Vectors Calaulus 1.21

Center of Mass and Moment of Inertia Mass We saw before that the double integral over a region of the constant function 1 measures the area of the region.  If the region has uniform density 1, then the mass is the density times the area which equals the area.  What if the density is not constant.   Suppose that the density is given by the continuous function         Density  =  r(x,y) In this case we can cut the region into tiny rectangles where the density is approximately constant.  The area of mass rectangle is given by         Mass  =  (Density)(Area)  =  r(x,y) Dx Dy  You probably know where this is going.  If we add all to masses together and take the limit as the rectangle size goes to zero, we get a double integral. Mass Let r(x,y) be the density of a lamina (flat sheet) R at the point (x,y).  Then the total mass of the lamina is the double integral                Example A rectangular metal sheet with 2  <  x  < 5  and  0  <  y  <  3 has density function         r(x,y)  =  x + y Set up the double integral that gives the mass of the metal sheet. Solution We just have the integral         Moments and Center of Mass We have seen in first year calculus that the moments about an axis are defined by the product of the mass times the distance from the axis.          Mx  =  (Mass)(y)        My  =  (Mass)(x) If we have a region R with density function r(x,y), then we do the usual thing.  We cut the region into small rectangles for which the density is constant and add up the moments of each of these rectangles.  Then take the limit as the rectangle size approaches zero. This will give us the total moment.  Definition of Moments of Mass and Center of Mass Suppose that r(x,y) is a continuous density function on a lamina R.  Then the moments of mass are                                and if M is the mass of the lamina, then the center of mass is                      Example Set up the integrals that give the center of mass of the rectangle with vertices (0,0), (1,0), (1,1),  and (0,1) and density function proportional to the square of the distance from the origin.  Use a calculator or computer to evaluate these integrals. Solution The mass is given by          The moments are given by          These evaluate to          Mx  =  5k/12        and       My  =  5k/12 It should not be a surprise that the moments are equal since there is complete symmetry with respect to x and y.  Finally, we divide to get         (x,y)  =  (5/8,5/8) This tells us that the metal plate will balance perfectly if we place a pin at (5/8,5/8) Moments of Inertia We often call Mx and My the first moments.  They have first powers of y and x in their definitions and help find the center of mass.  We define the moments of inertia  (or second moments) by introducing squares of y and x in their definitions.  The moments  of inertia help us find the kinetic energy in rotational motion.  Below is the definition Moments of Inertia Suppose that r(x,y) is a continuous density function on a lamina R.  Then the moments of inertia are