FURTHER APPLICATIONS OF INTEGRATION 8

To work, our strategy is:  Break up the physical quantity into small parts.  Approximate each small part.  Add the results.  Take the limit.  Then, evaluate the resulting integral. APPLICATIONS TO PHYSICS AND ENGINEERING

Our main objective here is to find the point P on which a thin plate of any given shape balances horizontally as shown. MOMENTS AND CENTERS OF MASS

This point is called the center of mass (or center of gravity) of the plate. CENTERS OF MASS

We first consider the simpler situation illustrated here. CENTERS OF MASS

Two masses m 1 and m 2 are attached to a rod of negligible mass on opposite sides of a fulcrum and at distances d 1 and d 2 from the fulcrum. CENTERS OF MASS

The rod will balance if: CENTERS OF MASS Equation 2

This is an experimental fact discovered by Archimedes and called the Law of the Lever.  Think of a lighter person balancing a heavier one on a seesaw by sitting farther away from the center. LAW OF THE LEVER

Now, suppose that the rod lies along the x-axis, with m 1 at x 1 and m 2 at x 2 and the center of mass at. MOMENTS AND CENTERS OF MASS

Comparing the figures, we see that:  d 1 = – x 1  d 2 = x 1 – MOMENTS AND CENTERS OF MASS

CENTERS OF MASS So, Equation 2 gives: Equation 3

The numbers m 1 x 1 and m 2 x 2 are called the moments of the masses m 1 and m 2 (with respect to the origin). MOMENTS OF MASS

Equation 3 says that the center of mass is obtained by: 1.Adding the moments of the masses 2.Dividing by the total mass m = m 1 + m 2 MOMENTS OF MASS

In general, suppose we have a system of n particles with masses m 1, m 2,..., m n located at the points x 1, x 2,..., x n on the x-axis. Then, we can show where the center of mass of the system is located—as follows. CENTERS OF MASS Equation 4

The center of mass of the system is located at where m = Σ m i is the total mass of the system. CENTERS OF MASS Equation 4

The sum of the individual moments is called the moment of the system about the origin. MOMENT OF SYSTEM ABOUT ORIGIN

Then, Equation 4 could be rewritten as: m = M  This means that, if the total mass were considered as being concentrated at the center of mass, then its moment would be the same as the moment of the system. MOMENT OF SYSTEM ABOUT ORIGIN

Now, we consider a system of n particles with masses m 1, m 2,..., m n located at the points (x 1, y 1 ), (x 2, y 2 )..., (x n, y n ) in the xy-plane. MOMENTS AND CENTERS OF MASS

By analogy with the one-dimensional case, we define the moment of the system about the y-axis as and the moment of the system about the x-axis as MOMENT ABOUT AXES Equations 5 and 6

M y measures the tendency of the system to rotate about the y-axis. M x measures the tendency of the system to rotate about the x-axis. MOMENT ABOUT AXES

As in the one-dimensional case, the coordinates of the center of mass are given in terms of the moments by the formulas where m = ∑ m i is the total mass. CENTERS OF MASS Equation 7

Since, the center of mass is the point where a single particle of mass m would have the same moments as the system. MOMENTS AND CENTERS OF MASS

Find the moments and center of mass of the system of objects that have masses 3, 4, and 8 at the points (–1, 1), (2, –1) and (3, 2) respectively. MOMENTS & CENTERS OF MASS Example 3

We use Equations 5 and 6 to compute the moments: MOMENTS & CENTERS OF MASS Example 3

As m = = 15, we use Equation 7 to obtain: MOMENTS & CENTERS OF MASS Example 3

Thus, the center of mass is: MOMENTS & CENTERS OF MASS Example 3

Next, we consider a flat plate, called a lamina, with uniform density ρ that occupies a region R of the plane.  We wish to locate the center of mass of the plate, which is called the centroid of R. CENTROIDS

In doing so, we use the following physical principles. CENTROIDS

The symmetry principle says that, if R is symmetric about a line l, then the centroid of R lies on l.  If R is reflected about l, then R remains the same so its centroid remains fixed.  However, the only fixed points lie on l.  Thus, the centroid of a rectangle is its center. CENTROIDS

Moments should be defined so that, if the entire mass of a region is concentrated at the center of mass, then its moments remain unchanged. CENTROIDS

Also, the moment of the union of two non-overlapping regions should be the sum of the moments of the individual regions. CENTROIDS

Suppose that the region R is of the type shown here. That is, R lies:  Between the lines x = a and x = b  Above the x-axis  Beneath the graph of f, where f is a continuous function CENTROIDS

We divide the interval [a, b] into n subintervals with endpoints x 0, x 1,..., x n and equal width ∆x.  We choose the sample point x i * to be the midpoint of the i th subinterval.  That is, = (x i–1 + x i )/2 CENTROIDS

This determines the polygonal approximation to R shown below. CENTROIDS

The centroid of the i th approximating rectangle R i is its center. Its area is: f( ) ∆x So, its mass is: CENTROIDS

The moment of R i about the y-axis is the product of its mass and the distance from C i to the y-axis, which is. CENTROIDS

Therefore, CENTROIDS

Adding these moments, we obtain the moment of the polygonal approximation to R. CENTROIDS

Then, by taking the limit as n → ∞, we obtain the moment of R itself about the y-axis: CENTROIDS

Similarly, we compute the moment of R i about the x-axis as the product of its mass and the distance from C i to the x-axis: CENTROIDS

Again, we add these moments and take the limit to obtain the moment of R about the x-axis: CENTROIDS

Just as for systems of particles, the center of mass of the plate is defined so that CENTROIDS

However, the mass of the plate is the product of its density and its area: CENTROIDS

Thus,  Notice the cancellation of the ρ’s.  The location of the center of mass is independent of the density. CENTROIDS

In summary, the center of mass of the plate (or the centroid of R ) is located at the point, where CENTROIDS Formula 8

Find the center of mass of a semicircular plate of radius r.  To use Equation 8, we place the semicircle as shown so that f(x) = √(r 2 – x 2 ) and a = –r, b = r. CENTERS OF MASS Example 4

Here, there is no need to use the formula to calculate.  By the symmetry principle, the center of mass must lie on the y-axis, so. CENTERS OF MASS Example 4

The area of the semicircle is A = ½πr 2. Thus, CENTERS OF MASS Example 4

The center of mass is located at the point (0, 4r/(3π)). CENTERS OF MASS Example 4

Find the centroid of the region bounded by the curves y = cos x, y = 0, x = 0, x = π/2 CENTERS OF MASS Example 5

The area of the region is: CENTERS OF MASS Example 5

CENTROIDS Example 5 So, Formulas 8 give:

The centroid is ((π/2) – 1, π/8). CENTROIDS Example 5

Suppose the region R lies between two curves y = f(x) and y = g(x), where f(x) ≥ g(x). CENTROIDS

Then, the same sort of argument that led to Formulas 8 can be used to show that the centroid of R is, where CENTROIDS Formula 9

Find the centroid of the region bounded by the line x = y and the parabola y = x 2. CENTROIDS Example 6

The region is sketched here. We take f(x) = x, g(x) = x 2, a = 0, and b = 1 in Formulas 9. CENTROIDS Example 6

First, we note that the area of the region is: CENTROIDS Example 6

Therefore, CENTROIDS Example 6

The centroid is: CENTROIDS Example 6