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TRANSCRIPT
Lecture Handouts
on
VECTORS AND CLASSICAL MECHANICS
(MTH-622)
Virtual University of Pakistan
Pakistan
ii
About the Handouts
The following books have been mainly followed to prepare the slides and handouts:
1. Spiegel, M.R., Theory and Problems of Vector Analysis: And an Introduction to Tensor
Analysis. 1959: McGraw-Hill.
2. Spiegel, M.S., Theory and problems of theoretical mechanics. 1967: Schaum.
3. Taylor, J.R., Classical Mechanics. 2005: University Science Books.
4. DiBenedetto, E., Classical Mechanics: Theory and Mathematical Modeling. 2010:
BirkhÀuser Boston.
5. Fowles, G.R. and G.L. Cassiday, Analytical Mechanics. 2005: Thomson Brooks/Cole.
The first two books were considered as main text books. Therefore the students are advised to
read the first two books in addition to these handouts. In addition to the above mentioned books,
some other reference book and material was used to get these handouts prepared.
3
Contents Module No. 101............................................................................................................................ 10
Example of Non- Conservative Field ..................................................................................... 10
Module No. 102............................................................................................................................ 11
Introduction to Simple Harmonic Motion and Oscillator ................................................... 11
Module No. 103............................................................................................................................ 13
Amplitude, Time period, Frequency and Energy of S.H.M ................................................. 13
Module No. 104............................................................................................................................ 14
Example of S.H.M ................................................................................................................... 14
Module No. 105............................................................................................................................ 16
Example of Energy of S.H.M .................................................................................................. 16
Module No. 106............................................................................................................................ 18
Damped Harmonic Oscillator ................................................................................................ 18
Module No. 107............................................................................................................................ 19
Example of Damped Harmonic Oscillator ........................................................................... 19
Module No. 108............................................................................................................................ 21
Eulerâs Theorem â Derivation ............................................................................................... 21
Module No. 109............................................................................................................................ 23
Chaslesâ Theorem.................................................................................................................... 23
Module No. 110............................................................................................................................ 24
Kinematics of a System of Particles ...................................................................................... 24
Module No. 111............................................................................................................................ 25
The Concept of Rectilinear Motion of Particles, Uniform Rectilinear Motion, Uniformly
Accelerated Rectilinear Motion ............................................................................................. 25
Module No. 112............................................................................................................................ 27
The Concept of Curvilinear Motion of Particles .................................................................. 27
Module No. 113............................................................................................................................ 28
Example Related to Curvilinear Coordinates ...................................................................... 28
Module No. 114............................................................................................................................ 30
Introduction to Projectile, Motion of a Projectile ................................................................ 30
Module No. 115............................................................................................................................ 33
Conservation of Energy for a System of Particles ............................................................... 33
Module No. 116............................................................................................................................ 34
Conservation of Energy .......................................................................................................... 34
4
Module No. 117............................................................................................................................ 36
Introduction to Impulse â Derivation ................................................................................... 36
Module No. 118............................................................................................................................ 37
Example of Impulse ................................................................................................................ 37
Module No. 119............................................................................................................................ 38
Torque ...................................................................................................................................... 38
Module No. 120............................................................................................................................ 39
Example of Torque ................................................................................................................. 39
Module No. 121............................................................................................................................ 41
Introduction to Rigid Bodies and Elastic Bodies ................................................................. 41
Module No. 122............................................................................................................................ 42
Properties of Rigid Bodies ...................................................................................................... 42
Module No. 123............................................................................................................................ 43
Instantaneous Axis and Center of Rotation.......................................................................... 43
Module No. 124............................................................................................................................ 44
Centre of Mass & Motion of the Center of Mass ................................................................. 44
Module No. 125............................................................................................................................ 46
Centre of Mass & Motion of the Center of Mass ................................................................. 46
Module No. 126............................................................................................................................ 48
Example of moment of Inertia and Product of Inertia ........................................................ 48
Module No. 127............................................................................................................................ 49
Radius of Gyration .................................................................................................................. 49
Module No. 128............................................................................................................................ 50
Principal Axes for the Inertia Matrix ................................................................................... 50
Module No. 129............................................................................................................................ 51
Introduction to the Dynamics of a System of Particles ....................................................... 51
Module No. 130............................................................................................................................ 54
Introduction to Center of Mass and Linear Momentum .................................................... 54
Module No. 131............................................................................................................................ 55
Law of Conservation of Momentum for Multiple Particles ................................................ 55
Module No. 132............................................................................................................................ 57
Example of Conservation of Momentum .............................................................................. 57
Module No. 133............................................................................................................................ 60
Angular Momentum â Derivation ......................................................................................... 60
5
Module No. 134............................................................................................................................ 62
Angular Momentum in Case of Continuous Distribution of Mass..................................... 62
Module No. 135............................................................................................................................ 64
Law of Conservation of Angular Momentum ...................................................................... 64
Module No. 136............................................................................................................................ 66
Example Related to Angular Momentum ............................................................................. 66
Module No. 137............................................................................................................................ 68
Kinetic Energy of a System about Principal Axes â Derivation ......................................... 68
Module No. 138............................................................................................................................ 70
Moment of Inertia of a Rigid Body about a Given Line ...................................................... 70
Module No. 139............................................................................................................................ 72
Example of M.I of a Rigid Body About Given Line............................................................. 72
Module No. 140............................................................................................................................ 75
Ellipsoid of Inertia .................................................................................................................. 75
Module No. 141............................................................................................................................ 76
Rotational Kinetic Energy ...................................................................................................... 76
Module No. 142............................................................................................................................ 79
Moment of Inertia & Angular Momentum in Tensor Notation ......................................... 79
Module No. 143............................................................................................................................ 81
Introduction to Special Moments of Inertia ......................................................................... 81
Module No. 144............................................................................................................................ 83
M.I. of the Thin Rod â Derivation ......................................................................................... 83
Module No. 145............................................................................................................................ 85
M.I. of Hoop or Circular Ring â Derivation ......................................................................... 85
Module No. 146............................................................................................................................ 87
M.I. of Annular Disk - Derivation ......................................................................................... 87
Module No. 147............................................................................................................................ 89
M.I. of a Circular Disk - Derivation ...................................................................................... 89
Module No. 148............................................................................................................................ 91
Rectangular Plate â Derivation.............................................................................................. 91
Module No. 149............................................................................................................................ 92
M.I. of Square Plate â Derivation .......................................................................................... 92
Module No. 150............................................................................................................................ 94
M.I. of Triangular Lamina â Derivation .............................................................................. 94
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Module No. 151............................................................................................................................ 96
M.I. of Elliptical Plate along its Major Axis â Derivation ................................................... 96
Module No. 152............................................................................................................................ 98
M.I. of a Solid Circular Cylinder - Derivation ..................................................................... 98
Module No. 153.......................................................................................................................... 100
M.I. of Hollow Cylindrical Shell - Derivation .................................................................... 100
Module No. 154.......................................................................................................................... 102
M.I of Solid Sphere - Derivation .......................................................................................... 102
Module No. 155.......................................................................................................................... 104
M.I. of the Hollow Sphere â Derivation .............................................................................. 104
Module No. 156.......................................................................................................................... 107
Inertia Matrix / Tensor of solid Cuboid .............................................................................. 107
Module No. 157.......................................................................................................................... 110
Inertia Matrix / Tensor of solid Cuboid .............................................................................. 110
Module No. 158.......................................................................................................................... 113
M.I of Hemi-Sphere â Derivation ........................................................................................ 113
Module No. 159.......................................................................................................................... 116
M.I. of Ellipsoid âDerivation................................................................................................ 116
Module No. 160.......................................................................................................................... 120
Example 1 of Moment of Inertia.......................................................................................... 120
Module No. 161.......................................................................................................................... 122
Example 2 of Moment of Inertia.......................................................................................... 122
Module No. 162.......................................................................................................................... 124
Example 3 of Moment of Inertia.......................................................................................... 124
Module No. 163.......................................................................................................................... 126
Example 4 of Moment of Inertia.......................................................................................... 126
Module No. 164.......................................................................................................................... 129
Example 5 of Moment of Inertia.......................................................................................... 129
Module No. 165.......................................................................................................................... 131
Example 5 of Moment of Inertia.......................................................................................... 131
Module No. 166.......................................................................................................................... 133
Example 1 of Parallel Axis Theorem ................................................................................... 133
Module No. 167.......................................................................................................................... 135
Example 2 of Parallel Axis Theorem ................................................................................... 135
7
Module No. 168.......................................................................................................................... 138
Example 3 of Parallel Axis Theorem ................................................................................... 138
Module No. 169.......................................................................................................................... 140
Example 4 of Parallel Axis Theorem ................................................................................... 140
Module No. 170.......................................................................................................................... 141
Perpendicular Axis Theorem ............................................................................................... 141
Module No. 171.......................................................................................................................... 143
Example 1 of Perpendicular Axis Theorem........................................................................ 143
Module No. 172.......................................................................................................................... 145
Example 2 of Perpendicular Axis Theorem........................................................................ 145
Module No. 173.......................................................................................................................... 149
Example 3 of Perpendicular Axis Theorem........................................................................ 149
Module No. 174.......................................................................................................................... 152
Problem of Moment of Inertia ............................................................................................. 152
Module No. 175.......................................................................................................................... 154
Existence of Principle Axes .................................................................................................. 154
Module No. 176.......................................................................................................................... 157
Determination of Principal Axes of Other Two When One is known .............................. 157
Module No. 177.......................................................................................................................... 159
Determination of Principal Axes by Diagonalizing the Inertia Matrix ........................... 159
Module No. 178.......................................................................................................................... 162
Relation of Fixed and Rotating Frames of Reference........................................................ 162
Module No. 179.......................................................................................................................... 165
Equation of Motion in Rotating Frame of Reference ........................................................ 165
Module No. 180.......................................................................................................................... 168
Example 1 of Equation of Motion in Rotating Frame of Reference................................. 168
Module No. 181.......................................................................................................................... 171
Example 2 of Equation of Motion in Rotating Frame of Reference................................. 171
Module No. 182.......................................................................................................................... 174
Example 3 of Equation of Motion in Rotating Frame of Reference................................. 174
Module No. 183.......................................................................................................................... 177
Example 4 of Equation of Motion in Rotating Frame of Reference................................. 177
Module No. 184.......................................................................................................................... 181
Example 5 of Equation of Motion in Rotating Frame of Reference................................. 181
8
Module No. 185.......................................................................................................................... 183
General Motion of a Rigid Body .......................................................................................... 183
Module No. 186.......................................................................................................................... 185
Equation of Motion Relative to Coordinate System Fixed on Earth ............................... 185
9
10
Module No. 101
Example of Non- Conservative Field
Problem:
Show that the force field given by
ᅵᅵ = ð¥2ðŠð§ð â ð¥ðŠð§2ð + 2ð¥ð§ï¿œï¿œ
is non-conservative.
Solution:
We have
ᅵᅵ = ð¥2ðŠð§ð â ð¥ðŠð§2ð + 2ð¥ð§ï¿œï¿œ
To verify whether the force field F is conservative or not, we will check whether ð» à ᅵᅵ = 0 or
not.
â à ᅵᅵ = |
ð ð ᅵᅵð
ðð¥â ðððŠâ ð
ðð§â
ð¥2ðŠð§ âð¥ðŠð§2 2ð¥ð§
|
= ð [ð
ððŠ(2ð¥ð§) â
ð
ðð§(âð¥ðŠð§2)] â ð [
ð
ðð§(ð¥2ðŠð§) â
ð
ðð¥(2ð¥ð§)] + ᅵᅵ [
ð
ðð¥(âð¥ðŠð§2) â
ð
ððŠ(ð¥2ðŠð§)]
= (2ð¥ðŠð§)ð â (ð¥2ðŠ â 2ð§)ð + (ðŠð§2 + ð¥2ð§)ᅵᅵ
â 0
As we obtain â à ᅵᅵ â 0
We conclude that ᅵᅵ is non-conservative.
11
Module No. 102
Introduction to Simple Harmonic Motion and Oscillator
Simple Harmonic Motion (SHM) is a particular type of oscillation and periodic motion in which
restoring force of an object is directly proportional to the displacement of the object acting in
opposite direction of displacement.
Mathematically, the restoring force ð¹ is given by
ð¹ð = âðð¥
where the subscript ð represents the restoring force and ð is the constant of proportionality often
called the spring constant or modulus of elasticity.
In Newtonian mechanics, by Newton's second law we have eq. of S.H.M
ð¹ = ðð = ðð2ð¥
ðð¡2= âðð¥
or
ðᅵᅵ + ðð¥ = 0
This vibrating system is called a simple harmonic oscillator or linear harmonic oscillator.
This type of motion is often called simple harmonic motion.
The following physical systems are some examples of simple harmonic oscillator
Mass on a spring
An object of mass m linked to a spring of spring constant k represents the simple harmonic
motion in closed region.
12
The equation representing the period for the mass attached on a spring
ð» = ðð âð
ð
The above equation expresses that the time period of oscillation is independent of amplitude as
well as the acceleration.
Simple pendulum
The movement of the mass attached to a simple pendulum is considered as simple harmonic
motion.
The time period of a mass m attached to a pendulum of length l with gravitational
acceleration g is given by
ð» = ðð âð
ð
13
Module No. 103
Amplitude, Time period, Frequency and Energy of S.H.M
Amplitude
Amplitude is defined as the maximum distance covered by the oscillating body in one oscillation
or length of a wave measured from its mean position.
The amplitude of a pendulum is one-half the distance that the mass covered in moving from one
terminal to the other. The vibrating sources generate waves, whose amplitude is proportional to
the amplitude of the vibrating source.
Time Period
Time period is minimum time required by a oscillation system to complete its one cycle of
oscillation of the specific system.
It is denoted by T and measured in seconds.
Frequency
The frequency (f) of an oscillatory system is the number of oscillations pass through a specific
point in one second.
It is measure in hertz (Hz).
The frequency of S.H.M can be calculate by using the following relation
ð =1
ð
Energy of S.H.M
If T is the kinetic energy, V the potential energy and ðž = ð + ð the total energy of a simple
harmonic oscillator, then we have
ðŸ. ðž = ð =1
2ðð£2
and
ð =1
2ðð¥2
Then the total energy of S.H.M will be
ðž =1
2ðð£2 +
1
2ðð¥2
14
Module No. 104
Example of S.H.M
Problem Statement
A particle of mass 4 moves along ðŠ axis attracted towards the origin by a force whose magnitude
is numerically equal to 6ðŠ. If the particle is initially at rest at ðŠ = 10,
i. The differential equation and initial conditions describing the motion,
ii. The position of the particle at any time,
iii. The speed and velocity of the particle at any time,
iv. The amplitude, time period and frequency of the vibration.
Solution:
i. Let ð = ðŠ be the position vector of the particle. The acceleration of the particle is
ð2ð
ðð¡2=
ð2ðŠ
ðð¡2
The net force acting on the particle is â6ðŠ. Then by using the Newton's second law,
â6ðŠ = 4ð2ðŠ
ðð¡2
or ð2ðŠ
ðð¡2 +3
2ðŠ = 0
which is the required differential equation. The required initial conditions (I. Cs) are
ðŠ = 10, ððŠ/ðð¡ = 0 ðð¡ ð¡ = 0
ii. Since the general solution the diff. eq. is
When ð¡ = 0, ð¥ = 20 ð ð ð¡âðð¡ ðŽ = 20. Thus
ðŠ = ðŽ cosâ3/2ð¡ + ðµ sinâ3/2 ð¡ (1)
Using I. Cs
At ð¡ = 0, ðŠ = 10. Thus ðŽ = 10. So,
ðŠ = 10 cosâ3/2ð¡ + Bsin â3/2 ð¡ (2)
15
ððŠ
ðð¡= â10â3/2 sinâ3/2 ð¡ + â3/2Bcosâ3/2ð¡
(3)
Since at ð¡ = 0,â ððŠ
ðð¡= 0
Thus, we get ðµ = 0. Hence (2) becomes
ðŠ = 10 cosâ3/2ð¡ (4)
This gives the position at any time.
iii. Speed at any time?
From (6) ððŠ
ðð¡= â10â3/2 sinâ3/2ð¡
This gives the speed at any time.
iv. Amplitude, T, F ?
General form for the position of a particle moving to and fro is
ðŠ(ð¡) = A cos(2ðð¡/T)
Thus, ðŠ = 10 cosâ3/2ð¡
Amplitude = 10
Period = T = 2â2
3ð
Frequency = 1 ðâ
16
Module No. 105
Example of Energy of S.H.M
ENERGY OF A SIMPLE HARMONIC OSCILLATOR
Example:
Find the total energy of the force ᅵᅵ = â3ð¥ð acting on a simple harmonic oscillator, where ð
represents the direction.
Solution:
Since the total energy of SHM
ðž = ð + ð
By Newton's second law,
ð¹ = ðð
therefore
ð¹ = ððð£
ðð¡= â3ð¥
ðð£
ðð¡= â
3
ðð¥
Integrating with respect to ð¡, we have
ð£ = â3
2ðð¥2 + ð
ð can be calculated if the initial condition is given. Assuming ð£ = 0 initially, which gives ð = 0.
Thus ð£ = â3
2ðð¥2
So,
ð =9
8ðð¥4
Now, as the potential energy is given by V where ᅵᅵ = âð»ð
or ð¹ = â3ð¥ð = â(ðð
ðð¥ð +
ðð
ððŠð +
ðð
ðð§ð)
17
Then
ðð
ðð¥= 3ð¥ ,
ðð
ððŠ= 0,
ðð
ðð§= 0
By integrating, we obtain
ð =3
2ð¥2 + ð1
Assuming ð = 0, corresponding to ð¥ = 0, we get ð1 = 0.
So the potential energy is ð =1
2ðð¥2
Hence
ðž =9
8ðð¥4 +
1
2ðð¥2
18
Module No. 106
Damped Harmonic Oscillator
The forces acting on a harmonic oscillator are called damping forces which tend to decrease the
amplitude of the successive oscillations or simply force apposing the motion. Since the damping
force is proportional to the velocity. Thus, mathematically,
ð¹ð = âðð£
= âðᅵᅵ
where ð represents the damping force and ð is the damping coefficient.
The negative sign shows that that direction of ð¹ð is opposite to the velocity ð£.
As we know that for the restoring force:
ðᅵᅵ + ðð¥ = 0
Adding the restoring force with the damping force, the equation of motion of the damped
harmonic oscillator will be,
ðᅵᅵ + ðð¥ = âðᅵᅵ
or
ðᅵᅵ + ðᅵᅵ + ðð¥ = 0
Since the mass is not equal to zero, therefore ᅵᅵ +ð
ðᅵᅵ +
ð
ðð¥ = 0
let
ð
ð= 2ð, â
ð
ð= ð0
then the equation can be written as
ᅵᅵ + 2ðᅵᅵ + ð02ð¥ = 0
19
Module No. 107
Example of Damped Harmonic Oscillator
Problem Statement
A particle of mass 4 moves along ðŠ axis attracted towards the origin by a force whose magnitude
is numerically equal to 6ðŠ. If the particle is initially at rest at ðŠ = 10. Consider that the particle has
also a damping force whose magnitude is numerically equal to 6 times the instantaneous speed.
Find
i. Position of the particle
ii. Velocity of the particle at any time
Solution
i. Since we have by Newton's second law
ð¹ = ðð = 4ᅵᅵ
According to the given information, the damping force is â6ᅵᅵ.
So the net force is
â6ðŠ â 6ᅵᅵ
Hence,
4ᅵᅵ = â6ðŠ â 6ᅵᅵ
or
ᅵᅵ +2
3ᅵᅵ +
2
3ðŠ = 0
The solution of the above equation is
ðŠ = ðââ2 3â ð¡(ðŽ + ðµð¡)
and
ᅵᅵ = ðµðââ2 3â ð¡ â â2 3â (ðŽ + ðµð¡)ðââ2 3â ð¡
To find the values of constants, we use I. Cs
At ð¡ = 0, ðŠ = 10 and ððŠ/ðð¡ = 0; thus,
ðŽ = 10
and 0 = ðµ(1) â â2 3â (10)
ðµ = 10â2 3â
and the solution gives
ðŠ = ðââ2 3â ð¡(ðŽ + ðµð¡)
= 10ðââ2 3â ð¡(1 + â2 3â ð¡)
20
the position at any time t.
21
Module No. 108
Eulerâs Theorem â Derivation
The following theorem, called Euler's theorem, is fundamental in the motion of rigid bodies.
Theorem Statement
âA rotation of a rigid body about a fixed point of the body is equivalent to a rotation about a line
which passes through the point.â
The line referred to is called the instantaneous axis of rotation. Rotations can be considered as
finite or infinitesimal. Finite rotations cannot be represented by vectors since the commutative
law fails. However, infinitesimal rotations can be represented by vectors.
Proof:
Let O be the fixed point in the body, which we take as a sphere S. Further, we take O at the
center of the sphere. Let A, B be two distinct points on the sphere. As the body moves, the point
O (on the axis) remains foxed and A and B suffer displacement.
Let ðŽâ and ðµâ bet the new locations of the points A and B after an infinitesimal time interval ð¿ð¡
respectively.We join (A, B) and (ðŽâ, ðµâ) by great circular areas.
Also we join (ðŽ, ðŽâ) and (ðµ, ðµâ) by mean of great circular arcs. Let Aâ and Bâ draw axes at right
angles, which meat at the point C on the sphere.
We join C with A, B, ðŽâ, ðµâ by means of great circular arcs.
Consider the spherical triangles âð¶ðŽâ²ðŽ" and âð¶ðŽðŽ". Obviously
22
i. ð < ð¶ðŽ"A â CA"ðŽâ²
ii. ðŽðŽ"=A"ðŽâ²
iii. ð¶ðŽ" is common to triangle
Therefore âð¶ðŽðŽ" â âð¶ðŽâ²ðŽ" and it follows that ð¶ðŽ = ð¶ðŽâ
Similarly for the triangle âð¶ðŽðµ ððð âð¶ðŽâ²ðµâ², we have
ð¶ðµ = ð¶ðµâ, ð¶ðŽâ = ð¶ðŽ ððð ðŽðµ = ðŽâðµâ
The last relation is due to the fact that by the definition of rigid body, the distance between a pair
of particles remains unchanged. Hence
âð¶ðŽðµ â âð¶ðŽâ²ðµâ²
The portion of rigid body lying in âð¶ðŽðµ has moved to âð¶ðŽâ²ðµâ².
In this process the point O and C have remained fixed, Although the later was at rest only
instantaneously. Therefore the body has under gone a rotation about the axis OC.
Hence the theorem.
23
Module No. 109
Chaslesâ Theorem
Theorem Statement:
Chasleâs theorem states that the most general rigid body displacement can be produced by a
translation along a line (called its screw axis) followed (or preceded) by a rotation about that line.
Explanation:
A rigid body has six degrees of freedom.
By Eulerâs theorem, three of these are associated with pure rotation.
The remaining three must be associated with translation.
To describe the general motion of a rigid body, think of the general motion as translation
of a fixed point ð in the body to a point ðâ² followed by the rotation about an axis through
ðâ².
24
Module No. 110
Kinematics of a System of Particles (Space, time & matter)
Kinematics is the branch of mechanics deals with the moving objects without reference to the
forces which cause the motion.
In other words we can say those kinematics are the features or properties of motion of concerned
with system of particles (rigid bodies).
Here some features of rigid body motion are
Displacement
Position
Velocity
Linear Velocity & Angular Velocity
Linear Acceleration & Angular Acceleration
Motion of a Rigid Body (Translation & Rotation)
From everyday experience, we all have some idea as to the meaning of each of the following terms
or concepts. However, we would certainly find it difficult to formulate completely satisfactory
definitions. We take them as undefined concepts.
I. Space. This is closely related to the concepts of point, position,' direction and displacement.
Measurement in space involves the concepts of length or distance, with which we assume
familiarity. Units of length are feet, meters, miles, etc.
II. Time. This concept is derived from our experience of having one event taking place after, before
or simultaneous with another event. Measurement of time is achieved, for example, by use of
clocks. Units of time are seconds, hours, years, etc.
III. Matter. Physical objects are composed of "small bits of matter" such as atoms and molecules.
From this we arrive at the concept of a material object called a particle which can be considered as
occupying a point in space and perhaps moving as time goes by. A measure of the "quantity of
matter" associated with a particle is called its mass. Units of mass are grams, kilograms, etc. Unless
otherwise stated we shall assume that the mass of a particle does not change with time.
25
Module No. 111
The Concept of Rectilinear Motion of Particles, Uniform Rectilinear Motion,
Uniformly Accelerated Rectilinear Motion
When a moving particle remains on a single straight line, the motion is said to be rectilinear. In
this case, without loss of generality we can choose the ð¥-axis as the line of motion.
The general equation of motion is then
ð¹ = ðð â¹ ð¹(ð¥, ᅵᅵ, ᅵᅵ) = ðᅵᅵ
Rectilinear motion for a particle:
Rectilinear motion of a body is defined by considering the two point of a body covered the same
distance in the parallel direction. The figures below illustrate rectilinear motion for a particle and
body.
Rectilinear motion for a body:
In the above figures, ð¥(ð¡) represents the position of the particles along the direction of motion, as
a function of time t.
An example of linear motion is an athlete running g along a straight track.
The rectilinear motion can be of two types:
i. Uniform rectilinear motion
ii. Non uniform rectilinear motion
Uniform Rectilinear Motion:
Uniform rectilinear motion is a type of motion in which the body moves with uniform velocity or
zero acceleration.
26
In contrast, Non uniform rectilinear motion is such type of motion with variable velocity or non-
zero acceleration.
Uniformly accelerated rectilinear motion:
Uniformly accelerated rectilinear motion is a special case of non-uniform rectilinear motion along
a line is that which arises when an object is subjected to constant acceleration. This kind of motion
is called uniformly accelerated motion.
Uniformly accelerated motion is a type of motion in which the velocity of an object changes by an
equal amount in every equal intervals of time.
An example of uniformly accelerated body is freely falling object in which the amount of
gravitational acceleration remains same.
ð¹ = ðð
27
Module No. 112
The Concept of Curvilinear Motion of Particles
The motion of a particle moving in a curved path is called curvilinear motion. Example: A stone
thrown into the air at an angle.
Curvilinear motion describes the motion of a moving particle that conforms to a known or fixed
curve. The study of such motion involves the use of two co-ordinate systems, the first being
planar motion and the latter being cylindrical motion.
Tangential and normal unit vectors are usually denoted by ðð¡ and ðð respectively.
Velocity of Curvilinear motion
If the tangential and normal unit vectors are ðð¡ and ðð respectively, then the velocity will be
ᅵᅵ =ðð
ðð¡ðð¡
You have already learnt that
ð£ = vT
Acceleration of Curvilinear motion
If the tangential and normal unit vectors are ðð¡ and ðð respectively, then the acceleration will be
ᅵᅵ =ð2ð
ðð¡2ðð¡ +
(ðsðð¡â )
2
ððð
You have already learnt that
ð = Tðv
ðð¡+
v2
rN
Example
A stone thrown into the air at an angle.
A car driving along a curved road.
Throwing paper airplanes or paper darts is an example of curvilinear motion.
28
Module No. 113
Example Related to Curvilinear Coordinates
Example 1
Problem Statement
Prove that a cylindrical coordinate system is orthogonal.
Solution
The position vector of any point in cylindrical coordinates is
ð = ð¥ð + ðŠð + ð§ï¿œï¿œ
We studied that the in cylindrical coordinates
ð¥ = ð cosð , ðŠ = ð sin ð, ð§ = ð§
this implies
ð = ð cosð ð + ð sinð ð + ð§ï¿œï¿œ
The tangent vectors to the Ï, Ï and г curves are given respectively by ðð
ðð ,
ðð
ðð,ðð
ðð§ and where
ðð
ðð= cosð ð + sinð ð
ðð
ðð= âð sinð ð + ð cosð ð
ðð
ðð§= ᅵᅵ
The unit vectors in these directions are
ð1 = ðð =ðð ððâ
|ðð ððâ |=
cosð ð + sinð ð
|cos ð ð + sinð ð|
=cosð ð + sin ð ð
âcos2 ð + sin2 ð = cosð ð + sinð ð
ð2 = ðð =ðð ððâ
|ðð ððâ |=
âð sinð ð + ð cosð ð
|âð sinð ð + ð cosð ð|
=ð(â sin ð ð + cosð)
âð2 cos2 ð + ð2 sin2 ð = âsinð ð + cosð ð
ð3 = ðð§ =ðð ðð§â
|ðð ðð§â |= ᅵᅵ
Then
29
ð1. ð2 = (cosð ð + sinð ð). (â sinð ð + cosð ð)
= âsinð cosð +sinð cosð = 0
ð2. ð3 = (âsinð ð + cosð ð). ᅵᅵ = 0
ð3. ð1 = ᅵᅵ. (cosð ð + sinð ð) = 0
and so ð1, ð2 and ð3 are mutually perpendicular and the coordinate system is orthogonal.
Example 2
Problem Statement
Prove
ððð
ðð¡= ððð
ððð
ðð¡= âððð
where dots denote differentiation with respect to time t.
Solution
We have
ðð = cosð ð + sinð ð
and
ðð = âsinð ð + cosð ð
Then
ððð
ðð¡=
ð
ðð¡(cosð ð + sinð ð)
= âsin ð ᅵᅵð + cosð ᅵᅵð = ᅵᅵ(â sinð ð + cosð ð) = ᅵᅵðð
ððð
ðð¡=
ð
ðð¡(â sinð ð + cosð ð)
= âcosð ᅵᅵð â sin ð ᅵᅵð = âᅵᅵ(cosð ᅵᅵð + sinð ᅵᅵð) = âᅵᅵðð
30
Module No. 114
Introduction to Projectile, Motion of a Projectile
Introduction to Projectile
If a ball is thrown from one person to another or an object is dropped from a moving plane, then
their path of traveling/motion is often called a projectile. If air resistance is negligible, a projectile
can be considered as a freely falling body so the Eq of motion will be
md2r
dt2= âmg
or
d2r
dt2= âg
with appropriate I.Cs
Motion of a Projectile with Resistance
Earlier, we studied the motion of a projectile under the assumption that air resistance is negligible.
If we further assume that the motion takes place in the vertical plane, (ðð-plane), and the only
force acting on the projectile is the gravity, then the equation of motion will be
d2r
dt2= g = âgj
Here we will discuss the problem of projectile motion when air resistance is taken into account.
Assuming the model of retarding force in which
ð¹ â ð£
or
ð¹ = âð1ð£
Let the initial velocity of the projectile be ð£0 and angle of projection be ð
The initial conditions can be taken as
ð¥(ð¡ = 0) = ðŠ(ð¡ = 0) = 0
ᅵᅵ(ð¡ = 0) = ð£0 cos ð = ð¢1
ᅵᅵ(ð¡ = 0) = ð£0 sin ð = ð£1
The equation of motion in the horizontal and vertical direction can be written as
ðᅵᅵ = âð1ᅵᅵ
or ᅵᅵ = âð1
ðᅵᅵ = âðᅵᅵ (1)
ðᅵᅵ = âðᅵᅵ â ðð (2)
By the substitution ᅵᅵ = ð§ we can solve eq (1) as
31
ð§ = ᅵᅵ = ðŽðâðð¡
With the initial condition ᅵᅵ = ð¢1, when ð¡ = 0, we obtain the constant of integration ðŽ = ð¢1
so that
ᅵᅵ = ð¢1ðâðð¡
Another integration gives
ð¥ = âð¢1ð
âðð¡
ð+ ðµ
The initial condition ð¥(0) = 0 gives ðµ =ð¢1
ðâ . Hence the solution of equation (1) can be
written as
ð¥ =ð¢1
ð(1 â ðâðð¡) (3)
We can solve equation (2) in the same manner and obtain
ðŠ = âðð¡
ð+
ðð£1+ð
ð2(1 â ðâðð¡) (4)
The path of motion can be obtained from (3) and (4) by eliminating ð¡
ðŠ =ð
ð2ln (1 â
ðð¥
ð¢1) +
ðð£1 + ð
ð2
ðð¥
ð¢1
which is no longer a parabola.
We can solve equation (2) in the same manner and obtain
ðŠ = âðð¡
ð+
ðð£1+ð
ð2(1 â ðâðð¡) (4)
The path of motion can be obtained from (3) and (4) by eliminating ð¡
ðŠ =ð
ð2ln (1 â
ðð¥
ð¢1) +
ðð£1 + ð
ð2
ðð¥
ð¢1
which is no longer a parabola.
32
Time of Flight ð Time of flight can be found by putting in equation (4) ðŠ = 0 and ð¡ = ð
0 = âðð
ð+
ðð£1 + ð
ð2(1 â ðâðð)
ð =ðð£1+ð
ðð(1 â ðâðð) (5)
This equation can be solved exactly for ð. To find an approximate solution, we expand the
exponential factor in equation (5), using
ðð¥ = 1 + ð¥ +ð¥2
2!+
ð¥3
3!+ â¯âŠâŠ ..
Form equation (5)
ð =ðð£1 + ð
ðð(ðð â
ð2ð2
ðð+
ð3ð3
3!+ ⯠. . )
1 =ðð£1 + ð
ðð(ð â
ð2ð
ðð+
ð3ð2
3!+ ⯠. . )
which on simplification gives
ð =2ð£1
ðð£1 + ð+
1
3ðð2 + ⯠. .
In the limit of small air resistance (i.e. ð â 0), the above expression reduces to
ð =2ð£1
ðð£1+ð+
ð
3ð2 (6)
When there is no resistance, (ð = 0), obtain from (6)
ð0 =2ð£1
ð=
2ð£0 sin ð
ð
If ð is non-zero small number, then the time of flight ð will be very close to ð0. Substituting ð0 for
ð on the R.H.S of (6), and simplifying we get
ð =2ð£1
ð(1 â
ðð£1
3ð)
The above equation gives a formula for approximate time of flight in the presence of a weak
retarding force.
33
Module No. 115
Conservation of Energy for a System of Particles
Statement
âThe law of conservation of energy describes that the net energy of an isolated system remains
conserved. Energy can neither be created nor destroyed; rather, it transforms from one form to
another.â
Theorem Statement
(Principle of conservation of energy)
In case of conservative force field, the total energy is a constant. i.e.,
If ð is for kinetic energy and ð is for potential energy, then the total energy ðž is
ðž = ð + ð = constant
Explanation
For a conservative force field, we have already learned that the work done by the system of
particles is
Work done = change in kinetic energy = ð = ð2 â ð1
Also Work done = change in potential energy = ð = ð1 â ð2
By comparing, we get
ð2 â ð1 = ð1 â ð2
or
ð1 + ð1 = ð2 + ð2
which can be written as
1
2ðð£1
2 + ð1 =1
2ðð£2
2 + ð2
.
34
Module No. 116
Conservation of Energy
Example
Consider the force field ð¹ = âðð3ð
i. Find whether the given field is conservative or not.
ii. Find the potential energy of the given force field of part (i).
iii. For the particle moving in ð¥ðŠ â plane, find the work done by the force in moving the
particle from A to B, where A is the point where ð = ð, and B where ð = ð.
iv. If the particle of mass ð moves with velocity ð£ =ðð
ðð¡ in this field, show that if ðž is the
constant, total energy then
v. 1
2 ð (
ðð
ðð¡)2
+1
5 ðð5 = ðž
Solution
i. As ᅵᅵ = âðð3ð = âð(ð¥2 + ðŠ2)3/2(ð¥ð + ðŠð)
â à ᅵᅵ = |
ð ð ð
ð ðð¥â ð ððŠâ ð ðð§â
âð(ð¥2 + ðŠ2)3/2ð¥ âð(ð¥2 + ðŠ2)3/2ðŠ 0
|
= ð [ð
ððŠ(0) â
ð
ðð§(âð(ð¥2 + ðŠ2)3/2ðŠ)] + ð [
ð
ðð§(âð(ð¥2 + ðŠ2)3/2ð¥) â
ð
ðð¥(0)]
+ ð [ð
ðð¥(âð(ð¥2 + ðŠ2)3/2ðŠ) â
ð
ððŠ(âð(ð¥2 + ðŠ2)3/2ð¥)]
= â3ð(ð¥2 + ðŠ2)12ð¥ðŠ + 3ð(ð¥2 + ðŠ2)
12ð¥ðŠ = 0
Thus the given field is conservative.
ii. Since the field is conservative there exists a potential V such that
ð¹ = âð»ð
ð¹ = âðð3ð = âð(ð¥2 + ðŠ2)32(ð¥ð + ðŠð)
= âð(ð¥2 + ðŠ2)32ð¥ð â ð(ð¥2 + ðŠ2)
32ðŠð = ââð
= âðð
ðð¥ð â
ðð
ððŠð â
ðð
ðð§ð
35
By comparing, we get
ðð
ðð¥= ð(ð¥2 + ðŠ2)3/2ð¥,
ðð
ððŠ=ð(ð¥2 + ðŠ2)3/2ðŠ and
ðð
ðð§= 0
From which, by omitting the constants, we get
ð =1
5 ð(ð¥2 + ðŠ2)5/2 =
1
5 ðð5
is required potential.
iii. The velocity potential at point A
=1
5 ðð5
and the velocity potential at point B
=1
5ðð5
So, the work done from A to B is
= Potential at A - Potential at B
=1
5 ðð5 â
1
5 ðð5 =
1
2 ð(ð5 â ð5)
which is required work done by the given force field.
iv. Since the kinetic energy of a particle of mass ð moving with velocity v =ðð
ðð¡ is
ð =1
2 ðð£2 =
1
2 ð(
ðð
ðð¡)2
From part (ii), we have the potential energy
ð =1
5 ðð5
Thus the total energy ðž will be
ðž = ð + ð
=1
2 ð (
ðð
ðð¡)2
+1
5 ðð5
Hence proved.
36
Module No. 117
Introduction to Impulse â Derivation
Impulse is a special type of force defined by applying the integral of a force ð¹, over the time
interval, t, for which it acts on the body.
Impulse is a directional (vector) quantity in the same direction of force as force is also a directional
quantity. When Impulse is applied to a rigid body, it results a corresponding vector change in its
linear momentum along the same direction. The SI unit of impulse is the newton second (ð · ð ),
and the dimensionally equivalent unit of momentum is the kilogram meter per second (ðððð â1).
The particle is located at ð1 and ð2 at times ð¡1 and ð¡2 where it has velocities ð£1 and ð£2 respectively.
The time integral of the force F given by
â« ð¹ðð¡
ð¡2
ð¡1
is called the impulse of the force ð¹.
Theorem Statement
The impulse is equal to the change in momentum; or, in symbols,
â« ð¹ðð¡ = ðð£2 â ðð£1 = ð2 â ð1
ð¡2
ð¡1
Proof
We have to prove that the impulse of a force is equal to the change in momentum.
By definition of impulse and Newton's second law, we have
â« ð¹ðð¡ =
ð¡2
ð¡1
â« ððð£
ðð¡ðð¡ =
ð¡2
ð¡1
â« ððð£ = ð|ð£|ð¡1ð¡2 =
ð¡2
ð¡1
ðð£2 â ðð£1 = ð2 â ð1
where we use the conditions
ð£(ð¡1) = ð£1 and ð£(ð¡2) = ð£2
The theorem is true even when the mass is variable and the force is non-conservative.
37
Module No. 118
Example of Impulse
Example:
What is the magnitude of the impulse developed by a mass of 200 gm which changes its velocity
from 5ð â 3ð + 7ð m/sec to 2ð + 3ð + ð m/sec?
Solution:
Since we have the following information:
ð = 200 gm
ð£1 = 5ð â 3ð + 7ð
ð£2 = 2ð + 3ð + ð
As we know that
Impulse = ð2 â ð1
= ðð£2 â ðð£1
= ð(ð£2 â ð£1)
Substituting the values, we get
Impulse = 200(2ð + 3ð + ð â (5ð â 3ð + 7ð))
= 200(â3ð + 6ð â 6ð)
The magnitude of the Impulse will be
Impulse magnitude = 200â9 + 36 + 36
= 200â81
= 200(9)
= 1800mgm/sec
= 1.8 N sec
38
Module No. 119
Torque
Definition
Torque is defined as the turning effect of a body. It is trend of an acting force due to which the
rotational motion of a body changes. It is also called twist and rotational force on an object.
Mathematically, torque is defined as the cross product of the force vector to the distance vector,
which causes rotational motion of the body.
ð = ᅵᅵ à ᅵᅵ
The magnitude of torque depends upon the applied force, the length of the lever arm connecting
the axis to the point where the force applied, and the angle between the force vector and the length
of lever arm. Symbolically we can write it as:
ð = |ð||ð| ð¬ð¢ð§ ðœ
Torque is a vector quantity implies that it has direction as well as magnitude.
The SI unit for torque is the newton meter (Nm).
The direction of torque can be approximate using Right Hand Rule.
Theorem:
The torque acting on a particle equals the time rate of change in its angular momentum, i.e.,
ð =ðΩ
ðð¡
where Ω = r à p is defined angular momentum of the system.
Proof:
As we know torque is defined as
ð = ð Ã ð¹ = ð Ã ðð
as m is constant, therefore
= ð(ð Ã ð)
= ð(ð Ãðð£
ðð¡)
= ðð
ðð¡(ð à ð£)
=ð
ðð¡(ð à ðð£)
=ð
ðð¡(ð Ã ð)
where ð = ðð£ is defined as linear momentum and hence the theorem.
39
Module No. 120
Example of Torque
Example
A particle of mass ð moves Along a space curve defined by ð = acosðð¡ ð + ð sinðð¡ ð.
Find
(i) The Torque
(ii) The angular momentum about the origin.
Solution
(i) As we know that the torque acting on a particle is equal to the time rate of change of its
angular momentum, i.e.
ð =ðΩ
ðð¡=
ð
ðð¡(ð Ã ð)
where ð = ðð£.
As ð = acosðð¡ ð + ð sinðð¡ ð. Therefore
ð£ =ðð
ðð¡
ð£ = âðð sinðð¡ ð + ðð cosðð¡ ð
So,
ð = âððð sinðð¡ ð + ððð cosðð¡ ð
Now
ð Ã ð
= |ð ð ᅵᅵ
acosðð¡ ð sinðð¡ 0âððð sinðð¡ ððð cosðð¡ 0
|
= [ððððcos2ðð¡ + ððððsin2ðð¡]ᅵᅵ
= ððððᅵᅵ
40
ð =ðΩ
ðð¡=
ð
ðð¡(ð Ã ð) = 0
(ii) From part (i), we already have
Ω = ð à ð
= ððððᅵᅵ
is the required angular momentum.
41
Module No. 121
Introduction to Rigid Bodies and Elastic Bodies
Definition of Rigid Bodies
When a force is applied to an object/ system of particles, and if the object maintains its
overall shape, then the object is called a rigid body.
Gap between two fixed points on the rigid body remains same regardless of external
forces exerted on it.
We can neglect the deformation of such bodies.
A rigid body usually has continuous distribution of mass.
Definition of Elastic Bodies
When a force is applied to a system of particles, it changes the distance between
individual particles. Such systems are often called deformable or elastic bodies.
Examples
A spring and rubber band are some common examples of elastic bodies.
A wheel is a common example of rigid body.
42
Module No. 122
Properties of Rigid Bodies
Following are some of the properties of the rigid bodies.
Degree of freedom
The number of coordinates required to specify the position of a system of one or more particles is
called the number of degrees of freedom of the system.
For example a particle moving freely in space requires 3 coordinates, e.g. (x, y, z), to specify its
position. Thus the number of degrees of freedom is 3.
Similarly, a system consisting of N particles moving freely in space requires 3N coordinates to
specify its position. Thus the number of degrees of freedom is 3N.
Translations
A displacement of a rigid body is a direct change of position of its particles. Translational motion
is the displacement of all particles of the body by the same amount and the line segment joining
the initial and the final position of the particles represented by parallel vectors.
Examples of translational motion are particles freely falling down to earth and the motion of a
bullet fired from a gun.
Rotations
Circular motion of a body about a fixed point or axis is called rotation. If during a displacement
the points of the rigid body on some line remains fixed and all other are displaced through the
same angle, then this displacement is called rotation. A rigid performs rotations around an
imaginary line called a rotation axis.
If the axis of rotation passes through the center of mass of the rigid body then body is said to be
spin or rotate upon itself. If a body rotates about some external fixed point is called revolution or
orbital motion of the rigid body. The example of revolution is the rotation of earth around sun
and motion of moon around sun.
Rotational motion concerns only with rigid bodies. The reverse rotation of a body (inverse
rotation) is also a rotation.
A wheel is common examples of rotation.
43
Module No. 123
Instantaneous Axis and Center of Rotation
Introduction to General Plane Motion
The general plane motion of a rigid body can be considered as:
Translational motion along the given fixed plane and rotational motion about a suitable
axis perpendicular to the plane.
This fixed axis is specifically chosen to pass through the center of mass of the rigid body.
Instantaneous Axis of Rotation
The axis about which the rigid body rotates is called instantaneous axis of rotation, where
this axis is perpendicular to the plane.
Instantaneous Center of Rotation
The point where instantaneous axis meets the fixed plane along which the body performs
translation motion is described as the instantaneous center of rotation.
44
Module No. 124
Centre of Mass & Motion of the Center of Mass
The center of mass (c.m.) or centroid of system of particles is a hypothetical particle such that if
the entire mass of the system were concentrated there, the mechanical properties would remain the
same. In particular expression of linear momentum, angular momentum and kinetic energy assume
simpler or more convenient forms when referred to the coordinated of this hypothetical particle
and the equation of motion can be reduced to simpler equation of a single particle.
Let ð1, ð2, ð3, âŠâŠâŠ . , ððbe the position vectors of a system of N particles of masses
ð1, ð2, ð3, âŠâŠâŠ . ,ðð respectively [see Fig.].
The center of mass or centroid of the system of particles is defined as that point C having position
vector
ðð =â ððððð
â ððð
When the system is moving the position vector will depend on time t and therefore ððwill also be
a function of t. the velocity and acceleration of center of mass will be then given by
ᅵᅵð = ðð =â ðððᅵᅵð
â ððð
ᅵᅵð = ðð =â ðððᅵᅵð
â ððð
Motion of Center of Mass Motion of center of mass can be examined by considering the following points:
45
1. If a system experiences no external force, the center-of-mass of the system will remain at
rest, or will move at constant velocity if it is already moving.
2. If there is an external force, the center of mass accelerates according to ð¹ = ðð.
3. Basically, the center-of-mass of a system can be treated as a point mass, following
Newton's Laws.
4. If an object is thrown into the air, different parts of the object can follow quite complicated
paths, but the center-of-mass will follow a parabola.
5. If an object explodes, the different pieces of the object will follow seemingly independent
paths after the explosion. The center of mass, however, will keep doing what it was doing
before the explosion. This is because an explosion involves only internal forces.
46
Module No. 125
Centre of Mass & Motion of the Center of Mass
The moment of inertia of a rigid body is a property which depends upon its mass and shape, (i.e.
the mass distribution of the body) and determines its behavior in rotational motion.
In rotational motion, the moment of inertia plays the same role as the mass in linear motion.
Definition of Moment of Inertia
Formally the moment of inertia ðŒ of the particle of mass ð about a line is defined by
ðŒ = ðð2
where ð is the perpendicular distance between the particle and the line (called the axis).
Moment of Inertia of System of particles
The moment of inertia of a system of particles, with masses ð1, ð2, ð3, ⊠,ðð about the line or
axis AB is defined as
ðŒ = â ðð
ð
ð=1ðð
2
= ð1ð12 + ð2ð2
2 + â¯+ ðððð2
In dimensions, the moment of inertia can be expressed as
[ðŒ] = [ð][ð¿2]
Moment of Inertia in Coordinate System
The moment of inertia of a particle of mass ð with coordinates (ð¥, ðŠ, ð§) relative to the
orthogonal Cartesian coordinate systemðððð about ð, ð, ð axes will be
ðŒð¥ð¥ = ð(ðŠ2 + ð§2)
ðŒðŠðŠ = ð(ð§2 + ð¥2)
ðŒð§ð§ = ð(ð¥2 + ðŠ2)
Product of Inertia
The product of inertia for the same particle w.r.to the pair of coordinate axes are defined as
ðŒð¥ðŠ = âðð¥ðŠ
ðŒðŠð§ = âððŠð§
ðŒð§ð¥ = âðð§ð¥
These definitions can be easily generalized to a system of particle and a rigid body.
47
48
Module No. 126
Example of moment of Inertia and Product of Inertia
Problem Statement:
Find the moments of inertia of a ring of radius ð about an axis through center.
Solution:
Let ð be the mass and ð the radius of the ring. Then the mass per unit length will be ð 2ððâ .
We regard this ring to be composed of small elements of mass (ð¿ð) each of length ð¿ð ,
we can write it as
ð¿ð
ð¿ð =
ð
2ððâ¹ ð¿ð =
ð
2ððð¿ð
Moment of inertia of the element about an axis through center O and the perpendicular to the
plane of the ring equals ð
2ððð¿ð ð2.
Therefore the M.I of the whole ring will be
ðŒðððð =ð
2ððð2 â ð¿ð
ððððððð¡ð =
ðð
2ðâ«ðð =
ðð
2ðà 2ðð = ðð2
Hence ðð2 is the required moment of inertia of the ring.
49
Module No. 127
Radius of Gyration
Radius of gyration specifies the distribution of the elements of body around the axis in terms of
the mass moment of inertia, As it is the perpendicular distance from the axis of rotation to a point
mass m that gives an equivalent inertia to the original object m The nature of the object does not
affect the concept, which applies equally to a surface bulk mass.
Mathematically the radius of gyration is the root mean square distance of the object's parts from
either its center of mass or the given axis, depending on the relevant application.
Let ðŒ = â ðððð=1 ðð
2 be the moment of inertia of a system of particles about AB, and
ð = â ðððð=1 be the total mass of the system.
Then the quantity K such that
ðŸ2 =ðŒ
ð=
â ðððð=1 ðð
2
â ðððð=1
or
ðŸ = âðŒ
ð
is called the radius of gyration of the system AB.
Example:
Find the radius of gyration, K, of the triangular lamina of mass M and moment of inertia ðŒ =1
6â ðâ2.
Solution:
Since formula for radius of gyration is given by
ðŸ2 =ðŒ
ð
by substituting value in above , we get
ðŸ2 =1
6â ðâ2
ð= 1
6â â2
or ðŸ =â
â6
50
Module No. 128
Principal Axes for the Inertia Matrix
Principal Axes
When a rigid body is rotating about a fixed point O, the angular velocity vector ð and the
angular momentum vector L (about O) are not in general in the same direction. However
it can be proved that at each point in the body there exists distinct directions, which are
fixed relative to the body, along which the two vectors are aligned i.e. coincident. Such
directions are called principal directions and the axes along them are referred to as
principal axes of inertia. The corresponding moments of inertia are called principal
moments of inertia.
Orthogonality of Principal Axes
If the principal axes at each point of the body exist, then their orthogonality can be proved by
stating that axes relative to which product of inertia are zero are the principal axes.
51
Module No. 129
Introduction to the Dynamics of a System of Particles
Dynamics
Dynamics is the branch of mechanics deals with forces and relationship fundamentally to the
motion but sometimes also to the equilibrium of bodies.
The Dynamics of a System of Particles
Suppose we have a system of N particles in motion under forces. These will be two types of
forces, internal and external.
Internal forces act between particles of the system; all other are external. We suppose further that
the internal forces satisfy Newtonâs third law of motion. i.e. Internal forces between a pair of
particles are equal and opposite If ð¹ðð(ððð¡)
denotes the internal force on the ith particle due to the
jth particle, then in the view of this assumption we can write
ð¹ðð(ððð¡) = âð¹ðð
(ððð¡) (1)
The equation of motion for the ith particle of system can therefore be written as
ð¹ð(ðð¥ð¡)
+ â ð¹ðð(ððð¡)ð
ð=1 = ððᅵᅵð (2)
where ð¹ð(ðð¥ð¡)
denotes the external force on the ith particle, and ᅵᅵð, the acceleration of the ith
particle. It is clear from equation (1) that the internal force on a given particle due to itself is zero
and therefore the term ð = ð does not contribute to the sum.
To obtain an equation of motion for the whole system we sum over i, (from 1 to N), on the both
sides of equation (2), and obtain
âð¹ð(ðð¥ð¡)
ð
+ ââð¹ðð(ððð¡)
ð
= âððᅵᅵð
ðð
or
ð¹(ðð¥ð¡) + â ð¹ðð(ððð¡)
ð,ð = â ððᅵᅵðð (3)
Where ð¹(ðð¥ð¡) denotes the total external force on the system. Now we will show that because of
condition (1), the second term on L.H.S of (3) is zero.
âð¹ðð(ððð¡)
ð,ð
=1
2â(ð¹ðð
(ððð¡) + ð¹ðð(ððð¡))
ð,ð
Since the indices I and j are dummy and vary over the same set of integers, we can interchange
them in the second sum on the R.H.S of the last equation. Therefore we have
52
âð¹ðð(ððð¡)
ð,ð
=1
2â(ð¹ðð
(ððð¡) + ð¹ðð(ððð¡))
ð,ð
=1
2â(ð¹ðð
(ððð¡) â ð¹ðð(ððð¡))
ð,ð
= 0
Therefore the equation (3) becomes
ð¹(ðð¥ð¡) = â ððᅵᅵðð (4)
In case of centroid, the relation becomes
âððᅵᅵð
ð
= ðᅵᅵð
where ᅵᅵð = ᅵᅵð is the acceleration of center of mass (c.m.) and M is the total mass of the system.
Hence equation (4) can be written as
ð¹(ðð¥ð¡) = ðᅵᅵð =ð
ðð¡(ðᅵᅵð) (5)
If ð¹(ðð¥ð¡) = 0 then ð
ðð¡(ðᅵᅵð) = 0, which implies that ðᅵᅵð =constant.
We conclude that if external forces are on a system of particles are zero, then its momentum will
be a constant of motions or a conserved quantity. Equation (5) describes the translational motion
of the system and may be referred to as the translational equation of motion which describes the
rotational motion of the system, we take the cross product of both sides of equation (2) with
ðð, and obtain
ððÃð¹ð(ðð¥ð¡)
+ â ðð à ð¹ðð(ððð¡)
ð
ð=1
= ðððð à ᅵᅵð
summing over i
â ððÃð¹ð(ðð¥ð¡)
+ â ðð à ð¹ðð(ððð¡)ð
ð,ð=1 =ð â ðððð à ᅵᅵðð (6)
Now we will show that in view of our assumption (1) about internal forces, the second term on
L.H.S of equation (6) is zero. On interchanging the dummy indices and using (1) we have
â ðð à ð¹ðð(ððð¡)
ð,ð
=1
2â(ðð à ð¹ðð
(ððð¡)
ð,ð
+ ðð à ð¹ðð(ððð¡))
=1
2â(ðð à ð¹ðð
(ððð¡) â
ð,ð
ðð à ð¹ðð(ððð¡))
=1
2â (ðð â ðð) à ð¹ðð
(ððð¡)ð,ð (7)
The vector ðð â ðð is in the direction of the line segment joining the particles I and j and therefore
it is parallel to ð¹ðð . Therefore the last term on R.H.S of equation (7) is zero.
Hence we finally obtain
âðð à ð¹ðð(ðð¥ð¡)
ð,ð
= âðððð à ᅵᅵð
ð
or
53
â ðºð = ðºðð¥ð¡ =
ð
âðððð à ᅵᅵð
ð
=ð
ðð¡(âðððð à ᅵᅵð
ð
)
=ð
ðð¡(âðð à ððᅵᅵð
ð
)
=ð
ðð¡âð¿ð =
ðð¿
ðð¡ð
which is usually written as
ðð¿
ðð¡= ðº(ðð¥ð¡)
where ðºð(ðð¥ð¡)
is the torque or momentum due to external force on the ith particle, and ðºð(ðð¥ð¡) is
the torque due to all external forces on the system.
Similarly L is the total angular momentum of the system.
âðð à ð¹ðð(ðð¥ð¡)
ð,ð
= âðºð = ðº
ð
Thus we have obtained the following two equations of motion for a system of particles. ð
ðð¡(ðᅵᅵð) = ð¹(ðð¥ð¡) (8)
and ðð¿
ðð¡= ðº(ðð¥ð¡) (9)
It follows from equation (8) that if the total external torque on the system is zero, then its angular
momentum will be a constant of motion or a conserved quantity.
54
Module No. 130
Introduction to Center of Mass and Linear Momentum
The rigid bodies are a system of particles in which the position of particles is relatively fixed. We
consider such a system of particles in this article and will discuss its center of mass and linear
momentum.
Center of Mass
Our general system consists of η particles of masses ð1, ð2, âŠâŠ . . , ððwhose position vectors
are, respectively, ð1, ð2, âŠâŠ . . , ðð. We define the center of mass of the system as the point whose
position vector ððð is given by
ððð =ð1ð1+ð2ð2+ð3ð3+â¯âŠâŠ.+ðððð
ð1+ð2+â¯âŠ..+ðð=
â ðððððð=1
ð (1)
where ð = â ðð ðð=1 is the total mass of the system.
The above definition of center of mass leads us to three equations
ð¥ðð =â ððð¥ð
ðð=1
ð, ðŠðð =
â ðððŠððð=1
ð, ð§ðð =
â ððð§ððð=1
ð
which are the rectangular coordinates of the center of mass of the system.
Linear Momentum
We define the linear momentum Ï of the system as the vector sum of the linear momentum of the
individual particles, namely,
ð = âðð = âððð£ð
ðð
From equation (1)
ððð =â ðððð
ðð=1
ð (1)
ᅵᅵðð = ððð =â ððð£ð
ðð=1
ð
it follows that
ð = ðððð
that is, the linear momentum of a system of particles is the product of the velocity of the center
of mass and the total mass of the system.
55
Module No. 131
Law of Conservation of Momentum for Multiple Particles
Statement
In the absence of the external force, the momentum of the system of particles will be conserved.
Proof
Suppose we have a system of N particles in motion under forces. These will be two types of
forces, internal and external.
Suppose now that there are external forces ð¹1ð¹2, . . . , ð¹ð, , . . . , ð¹ð acting on the respective particles.
In addition, there may be internal forces of interaction between any two particles of the system.
We denote these internal forces by ð¹ðð(ððð¡), meaning the force exerted on particle i by particle j.
Internal Forces act between particles of the system; all over the external. We suppose further that
the internal force satisfy Newtonâs third law of motion. i.e. Internal forces between a pair of
particles are equal and opposite.
If ð¹ðð(ððð¡) denotes the internal force on the ith particle due to the jth particle, then in the view of
this assumption we can write
ð¹ðð(ððð¡) = âð¹ðð
(ððð¡) (1)
The equation of motion for the ith particle of the system can therefore be written as
ð¹ð(ðð¥ð¡) + â ð¹ðð
(ððð¡)ðð=1 = ðððð = ðᅵᅵ (2)
where ð¹ð(ðð¥ð¡) denotes the total external force on the ith particle, and ðð the acceleration of the ith
particle.
To obtain an equation of motion for the whole system, we sum over I (from 1 to N), on both
sides of the equation (2).
âð¹ð(ðð¥ð¡)
ð
+ ââð¹ðð(ððð¡)
ð
ð=1ð
= âðððð
ð
= ðᅵᅵ
ð¹(ðð¥ð¡) + â ð¹ðð(ððð¡)
ð,ð = â ðððð = ðᅵᅵð (3)
where ð¹(ðð¥ð¡)denotes the total force of the system.
now we will show that because of equation 1, L.H.S of equation 3 is zero.
56
ðᅵᅵ = âð¹ðð(ððð¡)
ð,ð
= (1
2âð¹ðð
(ððð¡)
ð,ð
+1
2âð¹ðð
(ððð¡)
ð,ð
)
since the indices I and j are dummy and vary over the same set of integers, we can interchange
them in the second sum on R.H.S of the last equation. Therefore we have
ðᅵᅵ = âð¹ðð(ððð¡)
ð,ð
= (1
2âð¹ðð
(ððð¡)
ð,ð
+1
2âð¹ðð
(ððð¡)
ð,ð
)
using equation (1), we have
â ð¹ðð(ððð¡)
ð,ð = (1
2â ð¹ðð
(ððð¡)ð,ð â
1
2â ð¹ðð
(ððð¡)ð,ð )=0
therefore equation (3) becomes
ðᅵᅵ = ð¹(ðð¥ð¡) = âðððð
ð
for centroid, we know that â ððððð = ðððwhere ððis the acceleration of centroid.
Hence equation can be written as
ðᅵᅵ = ð¹(ðð¥ð¡) = ððð =ð
ðð¡(ðð£ð)
If ð¹(ðð¥ð¡) = 0 then ðᅵᅵ =ð
ðð¡(ðð£ð) = 0 ðð ð = ðð£ð = constant
We conclude that if external forces are on a system of particles are zero, then its momentum will
be a constant of motion or a conserved quantity.
57
Module No. 132
Example of Conservation of Momentum
Problem Statement
A particle ðŽ, of mass 6 kg, travelling in a straight line at 5 msâ1 collides with a particle ðµ, of
mass 4 kg, travelling in the same straight line, but in the opposite direction, with a speed of 3
msâ1. Given that after the collision particle ðŽ continues to move in the same direction at 1.5
msâ1, what speed does particle ðµ move with after the collision? (© mathcentre 2009)
Given Data
Mass of particle ðŽ = ð1 = 6kg
Velocity of ðŽ before collision = ð¢1 =5msâ1
Velocity of ðŽ after collision = ð£1 =1.5msâ1
Mass of Particle ðµ = ð2 = 4kg
Velocity of ðµ before collision = ð¢2 = 3msâ1
Required
Velocity of ðµ after collision = ð£2 = ?
Solution
It is always useful to depict the collision with the velocities both before and after.
Using the principle of conservation of momentum:
ð1ð¢1 + ð2ð¢2 = ð1ð£1 + ð2ð£2
6 à 5 + 4 à (â3) = 6 à 1.5 + 4 à ð£2
9 = 4 à ð£2
ð£2 = 9 4â = 2.3 ð ð â1
So after the collision particle B moves with a speed of 2.3 m sâ1, in the same direction as A.
Example 2
58
Problem Statement
A particle A, of mass 8 kg, collides with a particle B, of mass m2 kg. The velocity of particle
A before the collision was (â1ð + 4ð) ð ð â1 and the velocity of particle B before the collision
was (â0.8ð + 1.4ð) ð ð â1. Given the velocity of particle A after the collision was
(â2ð + 2ð) ð ð â1, and the velocity of particle B was 3ð ð ð â1, what was the mass of particle
B?
Given Data
Mass of particle A= ð1 = 8kg
Velocity of A before collision= ð¢1 = (â1ð + 4ð) ð ð â1
Velocity of A after collision= ð£1 = (â2ð + 2ð) ð ð â1
Velocity of B before collision= ð¢2 = (â0.8ð + 1.4ð) ð ð â1
Velocity of B after collision= ð£2 =?3ð ð ð â1
Required
Mass of Particle B= ð2 =?
Solution
It is always useful to depict the collision with the velocities both before and after.
Using the principle of conservation of momentum:
ð1ð¢1 + ð2ð¢2 = ð1ð£1 + ð2ð£2
6 à 5 + 4 à (â3) = 6 à 1.5 + 4 à ð£2
9 = 4 à ð£2
ð£2 = 9 4â = 2.3msâ1
So after the collision particle ðµ moves with a speed of 2.3 msâ1, in the same direction as ðŽ.
Example 2
Problem Statement
A particle ðŽ, of mass 8 kg, collides with a particle ðµ, of mass 2 kg. The velocity of particle ðŽ
before the collision was (â1ð + 4ð) msâ1 and the velocity of particle ðµ before the collision was
(â0.8ð + 1.4ð) msâ1. Given the velocity of particle ðŽ after the collision was (â2ð + 2ð) msâ1,
and the velocity of particle ðµ was 3ð msâ1, what was the mass of particle ðµ?
Given Data
59
Mass of particle ðŽ = ð1 = 8kg
Velocity of ðŽ before collision = ð¢1 = (â1ð + 4ð)msâ1
Velocity of ðŽ after collision = ð£1 = (â2ð + 2ð)msâ1
Velocity of ðµ before collision = ð¢2 = (â0.8ð + 1.4ð)msâ1
Velocity of ðµ after collision = ð£2 = 3ð msâ1
Required
Mass of Particle ðµ = ð2 = ?
Solution
By making use of principle of conservation of momentum:
ð1ð¢1 + ð2ð¢2 = ð1ð£1 + ð2ð£2
8(â1ð + 4ð) + ð2(â0.8ð + 1.4ð)
= 8(â2ð + 2ð) + ð2(3ð)
â8ð + 32ð + 16ð â 16ð + ð2(â0.8ð + 1.4ð) â ð2(3ð) = 0
8ð + 16ð + ð2(â0.8ð â 1.6ð) = 0
8ð + 16ð â ð2(+0.8ð + 1.6ð) = 0
8ð + 16ð = ð2(0.8ð + 1.6ð)
ð2 =0.8ð + 1.6ð
8ð + 16ð
By rationalizing and solving the above fraction, we obtain
ð2 = 10ðð
60
Module No. 133
Angular Momentum â Derivation
Angular momentum of a particle of mass ð, position vector ð and linear momentum ð is defined
as ð Ã ð. It is also called moment of momentum.
Let there be a system consists of N particles with positon vector ðð and the momentum ðð (ð =
1,2,3, âŠâŠ . , ð) then the total angular momentum L is given by
ð¿ = âðð à ðð = âðð à (ðð£ð)
ðð
To find the relation between L and ð, we proceed as follows:
ð¿ = âðð à (ððð£ð)
ð
= âðð à ðð(ᅵᅵ à ðð)
ð
= âðð[ðð à (ᅵᅵ à ðð)
ð
]
= â ðð[(ðð. ðð)ᅵᅵ â (ðð. ᅵᅵ)ððð ] (1)
Here (ðð. ðð)ᅵᅵ and (ðð. ᅵᅵ)ðð are not parallel vectors. If they were parallel then we could easily say
that L is parallel to ᅵᅵ.
We conclude that in general L and ᅵᅵ are not parallel to each other. To find the relationship
between the vectors L and ᅵᅵ we proceed as follows:
Since
ðð = ð¥ðð + ðŠðð + ð§ðᅵᅵ
therefore
ðð. ðð = ð¥ð2 + ðŠð
2 + ð§ð2
and
ᅵᅵ. ðð = ðð¥ð¥ð + ððŠðŠð + ðð§ð§ð
Therefore substituting these values in equation (1), we have
ð¿ = â ðð(ð ð¥ð2 + ðŠð
2 + ð§ð2)(ðð¥ð + ððŠð + ðð§ï¿œï¿œ) â (ðð¥ð¥ð + ððŠðŠð + ðð§ð§ð)(ð¥ðð + ðŠðð + ð§ðᅵᅵ)
(2)
Writing
ð¿ = ð¿ð¥ð + ð¿ðŠð + ð¿ð§ï¿œï¿œ â¡ ð¿1ð + ð¿2ð + ð¿3ᅵᅵ
and
ð = ðð¥ð + ððŠð + ðð§ï¿œï¿œ â¡ ð1ð + ð2ð + ð3ᅵᅵ
and comparing the coefficient of ð on both sides of equation (2), we obtain
61
ð¿ð¥ = âðð(
ð
ð¥ð2 + ðŠð
2 + ð§ð2)ðð¥ â (ðð¥ð¥ð + ððŠðŠð + ðð§ð§ð)ð¥ð
= âðð[
ð
ð¥ð2ðð¥ + (ðŠð
2 + ð§ð2)ðð¥ â ðð¥ð¥ð
2 â ððŠð¥ððŠð â ðð§ð¥ðð§ð]
= ðð¥ â ððð (ðŠð2 + ð§ð
2) â ðð¥ð¥ð2 â ððŠ â ððð ð¥ððŠð â ðð§ â ððð ð¥ðð§ð] (3)
As we studied the definitions of moment of inertia and product of inertia be defined as
ðŒð¥ð¥ = âðð(ðŠð2 + ð§ð
2)
ð
ðŒðŠðŠ = âðð(ð§ð2 + ð¥ð
2)
ð
ðŒð§ð§ = âðð (ð¥ð2 + ðŠð
2)
ð
and
ðŒð¥ðŠ = ââððð¥ððŠð
ð
ðŒðŠð§ = ââðððŠðð§ð
ð
ðŒð§ð¥ = ââððð¥ðð§ð
ð
Using these definitions and noting that ðŒð¥ðŠ = ðŒðŠð¥ etc. we can write
ð¿ð¥ = ðŒð¥ð¥ðð¥ + ðŒð¥ðŠððŠ + ðŒð¥ð§ðð§
Similarly we obtain
ð¿ðŠ = ðŒðŠð¥ðð¥ + ðŒðŠðŠððŠ + ðŒðŠð§ðð§
ð¿ð§ = ðŒð§ð¥ðð¥ + ðŒðŠð§ððŠ + ðŒð§ð§ðð§
62
Module No. 134
Angular Momentum in Case of Continuous Distribution of Mass
An actual rigid body consists of very large number of particles and therefore we may suppose that
there is a continuous distribution of mass. If ð(ð) denotes density of a volume element ðð,
surrounding the point ð, then the mass of this element will be ð(ð)ðð.
Hence we can write the expression of angular momentum in case of continuous distribution of
mass by
ðŒ11 â¡ ðŒð¥ð¥ = â«ð(ð)(ðŠ2 + ð§2)ðð
where
ð = (ð¥, ðŠ, ð§) â¡ (ð¥1, ð¥2, ð¥3), ðð = ðð¥ððŠðð§
Similarly the other two moment of inertia are defined as
ðŒ22 â¡ ðŒðŠðŠ = â«ð(ð)(ð§2 + ð¥2)ðð
and
ðŒ33 â¡ ðŒð§ð§ = â«ð(ð)(ð¥2 + ðŠ2)ðð
Similarly the product of inertia are defined as
ðŒ12 â¡ ðŒð¥ðŠ = ââ«ð(ð)ð¥ðŠðð,
ðŒ23 â¡ ðŒðŠð§ = ââ«ð(ð)ðŠð§ðð,
ðŒ31 â¡ ðŒð§ð¥ = ââ«ð(ð)ð§ð¥ðð
By using these definitions of moment of inertia, we can find the expression of angular momentum
by using the following result
ð¿ = ð¿ð¥ð + ð¿ðŠð + ð¿ð§ï¿œï¿œ
ð¿ = ðŒð¥ð¥ðð¥ + ðŒðŠðŠððŠ + ðŒð§ð§ðð§ + ðŒð¥ðŠ(ðð¥ + ððŠ) + ðŒðŠð§(ððŠ + ðð§) + ðŒð§ð¥(ðð§ + ðð¥)
where
ð¿ð¥ = ðŒð¥ð¥ðð¥ + ðŒð¥ðŠððŠ + ðŒð¥ð§ðð§
ð¿ðŠ = ðŒðŠð¥ðð¥ + ðŒðŠðŠððŠ + ðŒðŠð§ðð§
ð¿ð§ = ðŒð§ð¥ðð¥ + ðŒð§ðŠððŠ + ðŒð§ð§ðð§
and ðð¥, ððŠ, ðð§ are the components of angular velocity along (ð¥, ðŠ, ð§).
or
63
[
ð¿ð¥
ð¿ðŠ
ð¿ð§
] = [
ðŒð¥ð¥ ðŒð¥ðŠ ðŒð¥ð§
ðŒð¥ðŠ ðŒðŠðŠ ðŒðŠð§
ðŒð¥ð§ ðŒðŠð§ ðŒð§ð§
] [
ðð¥
ððŠ
ðð§
]
where we have used the property of products of inertia that
ðŒð¥ðŠ = ðŒðŠð¥, ðŒðŠð§ = ðŒð§ðŠ and
ðŒð¥ð§ = ðŒð§ð¥
The above matrix form can be written as
ð¿ = ðŒð
64
Module No. 135
Law of Conservation of Angular Momentum
Angular Momentum
The angular momentum of a single particle is defined as the cross product of linear momentum
and position vector of concerned particle.
Mathematically ð¿ = ð à ðð£.
Angular Momentum of System of Particles
The angular momentum L of a system of particles is defined accordingly, as the vector sum of the
individual angular momentum, namely,
ð¿ = âðð à ððð£ð (1)
ð
ð=1
Law of Conservation of Angular Momentum
The time rate change of angular momentum in the absence of some external forces is zero.
Mathematically, we can write
ð ð³
ð ð= ð â¹ ð³ = ððšð§ð¬ððð§ð
Let us calculate the time derivative of the angular momentum. Using the rule for differentiating
the cross product, we find
ðð¿
ðð¡=
ð
ðð¡(âðð à ððð£ð
ð
ð=1
) = â(ð£ð Ã
ð
ð=1
ððð£ð) + â(ðð Ã
ð
ð=1
ðððð)
Now the first term on the right vanishes, because,ð£ð à ð£ð = 0 and, because ðððð is equal to the
total force acting on particle i, we can write
ðð¿
ðð¡= â(ðð Ã
ð
ð=1
ðððð)
ðð¿
ðð¡= â [ðð Ã (ð
ð=1 â ð¹ð(ðð¥ð¡)
ð + â â ð¹ðð(ððð¡)ð
ð=1ðð=1 )] (2)
ðð¿
ðð¡= â ðð à ð¹ð
(ðð¥ð¡)ðð=1 + â â ð¹ðð
(ððð¡)ðð=1
ðð=1 ) (3)
where ð¹ð denotes the total external force on particle ð, and ð¹ðð denotes the (internal) force exerted
on particle i by any other particle ð. Now the double summation on the right consists of pairs of
terms of the form
65
(ðð à ð¹ð) + (ðð à ð¹ðð) (4)
Denoting the vector displacement of particle ð relative to particle ð by ððð, we see from the triangle
shown in Figure that
ððð = ðð â ðð (5)
Therefore, because ð¹ðð = -ð¹ðð, expression (3) reduces to
âððð à ð¹ðð (6)
which clearly vanishes if the internal forces are central, that is, if they act along the lines
connecting pairs of particles. Hence, the double sum in Equation (3) vanishes. Now the cross
product (ðð à ð¹ð)is the moment of the external force F. The sum â(ðð à ð¹ð) is, therefore, the
total moment of all the external forces acting on the system. If we denote the total external
torque, or moment of force, by N, Equation (3) takes the form
ðð¿
ðð¡= ð
That is, the time rate of change of the angular momentum of a system is equal to the total
moment of all the external forces acting on the system.
If a system is isolated, then ð = 0, and the angular momentum remains constant in both
magnitude and direction:
ð¿ = âðð à ððð£ð
ð
ð=1
= Constant vector (8)
This is a statement of the principle of conservation of angular momentum. It is a generalization
for a single particle in a central field.
66
Module No. 136
Example Related to Angular Momentum
Problem Statement
The moon revolves around the earth so that we always see the same face of the moon.
i. Find how the spin angular momentum and orbital angular momentum of the moon w.r.t.
the earth are related?
ii. Find the change in spin angular momentum of moon so that we could see the entire
moonâs surface during a month.
Solution
i. Let ð, ð ð denotes the mass and the radius of the moon respectively.
Then its spin angular momentum i.e. angular momentum about its axis of rotation is given by
ð¿ð = ðŒðð
where ðŒ is the moment of inertia and ðð angular velocity of the moon about its own axis
The moment of inertia of the sphere is
ðŒ =2
5ðð2
by generalizing it for moon, we obtain the moment of inertia for the moon
ðŒ =2
5ðð ð
2
then the angular momentum of moon will be
ð¿ð =2
5ðð ð
2ðð (1)
In addition to spinning about its own axis, the moon is also performing orbital motion about the
earth. If we denote the orbital angular momentum by ð¿0, then
ð¿0 = ð Ã (ðð) = ð Ã (ðð ð0)
= ðð 2ð0 (2)
⎠ð£ = ðð
where we have treated the moon as a particle and R is the distance between the moon and the
earth. Since we always see the same face of the moon, the moon makes on rotation about its axis
in the same time as it makes on revolution around the earth i.e. ðð = ð0. Form equation (1) and
(2) we have
ð¿ð
ð¿0=
25
ðð ð 2ðð
ðð 2ð0
67
=2
5(ð ð
ð )2
ii. If we have to see the entire moonâs surface during a month, then ð¿ð must either increase
or decrease by one-half of its present value.
68
Module No. 137
Kinetic Energy of a System about Principal Axes â Derivation
Kinetic Energy
The kinetic energy for a particle is given by the following scalar equation:
ð =1
2ðð£2
Where:
T is the kinetic energy of the particle with respect to ground (an inertial reference frame)
m is the mass of the particle
v is the velocity of the particle, with respect to ground
Kinetic Energy of a Rigid Body
For a rigid body experiencing planar (two-dimensional) motion, the kinetic energy is given by
the following general scalar equation:
ð =1
2ðð£ð
2 +1
2ðŒðð
2 (1)
where the first term in equation (1) shows the kinetic energy due to the motion of center of mass
and second term shows the rotational kinetic energy and subscript c denotes the center of mass
and ð£ð denotes the velocity of c.m and ðŒð denotes the inertia matrix w.r.t c.m.
Kinetic Energy of a Rigid Body w.r.t Origin
If the rigid body is rotating about a fixed point O that is attached to ground, we can express the
kinetic energy as:
ð =1
2ðŒ0ð
2
Where:
ðŒ0 is the moment of inertia of the rigid body about an axis passing through the fixed point O, and
perpendicular to the plane of motion and ð is the angular velocity.
Kinetic Energy of a Rigid Body about Principal Axes
69
If the center of mass is oriented along with origin then the coordinate axes (xyz axes) and the
principal axes of the rigid body are aligned. In this case, the inertia matrix reduces, i.e. products
of inertia are zero along the principal axes, (ðŒð¥ðŠ = ðŒðŠð§ = ðŒð§ð¥ = 0) .
For general three-dimensional motion, the kinetic energy of a rigid body about principal axes is
given by the following general scalar equation:
ð =1
2ðð£0
2 +1
2ðŒð¥ðð¥
2 +1
2ðŒðŠððŠ
2 +1
2ðŒð§ðð§
2 (3)
where ðŒð¥, ðŒðŠ, ðŒð§ are the moment of inertia along principal axes (xyz axes) also called principal
moment of inertia and products of inertia ðŒð¥ðŠ, ðŒðŠð§ , ðŒð§ð¥ are zero along the axes
Equation (3) is the required expression of Kinetic Energy along principal axes.
If the rigid body has a fixed point O that is attached to ground, we can give an alternate scalar
equation for the kinetic energy of the rigid body:
ð =1
2ðŒð¥ðð¥
2 +1
2ðŒðŠððŠ
2 +1
2ðŒð§ðð§
2
in this case, the kinetic energy due to the motion of c.m. vanishes.
70
Module No. 138
Moment of Inertia of a Rigid Body about a Given Line
Let M be the mass of the system and ᅵᅵ a unit vector along the line ð. Then ᅵᅵ = ðð + ðð + ðᅵᅵ,
where (ð, ð, ð)are direction cosines of the line.
If ðŒð denotes the record moment of inertia, then
ðŒð = âðððð2
ð
Form the figure
ðð = |ðð| sin ðð = |ðð| sin ðð = ðð sin ðð = |ðð Ã ðð|
Therefore
ðŒð = âðð|ðð à ðð|2
ð
(1)
Now
|ðð à ðð| = |ð ð ᅵᅵð ð ðð¥ð ðŠð ð§ð
|
= (ðð§ð â ððŠð)ð + (ðð¥ð â ðð§ð)ð + (ðð§ð â ðð¥ð)ᅵᅵ
|ðð à ðð|2 = (ðð§ð â ððŠð)
2 + (ðð¥ð â ðð§ð)2 + (ððŠð â ðð¥ð)
2
Hence on substitution in equation (1), we have
ðŒð = â ðð[
ð
(ðð§ð â ððŠð)2 + (ðð¥ð â ðð§ð)
2 + (ððŠð â ðð¥ð)2]
ðŒð = âðð[
ð
ð2ð§ð2 + ð2ðŠð
2 â 2ðððŠðð§ð + ð2ð¥ð2 + ð2ð§ð
2 â 2ððð¥ðð§ð + ððŠð + ðð¥ð â 2ððð¥ððŠð]
= âðð[
ð
ð2(ð¥ð2 + ð§ð
2) + ð2(ð¥ð2 + ðŠð
2) + ð2(ðŠð2 + ð§ð
2) â 2ððð¥ððŠð â 2ðððŠðð§ð â 2ððð¥ðð§ð
= âðð[
ð
ð2(ðŠð2 + ð§ð
2) + ð2(ð¥ð2 + ð§ð
2) + ð2(ð¥ð2 + ðŠð
2) â 2ððð¥ððŠð â 2ðððŠðð§ð â 2ððð¥ðð§ð
= ð2 âðð[
ð
(ðŠð2 + ð§ð
2) + ð2 âðð
ð
(ð¥ð2 + ð§ð
2) + ð2 âðð
ð
(ð¥ð2 + ðŠð
2) + 2ðð(ââðð
ð
ð¥ððŠð)
+ 2ðð(ââðð
ð
ðŠðð§ð) + 2ðð(ââðð
ð
ð¥ðð§ð)
71
or finally,
ðŒð = ð2ðŒð¥ð¥ + ð2ðŒðŠðŠ + ð2ðŒð§ð§ + 2ðððŒð¥ðŠ + 2ðððŒðŠð§ + 2ðððŒð§ð¥
which may also be written as
ðŒð = ð2ðŒ11 + ð2ðŒ22 + ð2ðŒ33 + 2ðððŒ12 + 2ðððŒ23 + 2ðððŒ31
72
Module No. 139
Example of M.I of a Rigid Body About Given Line
Problem Statement
Calculate the moment of inertia of a right circular cone about its axis of symmetry.
Solution
Let M be the mass, a the radius and h the height of right circular cone. We regard the cone as
composed of elementary circular discs of small thickness each parallel to the base of the cone.
We choose the z-axis along the axis of symmetry, and consider a typical disc of radius r and
width ð¿ð§ at a distance z from the base.
Mass of the disc is given by
ð¿ð = ððð2ð¿ð§
We regard the disk to be composed of concentric elementary circular rings of varying radii say ðâ²
then the mass ð¿ðâ² of one circular ring with height ð¿ð§ will be
ð¿ðâ² = ð2ððâ²ð¿ð§
Then the M.I of one circular ring will be
ðŒð.ð = ð2ððâ²ð¿ð§(ðâ²)2 = 2ðð(ðâ²)3ð¿ð§
Hence the M.I of the whole disk will be
ð¿ðŒ = â 2ðð(ðâ²)3ð¿ð§
ððððð
= â«2
ð
0
ð(ð¿ð
ðð2ð¿ð§)(ðâ²)3ð¿ð§ððâ²
=2ð¿ð
ð2â«(ðâ²)3ððâ²
ð
0
ð¿ðŒ =2ð¿ð
4ð2ð4 =
1
2ð¿ðð2
From the similar triangles,
we have
73
ð
ð=
â â ð§
â or ð = ð
â â ð§
â
Therefore
ð¿ð = ðð (ðâ â ð§
â)2
ð¿ð§
Since the M.I of the disk is
ð¿ðŒ =1
2ð¿ðð2
On substituting for ð¿ð and ð, the M.I of the cone about its axis of symmetry will be
ðŒ =1
2â« ðð (ð
â â ð§
â)2
(ðâ â ð§
â)2
â
0
ðð§
=1
2
ððð4â
â4â«(â â ð§)4ðð§
â
0
=1
2
ððð4â
â4â«(ð§ â â)4ðð§
â
0
Since the M.I of the disk is
ð¿ðŒ =1
2ð¿ðð2
On substituting for ð¿ð and ð, the M.I of the cone about its axis of symmetry will be
ðŒ =1
2â« ðð (ð
â â ð§
â)2
(ðâ â ð§
â)2
â
0
ðð§
=1
2
ððð4â
â4â«(â â ð§)4ðð§
â
0
=1
2
ððð4â
â4â«(ð§ â â)4ðð§
â
0
ðŒ =ððð4â
10â4|(ð§ â â)5|0
â
=ððð4â6
10â4=
ððð4â2
10
Since we know that
ð = density of the cone
74
=ð
(1 3â )ðð2â
So,
ðŒ =3
10ðð2
75
Module No. 140
Ellipsoid of Inertia
We have obtained the expression of moment of inertia of the given line l in terms of moment and
product of inertia w.r.t the coordinate axes OXYZ coordinate system whose origin O lies on the
line l.
ðŒð = ðŒ = ð2ðŒ11 + ð2ðŒ22 + ð2ðŒ33 + 2ðððŒ12 + 2ðððŒ23 + 2ðððŒ31
(1)
For the ellipsoid of inertia, we chose a point P such that |ðð| = 1âðŒ
â . If (ð¥, ðŠ, ð§) are coordinates
of point P then
ð =ð¥
|ðð|= ð¥âðŒ, ð =
ðŠ
|ðð|= ðŠâðŒ , ð =
ð§
|ðð|= ð§âðŒ
On eliminating ð, ð, ð from equation (1), we obtain
ðŒ = (ð¥âðŒ)2ðŒ11 + (ðŠâðŒ)2ðŒ22 + (ð§âðŒ)2ðŒ33 + 2(ð¥ðŠâðŒâðŒ)ðŒ12 + 2(ðŠð§âðŒâðŒ)ðŒ23 + 2(ð¥ð§âðŒâðŒ)ðŒ31
ðŒ = ðŒ[ð¥2ðŒ11 + ðŠ2ðŒ22 + ð§2ðŒ33 + 2ð¥ðŠðŒ12 + 2ðŠð§ðŒ23 + 2ð§ð¥ðŒ31]
1 = ð¥2ðŒ11 + ðŠ2ðŒ22 + ð§2ðŒ33 + 2ð¥ðŠðŒ12 + 2ðŠð§ðŒ23 + 2ð§ð¥ðŒ31
which can also be written as
ðŒ11ð¥2 + ðŒ22ðŠ
2 + ðŒ33ð§2 + 2ðŒ12ð¥ðŠ + 2ðŒ23ðŠð§ + 2ðŒ31ð§ð¥ = 1
Since ðŒ11, ðŒ22, ðŒ33 are all positive, equation (2) represents an ellipsoid. This ellipsoid is called
ellipsoid of inertia or momental ellipsoid. The momental ellipsoid contains information about
moments or product of inertia at given point. If P is any point on the ellipsoid of inertia, then
|ðð| = 1âðŒ
â or ðŒ = 1ðð2â i.e. the moment of inertia of a rigid body about any line ðð is equals
to the reciprocal of the square of the length of |ðð |.
76
Module No. 141
Rotational Kinetic Energy
Introduction to Kinetic Energy
Kinetic energy is the energy produced in any body during its motion. It is equal to the half of the
product of mass and square of the velocity of the moving body;
ðŸ. ðž.= ð =1
2ðð£2
We consider a rigid body in a general state of motion in which it has both translation and rotation
w.r.t a fixed coordinate system. We suppose that it has an instantaneous angular velocity ð about
a reference point C. We will use a model of a rigid body in which it is considered a collection of
large number N of particles which satisfy the constraint of rigidity.
If v is the velocity of C, then the velocity ð£ð of the ith particle is given by
ð£ð = ð£ + ð à ðð
If ðð is the mass and ð£ð the velocity of the ith particle, then the kinetic energy of the ith particle
will be
ðð =1
2ððð£ð
2
therefore the total kinetic energy of the system will be given by
ð = âðð
ð
ð=1
= â1
2ððð£ð
2
ð
ð=1
= â1
2ðð(
ð
ð=1
ð£ + ð à ðð)2
=1
2âðð[
ð
ð=1
ð£2 + 2ð£.ð à ðð + (ð à ðð)2]
=1
2(â ðð
ð
ð=1
)ð£2 +1
2âðð
ð
ð=1
2ð£.ð à ðð +1
2âðð(ð Ã ðð)
2
ð
ð=1
=1
2ðð£2 + ð£. (ð à âðððð) +
ð
ð=1
1
2âðð(ð Ã ðð)
2
ð
ð=1
=1
2ðð£2 + ð£.ð à â ðððð +ð
ð=11
2â ðð(ð Ã ðð)
2ðð=1 (1)
where â ðð = ð ðð=1 is the total mass of the system.
Now by using the definition of position vector center of mass (c.m.), we have
ðð =â ððððð
â ððð (2)
Now we will discuss the consequence of referring of the position vectors of particle of the system
to the c.m. If the position vector of the ith particle of the system w.r.t the c.s. , then
ðð = ðâ²ð + ðð
77
Then by substitution this expression in the definition of c.m.in (2) , we obtain
ðð =â ðð(ð ðâ²
ð + ðð)
â ððð
ðð =â ððð ðâ²
ð
â ððð+
ðððð
ðð =â ððð ðâ²
ð
â ððð+ ðð
or we can write
ðð =â ððð ðâ²
ð
ð+ ðð
which gives
â ððð
ðâ²ð = 0
Hence if the reference point C is identified with the center of the mass and origin is taken thereat,
the expression â ðððð = 0 ðð=1 and therefore
ð =1
2ðð£2 +
1
2â ðð(ð Ã ðð)
2
ð
ð=1
ð = ðð¡ð + ðððð¡
where ðð¡ð =1
2ðð£2 is the translational kinetic energy is also equals to the K.E of the center of the
mass and ð¡ððð¡ =1
2â ðð(ð Ã ðð)
2ðð=1 is the rotational kinetic energy of the system.
But we have
(ð Ã ðð)2 = (ð Ã ðð). (ð Ã ðð)
= ð. ðð Ã (ð Ã ðð)
therefore on substitution
78
ðððð¡ =1
2âðð[ð. ðð Ã (ð Ã ðð)] =
1
2ð. ð¿
ð
ð=1
Thus we have obtained the following formulas for translational and rotational K.E. of a rigid
body
ðð¡ð =1
2ðð£2
ðððð¡ =1
2ð. ð¿ (1)
As we studied the relation ð¿ = ðŒð
By using this value in relation (1) we obtain the rotational kinetic energy in terms of moment of
inertia.
ðððð¡ =1
2ð. ðŒð =
1
2ðŒð2 (2)
where ðŒ is the moment of inertia of the rigid body about the origin and equation (2) is the
required expression for Rotational Kinetic Energy.
79
Module No. 142
Moment of Inertia & Angular Momentum in Tensor Notation
To show that the 3 à 3 inertia matrix (ðŒðð ) is also a Cartesian tensor of rank 3, we proceed as
follows. Here we have to distinguish between the particle index and the component index, which
denotes particle number, will be denoted by Greek letter ðŒ, and will take the value
1,2, âŠâŠ . , ð. On the other hand, the component index, which denotes the component number
will be denotes by the Latin letters such as i and will take the value 1,2,3.
Using this notation, we can write the angular momentum of the system of N particles as
ð¿ = âððŒ à ððŒ =
ðŒ
âððŒ à (ððŒð£ðŒ)
ðŒ
where ð£ðŒ is the velocity of ðŒð¡â particle. Continuing we have
âððŒððŒ à ð£ðŒ =
ðŒ
âððŒððŒ à (ð à ððŒ) , ⎠ð£ðŒ = ð à ððŒðŒ
Recall that every particle of the rigid body has the same angular velocity at a given ð¡.
Simplification of the above reduces to
ð¿ = âððŒ((ððŒ. ððŒ)ð â (ð. ððŒ)ððŒ)
ðŒ
= âððŒ(ððŒ2ð â (ð. ððŒ)ððŒ)
ðŒ
But
ð. ððŒ = âððððŒ,ð =
3
ð=1
âððð¥ðŒ,ð =
3
ð=1
âððð¥ðŒ,ð
3
ð=1
where
ððŒ = (ð¥ðŒ , ðŠðŒ, ð§ðŒ) = (ð¥ðŒ,1, ð¥ðŒ,2, ð¥ðŒ,3) â¡ ð¥ðŒ,ð
Making substitution and taking the ith component of the expression of angular momentum,
we have
ð¿ð = âððŒ (ððŒ2ðð â (âðð
ð
ð¥ðŒ,ð)ð¥ðŒ,ð)
ðŒ
But ðð = â ððð ð¿ðð; where ð¿ðð = {0, ð â ð1, ð = ð
, thus
80
therefore
ð¿ð = âððŒ (ððŒ2 âðð
ð
ð¿ðð â âðð
ð
ð¥ðŒ,ðð¥ðŒ,ð)
ðŒ
ð¿ð = âððŒ â(ððŒ2ð¿ðð â ð¥ðŒ,ðð¥ðŒ,ð)ðð
ððŒ
ð¿ð = âðð
ð
âððŒ(ððŒ2ð¿ðð â ð¥ðŒ,ðð¥ðŒ,ð)
ðŒ
which can also be expressed as
ð¿ð = â ððð ðŒðð (1)
where
ðŒðð = â ððŒ(ððŒ2ð¿ðð â ð¥ðŒ,ðð¥ðŒ,ð)ðŒ (2)
Equation (1) shows that the component of angular momentum depends not only on the angular
velocity ð but also on the inertia tensor ðŒðð.
Since the angular velocity (ðð) and the angular momentum (ð¿ð) are known to be cartesion tensor
of rank 1, it follows from the quotient theorem ðŒðð that ðŒðð is a tensor of rank 2. It is called the
inertia tensor.
81
Module No. 143
Introduction to Special Moments of Inertia
In this module, we will study some special moment of inertia. In order to introduce the moment
of inertia of some specific shapes and special rigid bodies, we assume that the rigid body is of
uniform density of particles.
Solid Circular Cylinder
We assume the radius of cylinder is ð and mass ð about axis of cylinder.
ðŒ =1
2ðð2
Hollow Circular Cylinder
We assume the radius of cylinder is ð and mass ð about axis of cylinder.
We consider the thickness of Wall of cylinder is negligible.
ðŒ = ðð2
Solid Sphere
We assume the radius of sphere is ð and mass ð about a diameter.
ðŒ =2
5ðð2
Hollow Sphere
We assume the radius of sphere is ð and mass ð about a diameter.
We consider the thickness of sphere is negligible.
ðŒ = ðð2
Rectangular Plate
We consider sides of length ð and ð, and mass M about an axis perpendicular to the plate
through the center of mass.
ðŒ =1
12ð(ð2 + ð2)
Thin Rod
We assume the length of rod is ð and mass M about an axis perpendicular to the rod through the
center of mass.
82
ðŒ =1
12ðð2
Triangular Lamina
We assume the height ð and mass M of lamina.
ðŒ =1
6ðâ2
Right Circular Cone
We assume the radius of the circular cone is ð and mass M.
ðŒ =3
10ðð2
83
Module No. 144
M.I. of the Thin Rod â Derivation
Problem Statement
Calculate the moment of inertia of a uniform rod of length ð about an axis perpendicular to the
rod and passing through an end point.
Solution
Let the ð â ðð¥ðð be chosen along the length of the rod, with origin at one end point as shown in
the figure. Let ð and ð be the mass and length of rod respectively. We suppose the rod to be
composed of small elements.
Let ðð and ðð¥ be the mass and the length of the specific element of the rod at a distance ð¥ from
the end point O.
Then
ðð
ðð¥=
ð
ð
â¹ ðð = ð
ððð¥
Then the moment of inertia of this element about the given axis is
ðŒððððððð¡ = ððð¥2 =ð
ðð¥2ðð¥
84
Hence the moment of inertia of the whole rod will be
ðŒ = âð
ðððð ððððððð¡ð
ð¥2ðð¥
=ð
ðâ« ð¥2ðð¥
ð
0
=ð
ð. |
ð¥3
3|0
ð
=ð
ð
ð3
3
=1
3ðð2
85
Module No. 145
M.I. of Hoop or Circular Ring â Derivation
Problem Statement
Calculate the moment of inertia of a hoop of mass ð and radius ð about an axis passing through
its center.
Proof
Let ð be the mass and ð the radius of the hoop. Then we can define the density of the hoop by
ð =mass
area=
ð
2ðððð
We consider this hoop to be composed of small masses (ð¿ð) each of length ð¿ð .
We can write it as
ð =ð
2ðððð=
ð¿ð
ððð¿ð
â¹ ð¿ð =ð
2ððð¿ð
Moment of inertia of the small portion of the hoop of mass ð¿ð about an axis through center and
perpendicular to the plane of the ring equals
ðŒðððð¡ðððð = ð¿ðð2
=ð
2ððð¿ð ð2 =
ðð
2ðð¿ð
Therefore the ð. ðŒ of the
whole ring/hoop will be
86
ðŒ =ðð
2ðâ ð¿ð
ðððð¡ððððð
ðŒ =ðð
2ðâ«ðð
=ðð
2ð2ðð
= ðð2
Hence we obtain ðð2 as the moment of inertia of the hoop.
87
Module No. 146
M.I. of Annular Disk - Derivation Problem
Calculate the moment of inertia of annular disk of mass ð. The inner radius of the annulus is ð 1
and the outer radius is ð 2 about an axis passing through its center.
Solution
Subdivide the annulur disk into concentric rings one of which is shown in the fig.
Let the mass of the ring is ðð, and the radius be ð, then the moment of inertia of the ring will be:
ðŒðððð = ð2ðð
The Surface area of the ring is
Area = (2ðð)ðð = 2ðððð
Since the surface area of the
annulus is
ð(ð 22 â ð 1
2)
Therefore, we can have
88
ðð
ð=
2ðððð
ð(ð 22 â ð 1
2)
ðð =2ððð
(ð 22 â ð 1
2)ð
Since the moment of Inertia of the ring is:
ðŒðððð = ð2ðð
or
ðŒðððð = ð22ððð
ð 22 â ð 1
2 ð =2ðð3ðð
ð 22 â ð 1
2
Thus the total M.I of the annulur disk will be
ðŒ = â« ðŒðððð
ð 2
ð=ð 1
ðŒ = â«2ðð3ðð
ð 22 â ð 1
2
ð 2
ð=ð 1
=2ð
ð 22 â ð 1
2 â« ð3ðð
ð 2
ð=ð 1
=2ð
ð 22 â ð 1
2 |ð4
4|ð 1
ð 2
=2ð
ð 22 â ð 1
2
ð 24 â ð 1
4
4
ðŒ =1
2ð(ð 2
2 + ð 12)
89
Module No. 147
M.I. of a Circular Disk - Derivation
Problem
To find the moment of inertia of a circular disk of radius ð, and mass ð about the axis of the disk.
Solution
Subdivide the disk into concentric rings one of which is the element (ring) shown in the fig.
Let the mass the of the ring is ðð, then the moment of inertia of the ring will be:
ðŒðððð = ð2ðð
The Surface area this element is
Area = (2ðð)ðð = 2ðððð
Since we have
ðð
ð=
2ðððð
ðð2
ðð =2ððð
ð2ð
Since the moment of Inertia of the ring is:
90
ðŒðððð = ð2ðð
or
ðŒðððð = ð22ððð
ð2ð =
2ðð3ðð
ð2
Thus the total moment of inertia of the circular disk will be
ðŒððð ð = â« ðŒðððð
ð
ð=0
ðŒððð ð = â«2ðð3ðð
ð2
ð
ð=0
=2ð
ð2â« ð3ðð
ð
ð=0
=2ð
ð2|ð4
4|0
ð
=2ð
ð2
ð4
4
ðŒððð ð =1
2ðð2
which is M.I of a disk.
91
Module No. 148
Rectangular Plate â Derivation
Problem Statement
Calculate the inertia matrix of a uniform rectangular plate with sides ð and ð about its side.
Solution
We consider a rectangular plate (lamina) of sides of length a and b. We consider an element of
length ðð¥ and ððŠ.
The mass of selected element will be
ðð = ððð¥ððŠ
Its moment of inertia about the y-axis is
ððð¥ððŠð¥2 = ðð¥2ðð¥ððŠ.
Thus the total moment of inertia is
ðŒ = â« â« ðð¥2ðð¥ððŠ
ð
ðŠ=0
ð
ð¥=0
= ðð â«ð¥2
ð
0
= ðð |ð¥3
3|0
ð
=1
3ððð3
Since the total mass of rectangular plate is
ð = ððð
the moment o inertia will be
ðŒ =1
3ðð2
92
Module No. 149
M.I. of Square Plate â Derivation
Problem Statement
Calculate the moment of inertia of a uniform square plate with sides ð about any axis through its
center and lying in the plane of plate.
Solution
Consider a uniform square plate with length of its side to be ð, we have to find out M.I. of this
plate about any axis, through its center.
We consider this axis as ðŠ â ðð¥ðð .
Let ð be the density of plate and the total area of square plate is ð2.
So density will be
ð =ð
ð2
We assume that the plate has been divided into vertical strips. Let us consider a strip from the
whole square as shown in figure.
The strips are chosen in this way because each point on a particular strip is approximately the
same distance from axis of rotation i.e. y-axis, the mass of the strip is ð¿ð and the width of each
strip is ð¿ð¥, then the area of the strip will be ðð¿ð¥, so
ð¿ð = ððð¿ð¥
Let the distance of the strip from y-axis is ð¥, then the moment of inertia of strip will be
ð¿ðŒ = ð¿ðð¥2 = ð¥2ððð¿ð¥
So the moment of inertia of square is
ðŒð ðð¢ððð = â ð¥2ððð¿ð¥
ððð ð ð¡ðððð
93
ðŒð ðð¢ððð = â« ð¥2ðððð¥
ð2â
âð2â
= ðð â« ð¥2ðð¥
ð2â
âð2â
= ðð |ð¥3
3|âð
2â
ð2â
=ðð
3|ð3
8â
(âð3)
8|
=ðð
3(ð3
4) =
ðð4
12
Now substitute ð =ð
ð2, we obtain
ðŒð ðð¢ððð =ðð2
12
94
Module No. 150
M.I. of Triangular Lamina â Derivation
Problem Statement
Find the moment of inertia of a uniform triangular lamina of mass ð about one of its sides.
Solution
Let ABC be the lamina. We will find its M.I. about one of its side BC. We choose the X-axis and
the origin as shown in figure.
At a distance x from BC we consider a strip of lamina of width dx. If the mass of the strip is dm
then its moment of inertia about BC is = ð¥2ðð.
To find dm we note that the triangles ABC and ADE are similar. Therefore the ratios of sides are
the same.
Hence
ð·ðž
ðµð¶=
âðððâð¡ ðð ðŽð·ðž
âðððâð¡ ðð ðŽðµð¶=
â â ð¥
â
which gives
ð·ðž =â â ð¥
âà ðµð¶ =
â â ð¥
âà ð
where h is the height of the triangle ABC at base BC and a is the length of BC. Now
ð¿ð = ð(ðððð) = ðð·ðžð¿ð¥
= ð (â â ð¥
â)ð¿ð¥ð
95
where ð is the density.
Therefore the moment of inertia of triangular lamina about the side BC is
ðŒ = â«ð¥2ðð =â«ð (â â ð¥
â) ð¥2ðð¥ð
â
0
=ðð
ââ«ð¥2(â â ð¥)ðð¥
â
0
=ðð
ââ«(âð¥2 â ð¥3)ðð¥
â
0
=ðð
â|â
ð¥3
3â
ð¥4
4|0
â
=ðð
â(â4
3â
â4
4) =
ððâ3
12
Now substitute density= ð =ððð ð
ðŽððð=
ð
1 2â ðâ
we obtain
ðŒ =1
6ðâ2
Hence the required expression for moment of inertia of the triangular lamina is 1
6ðâ2.
96
Module No. 151
M.I. of Elliptical Plate along its Major Axis â Derivation
Problem Statement
Find the M.I. of a uniform elliptical plate with semi major axes and semi minor axes a, b
respectively about its major axes.
Solution
We consider the elliptical plate as
ð¥2
ð2 +ðŠ2
ð2 = 1 (1)
with semi major axes along x-axis as shown in figure.
From (1) we have
ðŠ = ±ð
ðâð2 â ð¥2
So, let ðŠ1 =ð
ðâð2 â ð¥2,
We consider a small element of plate of mass ð¿ð of the elliptical plate will be ððð which will be
equal to ððð¥ððŠ. The moment of inertia if this element along x-axis will be equal to ðŒ = ð¿ððŠ2.
Then the moment of inertia of whole plate will be
ðŒð¥ = â«(ð¿ð)ðŠ2 = â« ððð
ðððð¡ð
ðŠ2
97
= ð â«
(
â« ðŠ2ððŠ
ððâð2âð¥2
âððâð2âð¥2
)
ðð¥
ð
âð
= ð â«2ðŠ1
2
3ðð¥
ð
âð
=2ð
3â«
ð3
ð3(ð2 â ð¥2)
32â ðð¥
ð
âð
due to symmetry, we can write it as
=4ðð3
3ð3â«(ð2 â ð¥2)
32â
ð
0
ðð¥
By making use of polar coordinates, substitute ð¥ = ð sin ð, then ðð¥ = ð cos ð this integral
becomes
ðŒð¥ =4ðð3
3ð3â« ð3 cos3 ð (ð cos ð
ð2â
0
)ðð
=4ððð3
3â« cos4 ð
ð2â
0
ðð
=4ððð3
3
1 Ã 3
2 Ã 4Ã
ð
2
=1
4ððð3ð
=1
4ðð3ð Ã
ð
ððð=
1
4ðð2
where for elliptical plate, we have
ð =ð
ððð
is the required expression for M.I.
98
Module No. 152
M.I. of a Solid Circular Cylinder - Derivation
Problem Statement
To find the moment of inertia of a solid circular cylinder of radius ð, mass ð and the height of
the cylinder â about the axis of the cylinder.
Solution
Letâs subdivide the solid circular cylinder into concentric cylindrical shells/ hollow cylinders,
one of which is shown in the fig.
Let the mass of one shell isðð, height is same as â, thickness be ðð and the radius be ð then the
density of the shell will be
ð =ðð
2ððððâ
or
ðð = 2ðððððâ
Since we have
99
ð =ð
ðððŠðððððð=
ð
ðð2â=
ðð
2ððððâ
or
ðð =2ð
ð2ððð
Since the moment of inertia of the shell will be:
ðŒð âððð = ð2ðð
or
ðŒð âððð = ð22ð
ð2ððð =
2ð
ð2ð3ðð
Thus the total moment of inertia of the solid circular cylinder will be
ðŒððŠðððððð = â« ðŒð âððð
ð
ð=0
ðŒððŠðððððð = â«2ð
ð2ð3ðð
ð
ð=0
=2ð
ð2â« ð3ðð
ð
ð=0
=2ð
ð2|ð4
4|0
ð
=2ð
ð2
ð4
4
ðŒððŠðððððð =1
2ðð2
is the M.I of solid cylinder.
100
Module No. 153
M.I. of Hollow Cylindrical Shell - Derivation
Problem
To find the moment of inertia of a hollow open cylindrical shell of radius ð , thickness ðð , mass
ð and height of shell â about the axis of shell.
Solution
Letâs subdivide the hollow shell into small hoops/ rings, one of which is shown in the figure.
Let the mass of one ring isðð, height is ðâ, thickness be ðð and the radius be ð , then the mass
of one ring will be
101
ðð = ð2ðð ðð ðâ
Hence the moment of inertia of the ring of radius ð will be
ðŒðððð = ððð 2
or
ðŒðððð = (ð2ðð ðð ðâ)ð 2 = 2ððð 3ðð ðâ
In order to obtain the moment of inertia for the whole hollow cylindrical open shell, we will
integrate
ðŒ = â« ðŒðððð = â«2ððð 3ðð ðâ
â
0
ðŒ = 2ððð 3ðð â«ðâ
â
0
= 2ððð 3ðð â
Since the density of the hollow cylindrical shell is
ð =ð
2ðð ðð â
Therefore
ðŒ = ðð 2
is the M.I of hollow cylindrical open shell.
102
Module No. 154
M.I of Solid Sphere - Derivation
Problem
Find the moment of inertia of a uniform solid sphere of radius ð and mass ðŽ about an axis (the
z-axis) passing through the center.
Solution
For a uniform solid sphere, due to symmetry, we have
ðŒð¥ð¥ = ðŒðŠðŠ = ðŒð§ð§
In order to calculate the moment of Inertia of the sphere, we split the sphere into thin circular
discs, one of which is shown in Figure.
We have already derived the expression for the moment of inertia of a representative disc of
radius ð¥, which is
ðŒððð ð =1
2ð¥2ðð
of an elementary disc of mass ðð and the radius ð¥.
As we know the mass = (density)(Area of disc)
therefore
103
ðð = ððð¥2
Hence moment of inertia of the sphere along z-axis will be
ðŒð§ð§ = â«1
2ððð¥4ðð§
ð
âð
Now, to write âð¥âin
terms of ð§, we make
a triangle as shown in fig,
where
ð2 = ð¥2 + ð§2, â¹ ð¥2 = ð2 â ð§2
ðŒð§ð§ = â«1
2ðð(ð2 â ð§2)2ðð§
ð
âð
=8
15ððð5
Since the mass of the sphere is
ð =4
3ðð3ð
Therefore
ðŒð§ð§ =2
5ðð2
Also
ðŒð¥ð¥ = ðŒðŠðŠ = ðŒð§ð§ =2
5ðð2
is the required moment of inertia of the sphere.
104
Module No. 155
M.I. of the Hollow Sphere â Derivation
Problem
A thin uniform hollow sphere has a radius ð¹ and mass ðŽ. Calculate its moment of inertia about
any axis through its center.
Solution
In order to calculate the moment of inertia of the hollow sphere, we split the hollow sphere into
thin hoops (rings), as shown in Figure.
We have already derived the expression for the moment of inertia of a representative hoop of
radius ð¥, which is
ðŒ = ððð¥2
of an elementary ring of mass ðð and the radius ð¥.
The volume of the elementary ring is
ðð = 2ðð¥ð ðððð
As we know the mass = (density)(volume)
105
ðð = ððð
therefore
ðð = ð2ðð¥ð ðð ðð
Hence the moment of inertia of the small ring of radius ð¥ will be
ðŒðððð = ððð¥2
or
ðŒðððð = (2ðð¥ðð ðð ðð)ð¥2 = 2ððð ðð ð¥3ðð
which is the moment of inertia of ring of radius ð¥ chosen from the hollow sphere.
In order to obtain the moment of inertia for the whole hollow sphere, we will integrate
ðŒ = â« ðŒðððð = â« 2ððð ðð ð¥3ðð
ð2â
âð2â
due to symmetry, we can write
ðŒ = 4ððð ðð â« ð¥3ðð
ð2â
0
To solve the integral, we need to write ð¥ in terms of ð. From fig we have
ð¥ = ð cos ð
The integral becomes,
ðŒ = 4ððð ðð â« (ð cos ð)3ðð
ð2â
0
ðŒ = 4ððð 4ðð â« cos3 ð ðð
ð2â
0
ðŒ = 4ððð 4ðð â« cos ð cos2 ð ðð
ð2â
0
106
ðŒ = 4ððð 4ðð â« cos ð (1 â sin2 ð)ðð
ð2â
0
ðŒ = 4ððð 4ðð â« cos ð (1 â sin2 ð)ðð
ð2â
0
ðŒ = 4ððð 4ðð (â« cos ð ðð
ð2â
0
â â« sin2 ð cos ð ðð)
ð2â
0
ðŒ = 4ððð 4ðð |sin ð âsin3 ð
3|0
ð2â
= 4ððð 4ðð (1 â1
3)
= 4ððð 4ðð Ã2
3
=8
3ððð 4ðð
Since the density of the hollow sphere is
ð =ð
ð=
ð
4ðð 2ðð
Hence M.I of hollow sphere will be
ðŒ =2
3ðð 2
107
Module No. 156
Inertia Matrix / Tensor of solid Cuboid
Problem Statement
Calculate the inertia matrix / inertia tensor of uniform solid cube at one of its corners.
Solution
Let the length of the edges be ð of each side and let the axes be chosenalong the edges as shown
in the figure.
By definition
ðŒ11 â¡ ðŒð¥ð¥ = â«ð(ð)(ðŠ2 + ð§2)ðð
Since the box is made of uniform material, the density ð must be constant. Therefore
ðŒð¥ð¥ = ð â« â«â«(ðŠ2 + ð§2)ðð¥ððŠðð§
ð
0
ð
0
ð
0
= ð â« â«(ðŠ2 + ð§2)ððŠðð§
ð
0
ð
0
â«ðð¥
ð
0
= ðð â«â«(ðŠ2 + ð§2)ððŠðð§
ð
0
ð
0
108
= ðð [â« â«ðŠ2ððŠðð§
ð
0
ð
0
+ â« â«ð§2ððŠðð§
ð
0
ð
0
]
= ð (ð4
3+
ð4
3)
ðŒð¥ð¥ = ðð (2ð4
3) = ð (
2ð5
3)
Since we know that for the cube
ð =ð
ð3
We obtain
ðŒð¥ð¥ =2ð
3ð2
Similarly, due to symmetry, we can write
ðŒðŠðŠ =2ð
3ð2
ðŒð§ð§ =2ð
3ð2
Now for the product of inertia, we have
ðŒ12 = ââ«ðð¥ðŠðð
= âð â«â« â«ð¥ðŠðð¥ððŠðð§
ð
0
ð
0
ð
0
= âð â«ð¥ðð¥ â« ðŠððŠ â« ðð§
ð
0
ð
0
ð
0
= âððð2
2
ð2
2= âð
ð5
4
Again using
ð =ð
ð3
109
We obtain
ðŒ12 = âðð2
4
Similarly,
ðŒ12 = âðð2
4
ðŒ12 = âðð2
4
The required inertia matrix / inertia tensor will be
ðŒðð =
[ 2ð
3ð2 â
ðð2
4â
ðð2
4
âðð2
4
2ð
3ð2 â
ðð2
4
âðð2
4â
ðð2
4
2ð
3ð2
]
110
Module No. 157
Inertia Matrix / Tensor of solid Cuboid
Problem Statement
Calculate the inertia matrix of uniform solid cuboid (parallelepiped) at one of its corner.
Solution
Let the length of the edges be ð, ð, ð and let the axes be chosen along the edges as shown in the
figure.
By definition
ðŒ11 â¡ ðŒð¥ð¥
= â«ð(ð)(ðŠ2 + ð§2)ðð
Since the box is made of uniform material, the density ð must be constant. Therefore
ðŒð¥ð¥ = ð â«â«â«(ðŠ2 + ð§2)ðð¥ððŠðð§
ð
0
ð
0
ð
0
= ð â«â«(ðŠ2 + ð§2)ððŠðð§
ð
0
ð
0
â«ðð¥
ð
0
111
= ðð â«â«(ðŠ2 + ð§2)ððŠðð§
ð
0
ð
0
= ðð [â«â«ðŠ2ððŠðð§
ð
0
ð
0
+ â«â«ð§2ððŠðð§
ð
0
ð
0
]
= ðð (ðð3
3+ ð
ð3
3)
ðŒð¥ð¥ = ðððð
3(ð2 + ð2)
using relation
ð =ð
ððð
ðŒð¥ð¥ =ð
3(ð2 + ð2)
Similarly, due to symmetry, we can write
ðŒðŠðŠ =ð
3(ð2 + ð2)
ðŒð§ð§ =ð
3(ð2 + ð2)
Now for the product of inertia, we have
ðŒ12 = ââ«ðð¥ðŠðð
= âð â«â«â«ð¥ðŠðð¥ððŠðð§
ð
0
ð
0
ð
0
= âð â«ð¥ðð¥ â«ðŠððŠ â«ðð§
ð
0
ð
0
ð
0
âððð2
2
ð2
2= âð
ð2ð2ð
4
Again using
112
ð =ð
ððð
we get
ðŒ12 = âððð
4
Similarly,
ðŒ23 = âððð
4
ðŒ31 = âððð
4
The required inertia matrix / inertia tensor will be
ðŒðð =
[ ð
3(ð2 + ð2) â
ððð
4â
ððð
4
âððð
4
ð
3(ð2 + ð2) â
ððð
4
âððð
4â
ððð
4
ð
3(ð2 + ð2)]
113
Module No. 158
M.I of Hemi-Sphere â Derivation
Problem Statement
Find the moment of inertia of a uniform hemisphere of radius ð about its axis of symmetry.
Solution
We will use the spherical polar coordinates (ð, ð, ð). Their use makes computational work
simpler.
Their range of variation for hemisphere will be
0 †ð < ð
0 †ð †ð2â
0 †ð †2ð
ð¥ = ð sin ð cosð ,
ðŠ = ð sin ð sinð , ð§ = ð cos ð
We choose the z-axis as the axis of symmetry.
Hence
114
ðŒð§ð§ = â«ð(ð)(ð¥2 + ðŠ2)ðð
= ð â«(ð¥2 + ðŠ2)ðð
Now calculate ð¥2 + ðŠ2 in terms of sp. coordinates
ð¥2 + ðŠ2 = ð2(sin2 ð cos2 ð + sin2 ð sin2 ð)
= ð2 sin2 ð (cos2 ð + sin2 ð)
= ð2 sin2 ð
and the element of volume in spherical polar coordinates is given by
ðð = ðð(ððð)(ð sin ððð) = ð2 sin ð ðððððð
Therefore
ðŒð§ð§ = ð â« â« â« ð4 sin3 ð
2ð
0
ð2â
0
ð
0
ðððððð
= â« ð4ðð
ð
0
â« sin3 ð ðð
ð2â
0
â« ðð
2ð
0
= 2ððð5
5â« sin3 ð ðð
ð2â
0
= 2ððð5
5â«
3 sin ð â sin 3ð
4ðð
ð2â
0
= (2ððð5
5)1
4|â3 cos ð +
cos 3ð
3|0
ð2â
= ððð5
10[(0) â (â3 + 1/3)]
= ððð5
10(8
3)
115
= 4ððð5
15
By using the relation for density
ð =ð
23â ðð3
we obtain
ðŒð§ð§ =2
5ðð2
which is required expression for moment of inertia of hemisphere.
116
Module No. 159
M.I. of Ellipsoid âDerivation
Problem
Find the moment of inertia and product of inertia for the ellipsoid
ð¥2
ð2+
ðŠ2
ð2+
ð§2
ð2= 1
w.r.to its axes of symmetry.
Solution
By definition
ðŒ11 â¡ ðŒð¥ð¥
= â« ð(ð)(ðŠ2 + ð§2)ðð
ðððððð ððð
If we put
ð¥
ð= ð¥â²,
ðŠ
ð= ðŠâ²,
ð§
ð= ð§â²
then
ð¥ = ðð¥â², ðŠ = ððŠâ², ð§ = ðð§â²
â¹ ðð¥ = ððð¥â², ððŠ = ðððŠâ², ðð§ = ððð§â²
and
ðð = ðð¥ððŠðð§ = ððððð¥â²ððŠâ²ðð§â²
Now under the above transformation, the given ellipsoid is transformed into the unit sphere S:
ð¥â²2 + ðŠâ²2 + ð§â²2 = 1
The integration is now over the region enclosed by the unit sphere.
117
ðŒ11 = ð â«(ð2ðŠâ²2 + ð2ð§â²2)ððððð¥â²ððŠâ²ðð§â²
ð
or
ðŒ11 = ððð3ð â« ðŠâ²2ððâ² + ðððð3 â« ð§â²2ððâ²
ð ð
where ððâ² = ðð¥â²ððŠâ²ðð§â²
Now because of symmetry
â« ðŠâ²2
ð
ððâ² = â« ð§â²2
ð
ððâ²
Now we solve one integral.
We use the spherical polar coordinates (ð, ð, ð).
Their range of variation will be
0 †ð < 1, 0 †ð †ð, 0 †ð †2ð
ð¥ = ð sin ð cosð , ðŠ = ð sin ð sin ð , ð§ = ð cos ð
and
ððâ² = ðð(ððð)(ð sin ððð) = ð2 sin ð ðððððð
Thus
â« ðŠâ²2
ð
ððâ² = â«â« â« ð4 sin3 ð
2ð
0
ð
0
1
0
sin2 ð ðððððð
= â«ð4ðð
1
0
â« sin3 ð ðð
ð
0
â« sin2 ð ðð
2ð
0
=1
5
1
4|â3 cos ð +
cos 3ð
3|0
ð 1
2|ð â
sin 2ð
2|0
2ð
Thus
118
â« ðŠâ²2
ð
ððâ² =2ð
15
Therefore on substitution
ðŒ11 = ððð3ð2ð
15+ ðððð3
2ð
15
ðŒ11 = ðððð2ð
15(ð2 + ð2)
=ð
(4 3â )ððððððð Ã
2ð
15(ð2 + ð2)
ðŒ11 =ð
10(ð2 + ð2)
Similarly,
ðŒ22 =ð
10(ð2 + ð2)
ðŒ33 =ð
10(ð2 + ð2)
For product of inertia
ðŒ12 â¡ ðŒð¥ðŠ = â â« ð¥ðŠðð
ðððððð ððð
= âð â« ðð¥â²ððŠâ²(ððððð¥â²ððŠâ²ðð§â²
ð
)
= âðð2ð2ð âð¥â²ðŠâ²ðð¥â²ððŠâ²ðð§â²
ð
Using the polar coordinates (ð, ð, ð)
ðŒ12 = âðð2ð2ð â« â« â«ð sin ð cosð(
1
0
ð
0
2ð
0
rsin ð sinð)ð2 sin ð ðððððð
= 0
119
Similarly, we can obtain
ðŒ23 = 0
ðŒ31 = 0
120
Module No. 160
Example 1 of Moment of Inertia
Problem
Particles of masses 2ð, 3ð and 4ð are held in a rigid light framework at points (0,1,1), (1,1,0)
and (â1, 0, 1) resp.
Show that the M. I. of the system about the ð¥, ðŠ, ð§ â axis are 11, 13, 12 respectively.
Solution
Since the masses are 2, 3 and 4 i.e. ð1 = 2, ð2 = 3 and ð3 = 4, and the points are given by
ð1(0,1,1), ð2(1,1,0) and ð3(â1,0,1).
We are required to find out the moment of inertia about x-axis, y-axis and z-axis i.e. ðŒð¥ð¥, ðŒðŠðŠ, ðŒð§ð§ .
ðŒð¥ð¥ = âðð(
ð
ðŠð2 + ð§ð
2)
= ð1(ðŠ12 + ð§1
2) + ð2(ðŠ22 + ð§2
2) + ð3(ðŠ32 + ð§3
2)
= 2(1 + 1) + 3(1 + 0) + 4(1 + 0)
= 4 + 3 + 4 = 11
So,
ðŒð¥ð¥ = 11
ðŒðŠðŠ = âðð(
ð
ð¥ð2 + ð§ð
2)
= ð1(ð¥12 + ð§1
2) + ð2(ð¥22 + ð§2
2) + ð3(ð¥32 + ð§3
2)
= 2(0 + 1) + 3(1 + 0) + 4(1 + 1)
= 2 + 3 + 8 = 1
So,
ðŒðŠðŠ = 13
121
ðŒð§ð§ = âðð(ð¥ð2 +
ð
ðŠð2)
= ð1(ðŠ12 + ð¥1
2) + ð2(ð¥22 + ðŠ2
2) + ð3(ð¥32 + ðŠ3
2)
= 2(0 + 1) + 3(1 + 1) + 4(1 + 0)
= 2 + 6 + 4 = 12
So,
ðŒð§ð§ = 12
Hence Showed
122
Module No. 161
Example 2 of Moment of Inertia
Problem Statement
A boy of mass ð = 30kg is running with a velocity of 3 m/sec on ground just tangentially
to a merry-go-round which is at rest. The boy suddenly jumps on the merry-go-round. Calculate
the angular velocity acquired by the system. The merry-go-round has a radius of ð = 2m and a
mass ð = 120kg and its moment of inertia is 120 kgmâ2.
Solution
The merry-go-round rotates about an axis which we regard as passing through its c.m.
Letâs give the following notations:
M.I of boy = ðŒ1
M.I of merry go round = ðŒ2
Vel. of boy = ð£1
Vel. of merry go round = ð£2
Angular vel. of boy = ð1
Angular vel. of merry go round = ð2
The moment of inertia ðŒ1 of the boy about the axis of rotation can be found by
ðŒ1 = ðð2 = 30 à 22 = 120ðððâ2
The moment of inertia ðŒ2 of merry go round is given to be
ðŒ2 = 120ðððâ2
Since the velocity of the boy about the merry-go-round is ð£1 = 3msâ1, therefore his angular
velocity ð1 about the axis of rotation is therefore
ð1 =ð£
ð=
3
2= 1.5 radian per second
Initially the merry go round is at rest, therefore ð£2 = 0, and thus its angular velocity ð2 = 0.
We ignore friction and therefore there is no external torque on the system.
Hence by the law of conservation of angular momentum
(ðŒ1 + ðŒ2)ð = ðŒ1ð1 + ðŒ2ð2
ð =ðŒ1ð1 + ðŒ2ð2
(ðŒ1 + ðŒ2)
123
where ð is the angular velocity of the system when the boy jumps on the merry-go-round.
ð =120(5) + 0
120 + 120
= 0.75 radian per sec
124
Module No. 162
Example 3 of Moment of Inertia
Problem Statement[1]
Two particles of masses ð1 and ð2 are connected by a rigid massless rod of length ð and moves
freely in a plane. Show that the M. I. of the system about an axis perpendicular to the plane and
passing through the center of mass is ðð2 where
ð =ð1ð2
ð1 + ð2
Solution
Let ð1 be the distance of mass ð1 from center of mass ð¶. Then ð â ð1 is the distance of the mass
ð2 from ð¶.Since ð¶ is the center of the mass.
So,
ð1ð1 = ð2(ð â ð1)
â¹ ð1ð1 = ð2ð â ð2ð1
â¹ (ð1 + ð2)ð1 = ð2ð
â¹ ð1 =ð2ð
(ð1 + ð2)
125
â¹ ð â ð1 = ð âð2ð
(ð1 + ð2)
â¹ ð â ð1 =ð1ð + ð2ð â ð2ð
(ð1 + ð2)
â¹ ð â ð1 =ð1ð
(ð1 + ð2)
Thus the M.I. about an axis through ð¶ is
ð1ð12 + ð2(ð â ð1
2)2
= ð1 (ð2ð
ð1 + ð2)2
+ ð2 (ð1ð
ð1 + ð2)2
=ð1ð2ð
2
(ð1 + ð2)2(ð1 + ð2) =
ð1ð2
ð1 + ð2ð2
ðŒ = ðð2
where
ð =ð1ð2
ð1 + ð2
Hence showed.
126
Module No. 163
Example 4 of Moment of Inertia
Problem (© engineering.unl.edu)
Calculate the moment of inertia of the shaded area given in figure about y-axis.
Solution
Here, we have
ðŠ = ð¥2/3 (1)
We consider an elementary strip from the shaded area whose M.I is
ðŒð ð¡ððð =1
3ð¥2(ð¥ððŠ) =
1
3ð¥3ððŠ
127
Then by integration, we obtain the moment of inertia of the whole shaded area
ðŒðŠ = â3â«ðŠ3
2â ððŠ
4
1
= â32
5|ðŠ
52â |
1
4
= â32
5[(4)
52â â (1)
52â ]
=2â3
5[(4)
52â â (1)
52â ]
=2â3
5(32 â 1)
= 21.5 inch4
128
By using (1), we get
ðŒð ð¡ððð = â3ðŠ3
2â ððŠ
129
Module No. 164
Example 5 of Moment of Inertia
Problem
Determine the moment of inertia of the shaded area shown in figure with respect to each of the
coordinate axes.
Solution
Here we have
ðŠ = ðð¥2 (1)
From fig. we have
ð¥ = ð, ðŠ = ð, then
ð = ðð2 â¹ ð =ð
ð2
Substituting the value
of k in (1), we obtain
ðŠ =ð
ð2ð¥2 or ð¥ =
ð
âðâðŠ
Now the moment of inertia along x-axis will be
130
ðŒð¥ = â« ðŠ2ððŽ = â«ðŠ2(ð â ð¥)ððŠ
ð
0ðŽ
= â«ðŠ2 (ð âð
âðâðŠ)ððŠ
ð
0
= ð â«ðŠ2ððŠ âð
âðâ«ðŠ
52â ððŠ
ð
0
ð
0
= ð |ðŠ3
3|0
ð
âð
âð|2
7ðŠ
72â |
0
ð
=ðð3
3â
ð
âð
2
7ð
72â
=ðð3
3â
2ðð3
7
ðŒð¥ =ðð3
21
ðŒðŠ = â« ð¥2ððŽ = â« ð¥2ðŠðð¥
ð
0ðŽ
=ð
ð2|ð¥5
5|0
ð
=ð
ð2
ð5
5
ðŒðŠ =ð3ð
5
Hence
ðŒð¥ =ðð3
21
and
ðŒðŠ =ð3ð
5
are moment of inertia about x-axis and y-axis respectively.
131
Module No. 165
Example 5 of Moment of Inertia
Theorem Statement
The moment of inertia of a rigid body about a given axis is equal to the same about a parallel
axis through the centroid plus the moment of inertia due to the total mass placed at the centroid,
(the last quantity will be referred to as moment of inertia of the centroid).
Proof
By definition, if ðŒ denotes the M.I about the given axis, and ðð denotes the distance of ðð¡â particle
from the given axis, then
ðŒ = âðððð2
ð
If ð is the unit vector in the direction of the axis, then we have the following result
ðð2 = (ð Ã ðð)
2
Let ðð denotes the position vector of the centroid and
ððâ² the position vector of the ðð¡â particle w.r.t to the centroid, then
132
ðð = ðð + ððâ²
On making substitutions, we obtain
ðŒ = âðð [ð à (ðð + ððâ²)]
2
ð
= âðð [ð Ã ðð + ð Ã ððâ²]
2
ð
= âðð
ð
[(ð Ã ðð )2 + (ð Ã ðð
â²)2 + 2(ð Ã ðð ) . (ð Ã ððâ²)]
= âðð(ð Ã ðð )2 + âðð
ð
(ð Ã ððâ²)2
ð
+2(ð Ã ðð ) .âðð(ð Ã ððâ²)
ð
= â ðððð2 + âðð
ð
ððâ²2 + 2(ð Ã ðð ) . ð Ã âðððð
â²
ðð
But â ððððâ²
ð = 0 (We studied in earlier module).
Therefore
ðŒ = (âðð)ðð2 + âðð
ð
ððâ²2
ð
= ððð2 + ðŒâ²
where
ð = âðð
ð
is the total mass of the system.
or
ðŒ = ðŒ0 + ðŒâ²
where ðŒ0 denotes the moment of inertia of the centroid, and ðŒâ² denotes the M.I of the system
w.r.to a parallel axis through the centroid.
133
Module No. 166
Example 1 of Parallel Axis Theorem
Problem
Use the parallel axis theorem to find the moment of inertia of a solid circular cylinder about a
line on the surface of the cylinder and parallel to axis of cylinder.
Solution
Suppose the cross section of cylinder as in figure. Then the axis of the cylinder is passing
through the point ð¶, while the line on the surface of cylinder is passing through ðŽ. So, we have to
find out M.I of circular cylinder about a line passing through the point ðŽ whose radius is ð
(radius of circular cylinder) and mass is ð.
By parallel axis theorem
ðŒðŽ = ðŒð + ðð2 (1)
Since ðŒð¶ which is the moment of inertia of a solid circular cylinder about an axis passing from
the center of mass is defined by
ðŒð¶ = 1
2 ðð2 (2)
where ð is the radius of a solid circular cylinder.
By substituting equation (2) in equation (1) we have
ðŒðŽ = 1
2 ðð2 + ðð2
134
â¹
ðŒðŽ = (1
2+ 1) ðð2
ðŒðŽ = 3
2ðð2
135
Module No. 167
Example 2 of Parallel Axis Theorem
Problem
Prove that the moment of inertia of a uniform right circular cone using parallel axis theorem of
mass ð, height â and semi vertical angle ðŒ about a diameter of its base is
ðâ2(3 tan2 ðŒ + 2) 20â
Solution
In the case of M.I about its diameter, we consider the elementary disc of mass ð¿ð whose
moment of inertia about a diameter will be
ð¿ðŒ0 =1
4ð2ð¿ð
We note that the diameter passes through the center (which is also the centroid) of the
elementary disc.
Hence by parallel axis theorem, the M.I. ð¿ðŒ of the elementary disc about a parallel axis (parallel
diameter) at the base is given by
136
ð¿ðŒ = ð¿ðŒ0 + (ð¿ð)ð§2
=1
4ð2ð¿ð + ð¿ðð§2 = ð¿ð(
1
4ð2 + ð§2)
= ððð2ð¿ð§(1
4ð2 + ð§2) = ðð (
1
4ð4 + ð2ð§2) ð¿ð§
From the similar triangles,
we have
ð
ð=
â â ð§
â or ð = ð
â â ð§
â
Therefore = ðð [ð4
4â4 (â â ð§)4 +ð2
â2 (â â ð§)2ð§2] ð¿ð§
= ðð [ð4
4â4(â â ð§)4 +
ð2
â2(â2ð§2 â 2âð§3 + ð§4)] ð¿ð§
Therefore M.I of complete right circular cone about a diameter is given by
ðŒ = ðð â«{ð4
4â4(â â ð§)4 +
ð2
â2(â2ð§2 â 2âð§3 + ð§4)}
â
0
ð¿ð§
ðŒ = ðð (ð4
4â4
â5
5+
ð2
â2
â5
30)
Since we know that ð =ð
(1 3â )ðð2â
ðŒ =ð
20(3ð2 + 2â2)
137
Since the semi vertical angle of
the right circular cone is ðŒ,
So by right triangle AOB, we have
tanðŒ =ðŽð
ððµ=
ð
â
ð = â tanðŒ
Therefore
ðŒ =ð
20[3(htanðŒ)2 + 2â2]
or
ðŒ =ðâ2
20[3 tan2 ðŒ + 2]
Hence proved.
138
Module No. 168
Example 3 of Parallel Axis Theorem
Problem[1]
Find the moment of inertia of a uniform circular cylinder of length ð and radius ð about an axis
through the center and perpendicular to the central axis, namely ðŒð¥ or ðŒðŠ.
Solution
Consider the elementary disc of mass ðð and thickness ðð§ located at a distance ð§ from the ð¥ðŠ
plane. Then the moment of inertia about a diameter will be
ð¿ðŒ0 =1
4ð2ðð
Then, using the parallel-axis theorem, moment of inertia of the thin disc about the ð¥ â ðð¥ðð will
be
ððŒð¥ =1
4ð2ðð + ð§2ðð
where ðð = ððð2ðð§.
Thus
139
ðŒð¥ = ððð2 â« (1
4ð2 + ð§2)
ð2â
âð2â
ðð§
= ððð2 (1
4ð2ð +
1
12ð3)
but the mass of the cylinder ð is ððð2ð,
therefore
ðŒð¥ = ð (1
4ð2 +
1
12ð2)
Due to symmetry we have
ðŒð¥ = ðŒðŠ = ð (1
4ð2 +
1
12ð2)
140
Module No. 169
Example 4 of Parallel Axis Theorem
Problem
Calculate the moment of inertia ðŒðð for a uniform rod of length ð and mass ð rotating about an
axis through the center, perpendicular to the rod.
Solution
In order to calculate the moment of inertia through the center of mass c.m., we use parallel axes
theorem.
In a transparent notation
ðŒð = ðŒðð + ðð2 (1)
where ð is the distance between the origin and the center of mass and ð = ð2â .
Also ðŒð is the moment of inertia of rod about one of its end (which we calculated earlier).
Here
ðŒð =1
3ðð2 (2)
From (1), we have
ðŒðð = ðŒð â ðð2
Substituting value from (2) in (1)
ðŒðð =1
3ðð2 â ð (
ð
2)
2
=1
3ðð2 â
1
4ðð2
ðŒðð =1
12ðð2
which is required moment of inertia about its center of mass.
141
Module No. 170
Perpendicular Axis Theorem
Theorem Statement
The moment of inertia of a plane rigid body about an axis perpendicular to the body is equal to
the sum of the moment of inertia about two mutually perpendicular axis lying in the plane of the
body and meeting at the common point with the given axis.
Proof
We choose the coordinates such that ðð axes lie in the plane of the rigid body and the ð-axis is
perpendicular to it.
Then the theorem can be stated as
ðŒ33 = ðŒ11 + ðŒ22
ðr
ðŒð§ð§ = ðŒð¥ð¥ + ðŒðŠðŠ
where ðŒ11 = ðŒð¥ð¥ etc. are M.I about the ð¥-axis etc.
Since the ð§-axis has been
chosen to be perpendicular
to the laminar body, therefore
ðŒ33 = ðŒð§ð§ = â ðððð2
ð
142
where ðð is the distance of the ðð¡â particle (lying in the ð¥ðŠ-plane) from the ð§-axis.
If we denote the coordinate of this particle by (ð¥ð, ðŠð), then
ðð2 = ð¥ð
2 + ðŠð2
and therefore
ðŒ33 = ðŒð§ð§ = âðð(
ð
ð¥ð2 + ðŠð
2)
= âððð¥ð2
ð
+ âðððŠð2
ð
= ðŒð¥ð¥ + ðŒðŠðŠ
Hence the proof.
143
Module No. 171
Example 1 of Perpendicular Axis Theorem
Problem Statement
Find the moment of inertia of a uniform circular plate (or disc) about any diameter.
Proof
Since we have deduced that moment of inertia of a uniform circular disc is
ðŒððð ð = 1
2 Mð2
which is about a line passing through center and perpendicular to the plane.
Considering the same axis with same terms in 3-dim body (i.e. ð§ â ðð¥ðð passing through center
of mass and perpendicular to ð¥ðŠ â ððððð), we have using perpendicular axis theorem
ðŒð§ð§ = ðŒð¥ð¥ + ðŒðŠðŠ (1)
We have to find out M.I. of uniform circular plate (disc) about any of its diameters which are
along ð¥ â ðð¥ðð and ðŠ â ðð¥ðð (i.e. we have to find out ðŒð¥ð¥ ðð ðŒðŠðŠ both of them are equal to ðŒð)
Since
ðŒð¥ð¥ = ðŒð = ðŒðŠðŠ
So equation (1) becomes
ðŒð§ð§ = ðŒð + ðŒð
144
or
ðŒð§ð§ = 2ðŒð
ðŒð =1
2ðŒð§ð§
ðŒð = 1
2(1
2 Mð2)
(⎠ðŒð§ð§ = ðŒððð ð = 1
2 Mð2)
ðŒð =1
4ðð2
where ð is the mass of disc and a is its radius.
145
Module No. 172
Example 2 of Perpendicular Axis Theorem
Problem Statement
Find the M.I. of a uniform elliptical lamina with semi major axes and semi minor axes ð, ð
respectively about respective axes (x, y, z-axis) using perpendicular axis theorem.
Proof
We consider the elliptical plate as
ð¥2
ð2+
ðŠ2
ð2= 1 (1)
with semi major axes along x-axis as shown in figure.
From (1) we have
ðŠ = ±ð
ðâð2 â ð¥2
let ðŠ1 =ð
ðâð2 â ð¥2
We consider a small element
of mass ð¿ð of the elliptical
plate, then we will have
ð¿ð = ððð = ððð¥ððŠ
146
The moment of inertia of this element along x-axis will be equal to ðŒ = ð¿ððŠ2.
Then the moment of inertia of whole plate will be
ðŒð¥ð¥ = â«(ð¿ð)ðŠ2 = â« ððð ðððð¡ð
ðŠ2
= ð â« ( â« ðŠ2ððŠ
ðŠ1
âðŠ1
)ðð¥
ð
âð
= ð â«2ðŠ1
3
3ðð¥
ð
âð
=2ð
3â«
ð3
ð3(ð2 â ð¥2)
32â ðð¥
ð
âð
due to symmetry, we can write it as
=4ðð3
3ð3â«(ð2 â ð¥2)
32â
ð
0
ðð¥
By making use of polar coordinates,
substitute ð¥ = ð sin ð, then ðð¥ = ð cos ð this integral becomes
ðŒð¥ð¥ =4ðð3
3ð3â« ð3 cos3 ð (ð cos ð
ð2â
0
)ðð
=4ððð3
3â« cos4 ð
ð2â
0
ðð
=1
4ððð3ð
for elliptical plate, we have
ð =ð
ððð
Thus
ðŒð¥ð¥ =1
4ðð3ð Ã
ð
ððð
147
ðŒð¥ð¥ =1
4ðð2
Similarly, about y-axis
ðŒðŠðŠ = â«(ð¿ð)ð¥2 = â« ððð
ðððð¡ð
ð¥2
= ð â«( â« ð¥2ðð¥
ð¥1
âð¥1
)ððŠ
ð
âð
= ð â«2ð¥1
2
3ððŠ
ð
âð
=2ð
3â«
ð3
ð3(ð2 â ðŠ2)
32â ððŠ
ð
âð
due to symmetry, we can write it as
=4ðð3
3ð3â«(ð2 â ðŠ2)
32â
ð
0
ððŠ
Using polar coordinates, substitute ðŠ = ð sin ð, then ððŠ = ð cos ð ðð this integral becomes
ðŒðŠðŠ =4ðð3
3ð3â« ð3 cos3 ð (ð cos ð
ð2â
0
)ðð
=4ðð3ð
3â« cos4 ð
ð2â
0
ðð
=4ðð3ð
3
3
8Ã
ð
2
=1
4ðð3ðð
=1
4ð3ðð Ã
ð
ððð
ðŒðŠðŠ =1
4ðð2
now using perpendicular axis theorem, we obtain
148
ðŒð§ð§ = ðŒð¥ð¥ + ðŒðŠðŠ
=1
4ðð2 +
1
4ðð2
ðŒð§ð§ =1
4ð(ð2 + ð2)
which is M.I. of elliptical lamina along z-axis
149
Module No. 173
Example 3 of Perpendicular Axis Theorem
Problem
Find the moment of inertia of a rectangular plate with sides ð and ð about an axis perpendicular
to the plate and passing through a vertex using perpendicular axis theorem.
Solution
We consider a rectangular plate (lamina) of sides of length ð and ð. We consider an element of
length ðð¥ and ððŠ as shown in figure.
We find the M.I about ðŠ â ðð¥ðð .
The mass of selected element will be
ðð = ððð¥ððŠ
Itâs moment of inertia about the ðŠ â ðð¥ðð is
ðŒððððððð¡ = ððð¥ððŠð¥2 = ðð¥2ðð¥ððŠ.
where ð¥ is the perpendicular distance from the element to the ðŠ â ðð¥ðð .
150
Thus the total moment of inertia about y-axis is
ðŒðŠ = â« â« ðð¥2ðð¥ððŠ
ð
ðŠ=0
ð
ð¥=0
= ðð â«ð¥2
ð
0
= ðð |ð¥3
3|0
ð
=1
3ððð3
Since the density of the rectangular plate is
ð =ð
ðð
the moment of inertia will be
ðŒðŠ =1
3ðð2
In the similar manner, we will calculate the moment of inertia about x-axis
The total moment of inertia about x-axis is
ðŒð¥ = â« â« ððŠ2ðð¥ððŠ
ð
ðŠ=0
ð
ð¥=0
= ðð â«ðŠ2ððŠ
ð
0
= ðð |ðŠ3
3|0
ð
=1
3ððð3
Using the relation of the total mass of rectangular plate
ð = ððð
Then the moment of inertia will be
ðŒð¥ =1
3ðð2
Thus by using perpendicular axes theorem, we obtain the moment of inertia along z-axis
ðŒð§ = ðŒð¥ + ðŒðŠ
ðŒð§ =1
3ðð2 +
1
3ðð2
151
ðŒð§ =1
3ð(ð2 + ð2)
which are the required moments of inertia of rectangle along ð¥, ðŠ, ð§-axis.
152
Module No. 174
Problem of Moment of Inertia
Problem
Find the moment of inertia about the line of an apparatus (as shown in figure) consisting of a sphere
of mass ð and radius ð attached to a rod of length 2ð & mass ð.
Solution
Since from the figure it is clear that ð is the radius of the sphere whose mass is ð and is attached
to the rod of length 2ð whose mass is ð.
Therefore
ðŒð¿ = ðŒ1ð¿ + ðŒ2ð¿
where ðŒ1ð¿ is the M.I. of a rod of length 2ð about ð¿ and ðŒ2ð¿ is the M.I. of a sphere of radius ð
about the line ð¿, then
153
ðŒ1ð¿ = â« ðð¥2ðð¥
2ð
0
= ð |ð¥3
3|0
2ð
= ð8
3ð3
using relation for ð, we get
ðŒ1ð¿ =4
3ðð2 (2)
Now
ðŒ2ð¿ = ðŒâ² + ðð2
where ðŒâ² is the MI of sphere about its diameter
ðŒ2ð¿ =2
5ðð2 + ð(2ð + ð)2 (3)
Substituting equation (2) & (3) in (1), we get
ðŒð¿ =4
3ðð2 +
2
5ðð2 + ð(2ð + ð)2
which is the required M.I. of the apparatus.
154
Module No. 175
Existence of Principle Axes Introduction
The axes relative to which product of inertia are zero are called the principal axes and the
moment of inertia along these axes are called principal moment of inertia.
Theorem Statement
For a rigid body, there exist a set of three mutually orthogonal axes called principal axes
relative to which the product of inertia are zero and ᅵᅵ and ð¿ are considered along the
same direction.
Proof
We assume that there exists an axis through a point ð of the rigid body such that angular velocity
ᅵᅵ and angular momentum ð¿ of the body are parallel to this axis. Then we can write
ð¿ = ðŒï¿œï¿œ (1)
where ðŒ is a constant of proportionality.
Form equation (1), we have
ð¿1 = ðŒð1, ð¿2 = ðŒð2, ð¿3 = ðŒð3
Also from the general theory of angular momentum
ð¿ð = â ðŒððð ðð (2)
From equation (1) and (2), we have
ð¿1 = ðŒ11ð1 + ðŒ12ð2 + ðŒ13ð3 = ðŒð1
ð¿2 = ðŒ21ð1 + ðŒ22ð2 + ðŒ23ð3 = ðŒð2
ð¿3 = ðŒ31ð1 + ðŒ32ð2 + ðŒ33ð3 = ðŒð3
which can also be written as
(ðŒ11âðŒ)ð1 + ðŒ12ð2 + ðŒ13ð3 = 0
ðŒ21ð1 + (ðŒ22âðŒ)ð2 + ðŒ23ð3 = 0
ðŒ31ð1 + ðŒ32ð2 + (ðŒ33 â ðŒ)ð3 = 0 (3)
155
Equation (3) is a system of three homogeneous linear algebraic equation in the unknown
ð1, ð2, ð3.
This system will have a non-trivial solution, i.e. (ᅵᅵ â 0), if the determinant of the matrix of
coefficients is zero. i.e.
|ðŒ11 â ðŒ ðŒ12 ðŒ13
ðŒ21 ðŒ22 â ðŒ ðŒ23
ðŒ31 ðŒ32 ðŒ33 â ðŒ| = 0 (4)
This is a cubic equation in ðŒ and will in general have three roots. Equation (4) is called
characteristic equation of the matrix (ðŒðð).
The roots of equation (4) are called eigen values of the inertia matrix (ðŒðð).
The problem of finding principal moment of inertia and directions of inertia has been reduced to
that of finding the eigenvalues and eigenvectors of a symmetric 3 Ã 3 matrix.
The following results from the eigenvalue theory of matrices will deduce further results about the
principal moments and directions (axes) of inertia.
Related Theorems
Theorem 1
A 3 Ã 3 symmetrical matrix has three real eigenvalues, which may be distinct or repeated.
Theorem 2
The eigenvectors of a symmetrical matrix corresponding to distinct eigenvalues are orthogonal.
Theorem 3
It is always possible to find three mutually orthogonal eigenvectors for a 3 Ã 3 symmetric
matrix, whether the eigenvalues are distinct or repeated.
Results
In the light of above theorems, we deduce the following results about the inertia matrix ðŒððð¡ at
point ð of a rigid body.
i. The principal of moment of inertia are always real numbers. This is obviously physical
because the moment of inertia is defined as the quantity â ðððð2 ð where ðð and ðð are
both real.
156
ii. When all three principal moments ðŒ1, ðŒ2, ðŒ3 are distinct, then by the theorem 2, three
mutually orthogonal principal axes can be found.
iii. When ðŒ1 = ðŒ2 but ðŒ1 â ðŒ3, then by the theorem 3, we can still determine three mutually
orthogonal principal axes, because the inertia matrix is symmetric.
157
Module No. 176
Determination of Principal Axes of Other Two When One is known
In many instances a body possesses sufficient symmetry so that at least one principal axis can be
found by inspection, i.e;the axis can be chosen so as to make two of the three products of inertia
vanish.
We consider a plane rigid body i.e. a 2D body; for example a plate of uniform thickness. Such a
system can be regarded as a coplanar distribution of mass.
Since there are three mutually orthogonal principal axes, one of them must be perpendicular to
the plane of the body.
The other two axes will lie in the plane of lamina.
We choose ð and ð axes in the plane of lamina, and the ð axis perpendicular to its plane, as
shown in figure.
ðŒð¥ð§ = ðŒðŠð§ = 0 whereas ðŒð¥ðŠ â 0 (1)
Since we have obtained the following equations for principal axes.
(ðŒ11 â ðŒ)ð1 + ðŒ12ð2 + ðŒ13ð3 = 0
ðŒ21ð1 + (ðŒ22 â ðŒ)ð2 + ðŒ23ð3 = 0
ðŒ31ð1 + ðŒ32ð2 + (ðŒ33 â ðŒ)ð3 = 0 (2)
By making use of (1), and using the first two equations of (2), we obtain
(ðŒ11 â ðŒ)ð1 + ðŒ12ð2 = 0
ðŒ21ð1 + (ðŒ22 â ðŒ)ð2 = 0
Since the principal axes perpendicular to the plane of lamina is supposed to be known, and we
are interested in determining the principal axes in XY-plane, therefore we didnât consider third
eq. of (2).
Let ðð1, ðð2 denote the principal axes in the XY-plane and let ð denote the angle between the
ðð1 and the ð â ðð¥ðð .
158
We define
tanð = ð2 ð1â
which can also be written as
ð1
cosð=
ð2
sin ð= ð, (3)
where ð is any arbitrary constant.
Substituting for ð1, ð2 from equation (3) in (2) we have, after simplification
(ðŒ11 â ðŒ) cosð + ðŒ12 sinð = 0
ðŒ21 cosð + (ðŒ22 â ðŒ) sinð = 0
These equations can be put in the form
ðŒ11 â ðŒ = âðŒ12
sinð
cosð, ðŒ22 â ðŒ = âðŒ12
cosð
sinð
159
Module No. 177
Determination of Principal Axes by Diagonalizing the Inertia Matrix
Introduction
Suppose a rigid body has no axis of symmetry. Even so, the tensor that represents the moment of
inertia of such a body, is characterized by a real, symmetric 3 Ã 3 matrix that can be
diagonalized. The resulting diagonal elements are the values of the principal moments of inertia
of the rigid body.
The axes of the coordinate system, in which this matrix is diagonal, are the principal axes of the
body, because all products of inertia have vanished.
Thus, finding the principal axes and corresponding moments of inertia of any rigid body,
symmetric or not, is virtually the same as to diagonalzing its moment of inertia matrix.
Explanation
There are a number of ways to diagonalize a real, symmetric matrix. We present here a way that
is quite standard.
First, suppose that we have found the coordinate system (principal axes) in which all products of
inertia vanish and the resulting moment of inertia tensor is now represented by a diagonal matrix
whose diagonal elements are the principal moments of inertia.
Let ðð be the unit vectors that represent this coordinate system, that is, they point along the
direction along the three principal axes of the rigid body.
If the moment of inertia tensor is "dotted" with one of these unit vectors, the result is equivalent
to a simple multiplication of the unit vector by a scalar quantity, i.e.
ðŒðð = ððð (1)
The quantities ðð are just the principal M.I about their respective principal axes.The problem of
finding the principal axes is one of finding those vectors ðð that satisfy the condition
(ðŒ â ððŒ)ðð = 0 (2)
In general this condition is not satisfied for any arbitrary set of orthonormal unit vectors ðð. It is
satisfied only by a set of unit vectors aligned with the principal axes of the rigid body.
160
Any arbitrary ð¥ðŠð§ coordinate system can always be rotated such that the coordinate axes line up
with the principal axes. The unit vectors specifying these coordinate axes then satisfy the
condition in eq (2). This condition is equivalent to vanishing of the following determinant
|ðŒ â ððŒ| = 0 (3)
Explicitly, this equation reads
|ðŒ11 â ðŒ ðŒ12 ðŒ13
ðŒ21 ðŒ22 â ðŒ ðŒ23
ðŒ31 ðŒ32 ðŒ33 â ðŒ| = 0
It is a cubic in ð, namely,
â ð3 + ðŽð2 + ðµð + ð¶ = 0 (4)
in which ðŽ, ðµ, and ð¶ are functions of the ðŒ's. The three roots ð1, ð2 and ð3 are the three
principal moments of inertia.
We now have the principal moments of inertia, but the task of specifying the components of the
unit vectors representing the principal axes in terms of our initial coordinate system remains to
be solved.
Here we can make use of the fact that when the rigid body rotates about one of its principal axes;
the angular momentum vector is in the same direction as the angular velocity vector.
Let the angles of one of the principal axes relative to the initial ð¥ðŠð§ coordinate system be
ðŒ, ðœ and ðŸ and let the body rotate about this axis. Therefore, a unit vector pointing in the
direction of this principal axis has components (cos ðŒ, cos ðœ, cos ðŸ).
Using eq (1),
ðŒð1 = ð1ð1
where ð1, the first principal moment of the three (ð1, ð2, ð3), is obtained by solving eq (4).
In matrix form
[
ðŒ11 â ð1 ðŒ12 ðŒ13
ðŒ21 ðŒ22 â ð1 ðŒ23
ðŒ31 ðŒ32 ðŒ33 â ð1
] [
cos ðŒcos ðœcos ðŸ
] = 0
The direction cosines may be found by solving the above equations.
The solutions are not independent. They are subject to the constraint
161
cos2 ðŒ + cos2 ðœ + cos2 ðŸ = 1
In other words the resultant vector ð1 specified by these components is a unit vector.
162
Module No. 178
Relation of Fixed and Rotating Frames of Reference
In order to derive the relationship between fixed and rotating frames of reference, we will study
the following theorem[1,2].
Rotating Axes Theorem
Theorem Statement
If a time dependent vector function ðŽ is represented by ðŽð and ðŽð in fixed and rotating coordinate
system, then
(ððŽ
ðð¡)
ð
= (ððŽ
ðð¡)
ð
+ ð à ðŽð
where it is understood that the origins of the two systems coincide at ð¡ = 0
Proof
We denote the fixed and rotating coordinate systems by ðð0ð0ð0 and ðððð and denote the
associated unit vectors by {ð0, ð0, ð0} and {ð, ð, ð}.
Consider a vector ðŽ which is changing with time. To an observer fixed relative to ðððð system,
the time rate of change of ðŽ = ðŽ1ð + ðŽ2ð + ðŽ3ᅵᅵ will be
ð
ðð¡ðŽð =
ð
ðð¡ðŽ1ð +
ð
ðð¡ðŽ2ð +
ð
ðð¡ðŽ3ᅵᅵ
where ððŽð
ðð¡ denotes the time derivative of ðŽ relative to the rotating frame of reference.
However, the time rate of change of ðŽ relative to the fixed system ðð0ð0ð0 symbolized by the ððŽð
ðð¡ needs to be found.
To the fixed observer the unit vectors ð, ð, ð actually change with time.
Thus
163
ð
ðð¡ðŽð =
ð
ðð¡(ðŽ1ð + ðŽ2ð + ðŽ3ᅵᅵ)
=ððŽ1
ðð¡ð +
ððŽ2
ðð¡ð +
ððŽ3
ðð¡ï¿œï¿œ + ðŽ1
ðð
ðð¡+ ðŽ2
ðð
ðð¡+ ðŽ3
ðᅵᅵ
ðð¡
ððŽð
ðð¡=
ððŽð
ðð¡+ ðŽ1
ðð
ðð¡+ ðŽ2
ðð
ðð¡+ ðŽ3
ðᅵᅵ
ðð¡ (1)
Since ð is a unit vector, ðð/ðð¡ is perpendicular to ð. Then ðᅵᅵ
ðð¡ must lie in the plane of ð and ᅵᅵ.
Therefore
ðð
ðð¡= ðŒ1ð + ðŒ2ᅵᅵ (2)
Similarly,
ðð
ðð¡= ðŒ3ᅵᅵ + ðŒ4ð (3)
ðᅵᅵ
ðð¡= ðŒ5ð + ðŒ6ð (4)
Form ð. ð = 0, differentiation yields
ð.ðð
ðð¡+ ð.
ðð
ðð¡= 0 â¹ ð.
ðð
ðð¡= âð.
ðð
ðð¡
But from (2), we have
ð.ðð
ðð¡= ðŒ1 and ð.
ðð
ðð¡= ðŒ4
â¹ ðŒ4 = âðŒ1
Similarly form ð. ᅵᅵ = 0 we obtain
ð.ðᅵᅵ
ðð¡+ ᅵᅵ.
ðð
ðð¡= 0 â¹ ð.
ðᅵᅵ
ðð¡= âᅵᅵ.
ðð
ðð¡
From (3), we have
ᅵᅵ.ðð
ðð¡= ðŒ2 and ᅵᅵ.
ðð
ðð¡= ðŒ5
â¹ ðŒ5 = âðŒ2
164
and from ð. ᅵᅵ = 0 we obtain
ð.ðᅵᅵ
ðð¡+ ᅵᅵ.
ðð
ðð¡= 0 â¹ ð.
ðᅵᅵ
ðð¡= âᅵᅵ.
ðð
ðð¡
From (4), we have
ᅵᅵ.ðᅵᅵ
ðð¡= ðŒ2 and ð.
ðᅵᅵ
ðð¡= ðŒ5
â¹ ðŒ6 = âðŒ3
Then
ðð
ðð¡= ðŒ1ð + ðŒ2ᅵᅵ
ðð
ðð¡= ðŒ3ᅵᅵ â ðŒ1ð
ðᅵᅵ
ðð¡= âðŒ2ð â ðŒ3ð
follows that
ðŽ1
ðð
ðð¡+ ðŽ2
ðð
ðð¡+ ðŽ3
ðᅵᅵ
ðð¡= (âðŒ1ðŽ2 â ðŒ2ðŽ3)ð + (âðŒ1ðŽ1 â ðŒ3ðŽ3)ð + (âðŒ2ðŽ1 â ðŒ3ðŽ2)ᅵᅵ
where
ᅵᅵ = ð1ð + ð2ð + ð3ᅵᅵ
The vector quantity ᅵᅵ is the angular velocity of the moving system relative to the fixed system.
Thus from (1) and (5), we obtain
(ððŽ
ðð¡)
ð
= (ððŽ
ðð¡)
ð
+ ð à ðŽ
165
Module No. 179
Equation of Motion in Rotating Frame of Reference
There are two cases to be discussed in this article. Frist is when the origins of fixed and rotating
coordinate system coincide and other is when the origins of two system are distant.
Case I
In this case we consider the origins of the fixed and rotating coordinate system coincide. This
case was earlier discussed in detail, where it was found that:
To an observer fixed relative to ðððð system, the time rate of change of ð = ð1ð + ð2ð + ð3ᅵᅵ
will be
ð
ðð¡ðð =
ð
ðð¡ð1ð +
ð
ðð¡ð2ð +
ð
ðð¡ð3ᅵᅵ (1)
where ððð
ðð¡ denotes the time derivative of ð relative to the rotating frame of reference.
However, the time rate of change of ð relative to fixed system ðð0ð0ð0 symbolized by ððð
ðð¡ will
be
ððð
ðð¡= (
ðð
ðð¡)ð
= (ðð
ðð¡)ð+ ð Ã ð (2)
or in operator form, we can write
(ð
ðð¡)ð
= (ð
ðð¡)ð+ ð Ã
166
Differentiating both sides of (2) w.r.t the fixed coordinate system, we have
(ð
ðð¡)ðð£ð = (
ð
ðð¡)ð(ð£ð + ð à ð)
by applying operator, we have
(ð
ðð¡)ðð£ð = (
ð
ðð¡+ ð Ã)
ð(ð£ð + ð à ð)
= ðð + 2ð Ãðð
ðð¡+ ᅵᅵ à ð + ð à (ð à ð)
or
ðð = ðð + 2ð à ð£ð + ᅵᅵ à ð + ð à (ð à ð)
(3)
where
ðð =ðð£ð
ðð¡, ðð =
ðð£ð
ðð¡
are the acceleration in the fixed and rotating coordinate systems. The relation (3) is referred as
Coriolis theorem. The term 2ð à ð£ð is called Coriolis acceleration, whereas the term ð à (ð Ã
ð) is called centripetal acceleration.
The equations of motion in fixed and moving/ rotating coordinate system are
ð¹ = ððð , ð¹â² = ððð
where F and ð¹â² are the total forces in the fixed and the rotating coordinate systems.
On making substitution for ðð from (3) in equation ð¹ = ððð , we obtain
These forces are called fictitious or apparent or inertial forces. They do not have physical forces
and do not arise from interactions of the particles.
Case II
In this case we consider the origins of fixed and rotating coordinate systems are not coincident.
We have
(ð)ð = ð = (ð)ð + ð0
= ð¥ð + ðŠð + ð§ï¿œï¿œ + ð0
167
where ð0 is the position vector of the origin of the rotating system w.r.t the origin of the fixed
system.
Therefore
(ðð
ðð¡)ð
=ðð0ðð¡
+ð
ðð¡(ð¥ð + ðŠð + ð§ï¿œï¿œ)
ð£ð = ð£0 + (ᅵᅵð + ᅵᅵð + ᅵᅵᅵᅵ) + ð¥ðð
ðð¡+ ðŠ
ðð
ðð¡+ ð§
ðᅵᅵ
ðð¡
ð£ð = ð£0 + ð£ð + ð à ð
where ð â¡ (ð)ð.
Similarly the acceleration in the two systems will be related by
ðð = ð0 + ðð + 2ð à ð£ð + ᅵᅵ à ð + ð à (ð à ð)
168
Module No. 180
Example 1 of Equation of Motion in Rotating Frame of Reference
Problem Statement
The angular velocity of a rotating coordinate system ðððð relative to a fixed coordinate system
ðð0ð0ð0 is
ᅵᅵ = 2ð¡ð â ð¡2ð + (2ð¡ + 4)ᅵᅵ
where ð¡ is the time and {ð, ð, ᅵᅵ} unit vectors associated with ðððð. The position vector of a
particle at time ð¡ in the body system (ðððð system) is given by
ð = (ð¡2 + 1)ð â 6ð¡ð + 4ð¡3ᅵᅵ
To Find
i. The apparent and true velocities at time ð¡ = 1.
ii. The apparent and true acceleration at time ð¡ = 1
Solution
The apparent velocity is given by
ð£ð = (ðð
ðð¡)ð
=ð
ðð¡[(ð¡2 + 1)ð â 6ð¡ð + 4ð¡3ᅵᅵ]
= ðð
ðð¡(ð¡2 + 1) â ð
ð
ðð¡6ð¡ + ᅵᅵ
ð
ðð¡4ð¡3
= 2ð¡ð â 6ð + 12ð¡2ᅵᅵ
Therefore the apparent velocity at time ð¡ = 1 is given by
ð£ð(ð¡ = 1) = 2ð â 6ð + 12ᅵᅵ
The true velocity at any time t is given by
ð£ð = ð£ð + ᅵᅵ à ð
169
= 2ð â 6ð + 12ᅵᅵ + |ð ð ᅵᅵ
2ð¡ âð¡2 2ð¡ + 4ð¡2 + 1 â6ð¡ 4ð¡3
|
Therefore
ð£ð(ð¡ = 1) = 2ð â 6ð + 12ᅵᅵ + |ð ð ᅵᅵ2 â1 62 â6 4
|
= 34ð â 2ð + 2ᅵᅵ
ii. For apparent acceleration
ðð =ðð£ð
ðð¡=
ð
ðð¡(2ð¡ð â 6ð + 12ð¡2ᅵᅵ)
= 2ð + 24ð¡ï¿œï¿œ
Therefore
ðð(ð¡ = 1) = 2ð + 24ᅵᅵ
For true acceleration we have
ðð = ðð + 2ð à ð£ð + ᅵᅵ à ð + ð à (ð à ð)
(1)
now
ð(ð¡ = 1) = 2ð â ð + 6ᅵᅵ
and
ᅵᅵ = 2ð â 2ð¡ð + 2ᅵᅵ
ᅵᅵ(ð¡ = 1) = 2ð â 2ð + 2ᅵᅵ
Since ð = (ð¡2 + 1)ð â 6ð¡ð + 4ð¡3ᅵᅵ and
ð£ð = 2ð¡ð â 6ð + 12ð¡2ᅵᅵ, therefore
ᅵᅵ à ð = |ð ð ᅵᅵ2 â2 22 â6 4
| = 4ð â 4ð â 8ᅵᅵ
and
170
2ð à ð£ð = 2 |ð ð ᅵᅵ2 â1 62 â6 12
|
= 2(24ð â 12ð â 10ᅵᅵ)
Also since
ð à ð = |ð ð ᅵᅵ2 â1 62 â6 4
| = 32ð + 4ð â 10ᅵᅵ
Therefore
ð à (ð à ð) = |ð ð ᅵᅵ2 â1 632 4 â10
|
= â14ð + 212ð + 40ᅵᅵ
Hence on making substitution, we have
ðð = (2ð + 24ð¡ï¿œï¿œ) + (4ð â 4ð â 8ᅵᅵ) + 2(24ð â 12ð â 10ᅵᅵ) + (â14ð + 212ð + 40ᅵᅵ)
= 40ð + 184ð + 36ᅵᅵ
which is required true acceleration.
171
Module No. 181
Example 2 of Equation of Motion in Rotating Frame of Reference
Problem Statement
A coordinate system OXYZ rotates with angular velocity ᅵᅵ = 2ð â 3ð + 5ᅵᅵ relative to the fixed
coordinate system Oð0ð0ð0 both systems having the same origin. If ðŽ = sin ð¡ ð â cos ð¡ ð + ðâð¡ï¿œï¿œ
where ð, ð, ᅵᅵ refer to the rotating coordinate system OXYZ then find ððŽ
ðð¡ and
ð2ðŽ
ðð¡2 w.r.to
i. Fixed system
ii. The rotating system
Solution
i. To find ððŽ
ðð¡ and
ð2ðŽ
ðð¡2 in the fixed system, we have to apply the following formulas
(ðᅵᅵ
ðð¡)ð
= (ðᅵᅵ
ðð¡)ð+ ᅵᅵ à ðŽ (1)
(ð2ᅵᅵ
ðð¡2)ð
= (ð2ᅵᅵ
ðð¡2)ð+ 2ᅵᅵ Ã
ðᅵᅵ
ðð¡+
ðᅵᅵ
ðð¡Ã ðŽ + ᅵᅵ à (ᅵᅵ à ðŽ) (2)
(ððŽ
ðð¡)
ð
= [ððŽ
ðð¡]ð
=ððŽð
ðð¡
=ð
ðð¡(sin ð¡ ð â cos ð¡ ð + ðâð¡ï¿œï¿œ)
= cos ð¡ ð + sin ð¡ ð â ðâð¡ï¿œï¿œ
ᅵᅵ à ðŽ = |ð ð ᅵᅵ2 â3 5
sin ð¡ â cos ð¡ ðâð¡
|
ᅵᅵ à ðŽ = (â3ðâð¡ + 5 cos ð¡)ð â (2ðâð¡ â 5 sin ð¡)ð + (â2 cos ð¡ + 3 sin ð¡)ᅵᅵ
(ððŽ
ðð¡)
ð
= cos ð¡ ð + sin ð¡ ð â ðâð¡ï¿œï¿œ + (â3ðâð¡ + 5 cos ð¡)ð â (2ðâð¡ â 5 sin ð¡)ð
+ (â2 cos ð¡ + 3 sin ð¡)ᅵᅵ
172
= (cos ð¡ + 5 cos ð¡ â 3ðâð¡)ð + (5 sin ð¡ + sin ð¡ â 2ðâð¡)ð + (âðâð¡ â 2 cos ð¡ + 3 sin ð¡)ᅵᅵ
= (6 cos ð¡ â 3ðâð¡)ð + (6 sin ð¡ â 2ðâð¡)ð + (âðâð¡ â 2 cos ð¡ + 3 sin ð¡)ᅵᅵ
Now for solving equation (2), we proceed as follows
(ð2ðŽ
ðð¡2)
ð
=ð
ðð¡(ððŽ
ðð¡)
ð
=ð
ðð¡(cos ð¡ ð + sin ð¡ ð â ðâð¡ï¿œï¿œ)
= (âsin ð¡ ð + cos ð¡ ð + ðâð¡ï¿œï¿œ)
2ᅵᅵ à (ððŽ
ðð¡)
ð
= 2(2ð â 3ð + 5ᅵᅵ) à (cos ð¡ ð + sin ð¡ ð â ðâð¡ï¿œï¿œ)
= (4ð â 6ð + 10ᅵᅵ) à (cos ð¡ ð + sin ð¡ ð â ðâð¡ï¿œï¿œ)
= |ð ð ᅵᅵ4 â6 10
cos ð¡ sin ð¡ âðâð¡
|
2ᅵᅵ à (ððŽ
ðð¡)
ð
= (6ðâð¡ â 10 sin ð¡)ð + (4ðâð¡ + 10 cos ð¡)ð + (4 sin ð¡ + 6 cos ð¡)ᅵᅵ
ᅵᅵ = 2ð â 3ð + 5ᅵᅵ ⹠ᅵᅵ =ðᅵᅵ
ðð¡= 0
So,
ðᅵᅵ
ðð¡Ã ðŽ = 0 à (sin ð¡ ð â cos ð¡ ð + ðâð¡ï¿œï¿œ) = 0
ᅵᅵ à (ᅵᅵ à ðŽ) = (2ð â 3ð + 5ᅵᅵ) à [(2ð â 3ð + 5ᅵᅵ) à (sin ð¡ ð â cos ð¡ ð + ðâð¡ï¿œï¿œ)]
= (2ð â 3ð + 5ᅵᅵ) à [(â3ðâð¡ + 5 cos ð¡)ð â (2ðâð¡ â 5 sin ð¡)ð + (â2 cos ð¡ + 3 sin ð¡)ᅵᅵ]
= |ð ð ᅵᅵ2 â3 5
â3ðâð¡ + 5 cos ð¡ â2ðâð¡ + 5 sin ð¡ â2 cos ð¡ + 3 sin ð¡
|
= (6 cos ð¡ â 9 sin ð¡ â 25 sin ð¡ + 10ðâð¡)ð + (4 cos ð¡ â 6 sin ð¡ â 15ðâð¡ + 25 cos ð¡)ð + (4ðâð¡
â 10 sin ð¡ + 9ðâð¡ â 15 cos ð¡)ᅵᅵ
173
= (6 cos ð¡ â 34 sin ð¡ + 10ðâð¡)ð + (29 cos ð¡ â 6 sin ð¡ â 15ðâð¡)ð
+ (â4ðâð¡ + 10 sin ð¡ + 9ðâð¡ â 15 cos ð¡)ᅵᅵ
= (â sin ð¡ ð + cos ð¡ ð + ðâð¡ï¿œï¿œ) + (6ðâð¡ â 10 sin ð¡)ð + (4ðâð¡ + 10 cos ð¡)ð + (4 sin ð¡ + 6 cos ð¡)ᅵᅵ
+ 0 + (6 cos ð¡ â 34 sin ð¡ + 10ðâð¡)ð + (29 cos ð¡ â 6 sin ð¡ â 15ðâð¡)ð + (10 sin ð¡
â 13ðâð¡ + 15 cos ð¡)ᅵᅵ
(ð2ðŽ
ðð¡2)
ð
= (6 cos ð¡ â 45 sin ð¡ + 16ðâð¡)ð + (40 cos ð¡ â 6 sin ð¡ â 11ðâð¡)ð
+ (14 sin ð¡ â 12ðâð¡ + 21 cos ð¡)ᅵᅵ
ii. The rotating system
(ððŽ
ðð¡)
ð
= [ððŽ
ðð¡]ð
=ððŽð
ðð¡
=ð
ðð¡(sin ð¡ ð â cos ð¡ ð + ðâð¡ï¿œï¿œ)
= cos ð¡ ð + sin ð¡ ð â ðâð¡ï¿œï¿œ
(ð2ðŽ
ðð¡2)
ð
=ð
ðð¡(ððŽ
ðð¡)
ð
=ð
ðð¡(cos ð¡ ð + sin ð¡ ð â ðâð¡ï¿œï¿œ)
= (âsin ð¡ ð + cos ð¡ ð + ðâð¡ï¿œï¿œ)
174
Module No. 182
Example 3 of Equation of Motion in Rotating Frame of Reference
Problem Statement
A bead slides on a smooth helix whose central axis is vertical. The helix is forced to rotate about
its central axis with constant angular speed ð. Find the equation of motion of the bead relative to
the helix.
Solution
We choose a coordinate ðððð fixed in the helix such that ð-axis is coincident with the axis of
the helix.
Then the parametric eqs. of the helix are given by
ð¥ = ð cos ð , ðŠ = ð sin ð,
ð§ = ðð
In vector form these can be written as
ð = ð cos ð ð + ð sin ð ð + ððᅵᅵ
where ð, ð, ᅵᅵ point along the coordinate axis. Here ᅵᅵ is constant but ð, ð are variable.
175
The equation of motion in the rotating coordinate system is given by
ððð = ð¹ â 2ðð à ð£ð â ðᅵᅵ à ð â ðð à (ð à ð) (1)
In this problem, since ð is constant,
ð = ðᅵᅵ , ᅵᅵ = 0 (2)
Also
ð£ð = ᅵᅵ = (âð sin ð ð + ð cos ð ð + ðᅵᅵ)ᅵᅵ (3)
and
ðð = ᅵᅵ =ð
ðð¡[(âð sin ð ð + ð cos ð ð + ðᅵᅵ)ᅵᅵ]
ð
= (âð sin ð ð + ð cos ð ð + ðᅵᅵ)ᅵᅵ +(âð cos ð ð â ð sin ð ð)ᅵᅵ2 (4)
If ð¹ denotes the external force in the fixed (space) coordinate system, then
ð¹ = âððᅵᅵ + ð (5)
where ð is the reaction of the helix on the bead.
Before making substitution from equation (2-5) into equation (1), we obtain simplified
expression for the second, and the fourth terms of (1), (the third term being zero).
ð à ð£ð = ðᅵᅵ à (âð sin ð ð + ð cos ð ð + ðᅵᅵ)ᅵᅵ
= ðð(â sin ð ð â cos ð ð)ᅵᅵ (6)
and
ð à ð = ðᅵᅵ à (ð cos ð ð + ð sin ð ð + ððᅵᅵ)
= ðð(cos ð ð â sin ð ð)
ð à (ð à ð) = ð2ðᅵᅵ à (cos ð ð â sin ð ð)
= ð2ð(âcos ð ð â sin ð ð) (7)
Therefore on making substitutions from (4), (5), (6) and (7) into (1), we obtain
ð(âð sin ð ð + ð cos ð ð + ðᅵᅵ)ᅵᅵ â ðð(cos ð ð + sin ð ð) = âððᅵᅵ + ð + 2ððð(sin ð ð +
cos ð ð) + ððð2(cos ð ð + sin ð ð)
176
ð(âð sin ð ð + ð cos ð ð + ðᅵᅵ)ᅵᅵ â ðð(cos ð ð + sin ð ð) = âððᅵᅵ + ð + (2ððð +
ððð2) (sin ð ð + cos ð ð) (8)
Now if we write
ð£ð = (âð sin ð ð + ð cos ð ð + ðᅵᅵ)ᅵᅵ = ð¢ï¿œï¿œ
Then it is clear that the vector ð¢ is tangent to the helix. Since ð is normal to the helix,
ð . ð£ð = ð . ð¢ï¿œï¿œ = 0
Rewriting (8) in terms of ð¢ (after eliminating the common factor ð), we have
ð¢ï¿œï¿œ â ð(cos ð ð + sin ð ð)ᅵᅵ2 = âðð + ð + (2ðð + ðð2)(cosð ð + sin ð ð) (9)
Taking dot product of both sides of (9) with u, we have
ð¢. ð¢ï¿œï¿œ â ðð¢. (cos ð ð + sin ð ð)ᅵᅵ2 = âðð. ð¢ + ð . ð¢ + (2ðð + ðð2)(cosð ð + sin ð ð). ð¢ (10)
Now
ð¢. ð¢ = ð2 sin2 ð + ð2 cos2 ð + ð2 = ð2 + ð2
ᅵᅵ. ð¢ = ᅵᅵ. (âð sin ð ð + ð cos ð ð + ðᅵᅵ) = ð
ð . ð¢ = 0
(sin ð ð + cos ð ð). ð¢ = (sin ð ð + cos ð ð). (âð sin ð ð + ð cos ð ð + ðᅵᅵ)
= ð cos ð sin ð â ð cos ð sin ð
= 0
Therefore (10) reduces to
(ð2 + ð2)ᅵᅵ = âðð
or
ᅵᅵ = âðð
(ð2 + ð2)
177
Module No. 183
Example 4 of Equation of Motion in Rotating Frame of Reference
Problem Statement
Prove that the centrifugal force acting on a particle of mass ð on the earthâs surface is the vector
directed away from the earthâs center and perpendicular to the angular velocity vector ᅵᅵ.
i. Show that its magnitude is ðð2ðð cos ð.
ii. Determine the places on the surface of the earth where centrifugal force will be maximum
or minimum.
Solution
i. Since centrifugal force is âðᅵᅵ à (ᅵᅵ à ðð).
We have to show that itâs vector is directed away from center of the earth and perpendicular to
the angular velocity vector ᅵᅵ and its magnitude is ðð2ðð cos ð
where ð is the latitude as shown in figure and ðð is the radius of the earth.
It is quite clear that the particle is moving away from the center of earth and centrifugal force is
produced in the opposite direction of the path of the particle.
178
It can be observed from the given figure that the movement of the particle (i.e. ðᅵᅵ à (ᅵᅵ à ðð)) is
perpendicular to the angular velocity ᅵᅵ and vector ᅵᅵ à ðð , becomes centrifugal force in the
opposite direction of the particle, so its centrifugal force is also perpendicular to ᅵᅵ and ᅵᅵ à ðð .
Now we move onto the proof of magnitude of centrifugal force, since
ᅵᅵðð = âðᅵᅵ à (ᅵᅵ à ðð)
We need to show that
|ᅵᅵðð| = |âðᅵᅵ à (ᅵᅵ à ðð)| = ðð2ðð cos ð
Now we assume that the motion takes place in the ðð-plane.
Then,
ᅵᅵ = ᅵᅵ cos ð + ð sin ð
= ᅵᅵ cos(ð 2â â ð) + ð sin(ð 2â â ð)
ᅵᅵ = ᅵᅵ sin ð + ð cos ð
Consider the centrifugal force
= âðᅵᅵ à (ᅵᅵ à ðð)
If ᅵᅵ denotes a unit vector along the NS-axis, then we can write ᅵᅵ = ðᅵᅵ and the expression
becomes ᅵᅵ à (ᅵᅵ à ðð) = ð2ᅵᅵ à (ᅵᅵ à ðð)
âð2ᅵᅵ à (ᅵᅵ à ðð) = âð2[(ᅵᅵ. ðð)ᅵᅵ â (ᅵᅵ. ᅵᅵ)ðð]
= âð2[(ᅵᅵ. ðð)ᅵᅵ â ðð]
Now
ᅵᅵ. ðð = (ᅵᅵ sin ð + ð cos ð). ððᅵᅵ
= ðð cos ð = cos(ð 2â â ð) = ðð sin ð
So expression (1) becomes
âᅵᅵ à (ᅵᅵ à ðð) = âð2ðð sin ð (ᅵᅵ sin ð + ð cos ð) + ð2ððᅵᅵ
= (âð2ðð sin2 ð + ð2ðð)ᅵᅵ â ð2ðð sin ð cos ð ð
= (âð2ðð(1 â cos2 ð) + ð2ðð)ᅵᅵ â ð2ðð sin ð cos ð ð
179
|ᅵᅵðð| = |âðᅵᅵ à (ᅵᅵ à ðð)| = |âð((âð2ðð(1 â cos2 ð) + ð2ðð)ᅵᅵ â ð2ðð sin ð cos ð ð)|
= |âð(ð2ðð cos2 ð)ᅵᅵ + ðð2ðð sin ð cos ð ð)|
= ðâ(ð2ðð cos2 ð)2 + (ð2ðð sin ð cos ð)2
= ðâð4ðð2 cos4 ð + ð4ðð2 cos2 ð sin2 ð
= ðâð4ðð2 cos4 ð + ð4ðð2 cos2 ð (1 â cos2 ð)
= ðâð4ðð2 cos4 ð + ð4ðð2 cos2 ð â ð4ðð2 cos4 ð
= ðâð4ðð2 cos2 ð
|ᅵᅵðð| = ðð2ðð cos ð
Hence the required result.
The centrifugal force will be maximum at ð = 0, where cos ð = 1 (maximum) i.e. at equator
and minimum will be at NS-poles i.e. at ð = ð2â where cos ð = 0 (minimum).
180
181
Module No. 184
Example 5 of Equation of Motion in Rotating Frame of Reference
Problem Statement
A coordinate system ðððð is rotating with angular velocity ᅵᅵ = 5ð â 4ð â 10ᅵᅵ relative to fixed
coordinate system ðð0ð0ð0 both systems having the same origin. Find the velocity of the
particle at rest in the ðððð system at the point (3,1, â2) as seen by an observer in the fixed
system.
Solution
Since the given angular velocity is
ᅵᅵ = 5ð â 4ð â 10ᅵᅵ
and the point P(3,1, â2) at rest in ðððð system.
Then
ðð = ð = 3ð + ð â 2ᅵᅵ
Also, we know the equation of motion in case when the origins of the coordinate systems
coincide each other
ð£ð = ð£ð + ᅵᅵ à ð
But we have to find out the velocity of the particle at rest in ðððð system at P(3,1, â2).
i.e.
ð£ð = ð£0 + ᅵᅵ à ð
ð£ð = (0,0,0) + (5ð â 4ð â 10ᅵᅵ) à (3ð + 1ð â 2ᅵᅵ)
= (0ð + 0ð + 0ᅵᅵ) + |ð ð ᅵᅵ5 â4 â103 1 â2
|
= (0ð + 0ð + 0ᅵᅵ) + ((8 + 10)ð â (â10 + 30)ð + (5 + 12)ᅵᅵ)
182
= (0ð + 0ð + 0ᅵᅵ) + (18ð â 20ð + 17ᅵᅵ)
Hence
ð£ð = 18ð â 20ð + 17ᅵᅵ
is the required velocity of the system.
183
Module No. 185
General Motion of a Rigid Body
In general motion of the body, (i.e. no point of the body is fixed in space), let ððð¥ð¡ be the total
external force on the rigid body and ðºððð¥ð¡ the total external torque about its center of mass (i.e.
centroid). Then the equations of motion are
ððð =ððð¥ð¡ (1)
and
ð³ï¿œï¿œ = ð®ððð¥ð¡ (2)
where ðð is the acceleration of the c.m. and ð³ð is the total angular momentum about it.
Now we resolve the vectors ðð,ððð¥ð¡,ð®ððð¥ð¡and ð³ along the unit vector ð, ð, ð taken along the
principal axes at the mass center.
The triad of vectors ð, ð, ð may be inferred to as a principal triad. It will be assumed to be
permanently a principal triad.
Let Ω be its angular velocity. If the triad is fixed in the body then Ω = Ï, the angular velocity of
the body.
Now using the operator: (ð
ðð¡)ð
= (ð
ðð¡)ð+ ð Ã
(ðᅵᅵ
ðð¡)
ð
= (ðᅵᅵ
ðð¡)
ð
+ Ï Ã ï¿œï¿œ
which relates the rate of change of a vector in a fixed (i.e. inertial) frame and a rotating frame,
we have (on dropping the suffix r)
ðð â¡ (ðᅵᅵ
ðð¡)
ð
=ðᅵᅵ
ðð¡+ Ω à ᅵᅵ (⎠ðð = ð)
or ðð = ðð£
ðð¡ + Ω x ᅵᅵ (3)
where ð£ = ð£1ð + ð£2ð + ð£3ð is the velocity of the mass center (in the rotating coordinate system).
Substituting for ðð = ðð from equation (3) into (1), we obtain
184
ð(ðð
ðð¡+ Ω à ð) = ððð¥ð¡
which is equivalent to
ð(ð£1 + Ω2ð£3 + Ω3ð£2) = ð¹1
ð(ð£2 + Ω3ð£1 + Ω1ð£3) = ð¹2
ð(ð£3 + Ω1ð£2 + Ω2ð£1) = ð¹3
} (4)
From (2), on using
(ðð¿
ðð¡)ð =
ðð¿
ðð¡+ ðº x ð¿
and the relation ð¿ = ðŒ1ð1ð + ðŒ2ð2ð + ðŒ3ð3ᅵᅵ, we obtain the equations
ðŒ1Ï1 ð + ðŒ2Ï2 ð + ðŒ3Ï3 ᅵᅵ + (Ω2ð¿3 â Ω3ð¿2)ð + (Ω2ð¿1 â Ω1ð¿3)ð + (Ω1ð¿2 â Ω2ð¿1)ᅵᅵ = ᅵᅵ
From this vector equation we obtain the following three scalar equations
ðŒ1Ï1 + Ω2ð¿3 â Ω3ð¿2 = ðº1
ðŒ2Ï2 + Ω3ð¿1 â Ω1ð¿3 = ðº2
and
ðŒ3Ï3 + Ω1ð¿2 â Ω2ð¿1 = ðº3
where on using the results
ð¿1 = ðŒ1Ï1, ð¿2 = ðŒ2Ï2, ð¿3 = ðŒ3Ï3
we have
ðŒ1Ï1 + Ï3Ω2I3 â Ï2Ω3I2 = ðº1
ðŒ2Ï2 + Ï1Ω3I1 â Ï3Ω1I3 = ðº2
ðŒ3Ï3 + Ï2Ω1I2 â Ï1Ω2I1 = ðº3
} (5)
where ðŒ1, ðŒ2, ðŒ3 denote principal M.I at the centroid of the body.
The set of eqs (4) and (5) constitute six equations for the components of velocity of centroid and
the components of angular velocity of the body.
185
Module No. 186
Equation of Motion Relative to Coordinate System Fixed on Earth
We derived the equation of motion when the origins of fixed and rotating coordinate systems are
distant.
ᅵᅵð = ᅵᅵ0 + ᅵᅵð + 2ᅵᅵ à ᅵᅵð + ᅵᅵ à ð + ᅵᅵ à (ᅵᅵ à ð) (1)
From (1) the most general from of the equation of motion in a moving frame can be written as
ð (ð2ð
ðð¡2)ð
= ᅵᅵ â ðᅵᅵ à ð â 2ðᅵᅵ à ð£ð â ðᅵᅵ à (ᅵᅵ à ð) â ðᅵᅵ0 (2)
where the subscript ð refers to the rotating frame, ᅵᅵ is the total external force in the fixed
coordinate system and ᅵᅵ0 is the acceleration of the origin ð in moving (rotating) coordinate
system, ðððð. In the case of the earth, which is a rotating coordinate system, we choose the
origin as a point fixed on the earth.
We choose the fixed coordinate system at the center ð¶ of the earth and denote it by ð¶ð0ð0ð0 .
For a particle near the surface of the earth
ᅵᅵ = ðᅵᅵ (3)
where ᅵᅵ is the acceleration due to gravity.
ᅵᅵ0, the acceleration of the origin ð is the centripetal acceleration of ð due to the rotation of the
earth. It may be represented as
ᅵᅵ0 = ᅵᅵ à (ᅵᅵ à ðð) (4)
Therefore on substitution from (3) and (4) into (2), we obtain
ð (ð2ð
ðð¡2)ð
= ðᅵᅵ â ðᅵᅵ à ð â 2ðᅵᅵ à ð£ð â ðᅵᅵ à (ᅵᅵ à ð) â ðᅵᅵ à (ᅵᅵ à ðð) (5)
which gives the full equation of motion for a particle of mass ð w.r.t. a coordinate system fixed
on the earth.
Next we obtain a simple form of (5) taking into account the fact that the angular velocity ᅵᅵ of
the earth is nearly constant both in magnitude and direction, (which is along NS i.e. North-South
polar line), and of small magnitude. In fact
186
ð = |ᅵᅵ| =2ð radian
24 hours
=2ð
86400 radian per second
= 7.27 Ã 10â5rad/sec
Since ᅵᅵ may be taken as constant, ᅵᅵ = 0. Since the fourth term on R.H.S. of (5) has the
magnitude ðð2r (where r is the distance of the particle from the NS-axis), is negligibly small
because of ð2. However the fifth term ðᅵᅵ à (ᅵᅵ à ðð), the last term of equation (5) is not
negligible because |ðð| (the magnitude of the radius of the earth) is very large.
Hence equation (5) can be written as
ð (ð2ð
ðð¡2)
ð
= ðᅵᅵ â 2ðᅵᅵ à ð£ð â ðᅵᅵ à (ᅵᅵ à ðð)
which is a second order differential equation in ð and may also written as
ðð2ð
ðð¡2 + 2ðᅵᅵ Ãðð
ðð¡= ðᅵᅵ â ðᅵᅵ à (ᅵᅵ à ðð) (6)
where we have dropped the subscript ð, it being understood that the terms on L.H.S of (6) refer
to the rotating coordinate system.
If ᅵᅵ denotes a unit vector along the NS-axis, then we can write ᅵᅵ = ðᅵᅵ, and equation of motion
takes the form
ᅵᅵ + 2ðᅵᅵ à ð = ᅵᅵ â ð2ᅵᅵ à (ᅵᅵ à ðð)