Do objects always rotate around center of mass?

[A.T.]

but distance from P to c.g. will be same all the time..

If you look only at the distance from P to c.g., then you are talking about translation of the c.g., not about rotation, because the orientation of the plane plays no role in that.

 

[Dale]

draw line along aircraft in software and find around which point I see rotation.
…
What is name of this point/axis of rotation?

How do you find this point? If you literally find it just by looking and choosing a point then the proper word is “arbitrary”

 

[user079622]

How do you find this point? If you literally find it just by looking and choosing a point then the proper word is “arbitrary”

I know where is plane c.g. located, if I stop and pused video you can draw path of c.g. on the screen.
I draw line that connect c.g. and nose, in front of plane. Arrow is nose.
I think this 3 "rotation" is physically possible ?
Top= c.g. path is straight line then I know rotation is around c.g.
middle =c.g. path is curve ,line is tangent to curve path, plane "revolve" around point P and rotate around his c.g. (hm like a moon)
bottom= c.g. path is curve but line(nose) pointing into center of rotation P, he revolve around point P and rotate around his c.g.(like moon)

In last two cases he need centripetal force to keep him around point P

Hmm it seems he always rotate around his c.g.

 

[Dale]

I know where is plane c.g. located, if I stop and pused video you can draw path of c.g. on the screen.
I draw line that connect c.g. and nose, in front of plane. Arrow is nose.
I think this 3 "rotation" is physically possible ?

Yes.

Top= c.g. path is straight line then I know rotation is around c.g.

Yes, but there is also rotation around every other parallel axis.

line is tangent to curve path, plane "revolve" around point P and rotate around his c.g.

I think you are using “revolve” to refer to the instantaneous center as defined above. If so then you probably should mention it for the first case too. The com is not the instantaneous center in any of these cases.

Hmm it seems he always rotate around his c.g.

Sure. But it also always rotates around every other point too.

 

[user079622]

I think you are using “revolve” to refer to the instantaneous center as defined above. If so then you probably should mention it for the first case too. The com is not the instantaneous center in any of these cases.

If I didn't forget that moon rotate around itself I would never start this topic.
I realize this when draw this graph...

Do you think my top case is possible for airplane?

Sure. But it also always rotates around every other point too.

But revolve only around point P?

 

[Dale]

But revolve only around point P?

Yes. The instantaneous center of rotation is unique in a given inertial reference frame.

 

[user079622]

@Dale

Moon revolve in 27days/1 revolution and rotate 27/1 rotation

That mean if want to calculate centrifugal force at surface of moon, due to revolution or due to rotation, I must get same result?

 

[Dale]

@Dale

Moon revolve in 27days/1 revolution and rotate 27/1 rotation

That mean if want to calculate centrifugal force at surface of moon, due to revolution or due to rotation, I must get same result?

No. The centrifugal force also depends on the radius.

 

[user079622]

No. The centrifugal force also depends on the radius.

Yes I know it depend on radius.

Men that stand on moon dark side(outside),feel centrifugal force due to moon rotation and due to moon revolution.

Do we have to add this two results to get correct number?

 

[A.T.]

feel centrifugal force due to moon rotation

You don feel the centrifugal force. It's an inertial force that is determined by the choice of reference frame, not by how objects move.

 

[user079622]

You don feel the centrifugal force.

This is not important in my question.lets call it centripetal then

 

[jbriggs444]

So this is not that instant center? When I stop video at that time this is instant point around plane rotate..

Let us back up and define the "instantaneous center of rotation".

If you pick a frame of reference then every point on a rigid object will have some velocity as measured against that frame of reference.

If the object is not rotating then every point will have the same velocity. There is no rotation and no center of rotation. Some people would say that the center of rotation is somewhere at infinity in this case. Conversely, if every point on the object shares the same velocity then there is no rotation. But we do not care about the no rotation case. We want to talk about rotation.

If the object is rotating then the points on the object will have different velocities. Conversely, if the points on the rigid object have different velocities than there is rotation.

For a rigid body, those velocities will be related to one another in a strict pattern. For instance, if we are looking downward on an object rotating counter-clockwise then points farther east will have larger northward velocities than points farther west. Similarly, points farther north will have larger westward velocities.

We can pretend that we have a wire frame attached to the object that spans as far as it needs to.

If we hunt around, it is clear that we can find a line of points on the wire frame that are just right in the east-west direction to have a zero north-south velocity. Along this line we can find a point on the wire frame that is just right in the north-south direction to have a zero east-west velocity. We have discovered a point that, at this instant, has zero velocity relative to our chosen frame of reference.

This momentarily motionless point on the wire frame is the instantaneous center of rotation of the object against this reference frame.

 

[A.T.]

This is not important in my question. lets call it centripetal then

Then your question still makes no sense.

 

[user079622]

stick in free space,circle is com,all forces are equal in magnitude and act all the time, they are rocket engines connected to stick.

Top= pure rotation in clockwise direction around com, com zero translation
middle=net force is non zero so we have acceleration,stick rotate around point P like I draw at the right side? in com we have force=ma in opposite direction from right force ?
bottom=net force is zero so com dont accelerate in any direction, stick rotate clockwise around com because right force has longer lever arm?

If middle case has just impuls force ,hit with hammer instead const force ,com will translate up at my graph, and stick will rotate around com.

 

[user079622]

The first theorem also tell us that the rotation of a body will always be around its CM when there are no external forces applied, because in this case it will be $$0=\sum F=ma_{CM}\Rightarrow a_{CM}=0$$. If the rotation was around another point , then the CM would also rotate around that point which would mean that $a_{CM}\neq 0$, contradicting that $a_{CM}=0$.

But that mean object can not rotate around CM if net force≠0Moon or car in any turn has net force≠0, $a_{CM}\neq 0$ and CM "rotate" around point(earth) and around itslef(CM), so this statement is not 100% correct?

 

[A.T.]

stick in free space,circle is com,all forces are equal in magnitude and act all the time, they are rocket engines connected to stick.

Top= pure rotation in clockwise direction around com, com zero translation
middle=net force is non zero so we have acceleration,stick rotate around point P like I draw at the right side? in com we have force=ma in opposite direction from right force ?
bottom=net force is zero so com dont accelerate in any direction, stick rotate clockwise around com because right force has longer lever arm?View attachment 333245If middle case has just impuls force ,hit with hammer instead const force ,com will translate up at my graph, and stick will rotate around com.

top & bottom are OK, if you are asking about the instantaneous center of rotation in the initial rest frame of the stick (or the center of rotation that results in zero translation in that frame).

middle: the picture on the right is wrong. There is no centripetal force to create that circular motion.

 

[user079622]

top & bottom are OK, if you are asking about the instantaneous center of rotation in the initial rest frame of the stick (or the center of rotation that results in zero translation in that frame).

middle: the picture on the right is wrong. There is no centripetal force to create that circular motion.

I dont know what to do with middle case, net force is non zero that mean com must accelerate, it cant move in circle because we dont have centripetal force for do this, but how will com accelerate in straight line if force act all the time on stick, force change orientation all the time as stick rotate..
?

 

[Lnewqban]

I dont know what to do with middle case, net force is non zero that mean com must accelerate, it cant move in circle because we dont have centripetal force for do this, but how will com accelerate in straight line if force act all the time on stick, force change orientation all the time as stick rotate..
?

 

[user079622]

@Lnewqban

Hhahaha nice one!

So what is your answer?
It cant just rotate around com, because net force non zero, must somewhere accelerate

 

[Lnewqban]

That video is m

@Lnewqban

Hhahaha nice one!

So what is your answer?
It cant just rotate around com, because net force non zero, must somewhere accelerate

That video is my answer.
Yest, it can and it does.

Any other point of the machine describes a circle at the same time that rotates as much as the CM does.
That is why others have explained that not only the CM rotates.

The only reason the thing is landing some distance from where launched is that the plane of rotation was not perfectly vertical.
There was a little component of the tilted lift force that was forcing the CM to move sideways.

There is initial rotational acceleration, but only until the drag force equals the thrust of the propeller.
After that point, the rotational velocity is constant.
Then, that thrust decreases when the engine runs out of fuel, and sufficient lift is gone.

One question: Was your original question general or based on flying objects?

 

[user079622]

That video is m

That video is my answer.
Yest, it can and it does.One question: Was your original question general or based on flying objects?

I have lots of questions, every post other question fall out...

But how my middle case can do pure rotation if I have just one force?

 

[Lnewqban]

I have lots of questions, every post other question fall out...

But how my middle case can do pure rotation if I have just one force?

I have edited my previous post, sorry.
What about the inertia (ma) or natural resistance of the CM to be moved in the same direction in which the off-center force is initially applied?

In the video, the man-stick is helping to prevent the CM from any initial horizontal movement during the first instants.

 

[user079622]

What about the inertia (ma) or natural resistance of the CM to be moved in the same direction in which the off-center force is initially applied?

I dont know ,but I think one force(even it is continuos) cant cause pure rotation.

 

[Lnewqban]

I dont know ,but I think one force(even it is continuos) cant cause pure rotation.

 

[user079622]

Yes, that is all good

Monocopter prove that there is no translation of c.g. he rotate around c.g. like it has couple forces(two engines)
How can one continuos force only rotate him around c.g. if net force is non zero?

 

[vanhees71]

Around which point a rigid body rotates is up to your choice, and this choice must be well thought about. For a free rigid body (or a body freely falling in the constant gravitational field of the Earth) the choice of the center of mass of that body is the best choice though. For details, see

https://itp.uni-frankfurt.de/~hees/pf-faq/spinning-top.pdf

 

[pbuk]

Monocopter prove that there is no translation of c.g. he rotate around c.g

No it doesn't: you do not know where the center of mass is.

he rotate around c.g. like it has couple forces(two engines)

There is a couple of forces: what stops the monocopter from rotating faster and faster?

However the only way to find the instantaneous center of rotation is to draw a free body diagram containing all the forces and calculate it. You will need to know where the centre of mass is, and also the centre of lift which will vary according to the rate of rotation and the position of the centre of rotation (although if the 'copter's motion is in equilibrium there is something you know about the position of the centre of lift relative to the centre of mass).

This thread is just going in circles (pun intended), if you want to understand complicated motion you need to do the calculations (which start with a free body diagram), not just keep making guesses.

 

[user079622]

No it doesn't: you do not know where the center of mass is.
There is a couple of forces: what stops the monocopter from rotating faster and faster?

Aerodynamic drag force stop to going faster.
Don understand how they set center of rotation into CoM with just one force?
Maybe with shifting the drag force

At 5:20 he must put it on his c.g.

n (although if the 'copter's motion is in equilibrium there is something you know about the position of the centre of lift relative to the centre of mass).

I think gyro effect stop "roll" because here lift force is fore sure not align with c.g.

This thread is just going in circles (pun intended), if you want to understand complicated motion you need to do the calculations (which start with a free body diagram), not just keep making guesses.

I agree with you, I find I better learn if solve tasks then just talk about theory.
In future I will switch to homework questions.

 

[Dale]

Monocopter prove that there is no translation of c.g. he rotate around c.g. like it has couple forces(two engines)
How can one continuos force only rotate him around c.g. if net force is non zero?

I don’t know what a monocopter is, but it for sure doesn’t break the laws of physics as they have been taught to you here. And in this context “prove” would require detailed careful measurements and detailed careful math. Not vague handwaving assertions.

 

[user079622]

@Dale @A.T. @jbriggs444

So what is answer for my bottom case from post #119?

 

[Dale]

@Dale @A.T. @jbriggs444

So what is answer for my bottom case from post #119?

See post #121

 

[Dale]

For a free rigid body (or a body freely falling in the constant gravitational field of the Earth) the choice of the center of mass of that body is the best choice

@user079622 I think that this is the best summary. You can choose any point you like, the math does not constrain your choice. But for a free body the com is usually the best choice and for a body attached to a hinge or an axis the hinge or axis is usually the best point.

 

[user079622]

See post #121

Sorry I made mistake, I mean middle case, other cases are solved...

 

[Dale]

Sorry I made mistake, I mean middle case, other cases are solved...

You will have to solve that one mathematically. It doesn’t have an obvious simple motion. That is the whole reason that the math was developed.

 

[A.T.]

Don understand how they set center of rotation into CoM with just one force?

The video says the exact opposite of what you claim. The center of rotation is not the CoM when airborne (at 5:33):