Question

The center of mass of a non-uniform rod 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

Hint When the center of mass of a body cannot be found using the x-axis of symmetry, it can be found by integration. The thickness or length of the body can be divided into tiny strips, each with thickness $dx$ . Integrate mass of each element multiplied by its respective position concerning $x$ and divide by the integration of mass of each element. We will get the position of the center of mass.
Formula used:
${X_{CM}} = \dfrac{{\int\limits_0^L {xdm} }}{{\int\limits_0^L {dm} }}$

Complete step-by-step solution
An imaginary point in the body of matter, wherever the weighted relative position of all the distributed masses sums up to be zero. This is the point over which a force may be applied to cause some linear acceleration without an angular acceleration. For simple rigid bodies with uniform density, the center of mass is located at the centroid. The center of mass of any system is a point that signifies the mean location of the matter in a body or system.
The center of mass of a system can be calculated by taking the value of all the masses we are trying to find the center of a mass between and multiplying them by their positions. Now, we add these two and divide them by the sum of all the individual masses.
${X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + ....{m_n}{x_n}}}{{{m_1} + {m_2} + ....{m_n}}}$
In the case of a non-uniform rod, the position of the center of mass depends upon the density distribution of the rod.
The center of mass of a non-uniform rod is recognized by,
${X_{CM}} = \dfrac{{\int\limits_0^L {xdm} }}{{\int\limits_0^L {dm} }}$
where,
$L$ is the length of the rod
$x$ is the position of the mass element from one end of the rod
${X_{CM}} = \dfrac{{\int\limits_0^L {xdx} }}{{\int\limits_0^L {\rho dx} }}$
where, $\rho = \dfrac{{dm}}{{dx}}$
We are given,
$\dfrac{{dm}}{{dx}} = \dfrac{{k{x^2}}}{L}$
Therefore, $dm = \dfrac{{k{x^2}}}{L}dx$
${X_{CM}} = \dfrac{{\int\limits_0^L {x\dfrac{{k{x^2}}}{L}dx} }}{{\int\limits_0^L {\dfrac{{k{x^2}}}{L}dx} }}$
${X_{CM}} = \dfrac{{\dfrac{k}{L}\left[ {\dfrac{{{x^4}}}{4}} \right]_0^L}}{{\dfrac{k}{L}\left[ {\dfrac{{{x^3}}}{3}} \right]_0^L}}$
On further solving we get,
${X_{CM}} = \dfrac{{\dfrac{{{L^4}}}{4} - 0}}{{\dfrac{{{L^3}}}{3} - 0}}$
${X_{CM}} = \dfrac{{3L}}{4}$
The center of mass of a given rod is situated at a distance $\dfrac{{3L}}{4}$ from one end of the rod.

Hence, the correct option is (A) $\dfrac{{3L}}{4}$ .

Note If the rod has constant density $\rho $ , specified in terms of mass per unit length, then the mass of the rod is the product of the density and length of the rod. In the above case, we have specified the density of rod in the form of mass per unit length, so while integrating unit mass of rod, we took mass as the product of the density of rod and the position of the unit mass from one end of the rod.