Use Eq.(9.20) to calculate the moment of inertia of a uniform, solid disk with mass \(M\) and radius \(R\) for an axis perpendicular to the plane of the disk and passing through its center.

The formula for the moment of inertia is the "sum of the product of the mass" of each particle by "the square of its distance from the axis of rotation". The moment of inertia formula is expressed as

\(I = {\int r ^2}dm\) ….. (1)

Here,\(r\)and\(dm\)is the distance and mass respectively.

Determine the mass \(dm\) as follows:

\(dm = \rho dV\) ….. (2)

Here, \(\rho \) is the density, \(dm\) is the differential of mass, and \(dV\) is the differential of volume.

The differential of volume is,

\(dV = 2\pi rLdr\) ….. (3)

Here, \(r\) is the distance and \(L\) is the length.

Substitute \(2\pi rLdr\) for \(dV\) into equation (2).

\(dm = \rho 2\pi rLdr\) ….. (4)

Draw a uniform solid disk with a radius \(R\) for an axis perpendicular to the plan of the disk as follows:

Integrate equation(1) from \(r = 0\) to \(r = R\) as follows:

\(\begin{array}{c}I = {r^2}dm\\ = \int_0^R {{r^2}2\rho \pi rLdr} \\ = 2\rho \pi L\int_0^R {{r^3}dr} \end{array}\)

\(\begin{array}{c}l = 2\rho \pi L\left[ {\frac{{{r^4}}}{4}} \right]_0^R\\ = 2\rho \pi L\frac{{{R^4}}}{4}\end{array}\)

\(l = \rho \pi L\frac{{{R^4}}}{2}\) ….. (5)

It is known that \(V = AL\).

Determine the volume \(V\) as follows:

\(V = AL\)

\(V = \pi {R^2}L\) …..(6)

\(V = \frac{M}{\rho }\) ….. (7)

Equate the equation (7) and (8) to obtain the mass \(M\) as follows:

\(\begin{array}{l}\pi {R^2}L = \frac{M}{\rho }\\M = \rho \pi {R^2}L\end{array}\)

Substitute \(M\) for \(\rho \pi {R^2}L\) into equation (5) as follows:

\(\begin{array}{c}I = 2M\frac{{{R^2}}}{4}\\ = \frac{1}{2}M{R^2}\end{array}\)

Thus, the moment of inertia of a uniform solid disk is \(I = \frac{1}{2}M{R^2}\).