CH.6 PROPERTIES OF RIGID BODY
1- Center of gravity
A body is composed of an infinite number of particles of differential size,
then each of these particles will have a weight dW. These weights will
form a parallel force system, and the resultant of this system is the total
weight of the body W, which passes through a single point called the
center of gravity, G. So it is the point at which the gravity force (weight)
acts.
The location of the center of gravity, measured
from the y axis, is determined by equating the
moment of W about the y axis, to the sum of the
moments of the weights of all its particles about
the same axis, ๐๐ฎ๐พ = โซ ๐ ๐พ ๐. Therefore
๐๐ฎ =โซ ๐ ๐พ ๐
โซ ๐ ๐พ. So, the center of gravity can be
obtained with respect to the three axes by:
๐ฅ๐บ =โซ ๐๐ ๐ฅ
โซ ๐๐ ๐ฆ๐บ =
โซ ๐๐ ๐ฆ
โซ ๐๐ ๐ง๐บ =
โซ ๐๐ ๐ง
โซ ๐๐
2- Center of Mass
In order to study the accelerated motion of a body, it
becomes important to locate the bodyโs center of
mass Cm. This location can be determined by
substituting ๐๐ค = ๐ ๐๐ into the above equation.
Provided ๐ is constant, it cancels out, and the mass center is given by:
๐ฅ๐ถ๐ =โซ ๐๐ ๐ฅ
โซ ๐๐ ๐ฆ๐ถ๐ =
โซ ๐๐ ๐ฆ
โซ ๐๐ ๐ง๐ถ๐ =
โซ ๐๐ ๐ง
โซ ๐๐
3- Centroid
It is the geometrical center of the body, it is
determined by:
๐ฅ๐ถ =โซ ๐๐ ๐ฅ
โซ ๐๐ ๐ฆ๐ถ =
โซ ๐๐ ๐ฆ
โซ ๐๐ ๐ง๐ถ =
โซ ๐๐ ๐ง
โซ ๐๐
Important Notes
I. For homogeneous bodies ( = constant), the mass center coincides
with the centroid.
II. If a body has axis of symmetry, the centroid of
the body lies on this axis of symmetry. For
example, the cone in the figure has a centroid lies
on the y axis (y axis of symmetry) so that xc = zc
= 0. The location yc can be found by using a
single integration: ๐๐ช =โซ ๐ ๐ฝ ๐
โซ ๐ ๐ฝ
Example 6.1
Determine the location of the mass center of
the shown homogeneous cone with mass m,
base radius a and height h.
Solution:
The mass center coincides with the axis of symmetry, which is the y-axis.
Thus, xc = zc= o and
๐ฆ๐ถ =โซ ๐๐ ๐ฆ
โซ ๐๐
By choosing a differential element represented by a thin disk having a
thickness dy and radius r. Its volume is ๐๐ = ๐๐2๐๐ฆ
๐ฆ๐ถ =โซ ๐๐2๐ฆ ๐๐ฆ
โซ ๐๐2๐๐ฆ
From similarity of the triangles, then ๐
๐=
๐ฆ
โ
Substituting about r then,
๐ฆ๐ถ =โซ ๐ (
๐โ
๐ฆ)2
๐ฆ ๐๐ฆโ
0
โซ ๐ (๐โ
๐ฆ)2
๐๐ฆโ
0
=โซ ๐ฆ3 ๐๐ฆ
โ
0
โซ ๐ฆ2 ๐๐ฆโ
0
=(
๐ฆ4
4 ) โ0
(๐ฆ3
3 ) โ0
=3
4โ
z
y dy
r a
y
x
h
Mass Center of a Composite Body
A composite body consists of a series of connected โsimplerโ bodies.
Such a body can often be divided into number of parts provided the mass
and location of the mass center of these parts are known.
There is no need for integration to determine the center of mass for the
entire body, since we deal with finite number of simple bodies. We will
take the moment of masses about the 3 axes and equating each moment
with the moment of the mass of the whole body about the same axis.
๐ฅ๐ถ =โ ๐๐
๐=๐๐=1 ๐ฅ๐๐
โ ๐๐๐=๐๐=1
๐ฆ๐ถ =โ ๐๐
๐=๐๐=1 ๐ฆ๐๐
โ ๐๐๐=๐๐=1
๐ง๐ถ =โ ๐๐
๐=๐๐=1 ๐ง๐๐
โ ๐๐๐=๐๐=1
Where xCi , yCi , and zCi are the known coordinates of the mass center of
the part i of the composite body and n is the number of all parts.
It must be noted that in the case of holes (which are considered as parts of
the body), their masses must be considered negative because they must
subtracted from the body.
Example 6.2
Determine the mass center of the shown
pendulum consisting of a thin rod of mass
m1, and length L and a sphere of mass m2
and radius R.
Solution:
The x-axis is an axis of symmetry, the centroid of the body lies on this
axis. Thus, yC=0 and zC.
๐ฅ๐ถ =โ ๐๐
๐=2๐=1 ๐ฅ๐๐
โ ๐๐๐=2๐=1
๐ฅ๐ถ =๐1๐ฅ๐1 + ๐2๐ฅ๐2
๐1 + ๐2
๐ฅ๐1 = ๐ฟ/2
๐ฅ๐2 = ๐ฟ + ๐
๐ฅ๐ถ =๐1๐ฟ/2 + ๐2(๐ฟ + ๐ )
๐1 + ๐2
m2
m1 C1 R
O x
L/2 C2
L
4- Mass Moment of Inertia
The moment of inertia is a measure of the resistance of a body to the
angular acceleration in the same way that mass is a measure of the
bodyโs resistance to the linear acceleration.
We define the moment of inertia as the integral of the
โsecond momentโ about an axis of all the elements of mass
dm which compose the body. For example, the bodyโs
moment of inertia about the z axis is
๐ผ๐ง = โซ ๐2๐๐
Here the โmoment armโ r is the perpendicular distance from the z axis to
the arbitrary element dm. The value of I will depend on the axis about
which it is computed.
Since r is squared in the equation, the mass moment of inertia is always a
positive quantity. It is measured by kg.m2.
Radius of Gyration
Occasionally, the moment of inertia of a body about a specified axis is
reported in handbooks using the radius of gyration, k. This is a
geometrical property which has units of length. When the bodyโs moment
of inertia I and the bodyโs mass m are known, it is determined from the
equation:
๐๐ฅ = โ๐ผ๐ฅ
๐ ๐๐ ๐๐ฆ = โ
๐ผ๐ฆ
๐ ๐๐ ๐๐ฆ = โ
๐ผ๐ฆ
๐
Where ๐๐ฅ is the radius of gyration of the body with respect to x axis.
๐๐ฆ is the radius of gyration of the body with respect to y axis.
๐๐ง is the radius of gyration of the body with respect to z axis.
If the radius of gyration k and the mass of the body are known, the
moment of inertia is given by:
๐ผ๐ฅ = ๐ ๐๐ฅ2 ๐๐ ๐ผ๐ฆ = ๐ ๐๐ฆ
2 ๐๐ ๐ผ๐ง = ๐ ๐๐ง2
Parallel Axis Theorem
If the mass moment of inertia of a body is known
with respect to a centroidal axis Sc , it may be
easily determined with respect to any other axis
parallel to Sc. If the perpendicular distance
between the considered axes is d, then:
๐ผ๐ = ๐ผ๐ ๐ + ๐ ๐2
Dm
H
ro
d C
r o
d SC
S
Sc axis passing through its
mass center c
S axis parallel to Sc
Example 6.4
Determine the mass moment of inertia of a slender rod of length L and
mass m with respect to an axis perpendicular to the rod and passing
through:
a. one end of the rod.
b. the center of mass C of the rod.
Solution:
Choosing the differential element of
mass as shown, its mass equal
๐๐ =๐
๐ฟ ๐๐ฆ
๐ผ๐ง = ๐ผ๐ฅ = โซ ๐ฆ2๐๐ = โซ๐
๐ฟ๐ฆ2๐๐ฆ =
๐ฟ
0
๐
๐ฟ(
๐ฆ3
3)
๐ฟ
0=
๐๐ฟ2
3
Using parallel axis theorem
๐ผ๐ง๐ = ๐ผ๐ โ ๐ (๐ฟ
2)
2
=๐๐ฟ2
3โ
๐๐ฟ2
4=
๐๐ฟ2
12
z
y
x L
z
y dy y
x
Example 6.5
Determine the moment of inertia of the right circular cylinder with radius
R, height h and mass m, with respect the z axis. The mass density ฯ of
the materials is constant.
Solution:
This problem can be solved using the shell, its volume is
dV = 2rhdr, so that its mass is dm = dV = (2hr dr).
Since the entire element lies at the same distance r from
the z axis, the moment of inertia Iz is given by
๐ผ๐ง = โซ ๐2๐๐ = โซ ฯ(2ฯโ๐๐๐)๐
0
๐2 = 2๐โ๐ โซ ๐3๐
0
= 2๐โ๐๐ 4
4
=1
2๐๐ 4โ๐
Since ๐ = ๐๐ 2โ ๐, then
๐ผ๐ง =1
2 ๐ ๐ 2
Moment of Inertia of a Thin Plate
For a thin plate with uniform thickness h, that
locates in the x-y plane. The mass moments of
inertia of the plate with respect to coordinate axes
x, y and z are:
๐ผ๐ฅ = โซ ๐ฆ2๐๐ ๐ผ๐ฆ = โซ ๐ฆ2๐๐ ๐ผ๐ง = โซ ๐2๐๐
Noting that, ๐2 = ๐ฅ2 + ๐ฆ2, so
๐ผ๐ง = โซ ๐2๐๐ = ๐ผ๐ง = โซ (๐ฅ2 + ๐ฆ2) ๐๐
๐ผ๐ง = ๐ผ๐ฅ + ๐ผ๐ฆ
๐ผ๐ง is the moment of inertia about axis perpendicular to the plate, it may be
denoted by ๐ผ๐ (polar moment of inertia). Axis z is called polar axis. The
above relation, for a thin plate, is the same as that for area moments, of
inertia.
z
h
O y
r x
y dm
x
Example 6.7
Determine the mass moment of inertia of a
thin circular disk of mass m and radius R,
with respect to:
a) The disk axis of symmetry, z,
b) Two diameters x and y.
Solution
a) Considering example 6.5, it may be observed that the mass moment
inertia of the cylinder with respect to its axis, Iz=1/2 m R2, does not
depend on the height h. Hence, the mass moment of inertia of the
shown thin disk about its axis of symmetry
๐ผ๐ง =1
2 ๐ ๐ 2
b) From the symmetry of the disk, then Ix= Iy (x and y are diameters).
Applying the obtained relation for a thin plate located in x-y plane, we
have for the given circular disk;
๐ผ๐ง = ๐ผ๐ฅ + ๐ผ๐ฆ = 2๐ผ๐ฅ = 2๐ผ๐ฆ
Hence:
๐ผ๐ฅ = ๐ผ๐ฆ =๐ผ๐ง
2=
1
4๐ ๐ 2
z
R
y
x
Moment of Inertia of a Composite Body
If a body consists of a number of simple bodies such as disks, spheres,
and rods, the mass moment of inertia of this composite body is the sum of
the moments of inertia of its individual parts with respect to the same
axis. If a composite body has negative masses (such as holes), the
moment of inertia of these holes, must be considered negative quantity.
It must be noted that, the parallel axis theorem is needed for the
calculations if the center of mass of each composite part does not lie on
the axis. The moment of inertia for each part (such as rods and disks)
about axis passing through its mass center is determined from a table that
given in the book.
๐ผ๐ฅ = โ ๐ผ๐ฅ๐
๐=๐
๐=๐
= ๐ผ๐ฅ๐+ ๐ผ๐ฅ๐
+ ๐ผ๐ฅ๐+ โฏ + ๐ผ๐ฅ๐
The number of parts is (n)
Example 6.10
Determine the mass moment of inertia of
the shown pendulum with respect to the z-
axis. The pendulum consists of a slender
rod, of mass m1, and length L, and a disk
of mass m2 and radius R.
The body consists of two parts, then
๐ผ๐ง = ๐ผ๐ง1+ ๐ผ๐ง2
For part (1)
๐ผ๐ง1= ๐ผ๐งโฒ + ๐1 (
๐ฟ
2)
2
=๐1๐ฟ2
12+
๐1๐ฟ2
4=
๐1๐ฟ2
3
For part (2)
๐ผ๐ง2= ๐ผ๐งโฒโฒ + ๐2(๐ฟ + ๐ )2 =
๐2๐ 2
4+ ๐2(๐ฟ + ๐ )2
So
๐ผ๐ง =๐1๐ฟ2
3+
๐2๐ 2
4+ ๐2(๐ฟ + ๐ )2
z zโ m2 zโ
m1 C1 R
O x
L/2 C2
L
Example 6.11
Determine the mass moment of inertia of the shown
wheel with respect to its axis z. The density of the
wheel material is =300 kg/m3
Solution
The wheel may be considered as a solid disk (of radius R= 0.5 m and
thickness H=50 mm) from which two circular disks (of radius r=0.3 m
and thickness h=20 mm) are extracted. Thus:
Part (1) ๐1 = ๐๐๐ 2๐ป = 300๐(0.5)2(0.05) = 11.78 ๐๐
๐ผ๐ง1=
๐1๐ 2
2=
1
2(11.78)(0.5)2 = 1.4725 ๐๐. ๐2
Part (2) ๐2 = ๐๐๐2โ = 300๐(0.3)2(0.02) = 1.7 ๐๐
๐ผ๐ง2=
๐2๐ 2
2=
1
2(1.7)(0.3)2 = 0.0765 ๐๐. ๐2
๐ผ๐ง = ๐ผ๐ง1โ 2๐ผ๐ง2
= 1.4725 โ 2(0.0765) = 1.3195 ๐๐. ๐2
0.5 m
0.3m 10mm
z
2
1
50 mm