Question

A non uniform thin rod of length l lies along the x axis with one end at the origin. It has linear mass density λ kg/m, given by  λ= λ0 (1+x/l). Find the centre of mass of the rod

Answer 1

See diagram.

center of mass = $\frac{1}{M} \int\limits^L_0 {x} \, dm$

Total Mass = M =
$M = \int\limits^L_0 {} \, dm = \int\limits^L_0 {lambda\ } \, dx = lambda_0 \int\limits^L_0 {(1+\frac{x}{L})\ } \, dx \\ \\M= lambda_0\ [ x + \frac{x^2}{2L} ]_0^L = lambda_0\ [L + \frac{L^2}{2L}-0]\\ \\M=\frac{3\ lambda_0\ L}{2}$

Center of mass =
$= \frac{1}{M} \int\limits^L_0 {x} \, \ dm= \frac{1}{M} \int\limits^L_0 {x\ (density} \, \ dx)\\ \\= \frac{1}{M} * lambda_0 \int\limits^L_0 {x\ (1+\frac{x}{L})} \, \ dx \\ \\= \frac{1}{M} * lambda_0 * [\frac{x^2}{2} + \frac{x^3}{3L} ]_0^L \\ \\= \frac{1}{M} * lambda_0 * [\frac{L^2}{2} + \frac{L^3}{3L} - 0]\\ \\=\frac{1}{M} * lambda_0 * \frac{5}{6}L^2\\ \\=\frac{5\ lambda_0\ L^2}{6M}=\frac{5L^2}{6}*\frac{lambda_0}{M}$

center of mass =
$\frac{5L^2}{6}*\frac{2}{3L}=\frac{5L}{9}$

Center of mass = (5 L / 9, 0)

Answer 2

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