Using the parallel axis theorem, what is the moment of inertia of the rod of mass m about the axis shown below? (Use the following as necessary: m and L.) 2L 3 I =
Using the parallel axis theorem, what is the moment of inertia of the rod of mass m about the axis shown below? (Use the following as necessary: m and L.) 2L 3 I =
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Use of the Parallel Axis Theorem to Determine Moment of Inertia
**Problem Statement:**
Using the parallel axis theorem, what is the moment of inertia of the rod of mass \( m \) about the axis shown below? (Use the following as necessary: \( m \) and \( L \).)
**Diagram Description:**
The diagram displays a horizontal rod with a length of \( L \). The rod is horizontally aligned, and a vertical axis of rotation is shown passing through the left end of the rod. The sections along the rod are labeled as follows:
- The first section from the axis to a point \( \frac{L}{3} \) from the left end.
- The remaining section from the \( \frac{L}{3} \) point to the far right end, labeled \( \frac{2L}{3} \).
An arrow above the rod illustrates the rotational direction about the vertical axis through the left end.
**Equation:**
To solve the moment of inertia \( I \) for this scenario, use the parallel axis theorem, which is given by:
\[ I = I_{\text{CM}} + md^2 \]
Where:
- \( I_{\text{CM}} \) is the moment of inertia about the center of mass of the rod.
- \( m \) is the mass of the rod.
- \( d \) is the distance between the new axis and the axis through the center of mass.
Generally, for a uniform rod of mass \( m \) and length \( L \), the moment of inertia about its center of mass \( I_{\text{CM}} \) is:
\[ I_{\text{CM}} = \frac{1}{12}mL \]
For this problem, the distance \( d \) from the center of mass to the new axis at the left end is \( \frac{L}{2} \).
So,
\[ d = \frac{L}{2} \]
Plugging these values into the parallel axis theorem,
\[ I = \frac{1}{12}mL^2 + m\left(\frac{L}{2}\right)^2 \]
Simplify,
\[ I = \frac{1}{12}mL^2 + m\frac{L^2}{4} \]
\[ I = \frac{1}{12}mL^2 + \frac{1}{4}mL^2 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd0afef86-41bd-49f7-9deb-bb70dd2f5172%2Fc8619fea-b4f8-46e0-b101-9446bb2614a0%2Fbgoadtj.png&w=3840&q=75)
Transcribed Image Text:### Use of the Parallel Axis Theorem to Determine Moment of Inertia
**Problem Statement:**
Using the parallel axis theorem, what is the moment of inertia of the rod of mass \( m \) about the axis shown below? (Use the following as necessary: \( m \) and \( L \).)
**Diagram Description:**
The diagram displays a horizontal rod with a length of \( L \). The rod is horizontally aligned, and a vertical axis of rotation is shown passing through the left end of the rod. The sections along the rod are labeled as follows:
- The first section from the axis to a point \( \frac{L}{3} \) from the left end.
- The remaining section from the \( \frac{L}{3} \) point to the far right end, labeled \( \frac{2L}{3} \).
An arrow above the rod illustrates the rotational direction about the vertical axis through the left end.
**Equation:**
To solve the moment of inertia \( I \) for this scenario, use the parallel axis theorem, which is given by:
\[ I = I_{\text{CM}} + md^2 \]
Where:
- \( I_{\text{CM}} \) is the moment of inertia about the center of mass of the rod.
- \( m \) is the mass of the rod.
- \( d \) is the distance between the new axis and the axis through the center of mass.
Generally, for a uniform rod of mass \( m \) and length \( L \), the moment of inertia about its center of mass \( I_{\text{CM}} \) is:
\[ I_{\text{CM}} = \frac{1}{12}mL \]
For this problem, the distance \( d \) from the center of mass to the new axis at the left end is \( \frac{L}{2} \).
So,
\[ d = \frac{L}{2} \]
Plugging these values into the parallel axis theorem,
\[ I = \frac{1}{12}mL^2 + m\left(\frac{L}{2}\right)^2 \]
Simplify,
\[ I = \frac{1}{12}mL^2 + m\frac{L^2}{4} \]
\[ I = \frac{1}{12}mL^2 + \frac{1}{4}mL^2 \
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