# How does attaching a point mass to the CM of an object affect its moment of inertia about the CM?

How does attaching a point mass to the CM of an object affect its moment of inertia about the CM?

Intuitively, it seems to me that this would not affect the mass distribution about the center of mass, but the moment of inertia formulas contain a quantity "total mass", which would change in such a case.

We can test a simple case. Two point masses $$M$$ separated by a fixed distance $$2R$$. If the object is rotating, the angular momentum from the COM is $$L = RMv + RMv = 2MR\omega R = 2MR^2 \omega$$. So $$I = 2MR^2$$.

If there is another mass $$m$$ at the COM, the angular momentum doesn't change, and $$I$$ is the same.

But in the usual formulas for $$I$$ appears the total mass of the object. In the first case:

$$I = M_TR^2$$, where $$M_T = 2M$$

In the second case: $$I = (M_T-m)R^2$$, where $$M_T = 2M + m$$

So, the value of $$I$$ doesn't change when adding a mass at the COM, but the formula for its calculation changes.

For a point mass $$I=mr^2$$, where $$r$$ is the distance from the centre of rotation, so it is zero.

The total mass in other formulae comes from the shape the formula assumes - most formulae for shapes assume constant density for example, so the mass isn't exactly at the centre of mass.

Moment of inertia about points other than the centre of mass will increase by adding mass to the centre of mass, as there's then a non-zero distance.

I may be wrong, but if you're referring to these equations when you say "formulas," then your intuition is right that the moment of inertia doesn't change, despite the "total mass" factor. The reason for this is that all those formulas assume a uniform mass distribution, and when you put in that point mass, the distributions are no longer uniform, and thus those formulas are invalid.

In general, attaching a point mass on the axis of rotation of a system does not change the moment of inertia of that system.