Does every object rotate around its center of gravity?

[berkeman]

If you push stick at one end,it will translate and rotate,but pivot point(position which not change position in space) will be out of CG,can even be at point which is out of stick physical limits..

That's nonsense. Do you have any videos of objects rotating in the International Space Station?

 

[A.T.]

Because every human can see with his eyes that wheater vane is rotating about mast.

It's not your just eyes, but also your brain doing a lot of interpretation. But the interpretation your brain chooses doesn't have to be unique.

 

[John Mcrain]

It is the same weather vane regardless of what axis you choose to use for the analysis. It changes its orientation by the same amount regardless of what pair of body-fixed points you choose to use to measure deflection. If one does the analysis either way, the result will still come out predicting that that the center of pressure will equilibriate downwind from the mast.

I've already agreed with you that the position of the mast is a convenient choice for the "rotation axis". [It follows naturally from choosing to work in the ground frame]. What more do you wish?

I'm will not say that the weather vane "really" rotates about the mast because the norm is to reserve "real" as an adjective for things that are invariant. Not for things that are free choices.

If I want test aircraft in wind tunnel for wind gust/weather vane effect,where I must placed rod ?
At CG,infront CG,behind CG,at rudder,at propeller...?

 

[Ibix]

If I want test aircraft in wind tunnel for wind gust/weather vane effect,where I must placed attached point?
At CG,infront CG,behind CG,at rudder,at propeller...?

The point that @jbriggs444 is trying to make is that you are conflating two issues here.

The question you are asking in the post I quoted is: where do I attach a string to an object such that the string and the weight produce zero torque. That does have a unique, frame-invariant, answer.

That is not the question you asked originally. That question was which point the plane (or boat) rotates around, and the answer to that is frame dependent. Imagine that I run an axle through the nose of the plane and attach the axle firmly to a wall. Then I hold the plane horizontal and release. It will swing down to hang vertically from its nose (i.e. it will seek the position where the reaction force from the axle and its weight have zero torque, again). But where did the plane rotate around? In the lab frame it rotated about its nose, not its center of gravity. But in the rest frame of the center of gravity of the plane the axle moves and the plane rotates about its center of gravity. Neither description is more real than the other. From the point of view of the lab, though, it's easier to say that the plane rotated about the axle through its nose.

 

[russ_watters]

If I want test aircraft in wind tunnel for wind gust/weather vane effect,where I must placed rod ?
At CG,infront CG,behind CG,at rudder,at propeller...?

Anywhere will do. It's common to see the support under or behind the model.

 

[russ_watters]

The question you are asking in the post I quoted is: where do I attach a string to an object such that the string and the weight produce zero torque. That does have a unique, frame-invariant, answer.

Note also, that in these new questions @John Mcrain , the attachment point you are adding isn't just a point/axis of rotation. It applies a force, so in many cases it changes what happens. It isn't merely a convenient reference point.

 

[russ_watters]

If you push stick at one end,it will translate and rotate,but pivot point(position which not change position in space) will be out of CG,can even be at point which is out of stick physical limits..

If we want rotation at CG then we must apply two same force at both ends with opposite direction,equaly distance from CG.

That's nonsense. Do you have any videos of objects rotating in the International Space Station?

You don't need to be in space to test this, you just need to eliminate the influence of gravity. Set a pen or pencil on a table in front of you, oriented left to right (as opposed to pointing at you). Flick the right end away from you with your finger. It starts to spin and move away from you. Does it move straight away or off to the left or right at an angle?

 

[jbriggs444]

As I understand the claim to which @berkeman objected, it was that one can take an object (such as a thin rod) which is initially stationary, apply a momentary impulse at some point within the rod's physical extent and obtain a motion so that the rod has a fixed point (an instantaneous center of rotation) which falls outside the rod's physical extent.

That the rod begins rotating and that its center of mass begins translating, I think we all agree.

That the instantaneous center of rotation will be, at some times, outside the rod's physical extent, I would agree. [1/4 of a rotation after launch, the instantaneous center of rotation will be off to one side of the thin rod]

That the instantaneous center of rotation will be initially outside the rod's physical extent, I would disagree. For a uniform rod and a push at one tip, I seem to recall that the instantaneous center of rotation will be 1/3 of the way from the far end. Testing 5 seconds ago with a handy pen on desktop agrees. By no coincidence, this point is also known as the "sweet spot".

If you have a suitably concave object, one can certainly arrange matters so that the instantaneous center of rotation will initially be outside the object's physical extent. [Imagine a soviet sickle tapped on the handle end]. But for a convex object, I think not.

 

[berkeman]

And even weirder...

 

[A.T.]

As I understand the claim to which @berkeman objected, it was that one can take an object (such as a thin rod) which is initially stationary, apply a momentary impulse at some point within the rod's physical extent and obtain a motion so that the rod has a fixed point (an instantaneous center of rotation) which falls outside the rod's physical extent.
,,,,

That the instantaneous center of rotation will be initially outside the rod's physical extent, I would disagree. For a uniform rod and a push at one tip, I seem to recall that the instantaneous center of rotation will be 1/3 of the way from the far end. Testing 5 seconds ago with a handy pen on desktop agrees. By no coincidence, this point is also known as the "sweet spot".

Okay, but you have constrained to problem from "some point within the rod's physical extent" to "one tip".

 

[jbriggs444]

Okay, but you have constrained to problem from "some point within the rod's physical extent" to "one tip".

Oh, *doh*. You are right. Pushing near the center moves the instantaneous center away from the object.

 

[John Mcrain]

Oh, *doh*. You are right. Pushing near the center moves the instantaneous center away from the object.

Plane fly in wind tunnel test section,not attached to rod,free to move.
If gust hit aircraft from side, and lateral center of pressure is behind CG at fuselage ,then instantaneous center of rotation is somewhere infront CG,using ground as reference frame?

 

[berkeman]

Plane fly in wind tunnel test section,not attached to rod,free to move.

That sounds scary. Do you have any pictures or links?

 

[Richard R Richard]

Hi @John Mcrain I will respond to the initial point that is discussed in the thread.

The definitions that I will give are very perfectible, but let's just go to the main concept.

The center of mass of an object is an object's own characteristic, that is, of the distribution in space of the different densities of the materials that compose it.

The center of gravity is different because at each point of the object there is an acceleration towards the center of a gravitating object (Sun, earth, moon etc whatever), the sum of that mass by accelerations are forces that create a couple on any point reference. In this way, the center of gravity only coincides with the center of mass when there is a symmetry with respect to the main axes of inertia, we can say it in another way, the sum of these moments with respect to the point chosen as GC must be zero.

Well, but a ship, an airplane, are not only subjected to the force of gravity but also to hydrostatic thrust forces of the environment that surrounds it and to the force of aerodynamic or hydrodynamic friction. Therefore a boat is balanced when the Center of Thrust (the point of application of all external forces) is above the center of gravity, any angular imbalance creates a recuperative torque and the boat does not list. In the rocket (at low altitude) and in airplanes, friction is the force that causes the restoring moment while the center of application of this force is ahead in relation to the velocity vector with respect to the center of gravity.

 

[A.T.]

If gust hit aircraft from side, and lateral center of pressure is behind CG at fuselage, then instantaneous center of rotation is somewhere infront CG,using ground as reference frame?

Yes, and if the center of pressure is moved towards the CG, the instantaneous center of rotation moves towards infinity. Pushing exactly at the CG gives you pure translation, which doesn't have a well defined instantaneous center of rotation.

 

[Dale]

It is said that rocket,plane rotate about center of gravity ,why this is is not case for boats?

Do you have a reference for where this is said?

Plane fly in wind tunnel test section,not attached to rod,free to move.

I don’t think this is right.

I think a lot of the confusion in this thread is just you misunderstanding something you read. It would help if you would cite your sources so that we can have a more focused discussion.

 

[John Mcrain]

Yes, and if the center of pressure is moved towards the CG, the instantaneous center of rotation moves towards infinity. Pushing exactly at the CG gives you pure translation, which doesn't have a well defined instantaneous center of rotation.

Why we need 45 posts to answer what I asked?

What if plane flying in sky,a ground is reference frame,gust hit.
we have translation because of plane velocity over ground, but also have side translation cause by gust hit form side,and plane rotation becuase center of pressure is behind CG.
Where is then instantaneous center of rotation?

 

[John Mcrain]

Do you have a reference for where this is said?

I don’t think this is right.

In every video about rocket or plane stability ,they say it rotate about CG..

I put plane fly in test section ,only to avoid translation of airplane speed over ground.Because I didnt know how to deal with these 3 movement ,translation from speed,translation from gust hit and rotation from gust hit...

 

[PeroK]

In every video about rocket or plane stability ,they say it rotate about CG..

You could generalise this:

Every other point on a body moves relative to the CG.

Now, let's pick another point $P$ on the body:

Every other point on a body moves relative to point $P$.

Those are two ways to consider the overall motion:

1) Consider the motion of the CG, then the motion of the body relative to the CG.

2) Consider the motion of point $P$, then consider the motion of the body relative to point $P$.

Now, if we consider a rigid body rotating with angular frequency $\omega$.

1) Every other point on the body rotates about the CG with angular frequency $\omega$.

2) Every other point on the body rotates about the point $P$ with angular frequency $\omega$.

You can't have one without the other. If P moves relative to the CG, then the CG moves reciprocally relative to P.

 

[A.T.]

Why we need 45 posts to answer what I asked?

Because it took you so long to specify what center of rotation you mean.

What if plane flying in sky,a ground is reference frame,gust hit.
we have translation because of plane velocity over ground, but also have side translation cause by gust hit form side,and plane rotation becuase center of pressure is behind CG.
Where is then instantaneous center of rotation?

In front and upwind of the CM.

 

[Richard R Richard]

This is easily summarized, in the absence of gravity, any external force on a body has the effect of translating the center of mass and rotating about the center of mass. As near a planet you cannot avoid the force of gravity, it causes the resulting force in the same way and one moment, de say the objects will rotate with respect to the center of gravity, because in small objects, it practically coincides with the center of mass. Always the GC of a gravitationally stable object is closer to the Earth (in this case) than the MC.
When external forces act, they are distributed on the surface of the object, we can then add (integrate) all these forces to find a resultant, which will have the same effect, moving the MC and rotating on it. But as I said, you cannot avoid gravity, so only if the GC is displaced an appreciable distance from the MC, it will create a torsional moment trying to regain the previous stability.
The answer about which point broken is the MC.
If the density of the object is constant the MC coincides with the barycenter.
If the gravitational field is constant, the MC coincides with the GC, as no gravitational field is constant with the distance to the earth, then all stable objects have the GC lower than the MC, (it is a real but depreciable distance).
As long as the resultant of external forces (thrust, friction, etc.) passes through the same line of action as the resultant of gravity, then the object is translated in that frame of reference.
But if, in addition, the resultant of the external forces does not pass through the MC, it creates a torque that may or may not find equilibrium with the couple of the force of gravity, to find a new stable equilibrium or rotate with respect to the MC.

 

[John Mcrain]

In front and upwind of the CM.

Ok this is answer to my question.

So I just need write "instantaneous" center of rotation and then everything will be clear?

 

[A.T.]

So I just need write "instantaneous" center of rotation ...

If that's what you are interested in.

 

[John Mcrain]

Because it took you so long to specify what center of rotation you mean.

Dont put all blame at me.

Did everyone of you watched my first video at my first post?
It can be clear see what center of rotation I am looking for.

It is not my fault that everbody is lazy to watch video,so because of your laziness I turn out stupid and we are going in wrong direction all the thread..6:30-9:00

 

[jbriggs444]

The "apparent pivot point" described in the video is the same concept as "instantaneous center of rotation". It is that point on the vessel (or on a rigid wire frame one imagines extended from the vessel) that is momentarily stationary as the vessel moves.

Any time you see the word "stationary" in physics, you should immediately think "in what frame of refererence?"

The choice of reference frame affects the meaning of "stationary" and, accordingly, the position of the instantaneous center of rotation.

 

[A.T.]

Did everyone of you watched my first video at my first post?
It can be clear see what center of rotation I am looking for.

You also kept talking about using the CG. So you were told that different approaches are valid.

This is also what the video tells you at 7.22min:
- First the motion is decomposed into translation of the CG and rotation around the CG
- Then the same motion is described as a pure rotation around the instantaneous center of rotation.

 

[vanhees71]

The choice of the body-fixed reference frame in the theory of the rigid body is arbitrary and no physics depends on it in principle.

It's of course clear that in practice a clever choice of the reference frame is key to be able to get a treatable formulation.

E.g., if you describe a freely falling rigid body in the homogeneous gravitational field of the Earth it's a clever choice to make the center of mass of the body to the origin of the body-fixed reference frame and use the principle axes of the tensor of inertia around this point as basis.

If on the other hand you fix the body at a point, around which it can freely rotate ("spinning top" or "gyroscope") it's clever to choose this point as the origin of the body-fixed reference frame and the principle axes of the tensor of inertia around this point as Carstesian basis.

 

[Dale]

Dont put all blame at me.

Did everyone of you watched my first video at my first post?
It can be clear see what center of rotation I am looking for.

That is fair enough. I indeed had not watched it.

In terms of physics, it is important to recognize that you are free to specify the axis of rotation. When you analyze torque or angular momentum, you will find that the change in the angular momentum is equal to the torque about any axis. Because that holds true for any axis you must specify which axis you wish to analyze.

It is often convenient to specify an axis through the center of mass because in many cases the object's moment of inertia is constant about such an axis. Also, for a free object experiencing no net force the center of mass is the only point which must move in a straight line at constant speed. It is not necessary to choose that point, but it makes the math easier.

The ship pilot's discussion about a pivot point or an apparent pivot point is not something that is well defined in physics. He used the terminology many times, but never defined it. I am not sure what he means by that in terms of the motion of the ship.

The big difference between a ship and many other objects that you mentioned is that the ship is affected by large forces distributed all along the length of the hull. Those forces make it so that the center of mass of the ship is not moving inertially. The motion is more complicated than that of a free object experiencing no net force.

 

[John Mcrain]

That is fair enough. I indeed had not watched it.

In terms of physics, it is important to recognize that you are free to specify the axis of rotation. When you analyze torque or angular momentum, you will find that the change in the angular momentum is equal to the torque about any axis. Because that holds true for any axis you must specify which axis you wish to analyze.

It is often convenient to specify an axis through the center of mass because in many cases the object's moment of inertia is constant about such an axis. Also, for a free object experiencing no net force the center of mass is the only point which must move in a straight line at constant speed. It is not necessary to choose that point, but it makes the math easier.

The ship pilot's discussion about a pivot point or an apparent pivot point is not something that is well defined in physics. He used the terminology many times, but never defined it. I am not sure what he means by that in terms of the motion of the ship.

The big difference between a ship and many other objects that you mentioned is that the ship is affected by large forces distributed all along the length of the hull. Those forces make it so that the center of mass of the ship is not moving inertially. The motion is more complicated than that of a free object experiencing no net force.

Can I say that inertia is "responsible" why rotation happened when force act at object everywhere out of CG,because mass show reistance to change position ?

 

[Dale]

Can I say that inertia is "responsible" why rotation happened when force act at object everywhere out of CG,because mass show reistance to change position ?

Sorry, I don’t understand what you are asking here.