# Looking for proof that a free rigid body will rotate about its CM

1. Dec 15, 2009

### scoomer

Can anyone recommend a text book or web paper that proves that, in general, a free rigid body will act in the following way when the body is subjected to a force (or impulse) that is torque producing:
a) The center of mass (CM) moves in a straight line.
b) The body rotates about an axis through its CM.
(By "free" I mean unconstrained.)

It seems intuitive and many text books state the above. But the ones I've looked at don't seem to present a proof.

Thanks.

2. Dec 15, 2009

### A.T.

It follows from the conservation of total linear momentum, if you consider that the body consist of many small masses, each with it's own linear momentum. I don't have a formal proof, but that is the line it would along.

3. Dec 15, 2009

### Gerenuk

I also missed this very important concepts from mechanics books. So I worked it out by myself. I define coordinates (all relevant coordinates are vectors):
$$r_i=r_c+r_i'$$
$$v_i=v_c+v_i'=v_c+\omega\times r_i'$$
I suppose it is possible to show that the neccessary vector $\omega$ corresponds to the angular velocity about just any axis. I use center of mass coordinates since later I will need $\sum m_ir_i'=0$.
-----------------------------
Now
$$\sum F_i=\sum m_ia_i$$
$$\sum F_i=\frac{\mathrm{d}}{\mathrm{d}t}\sum m_iv_i$$
$$\sum F_i^\text{ext}=\frac{\mathrm{d}}{\mathrm{d}t}\left((\sum m_i)v_c\right)$$
$$F^\text{ext}=Ma_c$$
On the LHS I used that internal forces cancel $F_{ij}=-F_{ji}$ and on the RHS I used that $\sum m_iv_i'=\omega\times\sum m_i r_i'=0$. Now you have the linear law of motion.
--------------------------
For the angular part
$$\frac{\mathrm{d}}{\mathrm{d}t}\sum m_ir_i'\times v_i'=\sum m_i r_i'\times a_i'=\sum m_i r_i'\times (a_i-a_c)=\sum r_i'\times F_i=\sum r_i'\times F_i^\text{ext}$$
The internal forces cancel again since force are along the line of action $r_i\times F_{ij}+r_j\times F_{ji}=\Delta r\times F_{ij}=0$. So, the torque from external forces about the center of mass is equal to the angular momentum about the center of mass.

With some additional ideas one can show that therefore in the absence of external forces the center of mass velocity and the angular velocity are constant.

Last edited: Dec 15, 2009
4. Dec 16, 2009

### scoomer

thank you, that was very helpful

5. Dec 16, 2009

### Gerenuk

Actually my derivation doesn't specifically address the rigid body (so I don't need $\omega$). It is for a general set of interacting particles.

That's where I believe conservation of angular momentum comes from. At least I haven't met anyone who convincingly can use Noethers theorem, without including lots of hidden assumptions

6. Dec 20, 2009

### Mad_Eye

what?!
has my reply been deleted again..?
anyway... can you please prov the second law of newton for rotation?
thanks :D

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