Study Unit Ill Engineerin M Part4 an1cs By And.docx

Study Unit
Ill Engineerin M
Part4
an1cs
By
Andrew Pytel, Ph.D.
Associate Professor, Engineering Mechanics
The Pennsylvania State University
When you complete this study unit, you'll be able to
• Calculate the mass moment of inertia
• Calculate the kinetic energy of a body
• Determine the linear impulse and momentum of a body
• Analyze the equations and conditions used to determine the
forces involving rectilinear
translation
• Describe centripetal and centrifugal force
• Describe the forces that impact the rotation of a rigid body
without translation

• Explain the motion of a wheel, and calculate the magnitude of
the linear acceleration and
friction forces
• Analyze the work-energy method as it applies to the motion
and action of a body
iii
PRELIMINARY EXPLANATIONS PERTAINING TO
KINETICS .
FORCE-MASS-ACCELERATION METHOD .....
Translation of Rigid Body
Rotation of Rigid Body without Translation
General Plane Motion of Rigid Body
23
WORK-ENERGY METHOD . . . . . . . . . . . . . . . . . . . . . . . . .
53
Application of Method for Translation
Other Applications of Work-Energy Method
IMPULSE-MOMENTUM METHOD . . . . . . .
Rectilinear Translation of Single Body
Collision of Two Bodies
PRACTICE PROBLEMS ANSWERS
EXAMINATION . . . . . . . . . .

. ........... 77
93
95
Engineering Nlechanics, Part 4
PRELIMINARY EXPLANATIONS PERTAINING
TO KINETICS
Scope of This Text
1
1 • In the preceding texts on engineering mechanics, we have
discussed
separately the relations of forces in a system and the conditions
of mo-
tion of bodies. In this text, we shall consider the relation
between the
motion of a body and the force or forces acting on the body to
produce
the motion. The basis for the relationship between motion and
force
is Newton's second law of motion. However, there are three
different
methods of applying this law. These are commonly called the
force-
mass acceleration method, the work-energy method, and the
impulse-
momentum method. Each method is most useful for solving
certain

types of problems.
Statement of Newton's Second Law of Motion
2 • In Engineering Mechanics, Part 1, Newton's second law of
motion was
stated as follows:
If a resultant force acts upon a particle, the particle will be
accelerated
in the direction of the force. Furthermore, the magnitude of the
accel-
eration will be directly proportional to the magnitude of the
resultant
force and inversely proportional to the mass of the particle.
Newton's second law can be expressed mathematically by the
following
equation:
F
a=k-
m
in which a = magnitude of the acceleration of a particle
k = a numerical factor
F = magnitude of the force acting upon the particle
m = mass of the particle
(1)
The mass of a particle is a measure of the exact amount of
matter in
the particle. Any body is composed of a number of particles,
and the

mass of a body is the sum of the masses of all the particles in
the
body.
2 Engineering Mechanics, Part 4
The acceleration and the force are actually vector quantities,
because
their directions must be considered. However, since the
acceleration and
the force must have the same direction, it is permissible to
consider
only the magnitudes of these quantities in a particular problem.
In
Engineering Mechanics, Part 3, it was explained that these
magnitudes
can be indicated by the notations fdl and 111. The simpler
notations
a and F will be used in this text, but you must remember that
accelera-
tions and forces are really vector quantities. The mass of a
particle is a
scalar quantity.
The value of the factor k depends on the units in which the
other
quantities in the equation are expressed. However, it is the
usual prac-
tice to choose the units in such a manner that the value of k will
be 1.
When suitable units are used, the preceding equation becomes
F

a=-
m
This relation is generally given in the following form:
F=ma
Newton's Law of Universal Gravitation
(2)
(3)
3 • Another law that was formulated by Newton is known as the
law of
universal gravitation. According to this law, any two particles
of matter
are attracted to each other with a force which acts along the
straight
line between the two particles and which has a magnitude
expressed
by the following relationship:
in which F = magnitude of the force of attraction
K = a numerical factor
m1 and m2 = masses of the two particles
r = length of the straight line between the particles
(1)
The force F is actually a vector quantity. However, since it acts
in one
direction on one particle and it acts in the opposite direction on
the

other particle, and since the position of its line of action is
fixed by the
locations of the particles, it is permissible to consider only the
magni-
tude of the force in a particular problem. The factor K is
commonly
called the universal gravitation constant. The value used for this
constant
depends on the units in which the force, the masses, and the
distance
are expressed. These units must be consistent.
Engineering Mechanics, Part 4
The magnitude of the force of attraction between two bodies is
the sum
of the magnitudes of the forces of attraction between the
particles of
one body and the particles of the other body. In ordinary
practical
work, the only force of attraction that need be considered is the
force
between the earth and some other body. This force is commonly
called
the weight of the body. If W denotes the weight of a body, m
the mass
of the body, me the mass of the earth, and re the distance from
the center
of gravity of the earth to the center of gravity of the other body,
the
relation given by equation 1 becomes
(2)

Since the weight W is a force, it follows from equation 3,
Article 2, that
Km
the factor __ e in equation 2 must represent an acceleration.
This fac-
re
tor is really the acceleration due to gravity mentioned in
Engineering
Mechanics, Part 3, and is usually denoted by g. Therefore,
equation 2
is usually written in the following form:
W=gm (3)
The distance re is not exactly the same for every point on the
earth's
surface. Therefore, the value of g varies slightly in different
localities
on the earth. However, for ordinary practical purposes, it may
be
taken as 32.2 fps per sec (feet per second per second), or 9.81
meters
per sec per sec.
Relation Between Mass and Weight of a Body
3
4 • Although the amount of matter contained in a body is
indicated exactly
by the mass of the body, it is not practicable to determine the
mass of
a body directly. Since the weight of a body can usually be
measured

quite easily, the amount of matter in a body is generally
indicated by
its weight. When it is necessary to consider the mass of a body,
the
following relationship (obtained from equation 3, Article 3)
may be used:
w
m=-g
in which m =mass of a body
W =weight of the body
g = acceleration due to gravity
The units for mass will be discussed next.
Systems of Units
5 • Two general categories of systems of units are used in
problems in
which Newton's second law of motion is applied. Systems in
one cate-
gory are called absolute systems, and those in the other category
are
called gravitational systems. In a system in either category,
there are
three fundamental units. Also, in any system, two of these
fundamen-
tal units are a unit of length, or distance, and a unit of time.
However,
in an absolute system, the third fundamental unit is a unit of
mass;

whereas, in a gravitational system, the third unit is a unit of
force.
Acceleration is defined as a change in velocity in a unit of time;
velocity is
defined as a displacement in a unit of time. When Newton's
second law
is applied, it is necessary to use the same unit of time for the
change
in velocity and for the displacement. Therefore, acceleration
involves
a unit of distance and a unit of time. In an absolute system of
units,
there is a fundamental unit of mass, and the unit of force needed
to
make the factor kin equation 1, Article 2, equal to 1 depends on
the
units used for length, time, and mass. In a gravitational system
of
units, there is a fundamental unit of force, and the unit of mass
needed
to make the factor k equal to 1 depends on the units used for
length,
time, and force.
For English units, it is the common practice to use a
gravitational
system, known as the British Engineering System, in which the
unit
of length is the foot, the unit of time is the second, and the unit
of
force is the pound. Then the unit of mass is called the slug. To
deter-
mine the mass of a body in slugs from the weight of the body in
pounds by applying the relationship in Article 4, it is necessary
to di-

vide the weight by 32.2. For instance, if the weight of a body is
10 lb
(pounds), the mass of the body is 1%2.2 = 0.311 slug.
For metric units, there are three common systems. One such
system is
an absolute system, called the CGS (Centimeter-Gram-Second)
System,
in which the unit of length is the centimeter, the unit of time is
the sec-
ond, and the unit of mass is the gram. Then the unit of force is
called
the dyne. When equation 3, Article 2, is applied, the value of a
force in
dynes is obtained by multiplying the mass in grams by the
acceleration
in centimeters per second per second.
In a gravitational system for metric units, called the MKS
(Meter-
Kilogram-Second) System, the unit of length is the meter, the
unit of
time is the second, and the unit of force is the kilogram. Then
the unit
of mass is called the metric slug. To determine the mass of a
body in
metric slugs from the weight of the body in kilograms by
applying the
relationship in Article 4, it is necessary to divide the weight by
9.81.
The third system for metric units is an absolute system called
the
International System of Units, abbreviated "SI." In this system,
the
unit of length is the meter, the unit of time is the second, and

the unit
of mass is the kilogram. Then the unit of force is called the
newton.
When equation 3, Article 2, is applied, the value of a force in
newtons
is obtained by multiplying the mass in kilograms by the
acceleration
in meters per second per second. You must be careful to
distinguish
between a kilogram of mass used in the International System
and a
kilogram of force used in the MKS System.
Mass Moment of Inertia of Particle
5
6 • The mass moment of inertia of a particle with respect to a
certain
straight reference line is defined by the following relationship:
in which Ix = mass moment of inertia of a particle
m = mass of the particle
r x = perpendicular distance from the reference line to the
particle
In Figure lA are shown a horizontal reference line, a vertical
reference
line, and a particle. This particle is so located that the vertical
distance

from the horizontal reference line to the particle is r1 and the
horizontal
distance from the vertical reference line to the particle is r2. If
the mass
of the particle is m1r its mass moment of inertia with respect to
the
horizontal reference line is m1r/ and its mass moment of inertia
with
respect to the vertical reference line is m1rt In Figure lB are
shown a
selected reference line and a particle which has a mass m2 and
is so
located that the distance from the reference line to the particle
in a
direction at right angles to the line is r3• The mass moment of
inertia
of this particle with respect to the reference line is m1rt
VERTICAL
REFERENCE
LINE
r2 - f - - - -~ PARTICLE
90°
I r
I 1
I
90~1
I
A
HORIZONTAL

REFERENCE
LINE
FIGURE 1-Mass Moment of Inertia of Particle
' ' 90o,.)W"'
' r3
' ' :-
PARTICLE
B
The term moment of inertia has been adopted for convenience to
refer
to the quantity that would be obtained by multiplying the mass
of a
particle by the square of the distance from a line to the particle.
Since the
"moment" of the mass of a particle could be the value of the
quantity
mrx, the quantity mrx2 is often called a "second moment."
In the English system of units, the unit commonly used for mass
moment of inertia is slug-feet2 if the distance rx is in feet or
slug-inches2
if the distance r x is in inches.
In the metric system, the unit used for mass moment of inertia
depends
on the units used for mass and length. Common units are
kilogram-

meters2 and gram-centimeters2.
Formulas for Mass Moment of Inertia of Basic Bodies
7 • It is usually necessary to consider the mass moment of
inertia of a body,
rather than the mass moment of inertia of a single particle.
Since a
body is composed of a number of particles, the mass moment of
inertia
of a body with respect to any selected reference line is equal to
the sum
of the moments of inertia of all the particles in the body with
respect
to the same reference line. When a body has a simple regular
shape and
the reference line passes through the center of gravity of the
body, the
mass moment of inertia of the body can be computed by
applying a
simple formula. In this text, we shall consider only a few
special cases.
When the body is a sphere composed of uniform material and
the
reference line is any diameter of the sphere (a diameter must
pass
through the center of gravity of the sphere), the formula for the
mass
momentofinertiais
2
Ic=-mr
5 (1)

in which Ic = mass moment of inertia of a sphere with respect to
any
diameter
m = mass of the entire sphere
r =radius of the sphere
When the body is a right circular cylinder and the reference line
is the
line called the axis of the cylinder, as in Figure 2, the formula
for the
mass moment of inertia of the body is
(2)
in which Ic = mass moment of inertia of a right circular cylinder
with
respect to the axis of the cylinder
m = mass of the entire cylinder
r =radius of a circular cross section of the cylinder
1., 2. Diameters through center of gravity
FIGURE 2-Dimensions of Right Circular Cylinder
When the body is a right circular cylinder and the reference line
is any
diameter of the circular cross section through the center of
gravity of
the cylinder, as either line 1 or line 2 in Figure 2, the formula
for the

mass moment of inertia of the body is
1 2 1 2
Ic= 4mr + 12mz
in which Ic = mass moment of inertia of a right circular cylinder
(3)
7
with respect to any diameter through its center of gravity
m = mass of the entire cylinder
r =radius of a cross section of the cylinder
l = length of the cylinder
When the body is a rectangular solid, like that represented in
Figure 3,
and the reference line is the longitudinal line 1 through the
center of
gravity, the formula for the mass moment of inertia is
1. longitudinal line through center of gravity
2. 3. Traverse lines through center of gravity
FIGURE 3-Dimensions of Rectangular Solid
1 2 2 Ic = 12 m(a + b ) (4)
in which a and b are the dimensions of a cross section of the
body.

When the body is a rectangular solid and the reference line is
the
transverse line 2 in Figure 3 through the center of gravity, the
mass
moment of inertia can be computed by the formula
1 2 2 Ic = 12 m(l + a ) (5)
If the reference line is the transverse line 3 in Figure 3, the
formula for
a rectangular solid is
1 2 2 I =-- m(l +b)
G 12 (6)
Mass Moment of Inertia of Body with Respect to Any Line
8 • It is sometimes necessary to know the mass moment of
inertia of a body
with respect to a line that does not pass through the center of
gravity
of the body. Examples are the bodies represented in Figures 4
and 5. In
Figure 4, the body is a right circular cylinder, but the reference
line for
the moment of inertia is located at one end of the cylinder. In
Figure 5
is shown a side view of a body that consists of a sphere and a
right
circular cylinder which are welded together. The reference line
for
the moment of inertia of the entire body is located at the free
end of
the cylinder. There is a relatively simple relationship between
the
mass moment of inertia of a body with respect to any reference

line
and the mass moment of inertia of the same body with respect to
a par-
allel line through the center of gravity of the body. This
relationship,
which is given here_ without proof, is as follows:
10 =lc+md 2
in which 10 =mass moment of inertia of a body with respect to a
reference line through any point 0
Ic = mass moment of inertia of the body with respect to a
parallel line through the center of gravity of the body
m =mass of the body
d = perpendicular distance from the specified reference line
to the center of gravity of the body
It is important to remember that the line through the center of
gravity
of the body must be parallel to the specified reference line. The
method
of applying the relationship just given is illustrated in the
following
example problems.
Example Problems
FIGURE 4-cy/inder with Reference Line at One End

CYLINDER
FIGURE 5-connected Cylinder and Sphere
REFERENCE
LINE
At intervals throughout this text you will find one or more
example
problems solved to illustrate clearly the application of a
principle, a
rule, or a formula. Read each problem carefully and study the
solution
until you understand it thoroughly.
Problem 1: The diameter of the cylinder represented in Figure 4
is 1
ft (foot), and its length is 6ft. If the cylinder weighs 64.4lb,
what is its
mass moment of inertia with respect to the specified reference
line
which is perpendicular to the axis of the cylinder and passes
through
the centroid of the cross section at one end of the cylinder?
9
Solution
: In this problem, the value to be used for Ic in the formula for
computing 10 is the mass moment of inertia of the cylinder with

respect
to a horizontal line that passes through the center of gravity and
is
perpendicular to the axis. This line would correspond to the line
2 in
Figure 3, and Ic can be computed by applying formula 3 in
Article 7.
The mass m of the cylinder must be obtained by applying the
formula
in Article 4, in which W = 64.4lb and g = 32.2 fps per sec.
Hence,
64.4 2 1 m=--= sugs
32.2
Since r = ~ x 1 = 0.5 ft and 1 = 6 ft,
1 2 1 2 1 2 1 2 .2 Ic = 4mr + 12mz = 4 x 2 x 0.5 + 12 x 2 x 6 =
6.13 slug-ft
The distance d from the specified reference line to the center of

gravity
of the cylinder is one-half the length of the cylinder, as
indicated in
Figure 2. This distance is 1;2 x 6 = 3 ft. The required mass
moment of
inertia is
10 = Ic + md 2 = 6.13 + 2 x 32 = 24.1 slug-ft2
Problem 2: A body like that in Figure 5 consists of a sphere
whose
diameter is 2 ft, and a cylinder whose diameter is 6 in. and
whose
length is 5 ft. The weight of the sphere is 2200 lb and the
weight of the
cylinder is 500 lb. What is the mass moment of inertia of the
entire body
with respect to a vertical reference line through the centroid of
the cross
section of the cylinder at its right hand end?

Study Unit Ill Engineerin M Part4 an1cs By And.docx

Study Unit Ill Engineerin M Part4 an1cs By And.docx

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