Introduction
Hello and welcome to the second post on deriving the moment of inertia equations for different shapes. If you haven't yet read my post on deriving the moment of inertia for a slender rod, then please do. This post will build on the information that I explained in the first post and is essentially a more difficult derivation than the slender rod.
Moment of Inertia Derivation - Hollow/Solid Cylinder
I will now derive the respective equation for the moment of inertia for a hollow cylinder (this will also cover the moment of inertia for a solid cylinder, as you can simply set the inner radius to 0). I have defined the hollow cylinder to have an inner radius of R_1, mass, M, an outer radius of R_2 and a length of L. I have also included the central axis (the axis of rotation) as a dotted line, labelled x, that goes straight through the centre of the cylinder. I have included an image of this below:
Our general integral for the moment of inertia is still the same as in the previous derivation. I have included an image below to remind ourselves:
As before, we must rearrange dm in terms of r (and some other constants) so that we can integrate the expression. The mass of one infinitesimally thin, dm, is equal to the density of the cylinder, ρ, multiplied by the volume of the ring that we are considering, dV. This can expressed in the following expression, included below:
We can now express dV in terms of dA (the area of the top of the infinitesimally thin ring) and the length of the cylinder, L. This will result in the following expression, shown below:
We can work out dA in terms of dr. The area of the top of the ring is the area of the circle, with the infinitesimally small ring, minus the area of the circle without. This can be expressed like so:
Since dr referred to an infinitesimal distance, dr^2 is 0, as you multiplying an infinitesimally small number by another, which will yield an answer of 0. If you are not convinced, you can easily use differentiation to find out dA. We know that the area of a circle is πr^2 (I have actually proven this in my post on de Moivre’s theorem, check it out if you want). From this we can simply differentiate the expression with respect to dr and get the following answer:
We know have dA in terms of r and dr. We can therefore now express dV in terms of r and dr, as I have shown below:
Ergo, we can also express dm in terms of r and dr, which was our ultimate aim. This will allow us to substitute for dm into our previous integral, as I have shown in an image below:
The density, ρ, is a constant and can thus be taken out of the integral. The density of the cylinder, ρ, can be expressed as M/V, where M is the total mass of the cylinder and V is the total volume. We can also express V in terms of R_1 and R_2, resulting in the integral looking like so:
From here we just evaluate the integral, with limits R_2 and R_1. This will result in the following general equation for the moment of inertia for a hollow cylinder. I have an included an image of this below:
By setting R_1 = 0, we can therefore work out the specific moment of inertia equation for a solid cylinder. I have included an image of this below:
Moreover, in order to obtain the moment of inertia for a thin cylindrical shell (otherwise known as a hoop), we can substitute R_1 = R_2 = R, as the shell has a negligible thickness. This will result in the following equation (note that R_2 is just simply R).
I hope you have enjoyed this derivation! I really enjoy making these as I believe it gives a deeper physical understanding behind concepts, rather than blindly following an equation. I would very much like to derive many different equations, so please comment on which equation you’d like to see next! As for moment of inertia, I will be deriving the moment of inertia of a sphere soon.
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