Question

Consider a uniform wire of mass \(M\) and length \(L\). It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is:

• A. \(\frac{ML^2}{\pi^2}\)
• B. \(\frac{1}{2} \frac{ML^2}{\pi^2}\)
• C. \(\frac{1}{4} \frac{ML^2}{\pi^2}\)
• D. \(\frac{2}{5} \frac{ML^2}{\pi^2}\)

Hint:

If all mass is at a constant distance \(R\) from an axis, \(I = MR^2\) regardless of the shape (ring, arc, or point mass).

Answer 1

The Correct Option is A

Step 1: The wire of length \(L\) is bent into a semicircle of radius \(R\). The relationship is \(L = \pi R \implies R = \frac{L}{\pi}\).
Step 2: For a semicircle (or any arc) where all mass is at a distance \(R\) from the center, the moment of inertia \(I\) about the axis through the center perpendicular to the plane is: \[I = \int r^2 dm = R^2 \int dm = MR^2\]
Step 3: Substitute \(R = \frac{L}{\pi}\): \[I = M\left(\frac{L}{\pi}\right)^2 = \frac{ML^2}{\pi^2}\]