Question

The density of a non-uniform rod of length 1m is given by $\rho \left(x\right)=a(1+{bx}^2)$ 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. $\frac{4(2+b)}{3(3+b)}$
• C. $\frac{3(3+b)}{4(2+b)}$
• D. $\frac{4(3+b)}{3(2+b)}$

Answer 1

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

Consider a differential part at the rod at a distance $x$

Then,

$l=\frac{dm}{dx}=a(l+b{x}^2)$

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

We know,

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

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

$X_{COM}=\frac{\frac{1}{2}+\frac{b}{4}}{1+\frac{b}{3}}=\frac{3(2+b)}{4(3+b)}$

Option A is correct.