I've learned that the angular momentum of an object rotating about a fixed axis is $I \omega $. Also, in absence of external torques, $I_1 \omega_1 = I_2 \omega_2 $ (meaning, two different events).

I saw in the books that they use it sometime to solve problems. But, if for instance the mass and the axis of rotation were changed, how can it still be conserved around different axises? One of the first things we learned is that you should always calculate angular momentum about the same axis.

I'd really appreciate an explanation.

Edit: A more specific example:

Two blocks are connected by a string and spin around a fixed axis. Then, a 3rd block collides with one of the blocks and sticks to it (no external torques in the system of the 3 blocks). Isn't the angular momentum conserved?

  • $\begingroup$ Your question is unclear. What is "1" and "2"? What do you mean by changing the mass? How would you do that? How would you change the axis of rotation? Generally in either of these cases angular momentum would not be conserved. $\endgroup$ – garyp Feb 20 '15 at 14:50
  • $\begingroup$ @garyp thanks for your response. I've edited the question and added a more specific example. I'm not a native English speaker, obviously, so it's quite hard for me to explain myself, but I hope it's better now. $\endgroup$ – Yes Feb 20 '15 at 15:22
  • $\begingroup$ Please take a look at this meta post regarding good question titles. $\endgroup$ – DanielSank Feb 20 '15 at 17:56

First, angular momentum isn't measured about an axis. It's measured about a point.

Second, well, of course the angular momenta about different points will be different in general. But they will each be conserved -- there's no need for the point to be in the axis of rotation or even in the same galaxy as the rotating object you're interested in.

Now, about your example. The total angular momentum, on the three block system, is definitely conserved. Why wouldn't it be? What you're perhaps missing is the fact that there will be in general an angular momentum associated with the 3rd block as well, even if it's traveling on a straight line. For example, say you have a block with momentum $\vec{p}$ traveling in a straight line at constant velocity until it hits the block at one end of the string (at the point of closest approach to the center of mass). Say the string has length $2\ell$. Now let's calculate the angular momentum of this third block about the center of mass of the two rotating blocks when it's at a distance $r$ from that center of mass.

$$\vec{L} = \vec{r} \times \vec{p} = rp \sin \theta \hat{z} = rp \frac{\ell}{r}\hat{z} = rp \hat{z} $$

(The $\hat{z}$ is just my choice in how to orient the system in space).

Notice now that $L$ does not depend on the distance $\vec{r}$. It's a constant of the motion, as promised, so long as $\vec{p}$ itself is conserved. Notice that there was nothing special in this derivation about the center of mass: it could've been literally any point and the conclusion still would be valid.

Now you can work out what will happen in the collision, assuming that linear momentum is conserved, and you'll prove explicitly that angular momentum is conserved in this process.

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