Question

The center of mass of a non -uniform rods of length $L$ whose mass per unit length $\lambda = \dfrac{{k{x^2}}}{L}$ where $k$ is a constant and $x$ is the distance from the one end is:
(A) $\dfrac{{3L}}{4}$
(B) $\dfrac{L}{8}$
(C) $\dfrac{k}{L}$
(D) $\dfrac{{3k}}{L}$

Answer 1

The Center of mass of a rod and the geometric center of a rod coincides with each other, this only happens when the rod has a uniform density. But when there is a non-uniform rod then we have to calculate its center of mass using calculus.
Formula used :
${x_{com}} = \dfrac{{\int\limits_0^L {x.dm} }}{{\int\limits_0^L {dm} }}$
Where ${x_{com}}$ is the x-coordinate of the center of mass, $dm$ is the elemental mass at a distance $x$ from one end of the rod, and $L$ is the total length of the rod.

Complete step-by-step answer:
Let the elemental mass of an elemental length of the rod at a distance $x$ be $dm$.
We know that,
${x_{com}} = \dfrac{{\int\limits_0^L {x.dm} }}{{\int\limits_0^L {dm} }}$
Where ${x_{com}}$ is the x-coordinate of the center of mass, $dm$ is the elemental mass at a distance $x$ from one end of the rod.
We can say that for a body sum of all elemental mass will be the total mass of the body let it be $M$.
Therefore,
$\int\limits_0^L {dm} = M$
It is given that the total length of the rod is $L$and the equation of the distance varying linear density($\lambda$) is $\lambda = \dfrac{{k{x^2}}}{L}$where $k$ is a constant and $x$is the distance from the one end of the rod.
Hence the elemental mass $dm$ of the elemental length $dx$of the rod will be
$dm = \lambda dx$
As mass is the product of linear density and length.
$\Rightarrow dm = \dfrac{{k{x^2}}}{L}dx$
Hence using the formulas stated above,
$\Rightarrow {x_{com}} = \dfrac{{\int\limits_0^L {x\dfrac{{k{x^2}}}{L}dx} }}{{\int\limits_0^L {dm} }}$
$\Rightarrow {x_{com}} = \dfrac{{\int\limits_0^L {\dfrac{{k{x^3}}}{L}dx} }}{M}$
$\Rightarrow {x_{com}} = \dfrac{{k\int\limits_0^L {{x^3}dx} }}{{LM}}$
$\Rightarrow {x_{com}} = \dfrac{{k(\dfrac{{{x^4}}}{4})_0^L}}{{LM}}$
$\Rightarrow {x_{com}} = \dfrac{{k(\dfrac{{{L^4}}}{4} - 0)}}{{LM}}$
$\Rightarrow {x_{com}} = \dfrac{{k{L^3}}}{{4M}}$
We stated above that $\int\limits_0^L {dm} = M$and $dm = \dfrac{{k{x^2}}}{L}dx$
$\Rightarrow \int\limits_0^L {\dfrac{{k{x^2}}}{L}dx} = M$
$\Rightarrow (\dfrac{{k{x^3}}}{{3L}})_0^L = M$
$\Rightarrow \dfrac{{k{L^2}}}{3} = M$
We know that ${x_{com}} = \dfrac{{k{L^3}}}{{4M}}$
Hence
${x_{com}} = \dfrac{{3ML}}{{4M}}$
$\Rightarrow {x_{com}} = \dfrac{{3L}}{4}$

Therefore the correct answer to the above question is (A) $\dfrac{{3L}}{4}$

Note:
To solve the above question one must know the basics of calculus that $\int\limits_a^b {{x^n}dx} = (\dfrac{{{x^{n + 1}}}}{{n + 1}})_a^b$ . Here we did not calculate the y-coordinate of the center of mass as from the question we came to know that the mass is changing along the x-axis not along the y-axis so the center of mass along the y-axis will coincide with the y-coordinate of the geometric center.