Undergrad Do objects always rotate around center of mass?

[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.

 

[Dale]

And at any instant you can find a reference point O so that VP→ is zero. This point is what they call "instantaneous center of rotation" or "instant center of rotation".

Ah, that is a valid definition. This point may be on or off the object, and it will be different in different reference frames.

 

[nasu]

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

I did not watch all the video. but he seems to talk about a dynamics problem. When it comes to momentum and KE, the decomposition into rotation and translation works only for the COM. But this has nothing to do with the discussion in this thread.

 

[nasu]

Ah, that is a valid definition. This point may be on or off the object, and it will be different in different reference frames.

Yes, it does not have on be on the object. I meant to say this but I forgot.
For a body in pure translation, it is somewhere "at infinite".

 

[user079622]

@Dale

He forgot in my video to calculate mxa in case where net force in non zero?

 

[jbriggs444]

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?

This changes the question. We are no longer talking about the instantaneous center of rotation. Instead, we are talking about the behavior of the stick.

Naturally, two different force patterns will produce two different results.

Possibly you have in mind that a stick with a perpendicular rocket on the end that burns for a very long time will settle down into an accelerating spin. Like a pinwheel. You expect that the limiting motion will be a stick spinning extremely rapidly while the stick as a whole drifts at a small velocity in a direction somewhat spinward of the initial thrust. I expect the same.

A drift velocity arises because the initial thrust goes in the same direction for a long time. As the stick speeds up, subsequent thrusts last for a shorter and shorter time in any given direction. Eventually, it is almost averaging out. I've not done the math to determine whether the drift velocity approaches a limit or increases without bound, but I think it is the sum of an inverse square series which will converge. [Back of the envelope says that the impulse from each half-rotation will scale as a simple inverse. This would seem to be a harmonic series -- divergent. But then we are subtracting out the opposite impulse from the next half-rotation. The difference between the two will be approximately an inverse square. So we have a convergent situation].

Now then, you wonder whether the center of rotation will be at the center of mass.

Yes -- nearly so anyway. But only because we have naturally chosen a frame of reference where the stick's center of mass is motionless.

It is only "nearly so" as long as the rocket is still firing. Because it is firing, the center of mass is accelerating. It is moving in a spiral trajectory about a point that I would consider to be a better choice for a reference axis.

 

[A.T.]

But is instant axis of rotation through CoM?

No, because the instantaneous velocity of the CoM is not zero in the ground frame. Of course you can choose another frame as @jbriggs444 noted.

These are two separate choices:
- You can choose a frame of reference, which determines the instantaneous center of rotation
- For every chosen frame of reference you can choose a rotation center, which determines the decomposition into rotation and translation

 

[user079622]

This changes the question. We are no longer talking about the instantaneous center of rotation. Instead, we are talking about the behavior of the stick.

Naturally, two different force patterns will produce two different results.

In free space there is no resistance so one hit or const. force will get same result?

when plane is hit by wind, this is not equlibrium case so moment in all points are not the same.

 

[user079622]

No, because the instantaneous velocity of the CoM is not zero in the ground frame.

If plane rotate around com ,then from ground frame com travel in straight line. If plane rotate around any other point ,com will not travel in straight line.

I dont know how to express myself, I know when I spin a ball or frisbee, my eyes see that axis of rotation goes through center, I dont know how you call it, maybe I can call it "apparent pivot point"...

 

[A.T.]

I dont know how to express myself...

What is your point? Do you still have questions? Did you read the other thread that I linked?
https://www.physicsforums.com/threa...ays-rotate-about-their-centre-of-mass.990571/

 

[user079622]

What is your point? Do you still have questions? Did you read the other thread that I linked?
https://www.physicsforums.com/threa...ays-rotate-about-their-centre-of-mass.990571/

I didnt I will

 

[user079622]

Yes, that is all good

A.T. say it is not good because c.g. has velocity for ground frame

 

[jbriggs444]

In free space there is no resistance so one hit or const. force will get same result?

If the object is rotating or otherwise moving under that "constant" force, then you will have to specify details about how the force changes to match. It will not usually remain constant in direction, magnitude and point of application when the point of application is moving.

 

[Dale]

If plane rotate around com ,then from ground frame com travel in straight line. If plane rotate around any other point ,com will not travel in straight line.

If a rigid object rotates with angular velocity $\omega$ about one axis then it will rotate with angular velocity $\omega$ about any other axis.

The com travels in a straight line if there is no net force, regardless of rotation.

 

[user079622]

If a rigid object rotates with angular velocity $\omega$ about one axis then it will rotate with angular velocity $\omega$ about any other axis.

The com travels in a straight line if there is no net force, regardless of rotation.

Object rotate around com, but com translate in straight line.
Camera is ground frame.

 

[Dale]

Object rotate around com, but com translate in straight line.

Yes. The object also rotates around other points besides the com, but the other points do not translate in a straight line. This was covered also in the thread that @A.T. has linked

 

[user079622]

Yes. The object also rotates around other points besides the com, but the other points do not translate in a straight line. This was covered also in the thread that @A.T. has linked

Mathematicaly yes, phyiscly not, com is only position on object where ball inside "centrifugal device"(G-indicator) will stay at the middle.

So if airplane really rotate around axis through c.g. when gust hit him, ball inside g-indicator that is placed at c.g. must stay at the middle.
If plane rotate around any other axis, ball will move to side due to centrifugal force.

But ball will move to side due to downwind translation, because wind accelerate plane slightly downwind, later thurst from engine stop this movement.

Do you agree?

So basically my question is; will plane c.g. move in straight line during plane change his orientation when gust of wind hit him?

 

[nasu]

Mathematicaly yes, phyiscly not, com is only position on object where ball inside "centrifugal device"(G-indicator) will stay at the middle.So basically my question is; will plane c.g. move in straight line during plane change his orientation when gust of wind hit him?

Your first sentence is logically faulty. The special dynamic properties of the COM are completely irelevant to the discussion about the kinematics of the rotationg bodies. The motion of points on the object is what it is. There is no difference between "matematically" and "physically". They either move in a specific way or they don't.

You keep going around in circles (or more like on tangents) where there is nothing to argue about. There is no doubt that objects may and do rotate around the COM. Nobody denied this. But rotation is not an exclusivist club. Rotation around one point does not exclude rotation about any other point. It's not and either/or situation. Do you have same problem with just motion? If the moon moves relative to the Earth will you deny that it also moves relative to Jupiter or the Sun? If someone says that the Moon moves relative to the Sun will you say, "Hey, this should be wrong because Wiki says that it moves relative to the Earth!"?If the plane was already moving before the wind hits, the trajectory of the COM will follow a curve and not a straight line. It si the same as with a rock thrown horizontally and being deviated by gravity.
There is the special case where the plane's COM was already moving in a straight line and the wind direction is exactly along the straight line. But then there will be no change in orientation either.

 

[user079622]

If the plane was already moving before the wind hits the trajectory of the COM will follow a curve and not a straight line. It si the same as with a rock thrown horizontally and being deviated by gravity.
There is the special case where the plane's COM was already moving in a straight line and the wind direction is exactly along the straight line. But then there will be no change in orientation either.

Yes I agree with this, in nature there is not sharp turn, so when plane change his direction that must be done in curve.

But I focus at c.g. only at rotation phase, when he change his orientation into wind...

 

[Dale]

com is only position on object where ball inside "centrifugal device"(G-indicator) will stay at the middle.

That is true regardless of which point you consider the rotation axis.

So if airplane really rotate around axis through c.g. when gust hit him, ball inside g-indicator that is placed at c.g. must stay at the middle.
If plane rotate around any other axis, ball will move to side due to centrifugal force.

This is simply not true. The choice of rotation axis does not alter the motion of the ball.

So basically my question is; will plane c.g. move in straight line during plane change his orientation when gust of wind hit him?

No. The net force is not zero, so there will be acceleration. The motion of the com will therefore not be a straight line at a constant speed.

 

[nasu]

Yes I agree with this, in nature there is not sharp turn, so when plane change his direction that must be done in curve.

This has nothing to do with sharp turns. A gust of wind will not cancell the 500 mph speed of the airplane (including the COM). What happens with a point object moving in a straight line at 500 mph when a force start acting perpendicular to the initial trajectory? Does it move in a straight line? Does it move with constant velocity? If you understand the motion of point like objects, you should understand the motion of the COM. If not, maybe you need to review motion of point-like objects under the action of various forces, before looking at motion of rigid bodies.