# When to take the other body as point mass or apply parallel axis theorem on it?

What is the range of $$r$$ in the formula for the parallel axes theorem? Eg. if two discs are attached at a certain point on their circumference and both are in a plane, what is the moment of inertia of the system about the axis of one of the discs, so should I consider the other disc as a point mass located at $$2R$$ or apply parallel axes on it. In many such questions, they take the other body as a point mass and in some cases, they apply parallel axis theorem on it.

• Use the distance between the two centers of mass projected on the direction perpendicular to this axis of rotation. Commented May 9, 2022 at 19:17

The Parallel Axis Theorem does not have a range. It applies regardless of how far an object is from the axis of rotation.

However, as the distance $r$ from the axis of rotation increases, treating the distant object as a point particle becomes a better and better approximation. Whether you decide to use the Parallel Axes Theorem or treat the distant object as a point particle depends on what level of accuracy you want to obtain.

Suppose the moment of inertia through the centre of mass is $I_0=Mk^2$ where $k$ is the radius of gyration, which is on the same order of magnitude as the width/breadth/length $d$ of the object. Then the moment of inertia about a parallel axis at distance $r$ is
$$I=I_0+Mr^2=M(k^2+r^2)$$ If $r \gg d \approx k$ then $I \approx Mr^2$ which is the moment of inertia of a point particle. The fractional level of accuracy you can expect from using the approximation is about $(\frac{d}{r})^2$. So if $r \approx 10d$ then the approximation will be accurate to about 1%.

If the point at which you want to measure angular momentum or consider mass moment of inertia, is not at the center of mass, you need to consider the parallel axis theorem.

Most generally, to measure the mass moment of inertia at a point A not at the center of mass C in 3D to the following

$$\mathrm{I}_A = \mathrm{I}_C + m \begin{vmatrix} y^2+z^2 & -x y & - x z \\ - x y & x^2+z^2 & - y z \\ - x z & - y z & x^2+y^2 \end{vmatrix}$$

In shorthand notation the above is

$$\mathrm{I}_A = \mathrm{I}_C - m [\mathbf{r}\times] [\mathbf{r} \times]$$

where $[\mathbf{r}\times] = \pmatrix{0 & -z & y\\z & 0 & -x\\ -y&x&0}$ is the 3×3 skew symmetric cross product matrix.

The same notation is used for the general rotation matrix about an axis $\mathbf{z}$

$$\mathrm{R} = 1 + \sin \theta [\mathbf{z}\times] + (1-\cos \theta) [\mathbf{z}\times] [\mathbf{z}\times]$$

To find the rotational inertia of a solid object which will rotate about a new parallel axis which does not pass through the center of mass, first treat the object as a point mass at the CM which will resist being accelerated about the new axis, and then remember that the object will still resist acceleration about the old axis through the center of mass. (Both axes see the same angular acceleration.)