# Conservation of Angular Momentum for Rigid Bodies

I have question about Conservation of Angular Momentum of Rigid Bodies. I've been doing some examples from Hibbeler's book, and noticed that in this chapter about Conservation of Angular Momentum of Rigid Bodies, there are some examples where we sum all the angular momentums about some fixed point O, but when they are writing equation, they write $$I_G$$ (inertion in point G, center of mass). Why $$I_G$$ ? Why they didn't write $$I_O$$ (inertion in point O, the point about which we are writing this equation of conservation). So this confuses me. I mean, there are also examples where they do it like I expect (Inertion about point O).

Look at this example. As you can see, we are writing conservation of angular momentum about point A. So:

$$\sum{H_{A1}}$$ = $$\sum{H_{A2}}$$.

And as you can see they calculated $$I_G$$. Why? Why they didn't wrote about point A, and wrote $$I_A$$ = $$I_G + md^2$$ (where $$d$$ is distance of center of mass of body from point A).

So this confuses me. Whoever helps, thanks in advance!

• Please don't post screenshots of text. It breaks search functionality, and it doesn't work for blind users. – user4552 Jan 8 at 22:40
• Please credit the author of the problem, and give the source. (The publisher is not the author). This is one of the things that we ask you to do in our homework policy: physics.meta.stackexchange.com/questions/714/… – user4552 Jan 8 at 22:41
• I've added the homework-and-exercises tag. In the future, please add this tag to this type of problem. This is one of the things that we ask you to do in our homework policy: physics.meta.stackexchange.com/questions/714/… – user4552 Jan 8 at 22:41

$$\boldsymbol{H}_A = \mathbf{I}_G \boldsymbol{\omega} + \boldsymbol{c} \times \boldsymbol{p} = \mathbf{I}_G \boldsymbol{\omega} + \boldsymbol{c} \times ( m \,\boldsymbol{v}_G ) = \mathbf{I}_G \boldsymbol{\omega} - m \boldsymbol{c} \times (\boldsymbol{c} \times \boldsymbol{\omega}) = \mathbf{I}_A \boldsymbol{\omega}$$
This works because $$\require{cancel} \boldsymbol{v}_G = \cancel{ \boldsymbol{v}_A }+ \boldsymbol{\omega} \times \boldsymbol{c}$$, where $$\boldsymbol{c}$$ is the vector to the center of mass from A.
But since the bullet isn't rotating, it doesn't make sense to assign a rotation and calculate $$\mathbf{I}_A \boldsymbol{\omega}$$ for the bullet. It makes more sense to just use $$\boldsymbol{H}_A = \cancel{ \mathbf{I}_G \boldsymbol{\omega}} + \boldsymbol{c} \times m \boldsymbol{v}$$
• I know that. I have no problem with bullet. I am asking for rod and disk. Why $I_G$ and why not $I_A = I_G + md^2$ (for disk and rod). – pino123 Jan 9 at 20:13