A solid sphere of density $ρ$ and radius $R$ is centered at the origin. It has a spherical cavity in it that is of radius $R/4$ and which is centered at $(R/2, 0, 0)$, i.e. a small sphere of material has been removed from the large sphere. What is the the center of mass $R_{cm} = (x_{cm}, y_{cm}, z_{cm})$ of the large sphere, including the cavity?
My attempt:
$$\vec{R}=\frac{1}{M}\int \vec{r}dm$$
$M=\rho V=\rho (V_{total}-V_{cavity})=$total mass
$\vec{r}=x^2\hat x + y^2 \hat y+z^2 \hat z$
$dm= \rho dV$
$$\vec{R}=\frac{1}{\rho (V_{total}-V_{cavity})}\int_V \rho (x^2\hat x + y^2 \hat y+z^2 \hat z)dV$$
$$=\frac{1}{ (V_{total}-V_{cavity})}\int_0^x \int_0^y \int_0^{\sqrt{R-x^2-y^2}} (x^2\hat x + y^2 \hat y+z^2 \hat z)dxdydz$$
…but here I'm confused – how do I set up the integral to account for the cavity?
Best Answer
You dont have to complicate using such integrals. Just do this. Assume the larger sphere to be complete and find its centre of mass (which is its geometrical centre of course). Now for the cavity, assume it is made of negative mass(no such thing of course, just eases the calculation).The mass of the smaller sphere would then be $-M/64$ (where $M$ is the mass of the larger sphere) Find its centre of mass co-ordinates. Then use the formula for the centre of mass for two point masses located at two points(which are the co-ordinates of the centre of mass of larger sphere and that of the smaller sphere/cavity) giving $R_{COM}$ = $(-\frac{R}{126},0,0)$