Undergrad Do objects always rotate around center of mass?

[user079622]

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

Left object rotate around P1, ball in g indicator set at c.g. will move out,
Right object rotate around around c.g. ball will stay in center.

This is what I want to say..

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.

Lets assume engine is mount on tail so it cant make any torque around c.g..
In time t1 plane start change direction and change orientation, in time t2 plane settle new direction but still change his orientation toward wind, in time t3 plane is settle his new orientation

Does com travel in straight line from time t2 to t3?

Or I can assume momentum of plane is so high 200tones x high speed, so change in direction due to gust of 60km/h is neglibile, so we can analyze/focus only change in orientation..

 

[Dale]

Does com travel in straight line from time t2 to t3?

Is there a net force from t2 to t3?

 

[user079622]

Is there a net force from t2 to t3?

Is possible situation that net force=0 but net moment is non zero due to gust, so plane still changing just orientation?
If yes, then yes net force is zero

 

[jbriggs444]

Left object rotate around P1, ball in g indicator set at c.g. will move out,
Right object rotate around around c.g. ball will stay in center.

This is what I want to say..

Let me make sure that I am clear about the drawing. We have two depictions of the same scenario. In both cases, the rod is rotating at the same rate in the counter-clockwise direction.

In the first case, we place the instantaneous cebter of rotation well out north (toward the top of the page) from the rod. The instantaneous axis of rotation is supposed to be stationary. That means that the frame of reference which you have chosen is one in which the rotating rod is moving rapidly to the east. You might imagine the axis along with its rest frame moving rapidly to the west.

In the second case, we place the instantaneous center of rotation in the center of the rod. That means that the frame of reference which you have chosen is one which moves along with the rod but without rotation. Viewed against this frame, the rod is rotating in place.With this understanding in hand, your point is that the two depictions of the scenario yield physically distinguishable results. In one case there is a force detectable by an accelerometer at the center of mass. In the other case there is not. You conclude from this that one depiction is unambiguously right and the other is unambiguously wrong.

This turns out to be incorrect.

The reason it is incorrect involves what we mean by an "instantaneous center of rotation". It is the center only for an instant. It is not permanent. At any later time, there may be a different point that is the new "instantaneous center of rotation".

Because of this, knowing one "instantaneous center of rotation" plus the rotation rate only allows you to know the velocity of every point on the rigid object. Not their future and past trajectories. Not the curvatures of those trajectories. Not any accelerations.

In the case of the drawing you have provided with the spinning rod moving eastward as it spins counterclockwise about an instantaneous center of rotation to the north, the situation would be like a wheel rolling along a wall to the north. The instantaneous center of rotation would always be due north of the center of mass. The accelerometer would read zero.

Just like an accelerometer on the axle of a wheel on a car driving down the highway.

 

[Dale]

Is possible situation that net force=0 but net moment is non zero due to gust, so plane still changing just orientation?
If yes, then yes net force is zero

If the net force is zero then the com moves in a straight line. If the net force is not zero then the com does not move on a straight line

 

[user079622]

If the net force is zero then the com moves in a straight line. If the net force is not zero then the com does not move on a straight line

When net force =0 must be net moment also zero?for my case

 

[jbriggs444]

When net force =0 must be net moment also zero?

No. The net momentum need not be zero. Nor must the net torque be zero.

 

[user079622]

No. The net momentum need not be zero. Nor must the net torque be zero.

I try to draw forces for my case net force=0 and net moment non zero, but I struggle.

 

[A.T.]

Left object rotate around P1, ball in g indicator set at c.g. will move out,
Right object rotate around around c.g. ball will stay in center.

This is what I want to say..View attachment 333162

What you seem to mean is a pivot point that remains at rest in some inertial reference frame. This is very different from purely kinematic centers of rotation, which don't care about the reference frame being inertial.

Also note that if a gust hits a plane, it's CoM will not be inertial, so it's not the pivot you are looking for.

 

[user079622]

it's CoM will not be inertial,

What does it mean?

so it's not the pivot you are looking for.

And what pivot I am looking for?

 

[user079622]

If I record with camera(connected with earth) plane from top view, plane passes under camera when crosswind hit him.
After I can slow down video, draw line along aircraft in software and find around which point I see rotation.

Let say for this example I find that he rotate around point P.
Yes point P will move on my screen as plane fly,
but distance from P to c.g. will be same all the time..

What is name of this point/axis of rotation?

 

[nasu]

Any point at fixed distance from the com satisfies this condition. The plane rotates around it (if there is any rotation) and keeps the same distance from COM. The tip of the pilot's nose will do, as long as he does not move from his seat.

 

[hutchphd]

I looked up your question on CK-12 and this is what I got:
Yes, objects always rotate around their center of mass. This is because the center of mass is the point at which all of the mass of the object can be considered to be concentrated for the purpose of calculating rotational motion.
I hope this helps!

This answer restates the question IMHO. It seems to imply that objects can be considered point masses for rotational purposes.......which is not true. What is CK-12?

 

[A.T.]

I can slow down video, draw line along aircraft in software and find around which point I see rotation.

What is name of this point/axis of rotation?

It's called "the point around which user079622 sees rotation".

 

[user079622]

It's called "the point around which user079622 sees rotation".

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

 

[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?