Question

The density of a non-uniform rod of length $1m$ is given by $\rho \left(x\right)=a(1+b{x}^2)$ where $a$ and $b$ are constants and $0\le x\le 1$
The centre of mass of the rod will be at:

• A. $\frac{3(2+b)}{4(3+b)}$
• B. $\frac{4(3+b)}{3(2+b)}$
• C. $\frac{4(2+b)}{3(3+b)}$
• D. $\frac{3(3+b)}{4(2+b)}$

Answer 1

The correct option is A $\frac{3(2+b)}{4(3+b)}$

Formual Used:

$X_{COM}=\frac{\int \left(dm\right)x}{\int \left(dm\right)}$

Given,

Non uniform density,$\rho \left(x\right)=a(1+b{x}^2)$

Consider a small part at the rod at a distance $x.$ Assuming the rod to be of uniform cross section $\left(A\right)$, then

$\rho \left(x\right)=a(1+b{x}^2)$

$\Rightarrow dm=a(1+b{x}^2)$

We know,

$X_{COM}=\frac{{\int }_0^1\left(dm\right)x}{{\int }_0^1\left(dm\right)}=\frac{{\int }_0^{a}a(1+b{x}^2)Ax.dx}{{\int }_0^{1}a(1+b{x}^2)Adx}$

$X_{COM}=\frac{aA{\int }_0^1(x+b{x}^3)dx}{aA{\int }_0^1(1+b{x}^2).dx}\frac{{[\frac{x^2}{2}+\frac{b{x}^4}{4}]}_0^1}{{[x+\frac{b{x}^3}{4}]}_0^1}{X}_{COM}=\frac{\frac{1}{2}+\frac{b}{4}}{1+\frac{b}{}3}=\frac{3(2+b)}{4(3+b)}$

Alternative:
The answer can also be deducted by converting the given non uniform rod into a uniform rod i.e. if $b=0,$ density becomes uniform (being equal to $a$) along the length. Therefore, the centre of mass will be at the midpoint of the rod. Thus substituting $b=0,$ only option $\left(a\right)$ gives the position of $\text{COM}$ at $0.5m.$

Final Answer: Option (a)