Question

A rod of length L with linear mass density $\lambda = kx$ is placed along the x-axis with one end at origin. The distance of CM of rod form origin is
A) $\dfrac{L}{3}$
B) $\dfrac{{2L}}{3}$
C) $\dfrac{{3L}}{5}$
D) $\dfrac{L}{2}$

Answer 1

Hint: The Centre of mass of a body is the point where the whole mass of the body is concentrated. The point of the centre of mass is denoted using the coordinates of that point. Use the formula for the centre of mass of a rigid body to obtain the coordinate of the centre of mass. A rod is a one-dimensional body, hence the y and z coordinates of the centre of mass will be zero.

Formula used:
The formula for the centre of mass:
${X_{CM}} = \dfrac{{\int {xdm} }}{M}$
Where ${X_{_{CM}}}$stands for the coordinate of the centre of mass of the body, X stands for the coordinate of the small element under consideration, dm stands for the mass of the small element under consideration, and M stands for the mass of the rigid body.

Complete step by step solution:

Let L be the length of the rod. Consider a small element dx as shown in the figure.
It is given that the mass density of $\lambda = kx$
The mass of the small element dx is given by, $dm = \lambda dx$
Substituting the value of $\lambda $ in the above equation, we get
$dm = kxdx$…………………………………………………(1)
As the rod is a one dimensional rigid body, and is placed along the x axis, therefore the Y and Z coordinates will be zero.
$\therefore $The coordinates of centre of mass is $\left( {x,0,0} \right)$
The coordinate of centre of mass can be written as
${X_{_{CM}}} = \dfrac{{\int\limits_0^L {xdm} }}{M}$ ……………………………………………….(2)
The mass of the rigid body can be obtained by integrating the small mass dm
$M = \int {dm} $ …………………………………………………………….(3)
Substitute equation (3) in equation (2)
${X_{CM}} = \dfrac{{\int\limits_0^L {xdm} }}{{\int {dm} }}$
Substituting for dm from equation (1)
${X_{CM}} = \dfrac{{\int\limits_0^L {x(kx)dx} }}{{\int\limits_0^L {(kx)dx} }}$
Taking the constant outside the integral
${X_{CM}} = \dfrac{{k\int\limits_0^L {{x^2}dx} }}{{k\int\limits_0^L {xdx} }}$
Integrating the above equation, we get
${X_{CM}} = \dfrac{{\left[ {\dfrac{{{x^3}}}{3}} \right]_0^L}}{{\left[ {\dfrac{{{x^2}}}{2}} \right]_0^L}}$
Applying the limits we get,
${X_{CM}} = \dfrac{{\dfrac{{{L^3}}}{3}}}{{\dfrac{{{L^2}}}{2}}}$
Simplifying,
${X_{CM}} = \dfrac{{2{L^3}}}{{3{L^2}}} = \dfrac{{2L}}{3}$

The correct answer is (B), $\dfrac{{2L}}{3}.$

Note: Depending on the shape and dimension of the rigid body, the coordinates of the centre of mass varies. For one dimensional rigid body, there will be only one coordinate depending on which axis the body is aligned. There will be two and three coordinates for the two-dimensional and three-dimensional bodies. The centre of mass will greatly depend on the symmetry of the object. Therefore to find the centre of mass of bodies with irregular shapes we have to compare them with the centre of mass of bodies with regular shapes that have a close resemblance to the irregular body.