What is the moment of inertia of a hoop of mass M and radius R rotated about a
cylinder
axis, as shown below?

Fortunately, we do not need to use calculus to solve this problem. Notice that
all the mass is the same distance R from the axis of rotation. Thus we do not
need to integrate over a range, but can calculate the total moment of inertia.
Each small element dm has a rotational inertia of R^{2}dm, where r is constant.
Summing over all elements, we see that I = R^{2}dm = R^{2}M. The sum of all the
small elements of mass is simply the total mass. This value for I of MR^{2}
agrees with experiment, and is the accepted value for a hoop.

Problem :

What is the rotational inertia of a solid cylinder with length L and radius R,
rotated about its central axis, as shown below?

To solve this problem we split the cylinder into small hoops of mass dm, and
width dr:

This small element of mass has a volume of (2Πr)(L)(dr), where dr is the
width of the hoop. Thus the mass of this element can be expressed in terms of
volume and density:

dm = ρV = ρ(2ΠrLdr)

We also know that the total volume of the entire cylinder is given by: V = AL = ΠR^{2}L. In addition, our density is given by the total mass of the cylinder
divided by the total volume of the cylinder. Thus:

ρ = =

Substituting this into our equation for dm,

dm = = 2rdr

Now that we have dm in terms of r, we simply have to integrate over all possible
values of r to get our rotational inertia:

I

=

r^{2}dm

=

2r^{3}dr

=

[r^{4}/2]_{0}^{R}

=

Thus the rotational inertia of a cylinder is simply . Once
again, it has the form of kMR^{2}, where k is some constant less than one.