The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis passing from the edge of the disc and normal to the disc is

Step 1: Given data

1. Given a circular disc with radius $R$ and mass $M$
2. We know, the moment of inertia of a disc the center of mass and perpendicular to the plane is $I_{CM}=\frac{M{R}^2}{2}$
3. We have to find the moment of inertia through an axis passing from the edge of the disc and normal to the disc, i.e., through $AA\text{'}$

Step 2: Theorem used

Parallel axis theorem- The moment of inertia of a body about an axis parallel to the axis passing through the center of it is equal to the sum of the moment of inertia of the body through the center and the product of the mass of the body and the square of the distance between them.

'$I\text{'}=I+M{R}^2$

Where $I$ is the moment of inertia through any axis passing through the center of the body and $I\text{'}$ is the moment of inertia of an axis parallel to $I$, $M$ is mass of the body, and $R$ is the distance between these two axes

Step 3: Calculation

By parallel axis theorem, the moment of inertia along $AA\text{'}$ ,

$I=I_{CM}+M{R}^2$

$\Rightarrow I=\frac{M{R}^2}{2}+M{R}^2$

$\Rightarrow I=\frac{3}{2}M{R}^2$

Thus, the moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis passing from the edge of the disc and normal to the disc is $\frac{3}{2}M{R}^2$.