---
title: "A uniform thin rod of length \\(L\\) and mass \\(M\\) is attached to a frictionless pivot at one end. The rod is released from rest in a horizontal position and allowed to rotate downward. Let \\(\\alpha_0\\) be the magnitude of the angular acceleration of the rod at the instant of release. Let \\(\\alpha_{45}\\) be the magnitude of the angular acceleration of the rod when it has rotated through an angle of \\(45^\\circ\\) below the horizontal. What is the ratio \\(\\dfrac{\\alpha_{45}}{\\alpha_0}\\)?"
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url: "https://nerd-notes.com/ubq/109828/"
date_modified: "2026-03-26T05:52:44+00:00"
---

# A uniform thin rod of length \(L\) and mass \(M\) is attached to a frictionless pivot at one end. The rod is released from rest in a horizontal position and allowed to rotate downward. Let \(\alpha_0\) be the magnitude of the angular acceleration of the rod at the instant of release. Let \(\alpha_{45}\) be the magnitude of the angular acceleration of the rod when it has rotated through an angle of \(45^\circ\) below the horizontal. What is the ratio \(\dfrac{\alpha_{45}}{\alpha_0}\)?

A uniform thin rod of length \(L\) and mass \(M\) is attached to a frictionless pivot at one end. The rod is released from rest in a horizontal position and allowed to rotate downward. Let \(\alpha_0\) be the magnitude of the angular acceleration of the rod at the instant of release. Let \(\alpha_{45}\) be the magnitude of the angular acceleration of the rod when it has rotated through an angle of \(45^\circ\) below the horizontal. What is the ratio \(\dfrac{\alpha_{45}}{\alpha_0}\)?

![A thin rod is pivoted at its left end. A dashed line shows the rod in its initial horizontal position. A solid line shows the rod rotated downward by 45 degrees. A vector labeled Mg acts downward from the center of the rod in both positions.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1774504364-BcRsbK.jpg)

- **A.** \(\dfrac{1}{2}\)
- **B.** \(\dfrac{\sqrt{2}}{2}\)
- **C.** \(1\)
- **D.** \(\sqrt{2}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/109828/*