Question

A rod of uniform thickness is placed along x-axis with one end at origin. If the length of rod is $L$ and its linear mass density is proportional to $x$, then find the distance of its center of mass from origin.

Answer 1

we will first consider a infinitesimal small length of rod having mass $dm$ that will be around the point $x$. We will first calculate the total mass of the rod by using the suitable formula. Then, we will calculate the distance of center of mass of the rod by using the formula given below.

Formula used:
The distance of center of mass of the rod can be calculated as shown below
$C.M. = \dfrac{1}{M}\int\limits_0^L {x\,dm}$
Here, C.M. is the distance of the center of mass of the rod, $M$ is the total mass of the rod and $x$ is the point on the rod.
Now, the total mass of the rod can be calculated as shown below
$M = \int\limits_0^L {dm}$
Here, $dm$ is the mass of an infinitesimal small rod.

Complete step by step answer:
Consider a rod that will be of uniform thickness and will be placed along x-axis with one end at origin. Let the length of the rod is $L$. Now consider a infinitesimal small rod of length $dl$ and mass $dm$ that will be around the point $x$. As given, the linear mass density is proportional to $x$. Therefore,
The density of rod is, $\lambda = {\lambda _0}x$
And the length of the rod $= L$
The distance of center of mass of the rod can be calculated as shown below
$C.M. = \dfrac{1}{M}\int\limits_0^L {x\,dm}$
Now, the total mass of the rod can be calculated as shown below
$M = \int\limits_0^L {dm}$
Now, if we put $m = \lambda x$
Then $dm = \lambda dx$
Putting this value in the equation of total mass of the rod, we get
$M = \int\limits_0^L {\lambda dx}$
$\Rightarrow \,M = \int\limits_0^L {{\lambda _0}xdx}$
$\Rightarrow \,M = {\lambda _0}\int\limits_0^L {xdx}$
$\Rightarrow \,M = {\lambda _0}\left[ {\dfrac{{{x^2}}}{2}} \right]_0^L$
$\Rightarrow \,M = {\lambda _0}\left[ {\dfrac{{{L^2}}}{2}} \right]$
$\Rightarrow \,M = {\lambda _0}\dfrac{{{L^2}}}{2}$
Therefore, the total mass of the rod is ${\lambda _0}\dfrac{{{L^2}}}{2}$.
Putting this value in the formula of center of mass as given below
$C.M. = \dfrac{1}{M}\int\limits_0^L {x\,dm}$
$\Rightarrow \,C.M. = \dfrac{2}{{{\lambda _0}{L^2}}}\int\limits_0^L {x\,dm}$
$\Rightarrow \,C.M. = \dfrac{2}{{{\lambda _0}{L^2}}}\int\limits_0^L {x\left( {\lambda dx} \right)}$
$\Rightarrow \,C.M. = \dfrac{2}{{{\lambda _0}{L^2}}}\int\limits_0^L {x\left( {{\lambda _0}xdx} \right)}$
$\Rightarrow \,C.M. = \dfrac{2}{{{L^2}}}\int\limits_0^L {x\left( {xdx} \right)}$
$\Rightarrow \,C.M. = \dfrac{2}{{{L^2}}}\int\limits_0^L {\left( {{x^2}dx} \right)}$
$\Rightarrow \,C.M. = \dfrac{2}{{{L^2}}}\left[ {\dfrac{{{x^3}}}{3}} \right]_0^L$
$\Rightarrow \,C.M. = \dfrac{2}{{{L^2}}}\left[ {\dfrac{{{L^3}}}{3}} \right]$
$\therefore \,C.M. = \dfrac{{2L}}{3}$

Therefore, the distance of center of mass from the origin is $\dfrac{{2L}}{3}$.

Note: The alternate way to solve the above equation is given below:now, if we draw a triangle depicting that the mass distribution of the mass will be along the x-axis. Here, the distance of the center of mass of the rod will be the centroid of the triangle. Also, we know that the centroid of the triangle will be equal to $\dfrac{2}{3}$ the length of the median (the midpoint of one side of the triangle to an opposite of vertex). Here, if we consider that the length of the median is $L$. Then the centroid of the triangle will $\dfrac{{2L}}{3}$.This is the required answer.