Body rotating about its center of mass

[fog37]

Hello,

As many know, when an external force acts on a rigid body and the force's line of action does not pass through the body's center of mass $c.m.$, the force will cause the body to both translate and rotate exactly about the $c.m.$. Otherwise, the body will solely translate without rotation.

How is this important result proven? How can we derived this result? I have been looking on several books but none mathematically justify this conclusion...

Thank you!

 

[mfb]

the force will cause the body to both translate and rotate exactly about the c.m

This is an arbitrary point of view. You can also let it rotate around any other point. If you do that its translation velocity will change over time (as the rotation now carries linear momentum which depends on the phase of the rotation), something that is usually unwanted. A rotation around the center of mass carries zero net momentum (you can derive this), which means the translation stays the same over time without external forces.

 

[phyzguy]

I think many books spell this out. I am familiar with Goldstein "Classical Mechanics", Chapter 4. This Wikipedia page also has some information. The idea is that after applying all of the constraints imposed by the fact that it is a "rigid body", we are left with only 6 degrees of freedom, which we break down into motion of the center of mass and rotation of the body about the center of mass. In this sense I think we define the rotation of the body to be about the center of mass, because then it makes the motion simpler to analyze.

 

[fog37]

thanks phyzguy and mfb.

I got a hold of Goldstein's book and looked at chapter 4. The book is advanced. I am slowly grasping the meaning of your replies. I understand that the rotation of an object is not an absolute concept. For example, the angular velocity vector is a free vector.

We start with a rigid body and a triad of unit vectors ( $\hat{i},\hat{j},\hat{k}$) that moves with the rigid body itself. The triad has an origin $\overrightarrow{O}=(x_O, y_O,z_O)$ that can be located at any arbitrary point in space. The angular velocity vector $\overrightarrow{\omega}$ is a free vector given by $$\omega_x = \frac {d \hat{j} }{dt } \cdot \hat{k}$$ $$\omega_y = \frac {d \hat{k} }{dt } \cdot \hat{i}$$ $$\omega_z = \frac {d \hat{i}}{dt } \cdot \hat{j}$$

• An arbitrary point on the rigid body has a position $\overrightarrow{P}=(x_P, y_P, z_P)$ relative to the external, fixed frame of reference (not relative to the triad).
• The triad has an origin $\overrightarrow{O} =(x_O, y_O, z_O)$ relative to the external, fixed frame of reference.
• The triad's origin $O$ has velocity $\overrightarrow{v_0}$ relative to the external, fixed frame of reference.
• An arbitrary point on the rigid body has a velocity $\overrightarrow{v_P}$ relative to the external, fixed frame of reference:

$$ \overrightarrow{v_P}= \overrightarrow{v_0}+ \overrightarrow{\omega} (\overrightarrow{P}-\overrightarrow{O})$$
So, it is clear that the arbitrary point instantaneous velocity $\overrightarrow{v_P}$ depends on the choice of the triad's origin $\overrightarrow{O}$ and the point rotates about an axis parallel to the vector $\overrightarrow{\omega}$ and passing this arbitrarily chosen point $\overrightarrow{O}$ with velocity $\overrightarrow{\omega} (\overrightarrow{P}-\overrightarrow{O})$.

So, this means that there is no absolute rotation axis for an object that is physically rotating. Therefore, when we throw an object in the air and the object is said to "rotate" about the center of mass c.m. in a plane perpendicular to the ground, the rotation is not physically happening about the center of mass as much as it is happening about any other arbitrary point $\overrightarrow{O} \neq c.m.$ as long as the point $\overrightarrow{O}$ is fixed (moves with the rigid body so an observer located at $\overrightarrow{O}$sees all the rigid body's point having zero velocity) with the rigid body (the point can be internal or even external to the rigid body).

In general, the point $\overrightarrow{O}$ is suitably chosen to be equal to the center of mass $c.m.$ for mathematical convenience.
Could you explain this mathematical convenience in more detail again? Does the total kinetic energy $KE$ of the rigid body have a simpler expression when $\overrightarrow{O} = c.m.$? What about the angular momentum expression for the rigid body?
Thanks!

 

[mfb]

Does the total kinetic energy $KE$ of the rigid body have a simpler expression when $\overrightarrow{O} = c.m.$? What about the angular momentum expression for the rigid body?

Yes, the total kinetic energy is a simple sum of the energy of the rotation and the translational kinetic energy - but only if you consider the rotation around the center of mass.

 

[fog37]

Thanks. So this whole concept of rotation is not absolute, i.e. the angular velocity ${\omega}$, etc. but when we see an object rotating it really seems to visually rotate about a specific axis of rotation. Why do we get that impression if the rotation can be defined about any arbitrary axis parallel with ${\omega}$?

 

[A.T.]

... but when we see an object rotating it really seems to visually rotate about a specific axis of rotation...

The brain often interprets motion subjectively.

 

[mfb]

The axis going through the center of mass is the only axis that doesn't change its velocity over time (or, if the rotation is more complex: The center of mass is the only point that doesn't change it).

 

[wrobel]

As many know, when an external force acts on a rigid body and the force's line of action does not pass through the body's center of mass c.m.c.m., the force will cause the body to both translate and rotate exactly about the c.m.c.m.. Otherwise, the body will solely translate without rotation.

No forces act on the body:

 

[Omega0]

No forces act on the body:

How do you come to the wrong conclusion that the movement seen is without external forces? It is obvious that the movement is influenced by outer sources. You shouldn't forget that there are higher order terms in the solution of the laplace equation for the gravitational field, see Phobos and Deimos, Mars. Gravitation is not = 1/r
Having said this, I am not sure if the experiments have been conducted without any magnetic field.

 

[wrobel]

How do you come to the wrong conclusion that the movement seen is without external forces? It is obvious that the movement is influenced by outer sources. You shouldn't forget that there are higher order terms in the solution of the laplace equation for the gravitational field, see Phobos and Deimos, Mars. Gravitation is not = 1/r
Having said this, I am not sure if the experiments have been conducted without any magnetic field.

The things are much simpler than you think:)
Do you know what the Euler top is? The rotations about the axes with max and min moment of inertia are stable while about the middle axis is unstable

 

[A.T.]

How do you come to the wrong conclusion that the movement seen is without external forces? It is obvious that the movement is influenced by outer sources. You shouldn't forget that there are higher order terms in the solution of the laplace equation for the gravitational field, see Phobos and Deimos, Mars. Gravitation is not = 1/r
Having said this, I am not sure if the experiments have been conducted without any magnetic field.

https://en.wikipedia.org/wiki/Tennis_racket_theorem

 

[Omega0]

The things are much simpler than you think:)
Do you know what the Euler top is? The rotations about the axes with max and min moment of inertia are stable while about the middle axis is unstable

I asked to delete my reply. It was a little bit distracting, perhaps... but how does it works that the unstable system is unstable?

 

[A.T.]

. but how does it works that the unstable system is unstable?

 

[Omega0]

The things are much simpler than you think:)
Do you know what the Euler top is? The rotations about the axes with max and min moment of inertia are stable while about the middle axis is unstable

I didn't now. Sorry. I never saw this effect before. It leads me to new questions. Thanks.