# How to find moment of inertia of complex objects?

Let's say we have a ring gear which is attached some mass and is supposed to rotate this mass. This gear is driven by another gear. How one can find the moment of inertia of the driven gear to calculate the torque required to rotate this gear system? Is it just the total mass attached multiply by the square of the radius of the gear or something more advanced?

• Show some effort, then people may help... – user207455 Jul 19 '19 at 8:53

In general, the moment of inertia is found by evaluating $$I = \iiint\limits_V{\rho r^2 dV},$$ where $$\rho$$ is the mass density and $$r$$ is the distance from the axis with respect to which you wish to calculate the moment of inertia (often the axis of symmetry). For complex objects, there usually will not be a simple formula for the moment of inertia and it may be necessary to evaluate this integral numerically, or approximate it with some simplifying assumptions.

If you think the gear can be approximated as a simple, uniform disk of radius $$R$$, its moment of inertia with respect to the axis of symmetry will be roughly $$\frac{1}{2}MR^2$$, where $$M$$ is the total mass. If the gear is uniform and has a constant thickness throughout, with a certain "inner radius" (not containing the teeth) and an "outer radius" (just containing the teeth), $$\frac{1}{2}MR^2$$ will still be valid, with $$R$$ having a value somewhere between these two radii.

The moment of inertia of any compound object made up of $$N$$ discrete masses is given by

$$I = \sum_{i=1}^{N} m_{i} r_{i}^{2}$$

where $$m_{i}$$ is the mass of the $$i$$th object and $$r_{i}$$ is the distance of the object of the $$i$$th mass from the axis of rotation. Now, for a continuous object, the summation becomes and integral as in Puk's response.

Now in the simplest case, a ring gear may be thought of as a ring of mass $$M$$, radius $$R$$ with uniform density (let mass per unit length = $$\lambda$$). Then the moment of inertia about an axis through the center of mass and perpendicular to the ring gear is given by:

$$I = \int_{0}^{2 \pi} R^2 \lambda R d\theta = 2 \pi \lambda R^3 = MR^2$$.

If you want a better model you need to account for the contributions of the teeth on the rim. Let the mass of a tooth be $$m_{t}$$ and let there be $$N$$ teeth. Then the moment of inertia is

$$I = MR^2 + Nm_{t}R^2$$

In all of this, we have assumed that $$R >>>>$$ thickness of the rim.

If the gear has masses attached, you need to add the moments of inertia due to the masses, but do take the MI from the same axis.