sum_{j \neq i} \vec{F}_{ij}(\vec{x}_i-\vec{x}_j).$$
Thanks to Newton's Lex III we have
$$\vec{F}_{ij}=-\vec{F}_{ji}, \quad F_{ii}=0.$$
This implies
$$\sum_i m_i \ddot{\vec{x}_i} = \sum_j \sum_{i \neq j} \vec{F}_{ij} = \frac{1}{2} \sum_j \sum_{i \neq j} [\vec{F}_{ij} +\vec{F}_{ji}]=0,$$
i.e., the center of mass,
$$\vec{R}=\frac{1}{M} \sum m_i \vec{x}_i, \quad M=\sum_i m_i$$
moves like a free particle,
$$M \ddot{\vec{R}}=0.$$
If you have in addition a homogeneous gravitational field like close to the Earth, then you add
$$m_i \ddot{\vec{x}}_i=\sum_{j \neq i} \vec{F}_{ij}+m_i \vec{g},$$
and then for the center of mass you have
$$M \ddot{\vec{R}}=\vec{g} \sum_i m_i = M \vec{g},$$
i.e., the center of mass is freely falling.

That's what can be said with really simple math about a Newtonian system of point particles that's closed or one in an external homogeneous gravitational field.

For the special case of a closed two-particle system, you can reduce the problem to the free motion of the center of mass and an effective one-body problem moving in an external force field by simply introducing $\vec{R}$ and the relative coordinates $\vec{r}=\vec{r}_1-\vec{r}_2$. Then you get
$$M \ddot{\vec{R}}=0, \quad \mu \ddot{\vec{r}}=\vec{F}_{12}(\vec{r}),$$
where $\mu=m_1 m_2/M$ is the reduced mass.

The case of a rigid body is treated in my above quoted manuscript:

https://itp.uni-frankfurt.de/~hees/pf-faq/spinning-top.pdf

using, however, the Lagrangian form of the Hamilton principle.

 

[John Mcrain]

For the tugboat there is no point which travels in a straight line at constant speed. So the motion is less straightforward. You have to deal with acceleration regardless of which point you choose for analyzing the rotation. Correctly judging that motion is not trivial and a simple stubborn "center of mass" doesn't necessarily reflect what is actually going on in the mind of the chief and worker.

Lets say CLR and cm of tanker are not in same place.

If tugboat start to push tanker perpendicualar to tanker at CLR point;
during acceleration phase(from 0-2mph) it will slightly rotate because c.m. is not at CLR, cm will rotate toward tugboat because of inertia.

after tanker reach constant speed (2mph) it will not rotate any more, because hydrodynamic moment, left and right from CLR is equal.
So tugboat and tanker travel at costant speed(2mph) in straight line..
(I neglected here lateral aerodynamic drag of tanker, because at 2mph water drag is 99% of total drag)

Dont agree?

 

[hutchphd]

Dont agree?

Don't care.
Not the original question.

/

 

[Dale]

@John Mcrain You are missing the actual issue. The issue is to explain what you mean when you say that “an object rotates around a point”.

Mathematically you can instantaneously decompose rigid body motion into a translation and a rotation about any point. So what rule do you use to determine which specific point to claim that something is rotating around?

Lets say CLR and cm of tanker are not in same place.

If tugboat start to push tanker perpendicualar to tanker at CLR point;
during acceleration phase(from 0-2mph) it will slightly rotate

About which point does it rotate? Please write down the actual mathematical expression for the motion here.

Whatever expression you write, and whatever point you claim is the center of rotation, I can write another expression that has the same rotation about a different point. The translation will be different, but the rotations will be the same.

 

[Vanadium 50]

I have to say I am not happy about the way this thread is going.

Bodies do not rotate about a point. They rotate about an axis.

Why does the axis contain the COM? It doesn't have to, but that's usually what we mwan by the word "rotation". The earth rotates about its axis (which contains the COM) but revolves around the sun. It's a convention to aid discussion.

Now one might say "yeah, but this is just terminology" and it is. But when trying to understand something, it's better to keep the potential miscommunications to a minimum.

 

[John Mcrain]

@John Mcrain You are missing the actual issue. The issue is to explain what you mean when you say that “an object rotates around a point”.

About wich point boat rotate? A,B,C,D or any point ?

My answer is A.