Do objects always rotate around center of mass?

[A.T.]

In my brain, rotate mean circular motion of object around axis that passes anywhere inside object geometry, if axis is out of object geometry than I call it revolve.

Why invent more names than needed based on where the axis is?

- rotation is change of orientation
- translation is change of position

They can be combined, but that requires choosing a reference point. Choosing the center of mass often makes the math of applying Newton's Laws simpler. That's why it's often used.

 

[Dale]

In my brain, rotate mean circular motion of object around axis that passes anywhere inside object geometry, if axis is out of object geometry than I call it revolve.
That seems logic to me.

Is this Ok?

That is fine, but the issue is that the axis is not unique. Consider a wheel. You can say that it rotates about the axle, or that it rotates about the contact point, or indeed about any other point whether on or off the wheel.

Problem is when airplane move elevator, he revolve around some axis far away from him and rotate around some axis that passes through airplane body, but it turn out that dont have to be through c.g.?

Correct. It can be considered about any point, it is not restricted to the cg

 

[user079622]

That is fine, but the issue is that the axis is not unique. Consider a wheel. You can say that it rotates about the axle, or that it rotates about the contact point, or indeed about any other point whether on or off the wheel.

Correct. It can be considered about any point, it is not restricted to the cg

But..
If we dont know that airplane rotate around cg, then we can put lift force in front of cg and plane will unstable and kill someone.

So we must know what is real axis around which will plane rotate when disturbance comes.
I dont know what is name of this point/axis but I know engineers must know this position.
Is this called instant center of rotation?

Plane fly in straight line above runway and when gust hit him from side, he rotate into wind.
Where is axis of rotation? It can be anywhere?

 

[Dale]

If we dont know that airplane rotate around cg, then we can put lift force in front of cg and plane will unstable and kill someone.

This is simply not true. The axis of rotation is part of the analysis. It has no bearing on the stability of the air frame.

So we must know what is real axis around which will plane rotate when disturbance comes

You claim this, but it is simply false.

I dont know what is name of this point/axis but I know engineers must know this position.
Is this called instant center of rotation?

I don’t know it either. If you do find the term it would be useful. I am sure that the engineers have a clear definition for it.

Where is axis of rotation? It can be anywhere?

Yes.

 

[Dale]

@user079622

This is just a mathematical fact of rigid body motion, purely kinematically. It is 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

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:

View attachment 319994

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:

View attachment 319996

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.

 

[Lnewqban]

quote from text:

"The side force (F) is applied through the center of pressure of the vertical stabilizer which is some distance (L) from the aircraft center of gravity. This creates a torque

T = F * L​

on the aircraft and the aircraft rotates about its center of gravity."So people who read this conclude that object always rotate around c.g./com....Then comes to this forum and ask questions. I think it has 1000 questions in this forum at this topic

The center of mass of the airplane must follow Newton’s laws.
It would follow a rectilinear trajectory, unless forced to deviate from that natural traslation.
The external force on the rudder is not that force; therefore, the plane initially yaws (rotates around the CM until finding a position of balance) and its CM continues moving on a straight line.
That does not last long, as other aerodynamic forces appear.

For example, the tail is higher than the CM, reason for which the plane also rolls (rotates sideways).
Once it is rolled and crabing, one wing receives more airstream than the other, and more inestability appears.
Both additional movements must be controlled by ailerons, which may induce additional yaw.

In order to follow a horizontal circular trajectory, the CM must be forced to do so by a centripetal horizontal force, which is normally a horizontal component of the lift vector.
That vector is tilted because the plane has been rolled and pitched up simultaneously.

As you can see, a sequence of several torques and rotations appear for a simple maneuver.

 

[nasu]

quote from text:

"The side force (F) is applied through the center of pressure of the vertical stabilizer which is some distance (L) from the aircraft center of gravity. This creates a torque

T = F * L​

on the aircraft and the aircraft rotates about its center of gravity."So people who read this conclude that object always rotate around c.g./com....Then comes to this forum and ask questions. I think it has 1000 questions in this forum at this topic

The fact that it rotates about one point does not exclude rotation about another point. There is no contradiction.
However, what the existence of a torque about the CG actually implies is that there will be some angular acceleration about this point. If there was no previous rotation about CG, obviously it will start to rotate.
And by the way, the same force will have a torque about many other points on the airplane. It's just easier to analyze the dynamics by using the COM.

 

[alan123hk]

Wikipedia says that for a free-floating (unattached) object, the axis of rotation is commonly around its center of mass. https://en.wikipedia.org/wiki/Rotational_energy

Does this mean there are exceptions where some objects' axis of rotation does not rotate around their center of mass in the case of free-floating?

 

[Dale]

Does this mean there are exceptions where some objects' axis of rotation does not rotate around their center of mass in the case of free-floating?

It means that there are exceptions where the people analyzing the scenario choose to analyze its rotation about a different axis. For instance, they might analyze the moon as rotating about the barycenter instead of its own axis.

 

[alan123hk]

Got it, thank you for your explanation.

 

[A.T.]

Wikipedia says that for a free-floating (unattached) object, the axis of rotation is commonly around its center of mass. https://en.wikipedia.org/wiki/Rotational_energy

What it actually means is that if you want to use the moment of inertia around the CoM (as it is commonly provided), then you also have to use the rotation around the CoM to compute the rotational KE. But you could use any other point, as long as you stay consistent. You will get the same total KE, but a different composition of linear vs. rotational KE.

 

[A.T.]

Plane fly in straight line above runway and when gust hit him from side, he rotate into wind.
Where is axis of rotation? It can be anywhere?

Yes, one can put the axis anywhere for analysis, but it is often not practical to do so. In this case, since the plane is moving and rotating, even the instantaneous axis of rotation is not the CoM, and might even be somewhere outside the plane.

 

[user079622]

Yes, one can put the axis anywhere for analysis, but it is often not practical to do so. In this case, since the plane is moving and rotating, even the instantaneous axis of rotation is not the CoM, and might even be somewhere outside the plane.

If force act at 90degress at the end of stick in free space, will he rotate around CoM and CoM will translate in straight line?

 

[user079622]

This is simply not true. The axis of rotation is part of the analysis. It has no bearing on the stability of the air frame.

You claim this, but it is simply false.

I don’t know it either. If you do find the term it would be useful. I am sure that the engineers have a clear definition for it.

Yes.

When gust of wind hit plane from side, it act at lateral center of pressure which is for sure behind c.g. because of big tail surface.
Plane will do some small side slip so we have another small aerodynamic force at center of lateral pressure but from opposite side.

thrust force from left and right engines are equal so they dont cause any toruque.

So we have aero side force from gust minus aero side force side slip and CoM which resist to accelarate if something want to do it.Free body diagram , top view
This side force will translate plane c.g. downwind and rotate plane into wind around instant axis that passes through c.g
Is it ok?

 

[A.T.]

If force act at 90degress at the end of stick in free space, will he rotate around CoM and CoM will translate in straight line?

That is one possible way to desctibe the motion, but not the only one.

 

[user079622]

That is one possible way to desctibe the motion, but not the only one.

But is instant axis of rotation through CoM?

 

[jbriggs444]

But is instant axis of rotation through CoM? In another words, can single force make object to rotate around com?

The "instantaneous axis of rotation" is the axis that is momentarily motionless.

However, "motionless" is always relative to a chosen frame of reference. Change the frame of reference and you change which axis is "motionless".

It is quite convenient to adopt the rest frame of the stick's center of mass. So convenient that we tend to do so without even thinking about it. That is why we tend to think of the stick as rotating about its center of mass.

 

[Dale]

Free body diagram , top view
This side force will translate plane c.g. downwind and rotate plane into wind around instant axis that passes through c.g
Is it ok?

Yes, that is all good

 

[user079622]

The "instantaneous axis of rotation" is the axis that is momentarily motionless.

However, "motionless" is always relative to a chosen frame of reference. Change the frame of reference and you change which axis is "motionless".

It is quite convenient to adopt the rest frame of the stick's center of mass. So convenient that we tend to do so without even thinking about it. That is why we tend to think of the stick as rotating about its center of mass.

Is any difference if my force is some rocket engine at the end of stik working all the time or if my force is short impulse ,when I hit end of stick with hammer?
In first case we have continous external force in second case just first "second" one hit with external force and after that no external force
Will stick behave the same in both case in free space?

 

[user079622]

Yes, that is all good

So I dont see any difference with stick in free space and my free body diagram of gust/plane, stick "rotate" around CoM and my plane "rotate" around c.g.

 

[user079622]

even the instantaneous axis of rotation is not the CoM, and might even be somewhere outside the plane.

Why do you think not at CoM? We have just one side force and plane CoM...

 

[Dale]

So I dont see any difference with stick in free space and my free body diagram of gust/plane, stick "rotate" around CoM and my plane "rotate" around c.g.

Me neither. The difference is in your claims, not in the physics.

In the one where I said it was ok you claimed that it rotates about the cg. That is fine, it does rotate about the cg.

The ones where I objected you were claiming that if it rotates about any other axis there is some problem. That is not fine, if it rotates about the cg then it also rotates about any other parallel axis. That is just a fact of rigid body motion.

The problematic claim is not that it rotates about the cg, the problematic claim is that it only rotates about the cg or that the cg is unique or that if it rotates about another axis then it is unstable, etc.

 

[user079622]

Me neither. The difference is in your claims, not in the physics.

In the one where I said it was ok you claimed that it rotates about the cg. That is fine, it does rotate about the cg.

The ones where I objected you were claiming that if it rotates about any other axis there is some problem. That is not fine, if it rotates about the cg then it also rotates about any other parallel axis. That is just a fact of rigid body motion.

The problematic claim is not that it rotates about the cg, the problematic claim is that it only rotates about the cg or that the cg is unique or that if it rotates about another axis then it is unstable, etc.

But that is not valid for instant center of rotation?
I am talking about instant center all the time..

 

[Dale]

But that is not valid for instant center of rotation?
I am talking about instant center all the time..

What is the precise scientific definition of “instant center of rotation”? I don’t know what that means.

 

[user079622]

What is the precise scientific definition of “instant center of rotation”? I don’t know what that means.

Nither I.
This is place in universe where in time t1, angular velocity is zero, every particle of this object rotate around this center.

 

[Dale]

angular velocity is zero, every particle of this object rotate around this center.

This is a self contradictory definition. If angular velocity is 0 then the object does not rotate around that center.

 

[Dale]

I am talking about instant center all the time..

Also, when I agreed with you above you were not talking about “instant center”. You were talking about the cg, which is well defined. You are probably better off sticking with the cg, which we both know, rather than instant center, which neither of us know.

 

[user079622]

Also, when I agreed with you above you were not talking about “instant center”. You were talking about the cg, which is well defined. You are probably better off sticking with the cg, which we both know, rather than instant center, which neither of us know.

dont understand you.. ,if I dont say instant center members repeat it rotate around any center, so I must choose this one

 

[Dale]

dont understand you.. ,if I dont say instant center members repeat it rotate around any center, so I must choose this one

You can choose any well defined point. The cg is a well defined point. This “instant center” is not, at least neither of us knows a valid definition of it.

When you use terms that neither you nor the other person understands then confusion is guaranteed

 

[user079622]

You can choose any well defined point. The cg is a well defined point. This “instant center” is not, at least neither of us knows a valid definition of it.

When you use terms that neither you nor the other person understands then confusion is guaranteed

How do you mean that neither you dont undrastand this term? You are professor of phyiscs..

I can choose any point to calculate moments only for body in equlibrium, net force=0 ?
Otherwise not?

If net force not zero then, I must choose c.g. as refernce point otherwise I will get wrong result?
That I learned it this video?

 

[Dale]

How do you mean that neither you dont undrastand this term? You are professor of phyiscs..

Yes, I don't understand that term. Yes, I am a physics instructor. When I don't understand a term it usually means that the term is nonsense, but occasionally it is simply a valid term that I am not familiar with.

I can choose any point to calculate moments only for body in equlibrium, net force=0 ?
Otherwise not?

You can choose any point to calculate moments. If the net force is 0 then the moment about any axis will be the same.

 

[user079622]

Yes, I don't understand that term. Yes, I am a physics instructor. When I don't understand a term it usually means that the term is nonsense, but occasionally it is simply a valid term that I am not familiar with.You can choose any point to calculate moments. If the net force is 0 then the moment about any axis will be the same.

Yes, I don't understand that term. Yes, I am a physics instructor. When I don't understand a term it usually means that the term is nonsense, but occasionally it is simply a valid term that I am not familiar with.You can choose any point to calculate moments. If the net force is 0 then the moment about any axis will be the same.

When net force is non zero every point has different moment but results are correct for that point?

 

[Dale]

When net force is non zero every point has different moment but results are correct for that point?

Yes, that is correct.

 

[user079622]

Yes, that is correct.

So men in video is wrong when he said if you choose any point that is not at cg you will get wrong results?

 

[nasu]

As already mentioned in several posts, the velocity of any point P of the rigid can be decomposed as a sum of two terms: the velocity (usualy called translation velocity) of another point, O and a rotation about this point O. $\vec{V_P}=\vec{V_O}+\vec{\omega}\times \vec{r_P}$ where $\vec{r_P}$ is the position vector of P relative to O and $\vec{\omega}$ is the same for all the points of the rigid and for any choice of the point O. When you change the point O, the translation changes but $\vec{\omega}$ does not. And at any instant you can find a reference point O so that $\vec{V_P}$ is zero. This point is what they call "instantaneous center of rotation" or "instant center of rotation". It is at rest at that instant. It will be just a coincidence if this point happens to be the COM or the CG at some specific time.

This works always for plan-parallel motion (or planar motion). For a more general motion, like an elicoidal one, I think that it does not work. I mean, there may not be any point instantaneously at rest.