The moment of inertia of a rotating solid disk about an axis through its CM is \(\frac{{\bf{1}}}{{\bf{2}}}{\bf{M}}{{\bf{R}}^{\bf{2}}}\) (Fig. 8–20c). Suppose instead that a parallel axis of rotation passes through a point on the edge of the disk. Will the moment of inertia be the same, larger, or smaller? Explain why.

The object's MOI can be defined as the quantity that measures the resistance to a change in rotation. Its value altered inversely according to the value of the angular acceleration.

It is given that the moment of inertial of the disk about the center of mass is \({I_{{\rm{CM}}}} = \frac{1}{2}M{R^2}\).

Draw a schematic figure of a disk.

The above figure shows a solid disk with mass \(M\) and radius \(R\). The axis y passes through the center of mass of the disk and is perpendicular to the disk, while another axis \(y'\) is parallel to the y-axis and touches the edge of the disk.

Now apply the parallel axis theorem to calculate the moment of inertia of the disk about the \(y'\)-axis.

\(\begin{aligned}{c}I' = {I_{{\rm{CM}}}} + M{R^2}\\I' = \frac{1}{2}M{R^2} + M{R^2}\\I' = \frac{3}{2}M{R^2}\end{aligned}\)

Thus, from the above equation, the moment of inertia of the disk about an axis that passes through a point on the edge of the disk is greater than its moment of inertia about the center of mass. That is, \(I' > {I_{{\rm{CM}}}}\).