Does every object rotate around its center of gravity?

[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.

 

[Dale]

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

I assume that you mean “How would you explain why object rotate about the CM when force is act anywhere that is not at CM?” We have already mentioned the importance of specifying what it is rotating about, so it is a little disheartening to see you continue to not include that.

As I said above, that is because the torque about the CM is zero when the force acts at the CM and nonzero when it acts elsewhere. Are you familiar with the concept of torque?

So inertia(m x a) is reason why plane start rotating upwind..

I have never seen inertia expressed as $ma$, which is force. I have seen it expressed as $m$ which is mass or as $mv$ which is momentum. Please be aware that we don’t permit just making things up here at PF. All posts must be consistent with the professional scientific literature. Inertia as $ma$ is not consistent.

 

[A.T.]

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

The above is not correct:
1) It's not the center of gravity but rather the center of mass
2) A force doesn't have to act at the center of mass to avoid angular acceleration, as long it acts through the center of mass.

Why? Because the center of mass is defined such that the above is true.

 

[John Mcrain]

As I said above, that is because the torque about the CM is zero when the force acts at the CM and nonzero when it acts elsewhere. Are you familiar with the concept of torque?

But this senteces don't explain why torque is non zero when force act elsewhere out of CM.

How can I call fictitius force what I feel as something push me to seat,when car accelerate?

How can you explain how I know that board will turn left,when accelerate in my post #64 ,without any calculation?

(Yes it is better to say about CM...Yes I know what is torque, force x lever arm...)

 

[A.T.]

...why torque is non zero when force act elsewhere out of CM ...Yes I know what is torque...

If you know what torque is you should know why it is non zero.

 

[John Mcrain]

A force doesn't have to act at the center of mass to avoid angular acceleration, as long it acts through the center of mass.

What is this?if force act CM -angular accelration is zero
if force act elsewhere out of CM- angular accelaration is non zero

 

[A.T.]

What is this?

What is what?

 

[jbriggs444]

But this senteces don't explain why torque is non zero when force act elsewhere out of CM.

If there is a non-zero net force on the object then the torque will depend on the chosen reference axis.

However, if one is interested in rotation, torque is only part of the answer. Torque only tells you how angular momentum changes over time. It does not tell you how the object rotates.

One can decompose angular momentum into two pieces. One piece accounts for the rotation of the body about is center of mass. The other piece accounts for the linear motion of the center of mass about the reference axis.

If we choose to put the reference axis at the center of mass then we eliminate one piece. This is very convenient.

 

[John Mcrain]

What is what?

I don't understand this construction.

 

[Dale]

Now I am thouroughly confused about what level to address my replies. First, you say this

But this senteces don't explain why torque is non zero when force act elsewhere out of CM.

Which makes me think, that I need to explain torque. But then you say this.

Yes I know what is torque, force x lever arm...)

Which makes me unsure how you can know what torque is and yet claim that my previous comment doesn't explain why torque is zero. I am not sure why you don't see that since you know the definition.

What is the lever arm about the CM for a force that goes through the CM?

Spoiler

lever arm = 0

Plug that into your definition and then what is the torque?

Spoiler

force x 0 = 0

How can I call fictitius force what I feel as something push me to seat,when car accelerate?

How can you explain how I know that board will turn left,when accelerate in my post #64 ,without any calculation?

I have no idea why you are bringing fictitious forces into the discussion. It seems wholly irrelevant. Regarding post #64, I don't know how you would do it without calculation. With calculation you simply calculate the torque about the CM and find that it is the torque for a left turn.

 

[A.T.]

I don't understand this construction.

I don't understand your question.

 

[PeroK]

What is this?if force act CM -angular accelration is zero
if force act elsewhere out of CM- angular accelaration is non zero

https://www.cambridge.org/gb/academ...nics-2nd-edition?format=HB&isbn=9780521198110

 

[Richard R Richard]

Any force that is not applied to an axis that does not pass through the MC creates a torque, when the force passes through the MC the length of the lever arm is zero and there is no torque, nor rotation.

In the video that you present, the force at the different ends of the boat creates different movements, as none passes through the MC, they create a moment or torque, which makes the boat rotate through its MC in the instant, but like the resistive forces of the water does not pass through the MC also create a moment with respect to the MC.
In case 1 the resistive forces oppose the turn but move the ship more, in the second case the resistive forces collaborate with the moment and the ship turns faster and moves slower.
When the ship has rotated, the resistive forces do not have to oppose the force pushing the ship in direction and direction, rather it will have a direction very close to the direction perpendicular to the ship's axis of symmetry, the unbalanced components also accelerate at ship in the $x$ direction, even though the force was applied in the $y$ direction.

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..

View attachment 276218

Check the drawing, the MC of a boat is always on or over the boat, (it is not a catamaran that is a special case of design), if the Cm is outside the boat it will lie down. Even if this does not happen, in a weak balance, accelerating the boat from the point of view of its occupants, has the same effect as changing the modulus and the direction of gravity (you can see the effect on the free surface of a glass of water). the resulting acceleration as a function of the acceleration with respect to the water $a_x$ and its angle $\theta$ remain
$$ a_R = \sqrt {g ^ 2 + a_x ^ 2} $$
$$ \tan \theta = \dfrac {g} {a_x} $$
In this way the direction of the resistive forces and the weight of the elements of the boat with respect to the MC changes, so the boat can capsize backwards if it accelerates against the position of the extra weight you have put on.

 

[John Mcrain]

Engines are same,have equal thrust...
Stick is is deep space,engine 1 is act at CM, engine 2 is turn OFF,stick travel with constant speed v.
In time t1 I turn OFF engine 1 and turn ON engine 2.

Will stick start rotating counterclock wise or it will keep going in straight line without rotation?

 

[A.T.]

Stick is is deep space,engine 1 is act at CM, engine 2 is turn OFF,stick travel with constant speed v.

If only rocket 2 is OFF, the stick will not travel with constant speed.

In time t1 I turn OFF engine 1 and turn ON engine 2.

Will stick start rotating counterclock wise

Yes.

 

[John Mcrain]

If only rocket 2 is OFF, the stick will not travel with constant speed.

I forget that in space there is no drag.
What then determine final speed if there is no drag and my engine has finite thrust?
stick accelarate to infinity?

 

[A.T.]

I forget that in space there is no drag.
What then determine final speed if there is no drag and my engine has finite thrust?
stick accelarate to infinity?

No, it will approach c.

 

[John Mcrain]

No, it will approach c.

¸What determine real rocket speed in space,how fast they move in space,if drag is zero?

 

[A.T.]

¸What determine real rocket speed in space,how fast they move in space,if drag is zero?

The speed of everything is determined by the reference frame you choose to describe its motion.

 

[John Mcrain]

The speed of everything is determined by the reference frame you choose to describe its motion.

imagine space is fill with fluid,what is speed in relation to fluid?similar as airspeed in planes

 

[A.T.]

imagine space is fill with fluid,what is speed in relation to fluid?

Look at:
https://en.wikipedia.org/wiki/Terminal_velocity
Just replace the constant gravity with your constant thrust.

 

[John Mcrain]

Look at:
https://en.wikipedia.org/wiki/Terminal_velocity
Just replace the constant gravity with your constant thrust.

But space don't have fluid,I just say fluid as example for reference point ..Fixed point in space

 

[A.T.]

I just say fluid as example for reference point ..Fixed point in space

There is no absolute rest, if that's what you mean.

 

[John Mcrain]

There is no absolute rest, if that's what you mean.

What frame use ussualy when people talk about rocket speed in space?

 

[John Mcrain]

There is no absolute rest, if that's what you mean.

I am in rocket when engine is turn off.I get out of rocket and rocket turn ON engine and going away from me..

What is rocket speed in relation to me?

 

[A.T.]

I am in rocket when engine is turn off.I get out of rocket and rocket turn ON engine and going away from me..

What is rocket speed in relation to me?

https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
https://en.wikipedia.org/wiki/Relativistic_rocket#Relativistic_rocket_equation

 

[jbriggs444]

Engines are same,have equal thrust...
Stick is is deep space,engine 1 is act at CM, engine 2 is turn OFF,stick travel with constant speed v.
In time t1 I turn OFF engine 1 and turn ON engine 2.

Will stick start rotating counterclock wise or it will keep going in straight line without rotation?

View attachment 276287

The stick will not rotate in this case. From the problem statement, we are given that the engine acts at the center of mass. That means that the engine is just a bit right of center on that huge massy blob. Just enough right so that things remain balanced.

Please do not present drawings that disagree with the accompanying text.

 

[jbriggs444]

What frame use ussualy when people talk about rocket speed in space?

We pick one that is convenient.

For a rocket on a sub-orbital trajectory aimed at a target, we might use the rotating frame in which the surface of the Earth is stationary.

For a rocket in orbit around the Earth, we might use the inertial frame in which the center of the Earth is stationary.

For a rocket making a trip to the moon, we might switch frames in the middle of the trip. [It would be foolhardy for Neil Armstrong to be using the earth-centered inertial frame when descending the last 30 meters to land on the moon].

For a rocket making a slingshot maneuver, we can shift frames and observe how a trajectory which conserves spacecraft mechanical (kinetic + potential) energy in one frame increases it in another. This might confuse us until we realize that energy is conserved but is not invariant.

 

[Dale]

Will stick start rotating counterclock wise or it will keep going in straight line without rotation?

Again, you need to specify the point of rotation you are asking about. “Will stick start rotating counterclock wise about the CG or it will keep going in straight line without rotation?” If you find that too cumbersome then the word “spin” refers to “rotation about the CG” in classical physics.

 

[jbriggs444]

Again, you need to specify the point of rotation you are asking about. “Will stick start rotating counterclock wise about the CG or it will keep going in straight line without rotation?” If you find that too cumbersome then the word “spin” refers to “rotation about the CG” in classical physics.

I am inclined to let this one pass.

The rate of change of orientation of a rigid object is an invariant. It does not depend on a choice of reference axis.

If we were asked about the rate of change of angular momentum, then the reference axis would [usually] matter.

 

[Dale]

I am inclined to let this one pass.

I am not. His questions are exceptionally sloppy. This is one topic that he has been specifically instructed about from the very first page. That this far into it he still is being sloppy indicates an unwillingness to learn or improve.