# The motion of a solid object

The motion of a solid object can be analyzed by thinking of the mass as concentrated at a single point, the center of mass. If the object has density at the point and occupies a region W, then the coordinates of the center of mass are given by where is the total mass of the body. Consider a solid is bounded below by the square , , and above by the surface the density of the solid be 1 g/cm$^3$, with x,y,z measured in cm. Find each of the following:

The mass of the solid   I am having troubles with starting and setting the integral

• FYI the "motion of a solid object..." is just color commentary which is irrelevant to the question you're posing. – Jonathan Mar 25 '14 at 2:48
• Also because in a real rigid motion you have to consider rotational momenta (which gives you the phase space $T^*(\mathbb{R}^3\times SO(3))$) and not only the center of mass ($T^*\mathbb{R}^3$). – Daniel Robert-Nicoud Sep 30 '15 at 5:17

## 2 Answers

1. The mass of the solid is $\int_W \rho \mathrm{d}V$ which in this case is equal to $\rho \int_W \mathrm{d}V$ as the solid has a uniform mass density. To find the mass consider a small rectangular element of sides $\mathrm{d}x$ and $\mathrm{d}y$ on the $xy$ plane. The volume of the cuboidal rod of solid with this element as the base is $z\mathrm{d}x\mathrm{d}y$. So the total volume of the solid will be $$\int_0^5\int_0^4(x+y+3) \mathrm{d}x \mathrm{d}y = 150$$ and therefore the mass = $1gm/cm^3\cdot150 cm^3 = 150 gm$

2. For the coordinates of center of mass the integrals are similar $$\overline{x} = \frac{1}{m}\int_0^5\int_0^4\rho x(x+y+3) \mathrm{d}x \mathrm{d}y$$ $$\overline{y} = \frac{1}{m}\int_0^5\int_0^4\rho y(x+y+3) \mathrm{d}x \mathrm{d}y$$ $$\overline{z} = \frac{1}{m}\int_0^5\int_0^4\rho \frac{(x+y+3)^2}{2} \mathrm{d}x \mathrm{d}y$$

The mass is

$$m = \int_0^4 \int_0^5 \int_0^{x+y+3}\,\rho \mathrm{d}z\mathrm{d}y\mathrm{d}x = 150 \rho$$

The center of mass is

$$\begin{pmatrix} \overline{x} \\ \overline{y} \\ \overline{z} \end{pmatrix} = \frac{1}{m} \int_0^4 \int_0^5 \int_0^{x+y+3}\,\rho \begin{pmatrix}x\\y\\z\end{pmatrix}\mathrm{d}z\mathrm{d}y\mathrm{d}x = \frac{1}{150 \rho} \begin{pmatrix} \frac{980 \rho}{3} \\ \frac{1250 \rho}{3} \\ \frac{1790 \rho}{3} \end{pmatrix} = \ldots$$

This is how you setup the integrals.