---
title: "A thin, uniform wire of total mass \\(M\\) and radius \\(R\\) is bent into a semicircle and lies in the \\(xy\\)-plane, centered symmetrically about the \\(y\\)-axis with its center of curvature at the origin. What is the rotational inertia of the wire about an axis perpendicular to the \\(xy\\)-plane that passes through the center of mass of the wire?"
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url: "https://nerd-notes.com/ubq/120826/"
date_modified: "2026-08-23T04:43:30+00:00"
---

# A thin, uniform wire of total mass \(M\) and radius \(R\) is bent into a semicircle and lies in the \(xy\)-plane, centered symmetrically about the \(y\)-axis with its center of curvature at the origin. What is the rotational inertia of the wire about an axis perpendicular to the \(xy\)-plane that passes through the center of mass of the wire?

A thin, uniform wire of total mass \(M\) and radius \(R\) is bent into a semicircle and lies in the \(xy\)-plane, centered symmetrically about the \(y\)-axis with its center of curvature at the origin. What is the rotational inertia of the wire about an axis perpendicular to the \(xy\)-plane that passes through the center of mass of the wire?

![A set of perpendicular coordinate axes labeled x horizontally and y vertically, intersecting at an origin labeled O. In the upper half-plane, a thick solid semicircular arc of radius R is centered at the origin, extending from the negative x-axis at (-R, 0) through the positive y-axis at (0, R) to the positive x-axis at (R, 0). A single straight dashed arrow extends from the origin O to a point on the arc at an angle in the first quadrant, labeled R. A filled circle sits on the positive y-axis between the origin and the top of the arc, labeled CM. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787460210-c12Ecf.jpg)

- **A.** \(MR^2\left(1 - \dfrac{2}{\pi}\right)\)
- **B.** \(\dfrac{1}{2}MR^2\)
- **C.** \(MR^2\left(1 - \dfrac{4}{\pi^2}\right)\)
- **D.** \(MR^2\left(1 + \dfrac{4}{\pi^2}\right)\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120826/*