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
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 locatedas follows. CENTERS OF MASS Equation 4
The center of mass of the system is located at where m = 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 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 i1 + 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 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 = r 2. Thus, CENTERS OF MASS Example 4 The center
of mass is located at the point (0, 4r/(3)). CENTERS OF MASS
Example 4 Find the centroid of the region bounded by the curves y =
cos x, y = 0, x = 0, x = /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 ((/2) 1, /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
