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The parallel axis theorem is a helpful tool in physics for finding the moment of inertia of a body about any axis, given that the moment of inertia about a parallel axis through the center of mass is known. In our situation, the theorem comes in handy for locating the moment of inertia of a hoop (hollow ring) when the axis is not at its center. The theorem can be mathematically described as:

\[ I = I_{cm} + Md^2 \]
where:

• M is the mass of the object.
• d is the distance between the center of mass axis and the new axis.

The theorem allows us to quickly compute the moment of inertia by using the known value of \(I_{cm}\) and simply adding the product of the mass and the square of the distance. This procedure works well with symmetrical objects, helping to simplify otherwise complex calculations.

A hollow ring, also known as a hoop, is a simple yet intriguing geometrical object commonly encountered in physics problems. The key features of a hollow ring include:

• It is thin-walled, meaning its thickness is considerably less than its radius.
• It has a uniform mass distribution around its circular shape.
• The entire mass is located at a constant distance from the center, which simplifies calculations.

In the realm of moments of inertia, the hollow ring represents an interesting case because its mass is concentrated at the same radius, leading to straightforward computations, especially when determining the inertia about axes parallel to its own center.

Calculating the moment of inertia for a hoop, specifically for an axis that is not central, requires an understanding of how mass distribution affects inertia. For a hoop:

• Its moment of inertia about an axis through its center and perpendicular to its plane is given by \(I_{cm} = MR^2\).
• When the axis is moved to the edge, as in the given problem, the parallel axis theorem tells us the inertia changes.

Let's take an example:- For a hoop with mass \(M\) and radius \(R\): - Starting with \(I_{cm} = MR^2\), - Utilizing the parallel axis theorem by adding \(Md^2\) (here \(d = R\)) on the side, we get: - \[ I = MR^2 + MR^2 = 2MR^2 \].This calculation shows the moment of inertia increases because of the mass's offset from the center, due to the repositioning of the rotation axis to the edge, effectively doubling the original central moment of inertia. Understanding these concepts allows us to tackle similar problems systematically.