Step 1: Understanding the Concept:
We need to find the moment of inertia of a semicircular ring about its diametric axis.
The axis XX' shown in the figure corresponds to the base diameter of the semicircle. Step 2: Key Formula or Approach:
For a complete uniform ring of mass \( m \) and radius \( R \), the moment of inertia about an axis passing through its center and perpendicular to its plane is \( I_z = mR^2 \).
By the perpendicular axis theorem for a planar object, \( I_z = I_x + I_y \).
Due to symmetry, the moment of inertia about any diameter is the same, so \( I_x = I_y = \frac{1}{2}mR^2 \).
A semicircular ring is essentially half of a full ring. Its moment of inertia about the diametric axis is the same as that of a full ring of the same mass and radius. Step 3: Detailed Explanation:
Let the mass of the semicircular ring be \( M \) and its radius be \( R \).
The moment of inertia of this semicircular ring about its diametric axis (XX') is given by:
\[ I_{XX'} = \frac{1}{2}MR^2 \]
We are given that the wire of length \( L \) is bent to form this semicircular ring.
The length of a semicircle is \( \pi R \).
Therefore, we have the relation:
\[ L = \pi R \implies R = \frac{L}{\pi} \]
Substitute this expression for \( R \) into the moment of inertia formula:
\[ I_{XX'} = \frac{1}{2} M \left(\frac{L}{\pi}\right)^2 \]
\[ I_{XX'} = \frac{1}{2} M \left(\frac{L^2}{\pi^2}\right) \]
\[ I_{XX'} = \frac{ML^2}{2\pi^2} \]
Note: While options (C) and (D) might not be explicitly visible in the cropped image, the derived answer \( \frac{ML^2}{2\pi^2} \) is the mathematically correct result for standard multiple-choice questions of this type. Step 4: Final Answer:
The moment of inertia about the axis XX' is \( \frac{ML^2}{2\pi^2} \).
