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Question: The density of a non – uniform rod of length 1 m is given by \(\rho(x) = a(1 + bx^{2})\) where a and...

The density of a non – uniform rod of length 1 m is given by ρ(x)=a(1+bx2)\rho(x) = a(1 + bx^{2}) where a and b are constants and 0≤x≤1.0 \leq x \leq 1.

The centre of mass of the rod will be at.

A

3(2+b)4(3+b)\frac{3(2 + b)}{4(3 + b)}

B

4(2+b)3(3+b)\frac{4(2 + b)}{3(3 + b)}

C

3(3+b)4(2+b)\frac{3(3 + b)}{4(2 + b)}

D

4(3+b)3(2+b)\frac{4(3 + b)}{3(2 + b)}

Answer

3(2+b)4(3+b)\frac{3(2 + b)}{4(3 + b)}

Explanation

Solution

Mass of a small element of length dx of the rod at a distance x from the one end of the rod is

dm=ρdx=a(1+bx2)dxdm = \rho dx = a(1 + bx^{2})dx

The centre of mass of the rod is

XCM=∫01xdm∫01dm∫01xa(1+bx2)dx∫01a(1+bx2)dxX_{CM} = \frac{\int_{0}^{1}{xdm}}{\int_{0}^{1}{dm}}\frac{\int_{0}^{1}{xa(1 + bx^{2})dx}}{\int_{0}^{1}{a(1 + bx^{2})dx}}

=∫01(x+bx2)dx∫01(1+bx2)dx=[x22+bx44]01[x+bx33]01=[12+b4][1+b3]=3(2+b)4(3+b)= \frac{\int_{0}^{1}{(x + bx^{2})dx}}{\int_{0}^{1}{(1 + bx^{2})dx}} = \frac{\left\lbrack \frac{x^{2}}{2} + \frac{bx^{4}}{4} \right\rbrack_{0}^{1}}{\left\lbrack x + \frac{bx^{3}}{3} \right\rbrack_{0}^{1}} = \frac{\left\lbrack \frac{1}{2} + \frac{b}{4} \right\rbrack}{\left\lbrack 1 + \frac{b}{3} \right\rbrack} = \frac{3(2 + b)}{4(3 + b)}