Animated Solution for Physics - Rotational Motion: A circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $\left(\frac{{\sigma }_0}{r}\right)$, then the radius of gyration of the disc about its axis passing through the centre is

This problem is a classic example of dealing with rigid bodies that possess a variable mass density. When the mass is not uniformly distributed, we cannot rely on our standard, pre-derived formulas for the moment of inertia or the center of mass. Instead, we must return to the fundamental principles of calculus.

We are given an annular disc (a disc with a hole in the center). The inner radius is $a$ and the outer radius is $b$. The crucial piece of information is that the surface mass density $\sigma$ varies inversely with the radial distance $r$, given by the relation $\sigma =\frac{{\sigma }_0}{r}$.

Our objective is to find the radius of gyration, $K$. The radius of gyration is defined by the equation $I=M{K}^2$, which can be rearranged as $K=\sqrt{\frac{I}{M}}$. Therefore, our roadmap is clear: we must calculate the total mass $M$ and the total moment of inertia $I$ of this specific disc.

Choosing the Element

Because the density varies only with the radial distance $r$, the most logical choice for a differential element is a thin circular ring of radius $r$ and infinitesimal thickness $dr$. All points on this ring are at the exact same distance from the center, meaning the density $\sigma$ is constant across this entire elemental ring.

The area of this elemental ring is $dA=2\pi rdr$.

The mass of this element, $dm$, is the product of its density and its area:

Notice the beautiful mathematical cancellation here! The $r$ in the denominator of the density cancels perfectly with the $r$ in the area formula, leaving us with a remarkably simple expression for the elemental mass:

To find the total mass $M$, we integrate $dm$ from the inner boundary $r=a$ to the outer boundary $r=b$:

Next, we calculate the moment of inertia of our elemental ring. The moment of inertia of a thin ring about its central axis is simply its mass multiplied by the square of its radius ($dI=dm\cdot r^2$).

To find the total moment of inertia $I$, we integrate $dI$ over the same limits:

The Radius of Gyration

Now we substitute our expressions for $I$ and $M$ back into the radius of gyration formula:

The constant term $2\pi {\sigma }_0$ cancels out completely:

To simplify this further, we recall the standard algebraic identity for the difference of two cubes: $b^3-a^3=(b-a)(b^2+a^2+ab)$. Substituting this into our numerator allows us to cancel the $(b-a)$ term:

Taking the square root yields our final, elegant result:

This problem beautifully demonstrates how a seemingly complex variable density can lead to clean, solvable integrals if the correct differential element is chosen.