Does every object rotate around its center of gravity?

[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.

 

[John Mcrain]

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

If force act at CG,we have only translation.
If force act anywhere else what is not at CG,we have translation plus rotation.

And now I say inertia is "responsible" why rotation happened when force act at object anywhere else that is not at CG,because mass show reistance to change position ?

 

[Dale]

If force act anywhere else what is not at CG,we have translation plus rotation.

If the force acts anywhere else we have translation plus rotation about the CG. It is important to specify what the rotation is about.

And now I say inertia is "responsible" why rotation happened when force act at object anywhere else that is not at CG,because mass show reistance to change position ?

Not really. It is because the torque about the CG is 0 when the force is applied at the CG. The torque is non-zero when the force is applied elsewhere.

I am not sure how inertia could enter into this. Perhaps you mean “moment of inertia” instead of “inertia”?

 

[Richard R Richard]

The product of the "moment of inertia" of the body with respect to the axis of rotation and the angular acceleration is equal to the sum of the torques produced by the external elements including gravity.
Every system of forces reduces a force as a result that is a vector that passes through the MC plus a moment that is another vector that can be applied at any point in the body.
Another way is to apply the resultant force on an axis of action parallel to the previous one but offset a distance. The torque thus created then turns out to be equal to the sum of all the torques created by each of the external forces.
In other words, before the same forces, an object of similar dimensions with a greater moment of inertia will influence that it will rotate with less angular acceleration.Centroid or barycenter

$$ \displaystyle \vec r_{\text {ce}} = \dfrac {\int \limits_V \vec r dV} {\int \limits_V dV} $$

Center of mass

$$ \displaystyle \vec r_{\text {mc}} = \dfrac {\int \limits_V \vec r \rho (V) dV} {\int \limits_V \rho (V) dV} $$

Gravity center

$$\displaystyle \vec r_{\text {gc}} = \dfrac {\int \limits_V \vec r \rho (V) g (V) dV} {\int \limits_V \rho (V) g (V) dV} $$

External force center, pressure or thrust

$$ \displaystyle \vec r_{\text {pc}} = \dfrac {\int \limits_V \vec r F_{ext}(V) dV} {\int \limits_V dV}$$

$$ \vec r = (r, \theta, \phi) $$

If I'm wrong, please correct me.

In the figures you can see that for there to be stability, the thrust forces must have an opposite consequence to the driving force or the weight, any imbalance creates a restorative torsional moment.

 

[John Mcrain]

If the force acts anywhere else we have translation plus rotation about the CG. It is important to specify what the rotation is about.

Not really. It is because the torque about the CG is 0 when the force is applied at the CG. The torque is non-zero when the force is applied elsewhere.

I am not sure how inertia could enter into this. Perhaps you mean “moment of inertia” instead of “inertia”?

I mean at accelerating phase,from time zero when force start to act to time when object stop accelerating..

Imagine we have boat with very very heavy metal ball at left side.Neglect aerodynamic forces at ball..
When boat is going at constant speed ,nothing happend,he is going in straight line.
But if you increase throttle,increase engine thrust,boat start accelerate,boat will turn to the left because mass of ball show resistance to change speed/position..

In similar sense ,when gust hit plane (time zero) plane start to translate downwind and rotate upwind(if center of lateral pressure is behind CG).
Upwind rotation of plane is because CG(mass) "show resistance" to change position..So inertia(m x a) is reason why plane start rotating upwind..

This how my brain "explain" why plane rotate upwind when center of pressure is behind CG.

Is this correct explantion?

 

[PeroK]

This is getting silly. That's because the "boat" is attached to something else, namely the large "ball". Now you have a system of boat+ball, whose CM is out on the boom arm somewhere.

The system accelerates and rotates because the force from the water is no longer directed through the CM of the system. The same would happen if you moved the engine to the side of the boat.

If you put an equivalent ball on the other side of the boat, there would be no rotation.

 

[John Mcrain]

If you put an equivalent ball on the other side of the boat, there would be no rotation.

for sure no rotation,because now CG will be in the middle of boat ,so engine thrust(force) act at CG...

 

[A.T.]

This how my brain "explain" why plane rotate upwind when center of pressure is behind CG.

Is this correct explantion?

It's not precise enough to tell if its correct or wrong. But the vague intuitions you have about resistance to acceleration are expressed formally by the concepts of:
- Center of mass
- Moment of inertia

 

[John Mcrain]

It's not precise enough to tell if its correct or wrong. But the vague intuitions you have about resistance to acceleration are expressed formally by the concepts of:
- Center of mass
- Moment of inertia

How would you explain why object rotate when force is act anywhere that is not at CG?

 

[vanhees71]

The description of the motion of a rigid body is among the more challenging but also among the most fascinating subjects of Newtonian mechanics. In general you have both translational motion as well as spin (rotation of the body around some point fixed within the body). Together the rigid body has 6 degrees of freedom (3 translational degrees of freedom describing the location of one fixed point within the body and 3 rotational degrees of freedom describing the rotation of a body-fixed right-handed cartesian basis system relative to space-fixed cartesian basis system (usually defining an inertial frame of reference), usually parametrized by the Euler angles).

A very good treatment can be found in

A. Sommerfeld, Lectures on Theoretical Physics, vol. 1.

 

[PeroK]

How would you explain why object rotate when force is act anywhere that is not at CG?

The "why" involves Newton's laws and the presence of (reciprocal) internal forces. The simplest model is a two point system, held together as a rigid body by a light connecting rod. If an impulse is applied to one mass, then an internal force arises that pulls that mass away from straight line motion and pulls the other mass along the direction of the connecting rod.

With a bit of algebra, you see that it's the CM that accelerates according to $F = ma$. Then, by introducing the concepts of torque, moment of inertia and angular momentum, you find that the body rotates according to $\tau = I \alpha$.

You can extend this to any rigid system of $N$ particles - which is usually covered in an undergraduate Classical Mechanics textbook.

Whether this is intuitively obvious or not is a moot point. Ultimately, it all rests on Newton's laws.