Calculate center of rotation of monocopter

[user079622]

Homework Statement

I stuck at beginning ,due to wrong reasoning. I need better explanation.

Relevant Equations

F=ma
M=Fxd
a=ω2r

I want to calculate center of rotation for monocopter, from inertial reference frame,earth.
One case when fly in the air in one steady height/place and second case is in free space,then has rocket engine instead fan.
I draw forces what I think that must exist.

m=2kg
Ft(thurst)= 100N
ω=500RPM

Can I find center with this data or I need more informations?

Axioms:

1. A pure force thorugh the center of mass (with no net torque about the center of mass) will purely translate a rigid body (any point on the body).
2. A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of mass.

Space case
(I am in space station near the stick wich is at rest, then I start rocket engine of stick and observing, neglect loss in fuel mass)
I have only one force and mass, so I know Fnet must be non zero.I know that CoM will not accelerate in straight line, because force dont going through CoM.
I also know that object cant rotate around CoM because of Fnet is non zero .
So Fnet is non zero implies, that CoM must orbit around some point but I don have centripetal force for this movement

I am stuck in my contradictions, obviously my reasoning is wrong because physics laws must be true.
What I am doing wrong?

 

[haruspex]

It is a somewhat complex dynamic system. The misalignment between the lift and mg leads to a torque, but the blade acts as a gyroscope, so the torque does not tip the system over in the obvious way.

 

[user079622]

It is a somewhat complex dynamic system. The misalignment between the lift and mg leads to a torque, but the blade acts as a gyroscope, so the torque does not tip the system over in the obvious way.

Yes, it is complex, gyro effect stop this moment. In this task I will not analyze why this moment not roll over monocopter.
Maybe is better to focus on first case in space.
What is your opinion for this case?

 

[haruspex]

Maybe is better to focus on first case in space.

Isn't it the same, just without the drag?

 

[user079622]

Isn't it the same, just without the drag?

Yes only drag.

Maybe is this case even easier.
I can delete left drag arrow and mathematicaly show that as shift in drag at wing to the right,
when drag equal thrust Fnet is zero, and we have pure rotation around c.g., zero translation.
So we manage to put center of rotation at c.g. that mean monocopter stay in place stable.

What do you think?

 

[nasu]

If the net force is zero is the acceleration of the COM that is zero, not its velocity. The center of mass may be moving with constant velocity. It could also be at rest or maybe the whole body is at rest at that instant so there is no rotation at all.

If you want to find the instantaneous center of rotation, you need to know how the object moves at that instant. You don't need the forces. The forces determine the future motion of the object not the motion at this instant.

 

[user079622]

If the net force is zero is the acceleration of the COM that is zero, not its velocity. The center of mass may be moving with constant velocity. It could also be at rest or maybe the whole body is at rest at that instant so there is no rotation at all.

If you want to find the instantaneous center of rotation, you need to know how the object moves at that instant. You don't need the forces. The forces determine the future motion of the object not the motion at this instant.

In space case, stick is at rest(not rotate, not translate) in relation to space station where I am observing.
In t1 I start rocket engine.

On earth everything the same , in t1 I start fan.

 

[nasu]

Then what is the point of your post? There is no center of rotation at that time, as described. The forces are irelevant.

 

[user079622]

If you want to find the instantaneous center of rotation, you need to know how the object moves at that instant. You don't need the forces.

But I cant know how will object move without forces..
I dont understand you

 

[nasu]

But you just said that is at rest. So you know. It is not moving, actually. For this situation there is no center of rotation, to speak of. If you want the center of rotation at a different time, you need to know the kinematic situation at that time. Not the forces but the distribution of velocities at that time.

For your last example, with zero net force, if the object was at rest to start with, the center of mass will remain at rest. If it was moving already, it will keep moving with constant velocity.
In the first case, the COM being at rest it will be the instantaneous center of rotation and it will be so as long as the net force is zero. For the second case (moving with constant velocity) the COM will not be the instantaneous center of rotation (ICR). The position of the ICR will be a function of time.

 

[user079622]

Left trajectory is how my case in space will look like.
As RPM increase, linear acceleration of CoM decrease, so point around which CoM orbit become smaller and smaller.
Or I can say intuitively when RPM increase force vector point in each dircetion, so stick going toward pure rotation.
Will it end up in pure rotation around CoM or CoM will orbit around center of smallest circle?

 

[nasu]

What is this last post about?

 

[user079622]

What is this last post about?

Fersnel integrals

 

[nasu]

How is this relevant to this thread?

 

[user079622]

How is this relevant to this thread?

Like this?

Space case
c=Icm/md=k^2/d

Com trajectory is Euler spiral, during path angle a is reducing, in final stage CoM rotate around point c, rod is tangent to that circle.

Angle a decrease

 

[nasu]

I thought that the thread is about the center of rotation.

 

[Lnewqban]

@user079622 , why does free rotation interest you so much?
Why do you choose this complex flying machine for your analysis?

When an irregular object is forced to simultaneously rotate on a vertical plane and fly, like a thrown knife or ax, it must experience an initial combined torque (for rotation) and pushing force (for flight trajectory).
That combined force is achieved by the hand holding the tool far from its center of mass, as you can see in the following video:





The same applies to a free horizontal rotation and rectilinear translation while supported on a horizontal low friction surface.

 

[user079622]

I thought that the thread is about the center of rotation.

My graph is wrong, c is not center or rotation. I made mistake.
I am giving up on this task, it looks trivial (one rod with just one force) but it turn out it is damn complex.