Find the center of mass of the lamina in the following...

Find the center of mass of the lamina in the following figure if the circular portion of the lamina has twice the density of the square portion of the lamina. (ar{x} (,)/(b)ar (y))=(◻) (Center of square is (0,a) and Radius of Circle is b) Find the center of mass of the lamina in the following figure if the circular portion of the lamina has twice the density of the square portion of the lamina 12 G) (x,y)

Transcript

00:01 In this problem we are going to compute the center of mass of the following shape on the xy plane.

00:09 So let us quickly draw the figure.

00:14 This is the x -axis, this is y.

00:17 We have a square of side length 18 here, right next to the y -axis, and right next to this square we have a circle, we have a disk.

00:31 Of radius 9 so it will fit into somewhere here okay let me redraw it because i didn't like that okay something like this so this is 9 this is minus 9 this point is 18 and this point is 36 at the center we have 27 7 but there is a, okay, let me first fill these areas with colors to indicate that we have solid objects instead of just the wires that goes around these shapes.

01:24 So this is the second shape.

01:30 Okay, there's a catch here or there's a trick here.

01:33 It's not a trick, there's a twist here.

01:35 It says this square is twice as dense.

01:40 As the this circle this disk is twice as dense as the square okay so let's keep it in mind if this square has danced raw then the circle this disk will have the density two times raw that's important okay now let us determine the center of mass of each of these shapes individually we have a square on the left, so the center of mass will be right at the center of this, right at the geometric center of this object, assuming that the mass distribution is uniform within this squares.

02:27 Since it is not given, we can always assume that for the sake of simplicity.

02:33 So this length is 9, and since it is at the, it is along the x -axis, we have y -equal to zero now let us write down the components and the coordinates of this point using the global x and y components we have nine as the x component and zero as the y component also i'm going to call this area a1 for future reference using which i'm going to compute the mass of this shape now let's do this the disk.

03:17 The center of mass of the disk will be at this geometric center, assuming again, that the mass distribution is uniform within this disk.

03:29 So somewhere here.

03:33 And this length is the radius simply, which is nine, and it is along the x -axis because this disk is centered along the x -axis.

03:45 Now let us write down.

03:46 The global coordinates of this point using this global frame of x and y.

03:53 We have this length given by 18 plus 9 so it's 27 and y is equal to 0.

04:05 Now i'm going to call this area a 2...