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Mathematics for Materials and Earth Sciences. Hilary Term. J. H. Woodhouse
2.2
Integration – Applications
One of the most frequent applications of integration is in finding continuous sums of various kinds. Here
we shall discuss the use of integration to find areas, volumes, arc length of curves, centroids of plane and solid
bodies, moments of inertia of plane and solid bodies. The notions of centroids and moments of inertia will be
defined; the main purpose of introducing them here is to provide examples of how integration is applied.
The use of centroids and moments of inertia in physical applications will come in other courses. Suffice it to
say that, for example, a complex molecule will spin about its centroid, and the Earth spins about its centroid.
The moment of inertia of a system governs the way in which a spinning body responds to external forces.
Thus it plays a role in the calculation of the specific heat of a molecular gas, and in determining the period
of the precession of the equinoxes due to the gravitational force of the Sun acting on the Earth’s bulge.
The idea in all these applications is the same, namely to regard the quantities to be determined as continuous
sums with respect to some parameter u, say, and to write the sum as a definite integral
Rb
(something as a function of u) du, where a and b define the range of the parameter u that occurs in a
a
particular application. The parameter summed over will not, generally be u, but could be x, y, z, ✓, etc.
depending on the application. An example of this has already been given in Example 2.1.2, in which we
found the volume of a cone by regarding it as a sum of the volumes of elementary disks. Here we shall
apply the same idea to a variety of other situations. This segment of the course will have succeeded if you
can generalize this idea to other continuous sums that you encounter in other areas of science, e.g finding
the electric or gravitational field due to a particular distribution of charge or mass etc..
2.2.1
Area
Example 2.2.1. Find the area of a semicircle of radius a.
First Method: This simple example can be done by findingpthe area
under the curve representing a circle (x2 + y 2 = a2 ): y = a2 x2 .
Subsequently we shall do it in two other ways. We regard
p the semicircle as consisting of thin strips of width dx and height a2 x2 (see
figure). The quantity to be summed is the area of the strip; since we
are here going to sum (i.e. integrate) over x we need to know the area
of the strip as a function of x and dx, and we need to know the range
of values of x that are to be summedZover.
a p
Based on the figure, we write A =
a2 x2 dx. This is the ana
Figure 2.2.1: The height of the elementary
p strip of width dx at position x is
a2 R x2 .pHence the area of the semicircle is aa a2 x2 dx
swer; it only remains to apply the techniques that we have learned to
evaluate it. The substitution (e.g.) x = a cos u suggests itself, thus
dx = a sin u du:
R x=a p
R x=a
A =
a2 a2 cos2 u a( sin u) du = a2 x= a sin2 u du
x= a
⇥
⇤⇡
R u=0
R⇡
=
a2 u=⇡ sin2 u du = a2 0 sin2 u du = a2 12 (u 12 sin 2u) 0 = a2 12 (⇡ 0) 12 (0 0) = 12 ⇡a2 .
Notice that (in line 2) we need to convert the limits for x into limits for the integration variable u. As x varies
from a to a, u varies from cos 1 ( 1) = ⇡ to cos 1 (1) = 0.
11
Example 2.2.2. Find the area of a semicircle of radius a.
Second method: We regard the semicircle as a collection of horizontal
strips of width (in the y direction) dy. The length of the strip
p in the x
direction (using the equation of the circle x2 + y 2 = a2 ) isp2 a2 y 2 .
Thus the quantity to be summed is the area of the strip: 2 a2 y 2 dy
and the range of y over which it is to be summed
Z a p is from y = 0 to
y = a. Based on the figure, we write A =
2 a2 y 2 dy. Except
0
for the factor of 2, and the limits, it is essentially the same integral as
in Example 2.2.1 and can be done in the same way. For variety we use
the substitution y = a sin u, thus dx = a cos u du:
R y=a p
R y=a
A =
2 a2 a2 sin2 u a cos u du = 2a2 y=0 cos2 u du
y=0
⇥
⇤⇡/2
R
u=⇡/2
= 2a2 u=0 cos2 u du = 2a2 12 (u + 12 sin 2u) 0
= 2a2 12 ( ⇡2 0) 12 (0 0) = 12 ⇡a2 .
Notice that we need to convert the limits for y into limits for the integration variable u. As y varies from 0 to a, u varies from sin 1 (0) = 0
to sin 1 (1) = ⇡2 .
Figure 2.2.2: The width of the elementary
p strip of height dy at position y is
2 a2 y 2 . p
Hence the area of the semiR
circle is 0a 2 a2 y 2 dy
Example 2.2.3. Find the area of a semicircle of radius a.
Third method: We regard the semicircle as a collection of narrow triangles, subtending an angle d✓ at the centre. The area of the elementary
triangle is 12 ⇥ base ⇥ height = 12 ⇥ a d✓ ⇥ a. Thus the quantity to be
summed is 12 a2 d✓ and the range of ✓ over which it is to be summed is
fromZ✓ = 0 to ✓ = ⇡. Based on the figure, we write
⇡
A=
0
1 2
a
2
2
1
d✓ = 12 a2 [✓]✓=⇡
✓=0 = 2 ⇡a .
Clearly this is the simplest of the three methods (Examples 2.2.1-2.2.3)
for finding the area of a semicircle in the form of an integral.
Figure 2.2.3: The semicircle is regarded as
a collection of narrow triangles. The ‘base’
of the triangle is a d✓, and the ‘height’ is a.
Example 2.2.4. Find the area of the surface of a sphere of radius a.
We introduce the angle ✓ which is the angle measured at the centre
of the sphere between the z-axis and any point on the sphere. If the
sphere represented the Earth, ✓ would represent ( ⇡2
latitude). The
‘N-pole’ is ✓ = 0, the ‘equator’ is ✓ = ⇡2 and the ‘S-pole’ is ✓ = ⇡. We
regard the spherical surface as a collection of narrow strips or ribbons
that lie on circles of ‘latitude’ (i.e. along lines of constant ✓), having
widths subtending an angle d✓ at the centre. Thus the area of the
’ribbon’, the quantity to be summed, is 2⇡a2 sin ✓ d✓ (see figure and
caption).
Z Thus the area of the sphere is
⇡
A=
2⇡a2 sin ✓ d✓. Notice that the limits are chosen so that the rib-
0
bons span the whole spherical surface from the ‘N-pole’ to the ‘S-pole’.
Thus we have
R⇡
A = 2⇡a2 0 sin ✓ d✓ = 2⇡a2 [ cos ✓]✓=⇡
✓=0
=
2⇡a2 ((1)
( 1)) = 4⇡a2 .
12
Figure 2.2.4: The spherical surface is regarded as a collection of narrow circular
strips or ‘ribbons’. The length of the ribbon is 2⇡ ⇥ (its radius) = 2⇡a sin ✓, and
the width of the ribbon is a d✓. Thus its
area is 2⇡a2 sin ✓ d✓
2.2.2
Arc Length
The same principle can be applied to find the arclength of a curve.
The quantity to be summed is the distance along the curve, ds say,
corresponding to an increment
dx (see figure). We have
q
p
2
2
2
ds = dx + dy = 1 + ( dy/ dx) dx.
Figure 2.2.5: The element of arc length ds
is given by dx/ cos ✓, where ✓ is the angle
that the curve makes with the x-axis. Since
tan ✓ p
= dy/ dx and sec2 ✓ = 1 + tan2 ✓,
ds = 1 + ( dy/ dx)2 dx
Example 2.2.5. Find the arc length of the curve y = 1b cosh bx in the interval x 2 [ a, a] (i.e. between x = a and
x = a).
Z a p
Z a
dy/dx = sinh bx. Therefore arc length =
1 + sinh2 bx dx =
cosh bx dx = [ 1b sinh bx]a a = 2b sinh ab.
a
a
(The curve y = (1/b) cosh bx is called a catenary; it is the shape naturally adopted by a hanging chain.)
2.2.3
The Centroid
Consider a collection of point masses mi
masses is defined by the relations:
P
P
X ( i mi ) =
X
i xi mi
P
P
or
Y ( i mi ) =
Y
i yi mi
P
P
Z ( i mi ) =
Z
i zi mi
at locations (xi , yi , zi ). The centroid (X, Y, Z) of the distribution of
P
P
= ( i xi mi ) / ( i mi )
P
P
= ( i yi mi ) / ( i mi ) .
P
P
= ( i zi mi ) / ( i mi )
The centroid (X, Y, Z) is sometimes described as the geometrical centre of a given mass distribution
(weighted according to mass). Thus the centroid of the galaxy can be thought of as the geometrical centre of the galaxy, about which all the stars of the galaxy rotate. The centroid of an atom is located in the
nucleus, since most of the mass is concentrated there, but is slightly displaced according to the distribution
of the electrons around the nucleus. The centroid of the solar system is located in the Sun, since most of
the mass is concentrated there, but is slightly displaced according to the distribution of the planets about
the Sun. The solar system rotates about this centroid, and thus the centre of the sun is slightly displaced
from the point around which the sun and the planets rotate. The centroid can be thought of as the average
location of a mass distribution. The centroid is sometimes called the centre of mass or the centre of gravity. A
continuous mass distribution also possesses a centroid. For example a plane figure (e.g. cut from a metal
sheet) possesses a centroid, which is the point in the plane on which the plane figure can be balanced under
gravity. If the body possesses an axis of symmetry, the centroid will lie on this axis.
The quantities (xi mi , yi mi , zi mi ) are called the x-, y- and z-moments of the point mass mi , and thus the
centroid can be defined as the point having the property that if all the mass is concentrated at the point
(X, Y, Z) then the x-, y- and z-moments are the same as the sum of the moments of the constituent masses:
!
!
x-, y- and z-moments of the total mass
concentrated at the centroid
=
Sum of the x-, y- and z-moments of the
constituent masses
We shall be concerned with situations in which the mass distribution is continuous. For a plane figure,
having mass-per-unit-area , we can write, symbolically
R
R
X
dA =
x dA
R
R
Y
dA =
y dA
13
where dA represents an element of area. For solid bodies, having mass-per-unit-volume ⇢, we can write
R
R
X ⇢ dV =
x⇢ dV
R
R
Y ⇢ dV =
y⇢ dV
R
R
Z ⇢ dV. =
z⇢ dV
where dV represents an element of volume. In the case that the mass-per-unit-area or the mass-per-unitvolume ⇢ are constant, these relations reduce to
R
R
X dA =
x dA
R
R
Y dA =
y dA
for two-dimensional bodies, and
R
R
X dV =
x dV
R
R
Y dV =
y dV
R
R
Z dV. =
z dV
for three-dimensional bodies, and the centroid becomes a purely geometrical property of the shape of the
body. Thus we can speak of the centroid of a semicircle or the centroid of a hemisphere.
Example 2.2.6. Find the centroid of the semicircle x2 + y 2 6 a2 , y > 0.
The area defined is as shown in the figure. By symmetry, the centroid
liesZ on the y-axis: X = Z
0. To find Y , we take y-moments:
a p
a
p
Y
2 a2 y 2 dy =
2y a2 y 2 dy.
0
0
Notice that the integral on the left side is the total area, in the form used
in Example 2.2.2. The integral on the right, which is the same, apart
from the extra factor y, is the sum of the y-moments of the elements.
We use horizontal (y = const.) strips because it is simple to take ymoments for such elements because y is the same for all constituent
particles within the strip.
The integral on the left side has been calculated in Example 2.2.2, and
in any case is known to be equal to 12 ⇡a2 , the area of the semicircle.
The integral on the right side is:

Z ap
y=a
(a2 y 2 )3/2
a2 y 2 d(y 2 ) =
= 23 a3 .
3/2
0
y=0
4
1
Thus Y = 23 a3
⇡a2 =
a ⇡ 0.42 a (seems reasonable).
2
3⇡
✓
◆
4
Thus the centroid of the semicircle is at (X, Y ) = 0,
a .
3⇡
14
Figure 2.2.6: The areapof the elementary
2
strip at position
y 2 dy. Its yp y is 2 a
moment is 2y a2 y 2 dy
Example 2.2.7. Find the centroid of the triangle having vertices (0, 0),
(a, 0), (e, h).
The equations of the sides AB and AC are (Exercise):
AB : y = xh/e
or x = ey/h
AC : y = h(x a)/(e a) or x = a + (e a)y/h.
(We can check that the coordinates of points A and B satisfy the first
pair of equations, and that the coordinates of points A and C satisfy
the second pair of equations X)
The length of the horizontal strip at position y (Fig. 2.2.7a) is a + (e
a)y/h ey/h = a(1 y/h). (This should be equal to a when y = 0
and equal to 0 when y = h X)
The equation for the coordinate Y of the centroid is:
Z h
Z h
Y
a(1 y/h) dy =
ay(1 y/h) dy
0
0
The integral on the left is:
⇥
Rh
a(1 y/h) dy =
a(y
0
⇤y=h
y 2 /2h) y=0
= a(h 12 h) 0 = 12 ah.
(This should be the area of the triangle: 12 ⇥base⇥height =
The integral on the right is:
⇥ 2
⇤y=h
Rh
ay(1 y/h) dy =
a(y /2 y 3 /3h) y=0
0
Thus Y =
1
ah2
6
= a( 12 h2
1
ah = 13 h.
2
1 2
h )
3
1
⇥a⇥h
2
X)
Figure 2.2.7: Finding the centroid of a triangle. If vertical strips are used the triangle
needs to be split into two parts.
0 = 16 ah2 .
Let us find the coordinate X of the centroid by using vertical strips. In
this case we need to write the integrals in two parts corresponding to the left and right pieces of the triangle as
in Fig. 2.2.7b. The equation for the coordinate X of the centroid can be written:
⇣R
⌘ ⇣R
⌘
Ra
Ra
e
e
X 0 xh
dx + e e h a (x a) dx = 0 x xh
dx + e x e h a (x a) dx
e
e
It can be shown (I am omitting some details here so that we can see the wood for the trees) that sum of the
integrals on the left reduces to 12 ah, as it must, and the sum of the integrals on the right reduces to 16 ah(a + e).
1
Thus X = 16 ah(a + e)
ah = 13 (a + e).
2
Another way to find X is to use horizontal strips, as in Fig. 2.2.7a, and to use the fact that the x-moment of the
horizontal strip
⇣ can be obtained
⌘ by concentrating its mass at its midpoint. The midpoint of the strip at position
y is at x =
y = h X)
Thus X ⇥
1
2
ey
h
1
ah
2
+a+
=
=
(e a)y
h
Rh
= 12 a +
( 12 a +
⇣
a2
h+
2
0
e a/2
y.
h
e a/2
y)(a
h
a2
h
+
ae
h
⌘
(This should be equal to 12 a when y = 0 and equal to e when
a
y) dy
h
h2
2
+
⇣
=
a2
2h2
R h h a2
0
ae
h2
⌘
2
+
h3
3
⇣
a2
h
+
= 12 a2 h
ae
h
⌘
y+
1 2
a h
2
⇣
a2
2h2
ae
h2
⌘
i
y 2 dy
+ 12 aeh + 16 a2 h
1
aeh
3
= 16 a2 h + 16 aeh = 16 ah(a + e),
giving X =
+ e), as previously.
Thus the centroid of the triangle is at (X, Y ) = 13 (a + e), 13 h .
1
(a
3
(This result can be used to show that the centroid of a triangle is at the common intersection of the three medians of the triangle,
which are the straight lines joining each vertex to the midpoint of the opposite side. The medians intersect at a point which is
one-third of the height above the base, whichever side is regarded as the base. In fact, having found the Y coordinate of the
centroid in this example, we could have argued that, since the shape of the triangle is general, the centroid must be one-third
of the height above the base when, for example, AB is regarded as the base, and used this to find X)
15
Example 2.2.8. Find the centroid of a uniform solid hemisphere of radius a.
By symmetry, the centroid lies on the z-axis. We regard the hemisphere
as a collection of disks at position z and having thickness dz. Since
the equation of the sphere is x2 + y 2 + z 2 = a2 , the equation of the
perimeter of the disk at somep
fixed position z is x2 +y 2 = a2 z 2 , which
represents a circle of radius a2 z 2 . Thus the volume of the disk is
⇡(a2 z 2 ) dz and the equation for the coordinate Z of the centroid is:
Z a
Z a
Z
⇡(a2 z 2 ) dz =
⇡ z(a2 z 2 ) dz
0
0
h
i
Ra
3 z=a
The integral on the left is: 0 ⇡(a2 z 2 ) dz = ⇡ a2 z z3
= 23 ⇡a3
z=0
(This should be the volume of the hemisphere X).
The integral on the right
h is: 2
iz=a
Ra
2
3
2z
z4
⇡(a
z
z
)
dz
=
⇡
a
= 14 ⇡a4
2
4
0
Therefore Z =
1
⇡a4
4
2
⇡a3
3
z=0
= 38 a
(seems reasonable).
Thus the centroid of the solid hemisphere is at (X, Y, Z) = (0, 0, 38 a).
2.2.4
Figure 2.2.8: The hemisphere
x2 + y 2 + z 2 6 a 2 , z > p
0. The radius
of the disk at position z is a2 z 2 . Its
volume is therefore ⇡(a2
z 2 ) dz and its
2
2 ) dz.
z-moment
is
⇡z(a
z
Therefore
R
Ra
2
Z= 0a ⇡z(a2 z 2 ) dz
z 2 ) dz .
0 ⇡(a
Moments of Inertia
Consider a collection of point masses mi , (i = 1, 2, 3, · · · ). The moment
of inertia I of the distribution of masses about a given axis is defined to
be
I=
X
i
mi b2i ,
where bi is the perpendicular distance of mass mi from the axis
As in the case of centroids, we can extend the definition to continuous
bodies, writing, symbolically
Z
I=
b2 dA,
= mass-per-unit-area
Figure 2.2.9: bi is the perpendicular distance
of the ith mass from the given axis.
for two dimensional bodies and
Z
I = ⇢ b2 dV,
⇢ = mass-per-unit-volume
for three dimensional bodies. We can also consider one dimensional bodies, such as a wire or a rod, considered to be of zero width, writing
Z
I = ⌧ b2 dl,
⌧ = mass-per-unit-length
where dl is an element of length. Of course, b in these expressions is not a constant, but varies according
to position within the body. In the examples considered later we shall take ⇢, , ⌧ to be constant, which
means that the body is uniform – e.g. a uniform sphere, a uniform circular plate or a uniform rod. However,
there are many important applications in which ⇢, , ⌧ are not constant. For example, when calculating the
moment of inertia of the Earth it is essential to account for the increase in density with depth.
The dimensions of moment of inertia are clearly mass ⇥ length2 , and so moments of inertia are often expressed in the form
I = (dimensionless number) ⇥ (total mass) ⇥ (characteristic length)2 .
For example, we shall later find that the moment of inertia of a uniform solid sphere of radius a, about an
axis through its centre, is 25 M a2 , where M is the mass of the sphere.
16
Example 2.2.9. Find the moment of inertia of a straight, uniform rod of mass
M and length 2a, about an axis passing through its centre, perpendicular to
the rod.
FromZ the definition of moment of inertia:
a
I=
x2 ⌧ dx
(see Figure 2.2.10 and caption)
a
where ⌧ is mass-per-unit-length, ⌧ = M/2a.
h 3 ix=a
Thus I = ⌧ x3
= 23 ⌧ a3 = 13 M a2 .
Figure 2.2.10: The distance ‘b’ of the mass
element ⌧ dx from the axis is equal to x. ⌧ is
the mass-per-unit-length, i.e. ⌧ = M/2a.
x= a
Example 2.2.10. Find the moment of inertia of a straight, uniform rod of mass M and length 2a, about an axis passing
through one end, perpendicular to the rod.
Exercise.
Answer: 43 M a2
The moment of inertia is much larger than in Example 2.2.9 because the mass distribution extends further from the axis
Example 2.2.11. Find the moment of inertia of a straight, uniform rod of
mass M and length 2a, about an axis perpendicular to the rod, at a perpendicular distance h from the centre of the rod.
FromZ the definition of moment of inertia:
a
I=
(h2 + x2 ) ⌧ dx,
(see Figure 2.2.11 and caption)
a
where ⌧ is mass-per-unit-length, ⌧ = M/2a.
h
i
3 x=a
Thus I = ⌧ h2 x + x3
= ⌧ 2h2 a + 23 a3 = M h2 + 13 a2 .
x= a
There are two ways for us to check whether this is looks right: (i) when
h = 0 the result should agree with the result in Example 2.2.9 X
(ii) when a = 0 the rod collapses to a point mass at a distance h from
the axis, and so the moment of inertia in this case should be M h2 X
Note: If you are tempted to say that the moment of inertia is the same
as if the mass of the body is concentrated at its centre, or centroid, resist
the temptation; this argument can be made only when calculating the xy- and z- linear moments for locating centroids.
Figure 2.2.11: The axis is perpendicular to
the plane of the figure. The distance ‘b’ of
thepmass element ⌧ dx from the axis is equal
to h2 + x2 . ⌧ is the mass-per-unit-length,
⌧ = M/2a.
Example 2.2.12. Find the moment of inertia of a a uniform, 2a ⇥ 2a square
plate of mass M , about an axis perpendicular to the plate and passing through
its centre.
We regard the plate as a collection of strips at distance y from the axis
and having mass 2a dy (see Figure 2.2.12 and caption), where is the
mass-per-unit-area of the plate, = M/4a2 . The moment of inertia of
each elementary strip is given by the result of Example 2.2.11, and thus
we do not need to refer to the definition of moment of inertia explicitly.
We just sum the moments of inertia of the strips. The moment of inertia
of each strip (from the result of Example 2.2.11, ‘M ’!2a dy, ‘h’!y)
is (2a dy) y 2 + 13 a2 , and consequently the moment of inertia of the
plate is:
Z a
h 3
iy=a
I=
2a y 2 + 13 a2 dy = 2a y3 + 13 a2 y
= 83 a4 .
y= a
a
Thus, using
= M/4a2 , the result is I = 23 M a2 .
Figure 2.2.12: The square can be regarded
as a collection of ‘rods’, at distance y from
the axis, of width dy. The mass of the rod
is 2a dy, where is the mass-per-unit-area
of the plate, = M/4a2 .
17
Example 2.2.13. Find the moment of inertia of a uniform, hollow spherical shell of radius a and mass M about an axis passing through its centre.
We define the angle ✓ as in Example 2.2.4 and, as in that example, consider the spherical shell to consist of a collection of circular strips or
‘ribbons’, having width a d✓ and radius a sin ✓. The moment of inertia
of the ribbon about the z-axis is equal to (its mass) ⇥ (its radius)2 , because all the mass particles that make up the ribbon are at the same
distance ‘b’= a sin ✓ from the axis. Let be the mass-per-unit-area
of the spherical shell, = M/4⇡a2 . Then the mass of the ribbon is
length ⇥ width ⇥ mass-per-unit-area = 2⇡a2 sin ✓ d✓, and its moment
of inertia is (a sin ✓)2 ⇥ 2⇡a2 sin ✓ d✓.
Z ⇡
Z ⇡
Thus I =
2⇡a4 sin3 ✓ d✓ = 12 M a2
sin3 ✓ d✓.
0
Figure 2.2.13: The angle ✓ ( ⇡2
defined as in Example 2.2.4.
0
The
can be
by standard ⇥methods e.g.:
⇤⇡
R ⇡ integral
R ⇡ evaluated
3
2
sin
✓
d✓
=
(1
cos
✓)
d(cos ✓) =
cos ✓ + 13 cos3 ✓ 0 = 1
0
0
and therefore I = 23 M a2 .
1
3
1+
1
3
‘latitude’) is
= 43 ,
Example 2.2.14. Find the moment of inertia of a uniform solid sphere of radius a and mass M about an axis passing
through its centre.
We can regard the sphere as consisting of a collection of spherical shells having varying radius r and thickness
dr. Let ⇢ be the mass-per-unit-volume: ⇢ = M 43 ⇡a3 . Then the mass of the spherical shell of radius r is
⇢ ⇥ (its area) ⇥ (its thickness) = ⇢ ⇥ 4⇡r2 ⇥ dr. Using the result of Example 2.2.13, its moment of inertia is
2
⇥ (its mass) ⇥ (its radius)2 = 8⇡
⇢ r4 dr.
3
3
 5 r=a
Z a
8⇡ 4
8⇡
r
8⇡ 5
8⇡ M 5
2
Thus I =
⇢ r dr =
⇢
=
⇢a =
a = M a2 .
3
3
5 r=0
15
15 43 ⇡a3
5
0
Notice that, as in Example 2.2.12, we do not need to refer to the definition of moment of inertia explicitly; we are
just using integration to add up the moments of inertia of the constituent parts of the body.
(If the density ⇢ were not constant, but was a function of r, we would still obtain I =
Ra
8⇡
4
0 3 ⇢ r dr, but now with ⇢ = ⇢(r).
4 dr, M =
2 dr. This can be used to study the density distributions and moments of inertia of the
I = 0 8⇡
⇢(r)
r
4⇡⇢(r)
r
0
3
Earth, other planets and stars.)
Ra
Ra
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2.2.5
An example in a different area (optional)
Example 2.2.15. Newton’s law of gravitation states that two point masses separated by a distance d exert attractive forces
on each other, directed along the line joining them, of magnitude Gm1 m2 /d2 , where m1 and m2 are the masses and where
G is the universal constant of gravitation (G ⇡ 6.6726 ⇥ 10 11 m3 kg 1 s 2 ). Find the force acting on a unit mass placed
on the rotational symmetry axis of a uniform circular plate of mass M and radius a, the point mass being at perpendicular
distance h from the centre of the plate.
(It would be essentially the same problem if we wished to calculate the electric field due to a uniformly charged disk at a point
on the axis, as the electric field due to a point charge also satisfies an inverse square law.)
We consider the plate to consist of rings having variable radius r and width dr. By symmetry, the force due to
the ring on the unit mass is directed along axis and therefore we need to know the projection of the force in the
direction of the axis. The angle between the axis and a line joining the point mass to the perimeter of the ring of
radius r is ✓ = tan 1 r/h. This angle is the same for each particle that makes up the ring of radius r; therefore
the projection onto the axis of the force due to the ring is
G(2⇡r dr)
2⇡G r dr
1
p
cos ✓ =
= 2⇡G hr(h2 + r2 ) 3/2 dr,
h2 + r 2
h2 + r 2
1 + r2 /h2
where
is the mass-per-unit-area of the plate,
p
1/ 1 + tan2 ✓.
= M/⇡a2 , and where we have used the identity cos ✓ =
Thus the total force acting on the unit point mass (which is the gravitational acceleration) is
 2
Z r=a
r=a
(h + r2 ) 1/2
f =
⇡G h(h2 + r2 ) 3/2 d(r2 ) = ⇡G h
1/2
r=0
r=0
✓
◆
✓
◆
1
1
2GM
h
p
= 2⇡G h
=
1 p
.
h
a2
h2 + a 2
h2 + a 2
(When h = 0 or, approximately, when h is small in comparison with a, there is a gravitational acceleration 2GM/a2 =
2⇡G , a fact that is used in the calculation of gravitational anomaly when it is required to estimate the gravitational
acceleration due to the rock mass above sea level; the relevant mass-per-unit-area is
= (mean rock density) ⇥
(height of topography above sea-level).)
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