Animated Solution for Physics - Rotational Motion: The figure shows two solid discs with radius $R$ and $r$, respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc around axis $AB$ (which is perpendicular to the plane of the disc and passing through its centre) of MI of smaller disc around one of its diameters lying on its plane?

Imagine you are holding two solid discs, one larger with a radius $R$ and one smaller with a radius $r$. The problem gives us a crucial piece of information: both discs have the exact same mass per unit area. Let's denote this surface mass density as $\sigma$.

This means that the mass of each disc isn't just a random variable; it is geometrically linked to its area. If you ignore this and just assume their masses are $M$ and $m$, you will miss the hidden $R^2$ and $r^2$ dependence, leading straight to an incorrect answer!

Since the mass per unit area $\sigma$ is constant, we can easily express the mass of any disc by multiplying $\sigma$ by its respective area.

For the smaller disc, the area is $\pi r^2$. Therefore, its mass $M^{\prime }$ is:

Similarly, for the larger disc, the area is $\pi R^2$. Its mass $M$ is:

Now, we need to recall our standard formulas for the moment of inertia of a uniform circular disc.

1. About a central perpendicular axis: The moment of inertia is $I_{\perp }=\frac{1}{2}M{R}^2$. The problem asks for the moment of inertia of the larger disc about axis $AB$, which is exactly this perpendicular axis. 2. About its diameter: Using the Perpendicular Axis Theorem ($I_z=I_x+I_y$), we know that the moment of inertia about any diameter is exactly half of the perpendicular one. So, $I_{\text{dia}}=\frac{1}{4}M{R}^2$. The problem asks for the moment of inertia of the smaller disc about its diameter $CD$.

Let's substitute our mass expressions into these formulas.

For the smaller disc about its diameter: Substituting $M^{\prime }=\sigma \pi r^2$:

For the larger disc about its perpendicular axis: Substituting $M=\sigma \pi R^2$:

Calculating the Ratio

We are almost there! The final step is to find the ratio of the moment of inertia of the larger disc to that of the smaller disc.

Notice how the constants $\pi$ and $\sigma$ appear in both the numerator and the denominator. They cancel out beautifully, leaving us with a clean algebraic fraction:

And there we have it! The ratio is $2R^4:r^4$.