A straight rod of length $L$ has one of its ends at the origin and the other at $\mathcal{x =}L$. If the mass per unit length of the rod is given by $Ax$ is constant, where is its mass centre?

Question

Dr. Abubacker · Accepted Answer

(b) Let the mass of an element of length $dx$ of rod located at a distance $x$ away from left end is $\frac{M}{L}\ dx$. The $x$-coordinate of the centre of mass is given by
$X_{CM} = \frac{1}{M}\ \int_{}^{}{x\ dm}$
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$= \frac{1}{M}\ \int_{0}^{L}{x\left( \frac{M}{L}dx \right)}$
=$\frac{1}{L}\left\lbrack \frac{x^{2}}{2} \right\rbrack_{0}^{L} = \frac{L}{2}$

Q. A straight rod of length $L$ has one of its ends at the origin and the other at $\mathcal{x =}L$. If the mass per unit length of the rod is given by $Ax$ is constant, where is its mass centre?

Explanation

Reviewed by Dr. Abubacker

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