Question

If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod this end is:

A. 2.5m
B. 1m
C. 1.5m
D. 2m

Answer 1

Hint: We must know how to use the distribution of mass formalism for an uniform body. The linear density of the uniform rod is $\left( \lambda \right)$, which is proportional to the distance from one end of the rod. Hence, \[\lambda \propto x\Rightarrow \lambda =kx\]. Further the centre of mass of the rod in this case can be found using: ${{x}_{CM}}=\dfrac{\int\limits_{0}^{L}{xdm}}{\int\limits_{0}^{L}{dm}}$. In this case, the centre of mass of the rod is at the same point as the centre of gravity of the rod.

Complete step by step answer:
Let’s start by making a diagram of the problem given to us.

An uniform rod of length (L) having one end named as A and the other end named as B. Given that the length of the rod is 3m, that is: L=3m. The linear mass density, which is the mass per unit length of the rod, is $\left( \lambda \right)$, which is proportional to the distance (x). This distance (x), is the distance from one end of the rod. We will consider it to be the end A. We will consider a small length of the rod $\left( dx \right)$.
We know that, \[\lambda \propto x\Rightarrow \lambda =kx\]. Here, (k) is the constant of proportionality.
Therefore, the mass of the small length of the rod (dx) will be given by: $dm=\lambda dx\Rightarrow dm=\left( kx \right)dx$.
To find the centre of the mass of this uniform rod, we will have to find the summation of all the individual small masses (dm) which are at different distances from the centre of mass and divide it by the summation of all the small masses (dm).

Converting the summation to integration, we find the formula of centre of mass of an uniform rod to be: ${{x}_{CM}}=\dfrac{\int\limits_{0}^{L}{xdm}}{\int\limits_{0}^{L}{dm}}$. The limits of the integration will be from zero to the net length of rod, which is 3m. Hence, substituting in the values from earlier we get:

\[{{x}_{CM}}=\dfrac{\int\limits_{0}^{L}{xdm}}{\int\limits_{0}^{L}{dm}}\Rightarrow {{x}_{CM}}=\dfrac{\int\limits_{0}^{3}{x\left( kxdx \right)}}{\int\limits_{0}^{3}{kxdx}}\Rightarrow {{x}_{CM}}=\dfrac{k\int\limits_{0}^{3}{{{x}^{2}}dx}}{k\int\limits_{0}^{3}{xdx}}\Rightarrow {{x}_{CM}}=\dfrac{\int\limits_{0}^{3}{{{x}^{2}}dx}}{\int\limits_{0}^{3}{xdx}}\].
Putting the values of integration, we get: \[{{x}_{CM}}=\dfrac{\left[ \dfrac{{{x}^{3}}}{3} \right]_{0}^{3}}{\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{3}}\Rightarrow {{x}_{CM}}=\dfrac{\left[ \dfrac{{{3}^{3}}}{3} \right]}{\left[ \dfrac{{{3}^{2}}}{2} \right]}\Rightarrow {{x}_{CM}}=\dfrac{{{3}^{2}}}{\left[ \dfrac{{{3}^{2}}}{2} \right]}\Rightarrow {{x}_{CM}}=2m\].

Hence, the centre of mass of the rod, from end A is at 2m. For an uniform rod, the centre of gravity and the centre of mass coincide. Hence, the centre of gravity of the rod will also be 2m, given by Option D.
Hence, the correct answer is option D.

Note:
The centre of mass of a body is defined as the point, at which the whole mass of the body acts at. That is, if a force acts on the body, then the motion of the body will be the same as the motion followed by the centre of mass of the body.
The centre of gravity of a body is defined as the point at which the net the gravitational force of the body seems to be acting at.
Only for uniform bodies having uniform density throughout, the centre of mass and centre of gravity of the body will coincide.