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Step-by-Step Solution
Step 1: Understand the Problem
A rod of length 3 m has a mass distribution such that the linear mass density is directly proportional to the distance x from one end of the rod. We want to find the position of its center of gravity (center of mass) from that same end.
Step 2: Express the Mass Element
Let the proportionality constant be $k$. At a distance $x$ from the chosen end, the small mass element for an infinitesimal length $dx$ is:
$ dm = k \, x \, dx $
Step 3: Set Up the Integral for Center of Mass
The center of mass (or center of gravity) $x_c$ is given by:
$ x_c = \frac{\int x \, dm}{\int dm} $
Since $ dm = k \, x \, dx $, we substitute in the integrals. For $x$ ranging from 0 to 3 m:
$ x_c = \frac{\int_{0}^{3} x \,(k \, x \, dx)}{\int_{0}^{3} k \, x \, dx}
= \frac{\int_{0}^{3} k \, x^{2} \, dx}{\int_{0}^{3} k \, x \, dx} $
Step 4: Solve the Integrals
Factor out the constant $k$ from numerator and denominator; it cancels out:
$ x_c = \frac{\int_{0}^{3} x^{2} \, dx}{\int_{0}^{3} x \, dx} $
Compute each integral separately:
Numerator: $ \int_{0}^{3} x^{2} \, dx = \left[ \frac{x^3}{3} \right]_{0}^{3} = \frac{3^3}{3} - 0 = \frac{27}{3} = 9 $
Denominator: $ \int_{0}^{3} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{3} = \frac{3^2}{2} - 0 = \frac{9}{2} $
Hence,
$ x_c = \frac{9}{\frac{9}{2}} = \frac{9}{1} \times \frac{2}{9} = 2 $
Step 5: State the Final Answer
The center of gravity of the rod is located at a distance of 2 m from the chosen end.
