Math 32B Discussion Session Session 3 Notes August 14, 2018

Transcription

1 Math 3B Discussion Session Session 3 Notes August 4, 8 In today s discussion we ll think about two common applications of multiple integrals: locating centers of mass and moments of inertia. Centers of Mass Let s consider the problem of finding the center of mass of a region in two or three dimensions. For a two-dimensional region, this is defined to be the point which splits the region s mass both horizontally and vertically. That is, half the region s mass is to the left of this point and half is to the right, and similarly half the mass is above/below this point. We find the center of mass by computing the average x- and y-values over our region. Assuming our region has a density function δ(x, y), this will actually be a weighted average. First we compute the x- and y-moments of our region. The moments are intended to give us the total signed distance of the region from the respective axes. Since x gives the signed distance of a point from the y-axis, the y-moment is obtained by integrating the function x over our region. We multiply this x by our density function to get a weighted total: M y = xδ(x, y)da, M x = yδ(x, y)da. The moments give the total (weighted) distance from their respective axes, and we want an average distance, so we divide the moments by the total mass of the region. The x coordinate of the center of mass should then be given by the y-moment divided by the total mass: x CM = M y y CM = M x where M = δ(x, y)da. For a region W 3, we can analogously define the moments about the three standard planes: M yz = xδ(x, y, z)dv, M xz = yδ(x, y, z)dv, M xy = zδ(x, y, z)dv. W The coordinates of the center of mass of W are then given by dividing by the total mass: W W x CM = M yz y CM = M xz z CM = M xy M. Example. Let be the region in bounded by the unit circle and satisfying y x, and assume that has uniform mass density δ(x, y) =. Find the center of mass of.

2 (Solution) The region in question satisfies the inequalities x + y and y x, and can be described in a vertically simple way by Here s a plot of : / x /, x y x. To compute the center of mass, we first compute the moments of this region. For the y-moment we have M y = xδ(x, y)da = xda. Notice that the function we re integrating is anti-symmetric about the y-axis. Since is symmetric about the y-axis, we ll enjoy some cancellation: M y = = / x x / x x xdydx + xdydx + x / x / x x xdydx xdydx =. So no matter what the mass M of the region is, our center of mass has x-coordinate x CM = M y /M =. For the y-coordinate we need to evaluate the integral M x = yδ(x, y)da = yda = / x / x ydydx. This last integral isn t impossible to compute, but it also isn t very pretty. Since admits such a simple description in polar coordinates, let s switch coordinate systems to evaluate this integral. We can quickly see that = {(r, θ) : r, θ 3}, so we have M x = 3 r sin θ rdrdθ = 3 [ ] sin θ 3 r3 dθ

3 = 3 = 3 3 sin θdθ = 3 [ ] = 3. [cos θ]3 Finally, we compute the mass of the region, again using polar coordinates: 3 3 [ ] 3 M = da = rdrdθ = r The y-coordinate of our center of mass is then given by y CM = M x M = 3 4 π = 4 3π. dθ = Here s a plot of our region, including the newly found center of mass: dθ = π 4. It might be worth noting that while we used polar coordinates to actually evaluate the integrals involved here, we can t make a naïve substitution of polar coordinates for rectangular ones when computing the center of mass. Namely, if we computed the (weighted) average r- and θ-values over our region, we would get the polar center of mass, pictured here (in blue): Since polar coordinates split up our region in a different way than cartesian coordinates, the polar center of mass doesn t quite match what we call the center of mass (which is marked in red). 3

4 Moments of inertia To rotate a mass about an axis we must apply a certain amount of torque. As you ve experienced when, say, opening a door, forces that are applied at a great distance from the axis of rotation yield more torque than those applied near the axis. In the same vein, mass that is further away from the axis is more difficult to rotate than mass that is near the axis. Just as torque is a sort of angular force, the moment of inertia is thought of as an angular mass. For instance, you ve probably seen the following formula for computing the kinetic energy of a moving object: KE = mv, where m is the mass of the object and v is its velocity. If ω is the angular velocity (measured in radians per second) of our object as it rotates about an axis and I is its moment of inertia, then rotational KE = Iω. Note: standard units for the moment of inertia is kg m. According to physicists, here s how we compute the moment of inertia: say we want to rotate a point mass m about an axis L. Then the moment of inertia of this point mass is given by m times the square of the distance to the axis L. For instance, if L is the x-axis and the point mass is located at (x, y, ), then the moment of inertia is given by my, since the distance from (x, y, ) to L is y. Moments of inertia are additive, so to find the moment of inertia of a region we sum up the moments of all of the regions constituent points that is, we integrate the moment of inertia function. For instance, in we have I x = D y δ(x, y) da and I y = D x δ(x, y) da, where δ(x, y) is the mass density function for the region D and I x, I y are the moments of inertia of D about the x- and y-axes, respectively. If we want to rotate D, a region in, about the z-axis then we have I = (x + y )δ(x, y) da, which we sometimes call the polar moment of inertia. D Example. Let D be the rectangle in described by x 4 and y 4 with uniform mass density function. Which is greater: the moment of inertia of D about the x-axis or about the y-axis? (Solution) We can take δ(x, y) = for all (x, y), so we compute I x = 4 4 y dxdy = 4 y dy = y = 4 kg m

5 and I y = 4 4 x dydx = 4 so the moment of inertia about the y-axis is greater. 4 3x dx = x 3 = 56 kg m, 5

6 Group Work. Find the centroid of the region W lying above the sphere x + y + z = 6 and below the paraboloid z = 4 x y. Hint: Argue by symmetry.. Consider the region W lying outside the sphere x + y + z =, inside x + y + z = 4, and above the xy-plane. Find the centroid of W. Is this point contained in W? Hint: Argue by symmetry. 3. Let D be the triangular domain bounded by the coordinate axes and the line y = 3 x, with mass density ρ(x, y) = y. Compute the following quantities: (a) Total mass (b) Center of mass (c) I x (d) I Answers ( ) 3.,, (7 6 6) (.,, 45 ), and no (a) 9/ ( 3 (b) 4, 3 ) (c) 43/ (d) 8/5 6