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Step-by-Step Solution
Step 1: Moment of Inertia About the Rod’s Center
For a thin rod of length $L$ and mass $M$, rotated about an axis passing through its center and perpendicular to the rod, the moment of inertia (denoted $I_{\text{center}}$) is given by:
$I_{\text{center}} = \frac{M L^2}{12}.$
Step 2: Identify the Desired Axis
We want the moment of inertia about an axis perpendicular to the rod and located at a distance $\frac{L}{3}$ from one end. Label the rod’s ends as A and B, so the axis is $\frac{L}{3}$ from end A.
Step 3: Distance from the Rod’s Center of Mass
For a uniform rod, the center of mass is at its midpoint, a distance $\frac{L}{2}$ from either end. The distance ($d$) between this midpoint axis and the new axis is:
$d = \left|\frac{L}{2} - \frac{L}{3}\right| = \frac{L}{6}.$
Step 4: Apply the Parallel Axis Theorem
The parallel axis theorem states:
$I_{\text{new}} = I_{\text{center}} + M d^2,$
where $d$ is the perpendicular distance between the two parallel axes. Substitute $I_{\text{center}} = \frac{M L^2}{12}$ and $d = \frac{L}{6}$:
$I_{\text{new}} = \frac{M L^2}{12} + M\left(\frac{L}{6}\right)^2.$
Step 5: Simplify the Expression
Compute the second term:
$M\left(\frac{L}{6}\right)^2 = M \cdot \frac{L^2}{36} = \frac{M L^2}{36}.$
Hence,
$I_{\text{new}} = \frac{M L^2}{12} + \frac{M L^2}{36}.$
To add these, express $\frac{M L^2}{12}$ with a common denominator of 36:
$\frac{M L^2}{12} = \frac{3 M L^2}{36}.$
So,
$I_{\text{new}} = \frac{3 M L^2}{36} + \frac{M L^2}{36} = \frac{4 M L^2}{36} = \frac{M L^2}{9}.$
Step 6: Final Answer
Therefore, the moment of inertia of the rod about the specified axis is:
$\displaystyle \frac{M L^2}{9}.$
