Animated Solution for Physics - Rotational Motion: A circular disc $D_1$ of mass $M$ and radius $R$ has two identical discs $D_2$ and $D_3$ of the same mass $M$ and radius $R$ attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis $O{O}^{\prime }$ passing through the centre of $D_1$, as shown in the figure will be

When you first look at the 2D diagram provided in the question, it can be a bit deceptive. It might seem like all three discs are lying flat in the same plane. However, the way they are drawn—with $D_1$ as an ellipse and $D_2$, $D_3$ as perfect circles—is a classic convention to represent a 3D structure on a 2D page.

Let's visualize this properly. Imagine disc $D_1$ is lying flat in the horizontal plane, much like a plate resting on a table. Meanwhile, discs $D_2$ and $D_3$ are standing upright in the vertical plane, attached to the edges of $D_1$ like the wheels of a car attached to an axle. The axis of rotation, $O{O}^{\prime }$, passes straight up vertically through the center of $D_1$.

A Straightforward Spin

Let's first tackle the central disc, $D_1$. The axis $O{O}^{\prime }$ is perpendicular to the plane of $D_1$ and passes right through its center.

From our standard formulas, the moment of inertia of a uniform disc about its central perpendicular axis is simply:

This part is straightforward and forms the foundation of our total calculation.

Parallel Axis Theorem to the Rescue

Now, what about the outer discs, $D_2$ and $D_3$? Notice how the vertical axis $O{O}^{\prime }$ relates to them. Because $D_2$ and $D_3$ are standing upright in vertical planes, the axis $O{O}^{\prime }$ is actually parallel to their vertical diameters.

To find their moment of inertia about $O{O}^{\prime }$, we must deploy the Parallel Axis Theorem. The theorem states that the moment of inertia about any axis is the sum of the moment of inertia about a parallel axis through the center of mass, plus $M{d}^2$, where $d$ is the perpendicular distance between the two axes.

For a disc rotating about its own diameter, the moment of inertia is:

Since $D_2$ and $D_3$ are attached at the opposite ends of $D_1$ (which has a radius $R$), the distance $d$ between their own central vertical axis and $O{O}^{\prime }$ is exactly $R$.

Let's set up the equation for $D_2$:

Adding them up, we get:

Because the setup is perfectly symmetrical, disc $D_3$ will have the exact same moment of inertia:

Finally, to get the total moment of inertia of the entire system, we simply add the individual moments of inertia together.

To make the addition easier, let's write $\frac{1}{2}$ as $\frac{2}{4}$:

This beautifully simplifies to exactly $3M{R}^2$. Always pay close attention to the spatial orientation of the objects relative to the axis of rotation—it makes all the difference!