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# The linear density of rod of length L and placed along the x-axis with the lighter and at origin, is given by $\lambda = B{x^2}$ where $B$ is a constant. The coordinates of centre of mass are:(A) $\left( {\dfrac{{3L}}{4},0} \right)$(B) $\left( {0,\dfrac{{3L}}{4}} \right)$(C) $\left( {\dfrac{{4L}}{3},0} \right)$(D) $\left( {0,\dfrac{{4L}}{3}} \right)$

Last updated date: 19th Sep 2024
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Answer
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Hint: We need to consider the area of cross section of the rod of length L and negligible width. Mass is the product of density and volume where volume is given by the length of an element.

Formula Used:
The formulae used in the solution are given here.
The x-coordinate of mass is given by ${X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$.

Complete Step by Step Solution: Centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particles appears to be concentrated.
It has been given that the linear density of rod of length L and placed along the x-axis with the lighter end at the origin, is given by $\lambda = B{x^2}$ where $B$ is a constant.
Density is a measure of mass per volume. The average density of an object equals its total mass divided by its total volume.
Consider an element of length $dx$ of the rod at a distance $x$ from the left end. Since the rod is a linear rod, it has negligible thickness and the area of cross section is $a$.
Area of the rod $a = dx$.
Thus, the mass of the element, $dm$ is given by $dm = \lambda dx = \left( {B{x^2}} \right)dx$
The x-coordinate of mass is given by ${X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}$.
It can also be written as, ${X_{CM}} = \dfrac{{\int {dmx} }}{{\int {dm} }} = \dfrac{{\int\limits_0^L {x\left( {B{x^2}} \right)adx} }}{{\int\limits_0^L {\left( {B{x^2}} \right)adx} }}$
Simplifying the equation, we get, $\dfrac{{B\dfrac{{\left[ {{x^4}} \right]_0^L}}{4}}}{{B\dfrac{{\left[ {{x^3}} \right]_0^L}}{3}}} = \dfrac{{B{L^4}}}{4} \times \dfrac{3}{{B{L^3}}}$
Thus, the x-coordinates of the centre of mass are: ${X_{CM}} = \dfrac{{3L}}{4}$.
The y-coordinate is equal to zero, because the rod is linear and has no vertical component.

Hence the correct answer is Option A.

Note: Motion of the point of centre of mass is identical to the motion of a single particle whose mass is equal to the sum of all individual particles of the system and the resultant of all the forces exerted on all the particles of the system by surrounding bodies (or) action of a field of force is exerted directly to that particle. This point is called the centre of mass of the system of particles. The concept of centre of mass (COM) is useful in analysing the complicated motion of the system of objects, particularly When two and more objects collide or an object explodes into fragments.