# Parallel axis theorem: adding mass

I'm going through a worked physics problem and have a question about the parallel axis theorem regarding only adding mass while not changing is axis of rotation.

Here is the problem:

A physics professor stands at the center of a frictionless turntable with arms outstretched and a $5.0 \text{ kg}$ dumbbell in each hand (Fig. 10.29). He is set rotating about the vertical axis, making one revolution in $2.0 \text{ s}$. Find his final angular velocity if he pulls the dumbbells in to his stomach. His moment of inertia (without the dumbbells) is $3.0 \text{ kg m}^2$ with arms outstretched and $2.2 \text{ kg m}^2$ with his hands at his stomach. The dumbbells are $1.0 \text{ m}$ from the axis initially and $0.20 \text{ m}$ at the end.

Focusing on the part with the parallel axis theorem, the book says $$I = 3.0\text{ kg m}^2 + 2(5\text{ kg})(1\text{ m})^2 = 13 \text{ kg m}^2$$ I have never seen the parallel axis theorem used this way. Since the $3\text{ kg m}^2$ is without the dumbbells how can you include it into the equation. From what I understand when the book derived the equation, the masses in both moments of inertia should be the same. Please help me figure out this part of the problem.

$$\mathscr{I}_{total}=\mathscr{I}_{man/arms}+\mathscr{I}_{dumbell}+\mathscr{I}_{dumbell}$$
The parallel-axis theorem deals with a shift in rotational axis (theoretical or actual) from an axis through the center of mass. The axis never shifts in this problem except for possibly the idea that the dumbbells shift. But they are simply treated as point masses which have $\mathscr{I}=mr^2$, where $r$ is the distance from the dumbbell point mass to the axis.