Solutions to Homework 3
1. Determine the centroid the region in √
the plane which lies above the
x-axis and below the semi-circle y = r2 − x2 .
Solution: The centroid is the same as the center of mass where the
density has a constant value. The convention is to set the density
equal to 1. Therefore, the mass M is the same as the area of the
region: M = 12 πr2 .
The moment about the y-axis My is obviously equal to zero since the
region is symmetric about the y-axis. (Challenge: Can you think of a
region which has My = 0 but which is not symmetric about the
y-axis?) You could compute an integral for My , but you should not
spend much time computing it:
Z r p
My =
x r2 − x2 dx.
−r
The integral
√ is equal to zero for reasons of symmetry: the integrand
φ(x) = x r2 − x2 is an odd function (meaning φ(x) = −φ(x), which
is easily verified for this function) and the limits of integration are
x = −r to x = r. If φ(x) is any odd function, the
Rfrom
a
φ(x)
dx = 0. (Challenge: Can you name some other commonly
−a
encountered odd functions?)
http://en.wikipedia.org/wiki/Even_and_odd_functions
The moment about the x-axis is not apparent. So, we calculate the
integral:
Z r √ 2
r − x2 + 0 p 2
· r − x2 dx
2
−r
Z
1 r 2
1 2
1 3 r
2
2
=
(r − x ) dx =
r x− x
= r3 .
2 −r
2
3
3
−r
Rr 2
R
r
(You might have also seen that −r (r − x2 ) dx = 2 0 (r2 − x2 ) dx
since φ(x) = r2 − x2 is an even function. Using its symmetry, the
value of the integral is doubled while the limits of integration are
changed from 0 to r intead of from −r to r. This step, while not
necessary, often leads to an easier step for evaluating the limits of
integration.)
Thus, x = My /M = 0 and y = Mx /M =
1
4
3π r.
2. The following principle is used frequently in physics: if system S of
point masses is given, then we can regard the system as a single
point mass located at the center of mass of S and having mass equal
to the total mass of S. For example, a large asymmetric object
might, for simplicity, be regarded as only occupying a single point in
space and that the mass of this large object might be regarded as
concentrated at this single point. The natural candidate for such a
point is the center of mass of the object. In this exercise, you will
provide a mathematical justification for why this is “natural”.
Let S be the system of point masses p1 , . . . , pk , and let T be the
system of point masses pk+1 , . . . , pn . Suppose that each point mass pi
has mass mi at that each that each point pi lies on the real line and
has x-coordinate equal to xi . Prove that the center of mass of the
system S ∪ T of consisting of all of the point masses, p1 , . . . , pn , is
equal to the center of mass of the system consisting of two special
point masses: one point mass pS which has a mass equal to the total
mass of the system S and which has x-coordinate equal to the center
of mass of the system S and a second point mass pT defined
analogously with respect to the system T .
Solution: Let xS denote the center of mass of the system S, and
analogously define xT and xS∪T . Let MS , MT , and MS∪T denote the
total mass of system S, system T , and the combined system S ∪ T ,
respectively. Thus, we have the following equations (by definition)
MS =
k
X
mi ,
n
X
MT =
i=1
xS =
MS
MS∪T =
Pn
,
xT =
i=k+1 mi xi
MT
mi
i=1
i=k+1
Pk
i=1 mi xi
mi ,
n
X
Pn
,
xS∪T =
i=1 mi xi
MS∪T
.
If we replace S and T by point masses and then compute the center
of mass of this two point system we obtain the value
x S MS + x T M T
.
MS + M T
If we distribute MS and MT across the expressions for xS and xT ,
respectively,Pthen we see that the numerator of the above expression
is equal to ni mi xi . Easier to
Pnsee is that the denominator of the
above expression is equal to i=1 mi . So, indeed, the expression
above is equal to xS∪T .
2
3. In this exercise you will use the principle above to determine the
centroid of pentagon which is shaped like a child’s drawing of a
house. Let A, B, C, D and E be points in the plane such that the
quadrilateral ABCD is a square with side length s and such that E
lies outside this square and such that AEB is a triangle such that
∠AEB is a right triangle; let P be the pentagon having the above
vertices. First determine the centroid of the square and the centroid
of the triangle; then use the principle in the previous exercise to
determine the centroid of P . (You will need to choose coordinates for
the vertices; try to make choices which simplify your calculations.
You should prefer arguments which take advantage of the symmetry
to arguments which use brute force calculations.)
Solution: The problem is ambiguous as stated: I should have said
that AEB is an isoceles right triangle (as would be the case if you
were making a child’s drawing of a house). Let’s make this
assumption here.
First, a system of coordinates need to be chosen. I’ll choose the
midpoint of AB as the origin and I’ll choose E to lie on the positive
y-axis and A = (−s/2, 0) and B = (s/2, 0). Because of our
assumption that AEB is isoceles E = (0, s/2). It is now clear from
symmetry that the centroid of the triangle lies on the positive y-axis
and that the centroid of the square is located at (0, −s/2). The only
unknown is how far above the origin on the y-axis does the centroid
of the triangle lie.
I’ll use the theorem from geometry which states that centroid of a
triangle lies at the intersection of the medians. The median from
vertex E to the side AB is a segment of the line x = 0. The median
from vertex A to the side EB passes through A = (−s/2, 0) and
M = (s/4, s/4) (the midpoint of E and B). The equation of the line
through A and M is (y − 0) = (1/3)(x + s/2) and so it intersects
x = 0 in the point (0, s/6). This is the centroid of the triangle.
To compute the centroid of the house we replace the square by its
centroid (0, −s/2) and assign a mass equal to the area of the square
s2 and we replace the triangle by its centroid (0, s/6) and assign a
mass equal to the area of the triangle s2 /4. The center of mass of
this two point system has x-coordinate equal to zero and has
3
y-coordinate equal to
(s2 )(−s/2) + (s2 /4)(s/6)
11
= − s.
2
2
(s ) + (s /4)
30
So, the centroid of the house lies 11s/30 units below the “gutters” of
the house.
4