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 from this end is
A.
$$2.5\,m$$
B.
$$1\,m$$
C.
$$1.5\,m$$
D.
$$2\,m$$
Answer :
$$2\,m$$
Solution :
Consider an element of length $$dx$$ at a distance $$x$$ from end $$A.$$
Here, mass per unit length $$\lambda $$ of rod
$$\eqalign{ & \lambda \propto x \Rightarrow \lambda = kx \cr & \therefore dm = \lambda dx = kx\,dx \cr} $$
Position of centre of gravity of rod from end $$A.$$
$$\eqalign{ & {x_{CG}} = \frac{{\int_0^L x dm}}{{\int_0^L d m}} \cr & \therefore {x_{CG}} = \frac{{\int_0^3 x \left( {kx\,dx} \right)}}{{\int_0^3 k x\,dx}} = \frac{{\left[ {\frac{{{x^3}}}{3}} \right]_0^3}}{{\left[ {\frac{{{x^2}}}{2}} \right]_0^3}} = \frac{{\frac{{{{\left( 3 \right)}^3}}}{3}}}{{\frac{{{{\left( 3 \right)}^2}}}{2}}} = 2\,m \cr} $$
Consider an element of length $$dx$$ at a distance $$x$$ from end $$A.$$
Here, mass per unit length $$\lambda $$ of rod
$$\eqalign{ & \lambda \propto x \Rightarrow \lambda = kx \cr & \therefore dm = \lambda dx = kx\,dx \cr} $$
Position of centre of gravity of rod from end $$A.$$
$$\eqalign{ & {x_{CG}} = \frac{{\int_0^L x dm}}{{\int_0^L d m}} \cr & \therefore {x_{CG}} = \frac{{\int_0^3 x \left( {kx\,dx} \right)}}{{\int_0^3 k x\,dx}} = \frac{{\left[ {\frac{{{x^3}}}{3}} \right]_0^3}}{{\left[ {\frac{{{x^2}}}{2}} \right]_0^3}} = \frac{{\frac{{{{\left( 3 \right)}^3}}}{3}}}{{\frac{{{{\left( 3 \right)}^2}}}{2}}} = 2\,m \cr} $$
