14.4 Center of Mass and Moment of Inertia
Suppose a planar lamina (a very thin slice of a plane) has a constant density r. Its mass would then be its area times r. For the region in the picture, we have the following equation for mass:
Suppose further that the lamina was not of uniform density, but had a function r(x,y) that represents its density at the point (x,y).
Def Mass of a Planar Lamina of Variable Density
If r(x,y) is a continuous density function on the lamina corresponding to a plane region R, then the mass m of the lamina is given by the double integral:
Ex 1 Find the mass of the planar lamina with density
at point (x,y) proportional the distance from the y axis: the density function
is 
.
or
Ex 2 Find the mass of the lamina that consists of
the semicircle 
in
the first and second quadrants where the density at (x,y) is proportional
to the distance from the origin.
This is a classic case of converting to polar coordinates!
Moments and Centers of Mass
For any object with mass m, the tendency for it to cause a rotation about
an axis is the product of its mass and the distance from the fulcrum. This
is called the moment of the mass about some rotation line, often the
x or y axis. If we think of a planar lamina with variable density, a small
enough rectangular portion of it can be thought to have a constant density
and we can assume all of its mass is concentrated on some interior point of
the rectangle. The moment of this small rectangular bit of mass (purple in
the picture) about the x axis is given by the formula
and the moment about the y axis is given by
Forming the Riemann Sum of all such products and taking the limit as max DAi approaches zero leads to the following:
Moments of a Variable Density Planar Lamina
Let r(x,y) be a continuous density function on the planar lamina R.
1. The moment of mass of R about the x axis is 
2. The moment of mass of R about the y axis is 
Center of Mass of a Variable Density Planar Lamina
If m is the mass of the lamina, then the center of mass is
If R is a simple plane region (density function = constant) , then 
is
called the centroid of the region.
Note: it is often possible to determine at least one of the coordinates of the center of mass by symmetry.
Ex 3 Find the center of mass for the triangular
lamina with vertices at the points (0,0), (0,1), and (1,0) and density function
. The mass of the
region is given by the integral
The limits of integration come from the equation of the
line through (0,1) and (1,0). The
density function is shown:
Ex 4 Find the center of mass of the lamina where the density at (x,y) is proportional to the distance from the y axis. We calculated the mass in Ex 1 as 64k.
Screen capture of a problem of finding the center of mass. 
Screen capture of a problem of finding center of mass of a cardioid.
The moments of mass are sometimes called the "first moments" of a mass about the x and y axes. The "second moment" is called the moment of inertia. It is a measure of the tendency of a mass to resist a change in rotational motion. If a particle of mass m is distance d away from its line of rotation, the moment of inertia about the line is I = m*d2. For that reason, the moments of inertia about the x and y axes are the following:
The sum of the moments of inertia Ix and Iy is called the polar moment of inertia and is denoted I0. (Some texts use Iz).
Ex 5 Find the moments of inertia and the polar moment
of inertia of the region given with density function 
.
The moment of inertia of a revolving lamina can be used to measure its kinetic energy. If a planar lamina is revolving about a line with an angular speed of w radians per second, the kinetic energy of the revolving lamina is given by
In other words, the kinetic energy of an object revolving about an axis is proportional to its moment of inertia.
If the mass m is moving in a straight line at velocity v, the kinetic energy is given by
Here, the kinetic energy is proportional to its mass.
Radius of Gyration
The radius of gyration 
of
a revolving mass m with moment of inertia I is defined to be
If the entire mass m is located 
units
from its axis of revolution, it would have the same moment of inertia, hence
the same kinetic energy.
Ex 6 Find the radius of gyration about the y axis
for the region given with density function 
.
Assignment Pg 1015 #1,5,7,13,21,27,29,33