Animated Solution for Physics - Rotational Motion: Let the moment of inertia of a hollow cylinder of length $30\text{ cm}$ (inner radius $10\text{ cm}$ and outer radius $20\text{ cm}$) about its axis be $I$. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also $I$, is

Imagine you have two objects made of the same amount of material (mass $M$). One is a thick, hollow pipe (a hollow cylinder), and the other is a very thin pipe (a thin cylinder). The problem asks us to find the radius of the thin pipe such that it is just as hard to spin around its central axis as the thick, hollow pipe. In physics terms, we want their moments of inertia to be exactly equal.

Let's look at the hollow cylinder first. It has an inner radius $R_1=10\text{ cm}$ and an outer radius $R_2=20\text{ cm}$. The moment of inertia of a hollow cylinder about its central longitudinal axis is given by the standard formula:

This formula tells us how the mass is distributed. Because it's hollow, the mass is spread out between $R_1$ and $R_2$, and the formula averages the squares of these radii.

Now, consider the thin cylinder. A thin cylinder is essentially a ring extended in 3D space. All of its mass $M$ is concentrated at a single distance $R$ from the axis of rotation. Therefore, its moment of inertia is simply:

The core condition of the problem is that these two moments of inertia are equal ($I_1=I_2$). Let's set up the equation:

The first beautiful thing that happens is that the mass $M$ cancels out from both sides. This means the answer doesn't depend on how heavy the cylinders are, only on their geometry!

Now, we substitute the given values $R_1=10$ and $R_2=20$:

Taking the square root gives us the radius $R$:

Looking at our options, the closest integer value is $16\text{ cm}$.

Radius of Gyration

Take a step back and look at what we just found. We found a distance $R$ where, if we concentrated all the mass $M$, it would have the same moment of inertia as the original hollow cylinder.

Does this sound familiar? Yes! This is the exact definition of the Radius of Gyration ($K$).

By equating $I=M{K}^2$ to the hollow cylinder's inertia, we effectively calculated its radius of gyration. The thin cylinder in this problem is just a physical manifestation of the radius of gyration concept!