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Step-by-Step Solution
Step 1: Understand the Problem
We have a thin rod of length $L$, with its linear mass density varying as
$\lambda(x) = \lambda_0\left(1 + \frac{x}{L}\right)$
where $x$ is measured from one end of the rod (point A). The total mass of the rod is $M$. We need to find the moment of inertia of this rod about an axis perpendicular to it and passing through point A.
Step 2: Express the Total Mass in Terms of $\lambda_0$
The total mass $M$ is obtained by integrating the linear mass density from $x=0$ to $x=L$:
$M \;=\; \int_{0}^{L} \lambda(x)\,dx
\;=\; \int_{0}^{L} \lambda_0 \left(1 + \frac{x}{L}\right)\,dx.$
Carrying out the integration:
$\displaystyle
M
= \lambda_0 \int_{0}^{L} \left(1 + \frac{x}{L}\right)\,dx
= \lambda_0 \left[\int_{0}^{L} 1\,dx + \int_{0}^{L} \frac{x}{L}\,dx\right].
$
$\displaystyle
= \lambda_0 \left[L + \frac{1}{L}\int_{0}^{L} x \,dx\right]
= \lambda_0 \left[L + \frac{1}{L} \cdot \frac{L^2}{2}\right]
= \lambda_0 \left[L + \frac{L}{2}\right]
= \lambda_0 \left(\frac{3L}{2}\right).
$
Thus,
$\displaystyle \lambda_0 = \frac{2M}{3L}.$
Step 3: Write the Expression for the Moment of Inertia
The moment of inertia $I$ of a rod about an axis at one end (point A) is given by integrating $x^2 dm$ over its length. Since $dm = \lambda(x)\,dx$, we have:
$\displaystyle
I = \int_{0}^{L} x^2 \lambda(x)\,dx = \int_{0}^{L} x^2 \lambda_0 \left(1 + \frac{x}{L}\right)\,dx.
$
Substituting $\lambda_0 = \frac{2M}{3L},$ we get:
$\displaystyle
I = \int_{0}^{L} x^2 \left(\frac{2M}{3L}\right) \left(1 + \frac{x}{L}\right)\,dx.
$
Factor out constants:
$\displaystyle
I = \frac{2M}{3L} \int_{0}^{L} x^2 \left(1 + \frac{x}{L}\right)\,dx
= \frac{2M}{3L} \int_{0}^{L} \left(x^2 + \frac{x^3}{L}\right)\,dx.
$
Step 4: Perform the Integration
Split the integral and integrate term by term:
$\displaystyle
I = \frac{2M}{3L} \left[\int_{0}^{L} x^2 \,dx + \int_{0}^{L} \frac{x^3}{L}\,dx\right].
$
$\displaystyle
= \frac{2M}{3L} \left[\int_{0}^{L} x^2 \,dx + \frac{1}{L}\int_{0}^{L} x^3 \,dx\right].
$
Compute these integrals:
$\displaystyle
\int_{0}^{L} x^2\,dx \;=\;\frac{L^3}{3},
\quad
\int_{0}^{L} x^3\,dx \;=\;\frac{L^4}{4}.
$
Substitute these results back in:
$\displaystyle
I = \frac{2M}{3L} \left[\frac{L^3}{3} + \frac{1}{L}\cdot \frac{L^4}{4}\right]
= \frac{2M}{3L} \left[\frac{L^3}{3} + \frac{L^3}{4}\right].
$
$\displaystyle
= \frac{2M}{3L} \left[\frac{4L^3}{12} + \frac{3L^3}{12}\right]
= \frac{2M}{3L} \cdot \frac{7L^3}{12}.
$
Simplify this expression:
$\displaystyle
I = \frac{2M}{3L} \cdot \frac{7L^3}{12}
= \frac{2M \cdot 7 L^3}{3 \cdot 12 \cdot L}
= \frac{14 M L^2}{36}
= \frac{7}{18} M L^2.
$
Step 5: State the Final Answer
The moment of inertia of the rod about an axis passing through A is
$\displaystyle \frac{7}{18} M L^2.$
