Calculation of center of mass of a bar with linear density by branches

Question

A query. If a bar of lengt $d$ has a linear density $\rho(x)=\begin{cases} x & x\in [a,b] \\ x^2 & x\in [b,c]\\ x^3 & [c,d] \end{cases}$ then its center of mass is $\overline{x}=\frac{\int_{a}^{b}x^2dx+\int_{b}^{c}x^ 3dx+\int_{c}^{d}x^4dx}{\int_{a}^{b}xdx+\int_{b}^{c}x^2dx+\int_{c}^{d}x^3dx}$? I want to verify if I am correct. Thank you very much.

Answer 1

yes. remember that the CoM of a continuous mass distribution $\mathbb M$ with mass density $\mathrm dm = \rho(\mathbf{\vec r}) \, \mathrm dV$ is defined as $$\mathbf{\vec R} = \dfrac1M \int_\mathbb M \mathbf{\vec r} \, \mathrm dm$$ where $M$ is the body's mass, $$M=\int_\mathbb M \mathrm dm$$

In the 1-D case over a body of length $L$, this becomes $$\bar x=\dfrac1M \int_L x \rho(x) \, \mathrm dx, \qquad M=\int_L \rho(x) \, \mathrm dx$$

Your result is obtained simply by integrating the function $\rho (x)$, which is integrated piecewise