1
$\begingroup$

Find the center of mass of the solid $$ R = {(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 \leq 1^2, 0 \leq z \leq \frac \pi2 }$$ if the density is given by $$\rho(x,y,z) = 5\bigg(\sqrt{x^2 + y^2} + sin(z)\bigg). $$

First i tried solving for the mass with the triple integral $$\int_0^{2\pi} \int_0^{\frac\pi2} \int_0^15r\big(r+sin(z)\big) = \frac{10\pi^2}{6} $$ Which gave me the $z$ coordinate $\frac{3+\pi}{\pi}$ Not sure if this is correct though. Also, i don't know how to find the other coordinates now seeing as the density is not uniform.

$\endgroup$
  • $\begingroup$ set up is ok but check the derivation, by wolfy the mass should be different wolframalpha.com/input/… $\endgroup$ – gimusi Mar 9 '18 at 15:01
1
$\begingroup$

Note that the method is correct since $\rho(x,y,z)$ is symmetric with respect to z axis the center of mass has coordinates $(0,0,z_G)$ with (I didn't check the detail of calculation):

$$z_G=\frac{\int_V z\cdot \rho(x,y,z)dV}{M}$$

$\endgroup$
  • $\begingroup$ How do you see that $\rho(x,y,z)$ is symmetric about the z-axis? Doesen't the $\sqrt{x^2 + y^2}$ term mean that it also depends on $x$ and $y$ ? $\endgroup$ – Pame Mar 9 '18 at 15:08
  • $\begingroup$ @Pame since $\rho(x,y,z)=\rho(-x,y,z)=\rho(x,-y,z)=\rho(-x,-y,z)$ $\endgroup$ – gimusi Mar 9 '18 at 15:10
  • $\begingroup$ @Pame or also note that for z=constant $\rho=\rho(r)$ $\endgroup$ – gimusi Mar 9 '18 at 15:17
  • $\begingroup$ Ok i see that it must be symmetric about the $z$-axis, but i don't quite get why it means the center of mass must lie on the $z$-axis. $\endgroup$ – Pame Mar 9 '18 at 15:31
  • 1
    $\begingroup$ @Pame for the mass integral the integrand is positive thus $\int_V \rho(x,y,z)dV>0$ whereas for $x_G=\frac{\int_V x\cdot \rho(x,y,z)dV}{M}$ we have a term x wich makes one half integral positive and one half equal but negative thus the sum is zero. $\endgroup$ – gimusi Mar 9 '18 at 16:20
1
$\begingroup$

In general, the coordinates of the center of mass of the region $\Omega$ are given by

$$ \overline{{\bf x}} = \frac{\int_\Omega{\rm d}{\bf x}~{\bf x}\rho({\bf x})}{\int_\Omega{\rm d}{\bf x}~\rho({\bf x})} \tag{1} $$

In your case $\Omega = \{(x,y,z) | x^2+y^2 < 1, 0 < z < \pi/2 \}$

We can then calculate

$\int_\Omega{\rm d}{\bf x}~\rho({\bf x})$

\begin{eqnarray} \int_\Omega{\rm d}{\bf x}~\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R\rho(R,\phi,z) \\ &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [5(R + \sin z)] \\ &=& 10\pi \int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R (R + \sin z) \\ &=& (\dots) \end{eqnarray}

$\int_\Omega{\rm d}{\bf x}~x\rho({\bf x})$

\begin{eqnarray} \int_\Omega{\rm d}{\bf x}~x\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~Rx\rho(R,\phi,z) \\ &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [R\cos\phi][5(R + \sin z)] &=& (\dots) \end{eqnarray}

$\int_\Omega{\rm d}{\bf x}~y\rho({\bf x})$

\begin{eqnarray} \int_\Omega{\rm d}{\bf x}~y\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~Ry\rho(R,\phi,z) \\ &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [R\sin\phi][5(R + \sin z)] &=& (\dots) \end{eqnarray}

$\int_\Omega{\rm d}{\bf x}~z\rho({\bf x})$

\begin{eqnarray} \int_\Omega{\rm d}{\bf x}~z\rho({\bf x}) &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~Rz\rho(R,\phi,z) \\ &=& \int_0^{2\pi}{\rm d}\phi\int_{0}^{\pi/2}{\rm dz}\int_0^1{\rm dR}~R [z][5(R + \sin z)] &=& (\dots) \end{eqnarray}

Everything together

When you calculate these integrals, just use Eq. (1)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.