The density of a non-uniform rod of length 1m is given by ρ(x)=a(1+bx2) where a and b are constants and 0≤x≤1. The center of mass of the rod is at k(2+b)4(3+b) , Then k=……..

ρx=a(1+bx2)dmdx=a(1+bx2)dm=a(1+bx2)dxxcm=∫01dm x∫01dm =∫01a(1+bx2) x dx∫01a(1+bx2) dx=∫01(x+bx3) dx∫01(1+bx2) dx=x22+bx4404x+bx3301=12+b41+b3=2+b43+b4=3(2+b)4(3+b)

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The density of a non-uniform rod of length 1m is given by $ρ$ (x)=a(1+bx²) where a and b are constants and $0 \leq x \leq 1$ . The center of mass of the rod is at $\frac{k(2+b)}{4(3+b)}$ , Then k=……..

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Detailed Solution

$ρ (x) = a (1 + b x^{2})$

$\begin{array}{l} \frac{d m}{d x} = a (1 + b x^{2}) \\ d m = a (1 + b x^{2}) d x \end{array}$

$x_{c m} = \frac{\int_{0}^{1} d m x}{\int_{0}^{1} d m} = \frac{\int_{0}^{1} {a(1+bx}^{2}) x dx}{\int_{0}^{1} {a(1+bx}^{2}) dx} = \frac{\int_{0}^{1} {(x+bx}^{3}) dx}{\int_{0}^{1} {(1+bx}^{2}) dx} = \frac{{[\frac{x^{2}}{2} + \frac{b x^{4}}{4}]}_{0}^{4}}{{[x + \frac{b x^{3}}{3}]}_{0}^{1}}$

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

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