# A rod of length L has non-uniform linear mass density

Question:

A rod of length $L$ has non-uniform linear mass density

given by $\rho(x)=a+b\left(\frac{x}{\mathrm{~L}}\right)^{2}$, where $a$ and $b$ are constants

and $0 \leq x \leq \mathrm{L}$. The value of $x$ for the centre of mass of the rod is at:

1. (1) $\frac{3}{2}\left(\frac{a+b}{2 a+b}\right) \mathrm{L}$

2. (2) $\frac{3}{4}\left(\frac{2 a+b}{3 a+b}\right) \mathrm{L}$

3. (3) $\frac{4}{3}\left(\frac{a+b}{2 a+3 b}\right) \mathrm{L}$

4. (4) $\frac{3}{2}\left(\frac{2 a+b}{3 a+b}\right) \mathrm{L}$

Correct Option: , 2

Solution:

(2) Given,

Linear mass density, $\rho(x)=a+b\left(\frac{x}{L}\right)^{2}$

$X_{C M}=\frac{\int x d m}{\int d m}$

$\int d m=\int_{0}^{L} \rho(x) d x$

$=\int_{0}^{L}\left[a+b\left(\frac{x}{L}\right)^{2}\right] d x=a L+\frac{b L}{3}$

$\int_{0}^{L} x d m=\int_{0}^{L}\left(a x+\frac{b x^{3}}{L^{2}}\right) d x=\left(\frac{a L^{2}}{2}+\frac{b L^{2}}{4}\right)$

$\therefore \quad X_{\mathrm{CM}}=\frac{\left(\frac{a L^{2}}{2}+\frac{b L^{2}}{4}\right)}{a L+\frac{b L}{3}}$

$\Rightarrow X_{\mathrm{CM}}=\frac{3 L}{4}\left(\frac{2 a+b}{3 a+b}\right)$

Leave a comment

Click here to get exam-ready with eSaral