
What is the centre of mass of a non-uniform rod of length \[L\] which has mass per unit length $\lambda = \dfrac{{k{x^2}}}{L}$ where $k$ is a constant and $x$ is the distance from the one end?
(A) $\dfrac{{3L}}{4}$
(B) $\dfrac{L}{8}$
(C) $\dfrac{k}{L}$
(D) $\dfrac{{3k}}{L}$
Answer
221.4k+ views
Hint As given that $\lambda = \dfrac{{k{x^2}}}{L}$
$\therefore dm = \dfrac{{k{x^2}}}{L}dx$
To find the centre of mass of non-uniform rod of length \[L\] so, we will use the formula-
${X_{CM}} = \dfrac{{\int\limits_0^L {xdm} }}{{\int\limits_0^L {dm} }} \cdots (1)$
where, ${X_{CM}}$ is the centre of mass of object
$L$ is the length of object
Complete step-by-step answer:
According to the question, it is given that the length of rod is $L$ so, we will take the limit for integration in the equation $(1)$ we get
${X_{CM}} = \dfrac{{\int\limits_0^L {xdm} }}{{\int\limits_0^L {dm} }}$
Now, putting the value of $dm$ in the above equation
${X_{CM}} = \dfrac{{\int\limits_0^L {x\dfrac{{k{x^2}}}{L}dx} }}{{\int\limits_0^L {\dfrac{{k{x^2}}}{L}dx} }}$
Integrating both sides with respect to $x$, we get
${X_{CM}} = \dfrac{{\dfrac{k}{L}\left[ {\dfrac{{{x^4}}}{4}} \right]_0^L}}{{\dfrac{k}{L}\left[ {\dfrac{{{x^3}}}{3}} \right]_0^L}}$
Cancelling $\dfrac{k}{L}$ on denominator and numerator we get
${X_{CM}} = \dfrac{{\left[ {\dfrac{{{x^4}}}{4}} \right]_0^L}}{{\left[ {\dfrac{{{x^3}}}{3}} \right]_0^L}}$
Now, using the limits in the above equation we get
$
{X_{CM}} = \dfrac{{\dfrac{{{L^4}}}{4} - 0}}{{\dfrac{{{L^3}}}{3} - 0}} \\
{X_{CM}} = \dfrac{{3L}}{4} \\
$
So, the centre of mass for non-uniform rod is $\dfrac{{3L}}{4}$
Hence, the correct option is (A)
Note The point at which the whole mass of the body appears to be concentrated is called the centre of mass of a body. In other words, the point where all the masses of the system of particles appear to be concentrated is called the centre of mass of the system of particles. Motion of this point is the same to the motion of a single particle whose mass is equal to addition of individual particles of the system and all the forces exerted on all the particles of the system by bodies which are surrounding it.
Internal force of a system can be defined as the force of mutual interaction between particles of the system.
$\therefore dm = \dfrac{{k{x^2}}}{L}dx$
To find the centre of mass of non-uniform rod of length \[L\] so, we will use the formula-
${X_{CM}} = \dfrac{{\int\limits_0^L {xdm} }}{{\int\limits_0^L {dm} }} \cdots (1)$
where, ${X_{CM}}$ is the centre of mass of object
$L$ is the length of object
Complete step-by-step answer:
According to the question, it is given that the length of rod is $L$ so, we will take the limit for integration in the equation $(1)$ we get
${X_{CM}} = \dfrac{{\int\limits_0^L {xdm} }}{{\int\limits_0^L {dm} }}$
Now, putting the value of $dm$ in the above equation
${X_{CM}} = \dfrac{{\int\limits_0^L {x\dfrac{{k{x^2}}}{L}dx} }}{{\int\limits_0^L {\dfrac{{k{x^2}}}{L}dx} }}$
Integrating both sides with respect to $x$, we get
${X_{CM}} = \dfrac{{\dfrac{k}{L}\left[ {\dfrac{{{x^4}}}{4}} \right]_0^L}}{{\dfrac{k}{L}\left[ {\dfrac{{{x^3}}}{3}} \right]_0^L}}$
Cancelling $\dfrac{k}{L}$ on denominator and numerator we get
${X_{CM}} = \dfrac{{\left[ {\dfrac{{{x^4}}}{4}} \right]_0^L}}{{\left[ {\dfrac{{{x^3}}}{3}} \right]_0^L}}$
Now, using the limits in the above equation we get
$
{X_{CM}} = \dfrac{{\dfrac{{{L^4}}}{4} - 0}}{{\dfrac{{{L^3}}}{3} - 0}} \\
{X_{CM}} = \dfrac{{3L}}{4} \\
$
So, the centre of mass for non-uniform rod is $\dfrac{{3L}}{4}$
Hence, the correct option is (A)
Note The point at which the whole mass of the body appears to be concentrated is called the centre of mass of a body. In other words, the point where all the masses of the system of particles appear to be concentrated is called the centre of mass of the system of particles. Motion of this point is the same to the motion of a single particle whose mass is equal to addition of individual particles of the system and all the forces exerted on all the particles of the system by bodies which are surrounding it.
Internal force of a system can be defined as the force of mutual interaction between particles of the system.
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