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Step-by-Step Solution
Step 1: Identify the linear mass density
The problem gives the rod’s linear mass density as
$ \rho(x) = a + b \left(\frac{x}{L}\right)^2 $.
Here, $a$ and $b$ are constants, $0 \le x \le L$, and $L$ is the total length of the rod.
Step 2: Express the mass element
An infinitesimal mass element $dm$ on a small segment $dx$ is given by
$ dm = \rho(x)\,dx = \left(a + b \frac{x^2}{L^2}\right) dx. $
Step 3: Calculate the total mass of the rod
The total mass $M$ of the rod is found by integrating $dm$ from $x = 0$ to $x = L$:
$ M = \int_0^L dm
= \int_0^L \left(a + b \frac{x^2}{L^2}\right)\,dx. $
Evaluate this integral:
$ \int_0^L a \,dx = aL. $
$ \int_0^L b \frac{x^2}{L^2} \,dx = b \frac{1}{L^2} \int_0^L x^2 \,dx
= b \frac{1}{L^2} \cdot \frac{L^3}{3}
= \frac{bL}{3}. $
So,
$ M = aL + \frac{bL}{3} = L \left(a + \frac{b}{3}\right). $
Step 4: Set up the center of mass formula
The $x$-coordinate of the center of mass, $X_{\text{com}}$, is given by:
$ X_{\text{com}} = \frac{\int_0^L x\,dm}{\int_0^L dm}. $
Since $dm = \left(a + b \frac{x^2}{L^2}\right)dx,$
$ \int_0^L x \, dm = \int_0^L x \left(a + b \frac{x^2}{L^2}\right) dx. $
Step 5: Evaluate the numerator for $X_{\text{com}}$
Calculate
$ \int_0^L x \left(a + b \frac{x^2}{L^2}\right) dx. $
$ \int_0^L a \,x \,dx = a \left[\frac{x^2}{2}\right]_0^L = \frac{aL^2}{2}. $
$ \int_0^L b \frac{x^3}{L^2} \,dx
= b \frac{1}{L^2} \int_0^L x^3 \,dx
= b \frac{1}{L^2} \cdot \frac{L^4}{4}
= \frac{bL^2}{4}. $
Thus,
$ \int_0^L x \, dm = \frac{aL^2}{2} + \frac{bL^2}{4} = L^2 \left(\frac{a}{2} + \frac{b}{4}\right). $
Step 6: Combine to find $X_{\text{com}}$
Now,
$ X_{\text{com}}
= \frac{\int_0^L x\,dm}{\int_0^L dm}
= \frac{L^2 \left(\frac{a}{2} + \frac{b}{4}\right)}{L \left(a + \frac{b}{3}\right)}
= \frac{L \left(\frac{a}{2} + \frac{b}{4}\right)}{\left(a + \frac{b}{3}\right)}.
$
Simplify the fraction:
$ \frac{\frac{a}{2} + \frac{b}{4}}{a + \frac{b}{3}}
= \frac{\frac{2a + b}{4}}{\frac{3a + b}{3}}
= \frac{2a + b}{4} \cdot \frac{3}{3a + b}
= \frac{3(2a + b)}{4(3a + b)}.
$
Hence,
$ X_{\text{com}} = L \cdot \frac{3(2a + b)}{4(3a + b)}
= \frac{3}{4} \left(\frac{2a + b}{3a + b}\right) L.
$
Therefore, the center of mass is located at
$ \displaystyle \frac{3}{4} \left(\frac{2a + b}{3a + b}\right) L. $
