Properties of surfaces II: Second moment of area Just as we have discussing first moment of an area and its relation with problems in mechanics, we will now describe second moment and product of area of a plane. In this lecture we look at these quantities as some mathematical entities that have been defined and solve some problems involving them. The usefulness of related quantities, called the moments of inertia and products of inertia will become clear when we deal with rotation of rigid bodies. Let us then consider a plane area in xy plane (figure 1). The second moments of the area A is defined as That is given a plane surface, we take a small area in it, multiply by its perpendicular distance from the x-axis and sum it over the entire area. That gives I XX. Similarly I YY is obtained by multiplying the small area by square of the distance perpendicular to the y-axis and adding up all contributions (see figure 2). 1
The product of area is defined as where x and y are the coordinates of the small area da. Obviously I XX is the same as I XY. Let us now solve a few examples. Example1: Let us start with a simple example of a square of side a with its center of the origin (see figure 2). 2
Figure 3 To calculate this, we choose the elemental area as shown in figure 4 and integrate. Then so that da = ady 3
Similarly for calculating I YY we choose a vertical elemental area and calculate Let us also calculate the product of inertia. Choose on elemental area dxdy and calculate (see figure 5) As noted earlier, I YX is equal to I XY and therefore it also vanishes. 4
A related problem is that of a rectangular area of size a x b. Its length of side a is parallel to the x-axis and the other side of length b is parallel to the y-axis. I leave this as an exercise for you to show that in this case,. Notice that due to the area being symmetrically distributed about the x- and y-axes, the product of the area comes out to be zero. Example 2 : Next let us consider a quarter of an ellipse as shown in figure 6 and calculate the moment and product of area for this area. 5
Equation of the ellipse whose quarter is shown in figure 6 is:. For calculating integral choose an area element parallel to the x-axis to calculate da=xdy and perform the Using the equation for ellipse, we get which gives This integral can easily be performed by substituting y = b sin θ and gives Similarly by taking a vertical strip to perform the integral, we calculate and get Next we calculate the product of area I XY. To calculate I XY, we take a small element ( shown in figure 7, multiply it by x and y and integrate to get ) as 6
For a given x, the value of y changes from 0 to so the integral is This integral is easily performed to get Thus for a quarter of an ellipse, the moments and products of area are If we put a = b, these formulas give the moments and products of area for a quarter of a circle of radius a. I will leave it for you to work out what will be for the full ellipse about its centre. Using the second moment of an area, we define the concept of the radii of gyration. This is the point which will give the same moment of inertia as the area under consideration if the entire area was concentrated there. Thus 7
define the radii of gyration k X and k Y about the x- and the y-axes, respectively. In the example of a rectangular area of size a x b with side a parallel to the x-axis, we had,. So for this rectangle, the radii of gyration are and. Having defined the moments and products of area, we now describe a relationship between the second moment of an area about a set of axes passing through the centroid of that area and another set of (x-y) axes which are parallel to those passing through the centroid. This is known a transfer theorem. Transfer theorem: Let the centroid of an area be at point ( x 0 y 0 ) with respect to the set of axes (xy). Let ( x' y' ) be a parallel set of axes passing through the centroid. Then But by definition which gives This is how the moment of area of a plane about an axis is related to the moment of the same area about another axis parallel to the previous one but passing through the centroid. Similarly it is easily shown that and. We now solve an example to show the application of this theorem. 8
Example 3 : Calculate the second moments and products of area of an ellipse with its centre at (x 0,y 0 ) In a previous exercise, you have already calculated the second moments and products of area of an ellipse about its centre, which is also its centroid. These are: We use these results now in applying the transfer theorem to obtain moments and products of area of the ellipse about a different origin (see figure 8). Thus Similarly 9
Transformation of moments and products of area from one system to another rotated with respect to the first one: We just learnt that if we translate an area so that its centriod moves to another point, how its second moments of inertia and products of inertia change when the axes passing through the centroid and the other set of axes are parallel. We now study how the moments and products are related when we calculate them about another set of axes that an rotated with respect to the first one. So we consider a set of area (xy) and another on (x'y') rotated with respect to the first one by an angle θ (see figure 1). We wish to relate that. In lecture 1, we have already learnt This gives changing 10
Similarly and = = Thus This gives the second moment and product about a set of axis (x'y') rotated about the other set (xy). Let us now discuss some examples. As expected for a circular area, no matter about which set of axes you calculate, it will always come out to be the same because the area looks the sum from all set axes. What is interesting, however, is that for a square also the moments and product of area are the same with respect to any set of axes passing through its centre. It happens because with respect to its centre, the I XX and I YY for a square are the same i.e. and. This is left as an exercise for you to show. 11
We now use for formulae derived above to obtain what we call the principal set of axes for a plane area. The principal set of axes at a point are those for which the product of inertia vanishes i.e. about the principal set of axes. Let us see how we determine these axes if we know about a given set of axes. In the following we refer to the principal set of axes as (1,2) where 1 refers to the x-axis and 2 to the y-axis. We know that we want where α is the angle the principal set of axes make with the (xy) set of axes. The equation above gives The principal set of axes has one more property: The moments of area is maximum one of the principal axis (say x-axis) and minimum about the other (y-axis). This is seen as follow: Since Let us find θ for which I X'X' is a maximum or a minimum. The condition gives This is the same angle a that makes I XY vanish. This means Thus 12
When α makes the function I XX a maximum, the angle makes I YY a minimum. I'll leave it for you to show that. Thus the principal set of axes are also those about which the II nd moment of area is maximum about one axis and minimum about the other. Notice that for a square, any set of axes passing through its centre is a principal set of axes. This follows from the exercise that you did above. As a related quantity, we also define polar moment of an area. This is calculated as Since r 2 is independence of the (xy) system chosen, I is the same about any set of axes passing through a point. Having defined these concepts, at the end I will point out that in a similar manner II nd moment of mass can also be defined. We will elaborate on that more in the later lectures on dynamics when we deal with the rotation of rigid bodies. 13