Why does an object rotate about its center of mass when thrown in the air?

[fisico30]

Hello everyone. Take an object in you hands. Throw it in the air. While launching it, your hands will push on the object in different points of it.
The object will ALWAYs rotate, free in the air, about an axis that passes for the center of mass.

Does anybody know why? I found that in that case there is stability.
Does the axis need to be one the principal axes or can it be any axis?
thanks!

 

[Doc Al]

The object will ALWAYs rotate, free in the air, about an axis that passes for the center of mass.

In general, the instantaneous axis of rotation can be anywhere, not necessarily passing through the center of mass. What is true, and maybe this is what you are thinking of, is that any arbitrary motion of a rigid body can be viewed as a rotation about the center of mass plus the translation of the center of mass.

Does anybody know why? I found that in that case there is stability.
Does the axis need to be one the principal axes or can it be any axis?

If you can start the object rotating about a principal axis, then the rotation will be stable about that axis.

 

[fisico30]

Thanks Doc Al.
I found a post somewhere else(allexperts.com) that goes:"Question
I can understand why an object accelerates in the direction of the force when subjected to it , and i can understand it rotating due to a torque. What i want to know is that , Why should the body (in the situation i mentioned); when subjected to a force not passing through its COM ; CHOOSE,so as to speak, to rotate about an axis passing through the COM only? Why not some other axis?
While thinking about this ques. i realized that out of all the infinite Moment of Inertias a body can have , the least one is would surely be about an axis through the COM. But why this should be the reason i seek,if at all, I have no clue.Here i hoped that u could shed some light."

The answer was:"ANYWAY, the reason that objects spin only around their CoM is because any attempt to spin around somewhere other than the CoM is inherently unstable. If an object tried to rotate about an axis other than one passing through its CoM, it would get so out of equilibrium that it would rapidly adjust its rotation to one passing through its CoM.

The following experiment (which you can do either for real or in your head) should show what happens. Take two buckets, fill them with water, and then hold one in each hand. Now start to spin yourself with enough speed that the buckets move away from body. You'll notice that you can do this indefinitely (or at least until you get dizzy); your spinning is stable.
Now let go of one of the buckets. Your spinning will immediately get very UNstable, because your CoM has changed so much that you can no longer maintain a spin around the axis that USED to be your CoM, but no longer is. The only way to return to a stable spin is to lean away from the remaining bucket of water; ie, to adjust your CoM so that your axis of spin again goes through your CoM.

I can't go into more detail about WHY such an attempted spin (ie, outside the CoM) is unstable, without introducing the concept of pseudo-forces within a rotating frame. But the above experiment should show you that attempting to spin around an axis outside the CoM is unstable".

 

[Doc Al]

That answer sounds reasonable to me. Any attempt to maintain a constant rotation about a non-principal axis requires a torque and thus is not stable. (And every principal axis passes through the center of mass, of course.)

 

[tiny-tim]

instantaneous axis of rotation

Hello everyone. Take an object in you hands. Throw it in the air. While launching it, your hands will push on the object in different points of it.
The object will ALWAYs rotate, free in the air, about an axis that passes for the center of mass.

Does anybody know why?

Hi fisico30!

(passes through the center of mass )

The instantaneous axis of rotation is the line all points of which are instantaneously stationary.

Its position relative to the object depends on the velocity of the object relative to the observer (it often lies well outside the object… the Moon's passes through the Earth ).

We normally prefer to choose an observer (a frame) with the same velocity as the centre of mass.

In that frame, if there is no force acting on the object, the centre of mass is instantaneously stationary, and so by definition must be on the axis of rotation.

So an object experiencing a torque force but no translational force will rotate about an axis through its centre of mass, relative to a frame with the same velocity as the object.

(stability is more complicated … )

 

[Topher925]

The way I like to think of it is that the sum of all centrifugal forces must equal zero, or else the object will not be going strait up. In math lingo thats:

\sum\frac{mV²}{r} = 0

EDIT: Latex doesn't want to work for this but you get the idea.

 

[atyy]

Would the parallel axis theorem or Chasles theorem be relevant here?

 

[Topher925]

Would the parallel axis theorem or Chasles theorem be relevant here?

I don't think so. Parallel axis relates to moments of inertia with regard to structural behavior, and Chasles theorem is just another way to evaluate kinematics (note, not kinetics) of rigid bodies in place of Eulers method. Neither of which really relate to determining centroids or principle axes.

 

[fisico30]

Cool answer Tiny-tim. I get why the axis need to pass through the CM now.
But do you know if that rotation axis depends on the initial launching conditions?
Once in the air, will the object continue to rotate around that particular axis or change to one of the principal axes, or change randomly to new axes?
thanks!

 

[tiny-tim]

my head's spinning …

Thanks fisico30!

I get very confused about that …

I vaguely recall that you can prove that the rotation axis of a freely rotating body gradually "precesses", like a spinning-top (or like the Earth's does every 25,800 years! ) …

and I know there's a straightforward explanation …

but I can't think what it is!

 

[fisico30]

Hello Everyone,
the topic/answer is found in one the posts on this website. Just type torque-free precession. I will study it and get back with a summary of what happens. Thank tiny-tim for the keyword "precession"

 

[fisico30]

So, in free torque precession, the angular momentum vector direction stays fixed, and the axis of rotation precess around the angular momentum vector itself.
This situation would happen if the initial launching spin was not one of the principal(symmetry) axes, but any other axis(which can be projected along the three principal axes). The situation can be thought of as all three principal axis rotating simultaneously around the angular momentum vector...
It is same thing as saying that the object is given a spin along more than one axis of principal direction.
The math is interesting... I will post it soon!