1. Given information

For a disc of mass $M$ and radius $R$:

• Moment of inertia about any diameter through the center is $I_{\text{diameter}}=\frac{MR^2}{4}.$

We need the moment of inertia about an axis:

• normal to the disc, and
• passing through a point on its edge.

It is given in the form $I=\frac{x}{2}MR^2.$

2. Find moment of inertia about the central axis normal to the disc

Using the perpendicular axis theorem for a planar lamina: $I_z = I_x + I_y$ where $I_x$ and $I_y$ are moments of inertia about two perpendicular diameters in the plane of the disc.

Since both diameters are equivalent, $I_x = I_y = \frac{MR^2}{4}.$

So, $I_z = \frac{MR^2}{4} + \frac{MR^2}{4} = \frac{MR^2}{2}.$

Thus, moment of inertia about the axis normal to the disc through its center is $I_{\text{center, normal}}=\frac{MR^2}{2}.$

3. Shift the axis to the edge using parallel axis theorem

The required axis is parallel to the central normal axis, but passes through a point on the rim.

Distance between these two parallel axes is $d=R.$

By the parallel axis theorem, $I_{\text{edge, normal}} = I_{\text{center, normal}} + Md^2.$

Therefore, $I_{\text{edge, normal}} = \frac{MR^2}{2} + MR^2.$

$I_{\text{edge, normal}} = \frac{3}{2}MR^2.$

4. Compare with the given form

Given, $I = \frac{x}{2}MR^2.$

So, $\frac{x}{2}MR^2 = \frac{3}{2}MR^2.$

Hence, $x=3.$

5. Comparison with stored answer

Stored correct answer: $3$

Our derived answer: $3$

They match.