# Moment of inertia about a non-fixed axis

Let’s say I have a cubical shape of mass M and sides s rotating about a vertical axis placed at distance D from the cube’s center of mass. Its moment of inertia I under these conditions is equal to:

I = (M × s²/6) + (M × D²)

Let’s now say that the axis is not fixed anymore but instead rotates itself as well while keeping a vertical position. Since now there’s a different motion pattern, and the overall velocity of the cube is altered, how can the modified moment of inertia (call it I’) be calculated?

• What's your proficiency in math? high-school? University? Can I talk about tensor? Commented Jun 3 at 17:54
• High school, but I’m trying to learn more. And sure, I think I’m familiar enough with tensors. Commented Jun 3 at 17:59
• which country (if Italy, which region/city/school) to be familiar with tensor at high-school? Commented Jun 3 at 18:03
• Italy, Lombardia to be more specific. I have a diploma from an agricultural school. Admittedly I had to refresh my knowledge once I concluded the school through other means (AKA online papers and the likes). Commented Jun 3 at 18:12
• I'll write my answer later, if nobody answers before. Meanwhile, could you add a sketch of the system you're dealing with? It could help to clarify your question Commented Jun 3 at 18:16

The modified moment of inertia $$I'$$ considering both the rotation of the cube about its axis and the rotation of the axis itself can be calculated as: $$I' = \frac{Ms^2}{6}+2MD^2$$

To explain it better, you have to combine the rotational motions with these two formulas:

1. The Moment of inertia about the cube's own center of mass: $$\boxed{I_{cm} = \frac{Ms^2}{6}}$$
2. The Moment of inertia about the axis at distance $$D$$: $$\boxed{I = I_{cm}+MD^2=\left( \frac{Ms^2}{6}\right) + MD^2}$$

Using both concepts you can understand that when the axis rotates, it contributes to an additional kinetic term due to the rotational inertia of the mass $$M$$ at the radius $$D$$.

Now, since your request focuses only on the combined moment of inertia without concentrating on the exact kinetic energy due to the combined angular velocities, the moment of inertia $$I'$$, if the axis is considered as a simple geometric line without additional mass, can be represented as (Where $$I_{\text{axis}}$$ is the moment of inertia of the rotating axis itself, also described as $$MD^2$$): $$I' = I_{cm} + MD^2 + I_{\text{axis}}$$

Going on with the equation you obtain: $$I' = \left( \frac{Ms^2}{6} \right) + MD^2 + MD^2$$ And, finally: $$\boxed{I' = \frac{Ms^2}{6} + 2MD^2}$$ $$\text{QED}$$.

• Very useful. Does this principle apply to whatever shape I might need to work with, correct? Also, since you mentioned it, how would this change to a non fixed axis of rotation affect the kinetic energy of the cube — assuming the cube maintains a velocity of 1 rad/s around the axis, while the latter spins at 2 rad/s (let’s call the two angular velocities ω1 and ω2 respectively)? Commented Jun 3 at 18:31
• To say that it would apply to "whatever shape you might need" would be too generic, but yeah, the general approach works with various shapes (i.e.: spheres, cylinders, rectangular prisms..). However, it's obvious that the formula will change if you change the kind of shape you're working with, so keep that in mind. Commented Jun 3 at 18:37
• Perfect, thanks for the info. What about the second question? Commented Jun 3 at 18:41
• To determine how the non-fixed axis of rotation affects the kinetic energy of the cube, you need to consider the contributions from both rotational motions. The rotational kinetic energy for an object rotating with angular velocity $\omega$ and moment of inertia $I$ is given by $K = \frac{1}{2}I\omega^2$. Unfortunately, I don't have enough characters left here to write the whole explanation, but I hope my comments and my answer were useful :) Commented Jun 3 at 18:42