Mass, Center-of-Mass, and Moment of Inertia

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Suppose that we have a thin plate, so thin that it's practically 2-dimensional. Such a plate is called a planar lamina. A lamina is described by the region $D$ in the $x$-$y$ plane that it covers, and by its mass density $\rho(x,y)$, which gives the mass per unit area. In the following video, we show how to get the mass and center-of-mass of a lamina by integration.

If the density were a constant, finding the total mass of the lamina would be easy: we would just multiply the density by the area. When the density isn't constant, we need to integrate instead. The mass of a little box of area $dA$ around the point $(x,y)$ is essentially $\rho(x,y) dA$. For the total mass of the lamina, we add up the boxes and take a limit to get $$M \ = \ \iint_D \rho(x,y) dA.$$ This integral can be done in rectangular coordinates, polar coordinates, or by whatever method you prefer.

Example 1: Find the mass of a lamina with density function $\rho(x,y)=6x$ covering the triangle $D$ bounded by the $x$-axis, the line $y=x$, and the line $y=2-x$.

Solution: We express $D$ as a Type II region, with $x$ running from $y$ to $2-y$ and $y$ running from $0$ to $1$.

Our mass is then \begin{eqnarray*}M & = & \iint_D \rho(x,y) dA \\ & = & \int_0^1 \int_{y}^{2-y} 6x\, dx\, dy \\ & = & \int_0^1 3[(2-y)^2-y^2] dy \\ & = & -[(2-y)^3+y^3]_0^1 \ = \ 6. \end{eqnarray*}

The center-of-mass of a body is a weighted average of the positions of the particles inside. Since a box of area $dA$ at position $(x,y)$ has a mass that's a fraction $\rho(x,y) dA/M$ of the total, the center-of-mass of our lamina is at position $(\bar x, \bar y)$, where \begin{eqnarray*} \bar x & = & \frac{1}{M} \iint_D x \rho(x,y) dA \\ \bar y & = & \frac{1}{M} \iint_D y \rho(x,y) dA \end{eqnarray*}

Example 2: Find the center-of-mass of the lamina of Example 1.

Solution: We have to compute two integrals, one for each coordinate: \begin{eqnarray*}\iint_D x \rho(x,y) & = & \int_0^1 \int_{y}^{2-y} 6x^2\, dx\, dy \\ & = & \int_0^1 2[(2-y)^3-y^3] dy \\ & = & - \left . \frac{(2-y)^4 + y^4}{2} \right |_0^1 \ = \ 7. \end{eqnarray*}

\begin{eqnarray*} \iint_D y \rho(x,y) & = & \int_0^1 \int_{y}^{2-y} 6xy\, dx\, dy \\ & = & \int_0^1 3y[(2-y)^2-y^2] dy \\ &=& \int_0^1 12y - 12y^2 dy \\ & = & \left . 6y^2 - 4y^3 \right |_0^1 \ =\ 2. \end{eqnarray*} Dividing by the mass gives the location $(\bar x, \bar y)$ of our center-of-mass: $$\bar x = \frac{7}{6}; \qquad \bar y = \frac{1}{3}.$$

The moment of inertia of an object indicates how hard it is to rotate. For a point particle, the moment of inertial is $I=mr^2$, where $m$ is the mass of the particle and $r$ is the distance from the particle to the axis of rotation. The moment of intertia of an object with many pieces is the sum of the moments of inertia of its pieces. The following video what the moment of inertia means physically, and how we can calculate it.

Let's imagine that we're rotating around the origin, so $r^2=x^2+y^2$. Since the moment of inertial of a little box of size $dA$ at position $(x,y)$ is $(x^2+y^2) \rho(x,y) dA$, the moment of inertia of the entire lamina is $$ I = \iint_D (x^2+y^2) \rho(x,y) dA.$$

Example 3: Find the moment of inertial of a lamina covering the inside of the unit circle, with density function $\rho(x,y) = 1-x^2-y^2$.

Solution: This problem is best worked in polar coordinates. Our integral is

\begin{eqnarray*} I \ = \ \iint_D r^2 \rho(x,y) dA & = & \int_0^{2 \pi} \int_0^1 r^2 (1-r^2) r dr d\theta \\ &=& \int_0^{2\pi} \int_0^1 (r^3-r^5) dr d\theta \\ & = & \int_0^{2\pi} \frac{d\theta}{12} \ = \ \frac{\pi}{6}. \end{eqnarray*}

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