# 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!

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– user4552
Jan 8, 2020 at 22:40
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– user4552
Jan 8, 2020 at 22:41
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– user4552
Jan 8, 2020 at 22:41

You can do it both ways, as the torque equation or conservation of angular momentum is valid for the center of mass of the system, or an inertial point. Here they did not use the moment of inertia directly about the point $$A$$ as to do that, you would have to find the distance of the center of mass of the system from the point $$A$$ ($$I c m + m d^{2}$$ ($$d$$ is required)) and that would unnecessarily involve one more step to the calculation and make it slightly more difficult for some people. So, in short, doing it by your method or by their method, you will get the same answer, they just chose the shorter way (at least that's what I think).

As long as point A isn't moving, you can do it either way. In vector form, angular momentum about A is

$$\boldsymbol{H}_A = \mathbf{I}_G \boldsymbol{\omega} + \boldsymbol{c} \times \boldsymbol{p} = \mathbf{I}_G \boldsymbol{\omega} + \boldsymbol{c} \times ( m \,\boldsymbol{v}_G ) = \mathbf{I}_G \boldsymbol{\omega} - m \boldsymbol{c} \times (\boldsymbol{c} \times \boldsymbol{\omega}) = \mathbf{I}_A \boldsymbol{\omega}$$

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). Jan 9, 2020 at 20:13
• The same applies to all bodies. Jan 9, 2020 at 22:42

Finding the rotational inertia of the disk about point A would involve a complex integration. Instead, they use the parallel axis theorem. The disk is treated like a point mass revolving about point A, but then you have to add its resistance to also rotating about its own center. Either approach would work for the rod. They are treating the bullet like a point mass.