Question

A rod of length $L$ has non-uniform linear mass density given by

$\rho \left(x\right)=\left(a+b\frac{x^2}{L^2}\right)$, where $a$ and $b$ are constants and

$0\le x\le L$

.The value of $x$ for the center of mass of the rod is at :

• A. $\frac{3L}{2}\left(\frac{2a+b}{3a+b}\right)$
• B. $\frac{3L}{4}\left(\frac{2a+b}{3a+b}\right)$
• C. $\frac{3L}{4}\left(\frac{a+b}{3a+b}\right)$
• D. $\frac{4L}{3}\left(\frac{a+b}{3a+b}\right)$

Answer 1

The correct option is B

$\frac{3L}{4}\left(\frac{2a+b}{3a+b}\right)$

Step 1: Given data,

Let the value of $x$ for the center of mass of the rod is $x_{CM}$.

Length of rod $=L$

Linear mass density $\rho \left(x\right)=a+b{\left(\frac{x}{L}\right)}^2$(Where, $a$ and $b$ are constants)

And here the limit is $0\le x\le L$

Step 2: To find the value of $x$

We can write, Number Density $N$ as,

$\frac{dN}{dx}=\rho$

$$\begin{gathered} dN=\rho dx \\ =\left(a+b{\left(\frac{x}{L}\right)}^2\right).dx \end{gathered}$$

To calculate the value of $x$ for the center of mass of the rod we can write $x_{CM}$,

$$\begin{gathered} x_{CM}=\frac{\int x.dm}{\int dm} \\ =\frac{\int x\rho dx}{\int \rho dx} \\ =\frac{{\int }_0^{1}x\left(a+\left({\displaystyle \frac{b{x}^2}{L^2}}\right)\right)dx}{{\int }_0^1\left(a+{\displaystyle \frac{b{x}^2}{L^2}}\right)dx} \\ =\frac{{\int }_0^1\left(ax+{\displaystyle \frac{b{x}^3}{L^2}}\right)dx}{{\int }_0^1\left(a+{\displaystyle \frac{b{x}^2}{L^2}}\right)dx} \\ =\frac{{\left[{\displaystyle \frac{a{x}^2}{2}}\right]}_0^1+{\displaystyle \frac{b}{L^2}}{\left[{\displaystyle \frac{x^4}{4}}\right]}_0^1}{a{\left[x\right]}_0^1+{\displaystyle \frac{b}{L^2}}{\left[{\displaystyle \frac{x^3}{3}}\right]}_0^1} \\ =\frac{{\displaystyle \frac{a_{}^{1^2}}{2}}+{\displaystyle \frac{b^{1^2}}{4}}}{aL+{\displaystyle \frac{bL}{3}}} \\ =\frac{\left(2a+b\right)L}{\left(3a+b\right)4}\times 3 \\ =\frac{3L}{4}\left(\frac{2a+b}{3a+b}\right) \end{gathered}$$

Hence the value of $x=\frac{3L}{4}\left(\frac{2a+b}{3a+b}\right)$

Therefore, the correct answer is Option $\text{'}B\text{'}$.