You can also find the moment of inertia of a disk (in fact a disk is a cylinder). You can use the moment of inertia of a disk to find the moment of inertia of a solid sphere.
First consider a cylinder of height \(h\) whose inner radius is \(R_1\) and outer radius is \(R_2\) as shown in Figure 1. The cylinder rotates about an axis which passes through the centres of the cross sections of the cylinder.
And we consider an infinitesimally small volume element of the cylinder of mass \(dm\) and thickness \(dr\). The inner radius of the element is \(r\) and outer radius is \(r + dr\). The volume of the element of the cylinder is \(dV = 2\pi r{\kern 1pt} dr{\kern 1pt} h\) and the mass of the element is \(dm = \rho dV = 2\rho \pi r{\kern 1pt} dr{\kern 1pt} h\). Here \(\rho \) is the density of the cylinder. So the moment of inertia of the element is
\[dI = dm{\kern 1pt} {\kern 1pt} {r^2} = (2\rho \pi r{\kern 1pt} dr{\kern 1pt} h){r^2} = (2\rho \pi h){r^3}dr\]
The total moment of inertia of the whole cylinder is obtained by integrating the moment of inertia \(dI\) from \(R_1\) to \(R_2\):
\[\begin{align*} I &= 2\rho \pi h\int\limits_{{R_1}}^{{R_2}} {{r^3}dr} = 2\rho \pi h\left( {\frac{{R_2^4}}{4} - \frac{{R_1^4}}{4}} \right)\\ {\rm{or,}}\quad I &= \frac{{\rho \pi h}}{2}(R_2^4 - R{}_1^4) \end{align*}\]
The total volume of the cylinder is \(V = \pi h(R_2^2 - R_1^2)\). If \(M\) is the total mass of the cylinder, the density is \(\rho = \frac{M}{V} = \frac{M}{{\pi h(R_2^2 - R_1^2)}}\). Thus, putting the value of \(\rho\) in the above expression for the total moment of inertia \(I\) we get,
\[I = \frac{{\pi hM(R_2^4 - R{}_1^4)}}{{2\pi h(R_2^2 - R_1^2)}} = \frac{1}{2}M(R_1^2 + R_2^2)\]
Case 1: If the cylinder is solid that is, \(R_1 = 0\) the moment of inertia is \(I = \frac{1}{2}MR_2^2 = \frac{1}{2}MR_{}^2\). There is only one radius, so \(R_2\) is written as \(R\). This is also the moment of inertia of a disk.
Case 2: When \(R_1 = R_2 = R\) the moment of inertia of the cylinder becomes \(I = MR_{}^2\). This is the same as the moment of inertia of a body of mass \(M\) about an axis at a distance \(R\) from the axis of rotation. This result is the same as the moment of inertia of a thin circular ring about the axis through its centre and perpendicular to its plane.
