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, 𝒙𝑮𝑾 = ∫ 𝒅𝑾 𝒙. Therefore

𝒙𝑮 =∫ 𝒅𝑾 𝒙

∫ 𝒅𝑾. So, the center of gravity can be

obtained with respect to the three axes by:

𝑥𝐺 =∫ 𝑑𝑊 𝑥

∫ 𝑑𝑊 𝑦𝐺 =

∫ 𝑑𝑊 𝑦

∫ 𝑑𝑊 𝑧𝐺 =

∫ 𝑑𝑊 𝑧

∫ 𝑑𝑊

2- Center of Mass

In order to study the accelerated motion of a body, it

becomes important to locate the body’s center of

mass Cm. This location can be determined by

substituting 𝑑𝑤 = 𝑔 𝑑𝑚 into the above equation.

Provided 𝑔 is constant, it cancels out, and the mass center is given by:

𝑥𝐶𝑚 =∫ 𝑑𝑚 𝑥

∫ 𝑑𝑚 𝑦𝐶𝑚 =

∫ 𝑑𝑚 𝑦

∫ 𝑑𝑚 𝑧𝐶𝑚 =

∫ 𝑑𝑚 𝑧

∫ 𝑑𝑚

3- Centroid

It is the geometrical center of the body, it is

determined by:

𝑥𝐶 =∫ 𝑑𝑉 𝑥

∫ 𝑑𝑉 𝑦𝐶 =

∫ 𝑑𝑉 𝑦

∫ 𝑑𝑉 𝑧𝐶 =

∫ 𝑑𝑉 𝑧

∫ 𝑑𝑉

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: 𝒚𝑪 =∫ 𝒅𝑽 𝒚

∫ 𝒅𝑽

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

𝑦𝐶 =∫ 𝑑𝑉 𝑦

∫ 𝑑𝑉

By choosing a differential element represented by a thin disk having a

thickness dy and radius r. Its volume is 𝑑𝑉 = 𝜋𝑟2𝑑𝑦

𝑦𝐶 =∫ 𝜋𝑟2𝑦 𝑑𝑦

∫ 𝜋𝑟2𝑑𝑦

From similarity of the triangles, then 𝑟

𝑎=

𝑦

ℎ

Substituting about r then,

𝑦𝐶 =∫ 𝜋 (

𝑎ℎ

𝑦)2

𝑦 𝑑𝑦ℎ

0

∫ 𝜋 (𝑎ℎ

𝑦)2

𝑑𝑦ℎ

0

=∫ 𝑦3 𝑑𝑦

ℎ

0

∫ 𝑦2 𝑑𝑦ℎ

0

=(

𝑦4

4 ) ℎ0

(𝑦3

3 ) ℎ0

=3

4ℎ

z

y dy

r a

y

x

h

Mass Center of a Composite Body

A composite body consists of a series of connected “simpler” 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.

𝑥𝐶 =∑ 𝑚𝑖

𝑖=𝑛𝑖=1 𝑥𝑐𝑖

∑ 𝑚𝑖𝑖=𝑛𝑖=1

𝑦𝐶 =∑ 𝑚𝑖

𝑖=𝑛𝑖=1 𝑦𝑐𝑖

∑ 𝑚𝑖𝑖=𝑛𝑖=1

𝑧𝐶 =∑ 𝑚𝑖

𝑖=𝑛𝑖=1 𝑧𝑐𝑖

∑ 𝑚𝑖𝑖=𝑛𝑖=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.

𝑥𝐶 =∑ 𝑚𝑖

𝑖=2𝑖=1 𝑥𝑐𝑖

∑ 𝑚𝑖𝑖=2𝑖=1

𝑥𝐶 =𝑚1𝑥𝑐1 + 𝑚2𝑥𝑐2

𝑚1 + 𝑚2

𝑥𝑐1 = 𝐿/2

𝑥𝑐2 = 𝐿 + 𝑅

𝑥𝐶 =𝑚1𝐿/2 + 𝑚2(𝐿 + 𝑅)

𝑚1 + 𝑚2

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

body’s resistance to the linear acceleration.

We define the moment of inertia as the integral of the

“second moment” about an axis of all the elements of mass

dm which compose the body. For example, the body’s

moment of inertia about the z axis is

𝐼𝑧 = ∫ 𝑟2𝑑𝑚

Here the “moment arm” 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 body’s moment

of inertia I and the body’s mass m are known, it is determined from the

equation:

𝑘𝑥 = √𝐼𝑥

𝑚 𝑜𝑟 𝑘𝑦 = √

𝐼𝑦

𝑚 𝑜𝑟 𝑘𝑦 = √

𝐼𝑦

𝑚

Where 𝑘𝑥 is the radius of gyration of the body with respect to x axis.

𝑘𝑦 is the radius of gyration of the body with respect to y axis.

𝑘𝑧 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:

𝐼𝑥 = 𝑚 𝑘𝑥2 𝑜𝑟 𝐼𝑦 = 𝑚 𝑘𝑦

2 𝑜𝑟 𝐼𝑧 = 𝑚 𝑘𝑧2

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:

𝐼𝑠 = 𝐼𝑠𝑐 + 𝑚 𝑑2

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

𝑑𝑚 =𝑚

𝐿 𝑑𝑦

𝐼𝑧 = 𝐼𝑥 = ∫ 𝑦2𝑑𝑚 = ∫𝑚

𝐿𝑦2𝑑𝑦 =

𝐿

0

𝑚

𝐿(

𝑦3

3)

𝐿

0=

𝑚𝐿2

3

Using parallel axis theorem

𝐼𝑧𝑐 = 𝐼𝑍 − 𝑚 (𝐿

2)

2

=𝑚𝐿2

3−

𝑚𝐿2

4=

𝑚𝐿2

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 ρ 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

𝐼𝑧 = ∫ 𝑟2𝑑𝑚 = ∫ ρ(2πℎ𝑟𝑑𝑟)𝑅

0

𝑟2 = 2𝜋ℎ𝜌 ∫ 𝑟3𝑅

0

= 2𝜋ℎ𝜌𝑅4

4

=1

2𝜋𝑅4ℎ𝜌

Since 𝑚 = 𝜋𝑅2ℎ 𝜌, then

𝐼𝑧 =1

2 𝑚 𝑅2

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:

𝐼𝑥 = ∫ 𝑦2𝑑𝑚 𝐼𝑦 = ∫ 𝑦2𝑑𝑚 𝐼𝑧 = ∫ 𝑟2𝑑𝑚

Noting that, 𝑟2 = 𝑥2 + 𝑦2, so

𝐼𝑧 = ∫ 𝑟2𝑑𝑚 = 𝐼𝑧 = ∫ (𝑥2 + 𝑦2) 𝑑𝑚

𝐼𝑧 = 𝐼𝑥 + 𝐼𝑦

𝐼𝑧 is the moment of inertia about axis perpendicular to the plate, it may be

denoted by 𝐼𝑜 (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

𝐼𝑧 =1

2 𝑚 𝑅2

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;

𝐼𝑧 = 𝐼𝑥 + 𝐼𝑦 = 2𝐼𝑥 = 2𝐼𝑦

Hence:

𝐼𝑥 = 𝐼𝑦 =𝐼𝑧

2=

1

4𝑚 𝑅2

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.

𝐼𝑥 = ∑ 𝐼𝑥𝒊

𝒊=𝒏

𝒊=𝟏

= 𝐼𝑥𝟏+ 𝐼𝑥𝟐

+ 𝐼𝑥𝟑+ ⋯ + 𝐼𝑥𝒏

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

𝐼𝑧 = 𝐼𝑧1+ 𝐼𝑧2

For part (1)

𝐼𝑧1= 𝐼𝑧′ + 𝑚1 (

𝐿

2)

2

=𝑚1𝐿2

12+

𝑚1𝐿2

4=

𝑚1𝐿2

3

For part (2)

𝐼𝑧2= 𝐼𝑧′′ + 𝑚2(𝐿 + 𝑅)2 =

𝑚2𝑅2

4+ 𝑚2(𝐿 + 𝑅)2

So

𝐼𝑧 =𝑚1𝐿2

3+

𝑚2𝑅2

4+ 𝑚2(𝐿 + 𝑅)2

z z’ m2 z”

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) 𝑚1 = 𝜌𝜋𝑅2𝐻 = 300𝜋(0.5)2(0.05) = 11.78 𝑘𝑔

𝐼𝑧1=

𝑚1𝑅2

2=

1

2(11.78)(0.5)2 = 1.4725 𝑘𝑔. 𝑚2

Part (2) 𝑚2 = 𝜌𝜋𝑟2ℎ = 300𝜋(0.3)2(0.02) = 1.7 𝑘𝑔

𝐼𝑧2=

𝑚2𝑅2

2=

1

2(1.7)(0.3)2 = 0.0765 𝑘𝑔. 𝑚2

𝐼𝑧 = 𝐼𝑧1− 2𝐼𝑧2

= 1.4725 − 2(0.0765) = 1.3195 𝑘𝑔. 𝑚2

0.5 m

0.3m 10mm

z

2

1

50 mm