Question

The mass per unit length of a rod of length $3$ $m$ is directly proportional to distance $x$ from its one end. The centre of mass of the rod from that end will be at

• A. $1.5\text{m}$
• B. $2\text{m}$
• C. $2.5\text{m}$
• D. $3\text{m}$

Answer 1

The correct option is B $2\text{m}$

In a case where mass density is unifoum,
COM= $\frac{L}{2}$

In the given situation, mass density is changing, as low at left side and incresing while moving towards right.

So, in this condition COM will be shifts towards right.
$x_{com}=\frac{m_1{x}_1+m_2{x}_2+....}{m_1+m_2+....}$
$x_{com}=\frac{\int xdm}{dm}$
where, $\frac{dm}{dx}=\lambda =kxdx=\left(kx\right)=dx$

then, $\frac{\int x\left(kx\right)dx}{kxdx}=\frac{\int x^{2}dx}{\int xdx}$
using the limit from $0$ to $L$,
$=\frac{{\int }_0^L{x}^{2}dx}{{\int }_0^{L}xdx}=\frac{[\frac{x^3}{3}{]}_0^L}{[\frac{x^2}{2}{]}_0^L}=\frac{2L}{3}$

So ,the total shift of COM = $\frac{2L}{3}\times 3=2L$
Hence, now the new COM is in 2L position of the rod.