Volume, mass and centre of mass of a bar

Text

A bar $3\ \mathrm m$ long has a circular cross-section of area $0.1\ \mathrm{m^2}$ and increasing density (mass per unit volume) $\rho(x)=x+1$, where $x$ is the distance from one end. Find: (a) the volume; (b) the mass; (c) the position of the centre of mass.

Solution

(a) Volume. The cross-section is constant, so $V=(\text{cross-section area})\cdot(\text{length})=0.1\cdot 3=0.3\ \mathrm{m^3}.$

(b) Mass. A slice of thickness $dx$ has volume $0.1\,dx$ and mass $dm=\rho(x)\cdot 0.1\,dx$: $m=\int_0^3 0.1(x+1)\,dx=0.1\left[\frac{x^2}{2}+x\right]_0^3=0.1\left(\frac92+3\right)=0.75\ \mathrm{kg}.$

(c) Centre of mass. $x_{cm}=\frac{\int_0^3 x\,\rho(x)\,0.1\,dx}{m}=\frac{0.1\int_0^3(x^2+x)\,dx}{0.75}=\frac{0.1\left(9+\tfrac92\right)}{0.75}=\frac{0.1\cdot 13.5}{0.75}=1.8\ \mathrm m.$ The centre of mass is shifted towards the denser end, as expected.

$\boxed{V=0.3\ \mathrm{m^3},\quad m=0.75\ \mathrm{kg},\quad x_{cm}=1.8\ \mathrm m}$