Question:
A rod of length $L$ has non-uniform linear mass density given by $\rho\left(x\right)=a+b\left(\frac{x}{L}\right)^{2},$ where a and b are constants and $0 \le x \le L.$ The value of $x$ for the centre of mass of the rod is a t :
A rod of length $L$ has non-uniform linear mass density given by $\rho\left(x\right)=a+b\left(\frac{x}{L}\right)^{2},$ where a and b are constants and $0 \le x \le L.$ The value of $x$ for the centre of mass of the rod is a t :
Updated On: Jul 8, 2024
- $\frac{3}{2}\left(\frac{2a+b}{3a+b}\right)L$
- $\frac{3}{2}\left(\frac{a+b}{2a+b}\right)L$
- $\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L$
- $\frac{4}{3}\left(\frac{a+b}{2a+3b}\right)L$
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The Correct Option is C
Solution and Explanation
$x_{cm} = \frac{\int xdm}{\int dm} =\frac{\int\left(\lambda dx\right)x}{\int dm}$
$= \frac{\int^{L}_{0}\left(a+\frac{bx^{2}}{L^{2}}\right)xdx}{\int ^{L}_{0}\left(a+\frac{bx^{2}}{L^{2}}\right)dx}$
$= \frac{\frac{aL^{2}}{2}+\frac{b}{L^{2}}. \frac{L^{4}}{4}}{aL+\frac{b}{L^{2}}. \frac{L^{3}}{3}}$
$= \frac{\left(\frac{4a+2b}{8}\right)L}{\frac{\left(3a+b\right)}{3}} = \frac{3}{4} \frac{\left(2a+b\right)L}{\left(3a+b\right)}$
$= \frac{\int^{L}_{0}\left(a+\frac{bx^{2}}{L^{2}}\right)xdx}{\int ^{L}_{0}\left(a+\frac{bx^{2}}{L^{2}}\right)dx}$
$= \frac{\frac{aL^{2}}{2}+\frac{b}{L^{2}}. \frac{L^{4}}{4}}{aL+\frac{b}{L^{2}}. \frac{L^{3}}{3}}$
$= \frac{\left(\frac{4a+2b}{8}\right)L}{\frac{\left(3a+b\right)}{3}} = \frac{3}{4} \frac{\left(2a+b\right)L}{\left(3a+b\right)}$
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