introduction & theory
Logic behind the moment of inertia: Why do we need this?
Definition for point bodies
I = mr2
It’s a scalar quantity (like its translational cousin, mass), but has unusual looking units.
Say it, kilogram meter squared and don’t say it some other way by accident.
For a collection of objects, just add the moments. It works like mass in this respect as long as you’re adding moments that are measured about the same axis.
I = ∑I = ∑mr2
For an extended body, replace the summation with an integral and the mass with an infinitesimal mass. You add up (integrate) all the moments of inertia contributed by the teeny, tiny masses (dm) located at whatever distance (r) from the axis they happen to lie.
In practice, for objects with uniform density (ρ = m/V) you do something like this…
|r2 dm =||⌠|
|r2 ρ dV =||⌠|
For objects with nonuniform density, replace density with a density function, ρ(r).
|r2 dm =||⌠|
|r2 ρ(r) dV|
The infinitesimal quantity dV is a teeny tiny piece of the whole body. In practice, this may take one of two forms (but it is not limited to these two forms). The infinitesimal box is probably the easiest conceptually. Imagine dicing the object up into cubes.
[photo of cubed potatoes]
The pieces are dx wide, dy high, and dz deep. The volume of each infinitesimal piece is…
dV = dx dy dz
When an object is essentially rectangular, you get a set up something like this…
|(x2 + y2 + z2)||m||dx dy dz|
|(x2 + y2 + z2) ρ(x, y, z) dx dy dz|
This is the way to find the moment of inertia for cubes, boxes, plates, tiles, rods and other rectangular stuff. Note that although the strict mathematical description requires a triple integral, for many simple shapes the actual number of integrals worked out through brute force analysis may be less. Sometimes, the integrals are trivial.
The other easy volume element to work with is the infinitesimal tube. Imagine a leek.
[photo of a leek]
Each layer of the leek has a circumference 2πr, thickness dr, and height h. The volume of each infinitesimal layer is then…
dV = 2πrh dr
For many cylindrical objects, you basically start with something like this…
|r2 ρ(r) 2πrh dr|
This method can be applied to disks, pipes, tubes, cylinders, pencils, paper rolls and maybe even tree branches, vases, and actual leeks (if they have a simple mathematical description).
When shapes get more complicated, but are still somewhat simple geometrically, break them up into pieces that resemble shapes that have already been worked on and add up these known moments of inertia to get the total.
Itotal = I1 + I2 + I3 +…
For slightly more complicated round shapes, you may have to revert to an integral that I’m not sure how to write. Something like for nested, cylindrical shells…
|Icylindrical shell(r) dr|
or this for stacked disks and washers
|Idisk or washer(r) dr|
These methods can be used to find the moment of inertia of things like spheres, hollow spheres, thin spherical shells and other more exotic shapes like cones, buckets, and eggs — basically, anything that might roll and that has a fairly simple mathematical description.
When you are done with all of this, you oftentimes end up with a nice little formula that looks something like this…
I = αmr2
where α is a simple rational number like 1 for a hoop, ½ for a cylinder, or ⅖ for a sphere.
What if an object isn’t being rotated about the axis used to calculate the moment of inertia? Apply the parallel axis theorem.
I = Icm + mL2
What can I say about the perpendicular axis theorem other than it’s interesting. It applies to laminar objects only. I haven’t needed to use it much.
Iz = Ix + Iy
The best way to learn how to do this is by example. Lots of examples.
|cause of acceleration||∑F||τ =||r × F||∑ τ|
|resistance to acceleration||m||I =||∑ ri2mi = ∫ r2 dm||I|
|newton’s second law||∑F =||ma||∑ τ =||Iα|