Obtain the position of centre of mass of a thin rod of uniform density.
Obtain the position of centre of mass of a thin rod of uniform density.
Homogeneous body means a body with uniformly distributed mass.
Let us consider a thin rod, whose width and breadth (in case of cross-section of the rod is rectangular) or radius (In case of the cross section of the rod is cylindrical) is much smaller than its length.
Taking the origin to be at the geometric centre of the rod and $X$-axis to be along the length of the rod, for every element $d m$ of the rod at $+x$ there is an element of the same mass $d m$ located at $-x$
The net contribution of every such pair to the integral and hence the integral $\int x d m$ itself is zero. $\therefore$ Centre of mass of rod,
$\mathrm{X}=\frac{1}{\mathrm{M}} \int x d m$
but $d m=\lambda d x$ where $\lambda=$ mass per unit length $=\frac{M}{L}$
$\quad=\frac{\mathrm{M}}{\mathrm{L}} d x \text { where } \mathrm{L}=\text { length of rod, } $
$\therefore \mathrm{X}=\frac{1}{\mathrm{M}} \int x \frac{\mathrm{M}}{\mathrm{L}} d x$
$=\frac{1}{\mathrm{~L}} \int_{0}^{\mathrm{L}} x d x$
$=\frac{1}{\mathrm{~L}}\left[\frac{x^{2}}{2}\right]_{0}^{\mathrm{L}}$
$=\frac{1}{\mathrm{~L}}\left[\frac{\mathrm{L}^{2}}{2}\right]$
$=\frac{\mathrm{L}}{2}$
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