To determine the centre of mass of a system with a hole, the hole should be treated as negative mass. If the geometry of the system is **symmetrical**, the centre of mass is also symmetrical in the x and y dimensions.
For each mass, its surface density $\sigma$ is equal to:
$$
\sigma = \frac{m}{A} \\
m = \sigma A
$$
Holes have negative mass, i.e., $m = -\sigma A$.
For a **one-dimensional** hole, the linear mass density uses a similar formula:
$$
\lambda =\frac{m}{L} \\
\lambda = \frac{dm}{dx}
$$
This means that a hole in a rod can use a different formula:
For a solid object, the centre of mass can be expressed as a Riemann sum and thus an integral:
$$r_{cm} = \frac{1}{M}\int_0^M r\cdot dm$$
In an **isolated system**, it is guaranteed that the centre of mass of the whole system never changes so long as only rigid bodies are involved.
## Rotational motion
### Moment of inertia
The moment of inertia of an object represents its ability to resist rotation, effectively the rotational equivalent of mass. It is equal to the sum of each point and distance from the axis of rotation.
$$I=\sum(mr)^2$$
For more complex objects where the distance often changes:
$$I=\int^M_0 R^2 dm$$
#### Common moment shapes
- Solid cylinder or disc symmetrical to axis: $I = \frac{1}{2}MR^2$
- Hoop about symmetrical axis: $I=MR^2$
- Solid sphere: $\frac{2}{5}MR^2$
- Thin spherical shell: $I=\frac{2}{3}MR^2$
- Solid cylinder about the central diameter: $I=\frac{1}{4}MR^2 + \frac{1}{12}ML^2$
Most translational variables have a rotational equivalent.
Although the below should be represented as cross products, this course only deals with rotation perpendicular to the axis, so the following are always true.