Question

Find the moment of inertia of a circular disk (uniform density) about an axis through its centre and perpendicular to the plane of the disk.

Short answer

Thus,the moment of inertia of the circular disk is $\frac{1}{2}M{R}^2$.

Step-by-step solution

Step 1:Determine the value of the mass density of strip.

The expression for the moment of inertia of an object of mass m is,

$I=m{r}^2$

Here, r is the distance of the object from the axis of rotation.

Moment of inertia of a circular disc:

Consider a disk of mass M ,radius R of mass density (mass per unit area)$\sigma$with centre O .

Let the disk is composed of small circular strips between the region O and R .

With centre O, assume a thin circular strip of mass dm ,radius x and thickness dx .

The mass density of strip is,

localid="1664336745532" $$\begin{gathered} \sigma =\frac{dm}{A} \\ =\frac{dm}{2\pi xdx} \\ dm=\left(2\pi xdx\right)\sigma \end{gathered}$$

The following figure shows the circular disk of radius R.

Determine the moment of inertia of the circular disk.

The moment of inertia of the strip about an axis through its centre and perpendicular to the plane of the strip is,

$$\begin{gathered} dI=\left(2\pi xdx\right)\sigma x^2 \\ =2\pi \sigma x^{3}dx \end{gathered}$$

Thus, the moment of inertia of the disk about an axis through its and perpendicular to the plane of the disk is,

$$\begin{gathered} I=\int dI \\ =\underset{0}{\overset{R}{\int }}2\pi \sigma x^{3}dx \\ =2\pi \sigma \underset{0}{\overset{R}{\int }}{x}^{3}dx \\ =2\pi \sigma {\left[\frac{x^4}{4}\right]}_0^R \end{gathered}$$

Further simplify,

$$\begin{gathered} 2\pi \sigma {\left[\frac{x^4}{4}\right]}_0^R=\frac{2\pi \sigma R^4}{4} \\ I=\frac{\pi \sigma R^4}{2}...\left(1\right) \end{gathered}$$

But, the mass of the disk is,

$M=\left(\pi R^2\right)\sigma$

Use this value in equation (1).

$$\begin{gathered} I=\frac{\left(\pi R^2\right)\sigma R^2}{2} \\ =\frac{1}{2}M{R}^2 \end{gathered}$$

Thus, the moment of inertia of the circular disk is $\frac{1}{2}M{R}^2$.