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Q.
If linear density of a rod of length 3 m varies as $\lambda $ =2+ $x$ , then the position of the centre of gravity of the rod is
NTA AbhyasNTA Abhyas 2020
Solution:
Let rod is placed along $x$ -axis. Mass of element $PQ$ of length $dx$ situated at $x=x$ is
$dm=\lambda \, dx=\left(2 + x\right) \, dx$
The CM of the element has coordinates ( $x$ , 0, 0).
Therefore, $x$ -coordinates of CM of the rod will be
$x_{C M}=\frac{\displaystyle \int _{0}^{3} x d m}{\displaystyle \int _{0}^{3} d m}$
$=\frac{\displaystyle \int _{0}^{3} x \left(2 + x\right) d x}{\displaystyle \int _{0}^{3} \left(2 + x\right) d x}$
$=\frac{\displaystyle \int _{0}^{3} \left(2 x + x^{2}\right) d x}{\displaystyle \int _{0}^{3} \left(2 + x\right) d x}$
$=\frac{\left[\frac{2 x^{2}}{2} + \frac{x^{3}}{3}\right]_{0}^{3}}{\left[2 x + \frac{x^{2}}{2}\right]_{0}^{3}}$
$=\frac{\left[\left(3\right)^{2} + \frac{\left(3\right)^{3}}{3}\right]}{\left[2 \times 3 + \frac{\left(3\right)^{2}}{2}\right]}=\frac{9 + 9}{6 + 9 / 2}$
$=\frac{18 \times 2}{21}=\frac{12}{7}$ m
