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 =

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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 \
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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