Engineering Mechanics is one of the fundamental branches of science which is important for the education of professional
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English Pages 175 [177] Year 2019
Table of contents :
Preface
Acknowledgments
Topic AK1
1.1 Application of Linear Momentum Principle to Determine the Velocity of Particle
1.2 Sample Problem
1.3 Solution
Topic AK2
2.1 Application of the Linear Momentum Principle and Motion of the Mass Center of a System of Particles
2.2 Sample Problem
2.3 Solution
Topic AK3
3.1 Application of the Angular Momentum Principle to Determine the Angular Velocity of the System of Particles
3.2 Sample Problem
3.3 Solution
Topic AK4
4.1 Determination of Velocity and Acceleration of Particle by Given Equations of Motion
4.2 Sample Problem
4.3 Solution
Topic AK5
5.1 Differential Equations of Motion of Rigid Body: Rotational Motion of Rigid Body
5.2 Sample Problem
5.3 Solution
Topic AK6
6.1 Study of Impact Dynamics between Rigid Bodies
6.2 Sample Problem
6.3 Solution
Topic AK7
7.1 Application of Virtual Displacements Principle to Problems on Balanced Forces Applied to Mechanical System with One Degree of Freedom
7.2 Sample Problem
7.3 Solution
Topic AK8
8.1 Projectile Motion Analyses with Quadratic Air Resistance
8.2 Sample Problem
Author Biography
Blank Page
Series ISSN: 25733168
BAKHTIYAROV
Synthesis Lectures on Mechanical Engineering
Solving Practical Engineering Mechanics Problems ADVANCED KINETICS
Engineering Mechanics is one of the fundamental branches of science which is important for the education of professional engineers regardless of major. Most of the basic engineering courses, such as mechanics of materials, fluid and gas mechanics, machine design, mechatronics, acoustics and vibrations, etc., are based on the Engineering Mechanics course. In order to absorb the materials of Engineering Mechanics, it is not enough to just consume theorems and theoretical laws. A student also must develop an ability to solve practical problems. Therefore, it is necessary to solve many problems independently. The books in this series are designed as supplements to the Engineering Mechanics course and can be used to apply the principles required for solving practical engineering problems in the following branches of Mechanics: Statics, Kinematics, Dynamics, and Advanced Kinetics. Each book contains several (between 6 and 8) topics of the branch. Each topic has 30 problems to be assigned as homework, tests, and midterm/final exams with the consent of the instructor. A solution of one similar sample problem from each topic is provided. This fourth book in the series contains eight topics of Advanced Kinetics, which is the branch of Mechanics that is concerned with the analysis of motion of both particles and rigid bodies with reference to the cause of the motion. This book is targeted to undergraduate students of the junior/senior level as well as graduate students majoring in science and engineering. The author welcomes all feedback/comments from the reader. Please feel free to contact him at [email protected]
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ABOUT SYNTHESIS This volume is a printed version of a work that appears in the Synthesis Digital Library of Engineering and Computer Science. Synthesis lectures provide concise original presentations of important research and development topics, published quickly in digital and print formats. For more information, visit our website: http://store.morganclaypool.com
SOLVING PRACTICAL ENGINEERING MECHANICS PROBLEMS: ADVANCED KINETICS
Sayavur I. Bakhtiyarov, New Mexico Institute of Mining and Technology
Solving Practical Engineering Mechanics Problems
ADVANCED KINETICS
Sayavur I. Bakhtiyarov
Synthesis Lectures on Mechanical Engineering
Solving Practical Engineering Mechanics Problems: Advanced Kinetics
iii
Synthesis Lectures on Mechanical Engineering Synthesis Lectures on Mechanical Engineering series publishes 60–150 page publications pertaining to this diverse discipline of mechanical engineering. The series presents Lectures written for an audience of researchers, industry engineers, undergraduate and graduate students. Additional Synthesis series will be developed covering key areas within mechanical engineering. Solving Practical Engineering Mechanics Problems: Advanced Kinetics Sayavur I. Bakhtiyarov June 2019 Natural Corrosion Inhibitors Shima Ghanavati Nasab, Mehdi Javaheran Yazd, Abolfazl Semnani, Homa Kahkesh, Navid Rabiee, Mohammad Rabiee, and Mojtaba Bagherzadeh May 2019 Fractional Calculus with its Applications in Engineering and Technology Yi Yang, Haiyan Henry Zhang March 2019 Essential Engineering Thermodynamics: A Student’s Guide Yumin Zhang September 2018 Engineering Dynamics Cho W.S. To July 2018 Solving Practical Engineering Mechanics Problems: Dynamics Sayavur I. Bakhtiyarov May 2018 Solving Practical Engineering Mechanics Problems: Kinematics Sayavur I. Bakhtiyarov April 2018 Mathematical Magnetohydrodynamics Nikolas Xiros January 2018 Design Engineering Journey Ramana M. Pidaparti January 2018 Introduction to Kinematics and Dynamics of Machinery Cho W. S. To December 2017
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Microcontroller Education: do it Yourself, Reinvent the Wheel, Code to Learn Dimosthenis E. Bolanakis November 2017 Solving Practical Engineering Mechanics Problems: Statics Sayavur I. Bakhtiyarov October 2017 Resistance Spot Welding: Fundamentals and Applications for the Automotive Industry Menachem Kimchi and David H. Phillips October 2017 Unmanned Aircraft Design: Review of Fundamentals Mohammad Sadraey September 2017 Introduction to Refrigeration and Air Conditioning Systems: Theory and Applications Allan Kirkpatrick September 2017 MEMS Barometers Toward Vertical Position Detection: Background Theory, System Prototyping, and Measurement Analysis Dimosthenis E. Bolanakis May 2017 Engineering Finite Element Analysis Ramana M. Pidaparti May 2017
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ISBN: 9781681735856 Paperback ISBN: 9781681735863 ebook ISBN: 9781681735870 Hardcover DOI 10.2200/S00924ED1V04Y201905MEC019 A Publication in the Morgan & Claypool Publishers series SYNTHESIS LECTURES ON MECHANICAL ENGINEERING, #19
Series ISSN: 25733168 Print 25733176 Electronic
Solving Practical Engineering Mechanics Problems: Advanced Kinetics Sayavur I. Bakhtiyarov
New Mexico Institute of Mining and Technology
SYNTHESIS LECTURES ON MECHANICAL ENGINEERING #19
M &C
MORGAN
& CLAYPOOL PUBLISHERS
viii
ABSTRACT
Engineering Mechanics is one of the fundamental branches of science which is important for the education of professional engineers regardless of major. Most of the basic engineering courses, such as mechanics of materials, fluid and gas mechanics, machine design, mechatronics, acoustics and vibrations, etc., are based on the Engineering Mechanics course. In order to absorb the materials of Engineering Mechanics, it is not enough to just consume theorems and theoretical laws. A student also must develop an ability to solve practical problems. Therefore, it is necessary to solve many problems independently. The books in this series are designed as supplements to the Engineering Mechanics course and can be used to apply the principles required for solving practical engineering problems in the following branches of Mechanics: Statics, Kinematics, Dynamics, and Advanced Kinetics. Each book contains several (between 6 and 8) topics of the branch. Each topic has 30 problems to be assigned as homework, tests, and midterm/final exams with the consent of the instructor. A solution of one similar sample problem from each topic is provided. This fourth book in the series contains eight topics of Advanced Kinetics, which is the branch of Mechanics that is concerned with the analysis of motion of both particles and rigid bodies with reference to the cause of the motion. This book is targeted to undergraduate students of the junior/senior level as well as graduate students majoring in science and engineering.
KEYWORDS
velocity, acceleration, force, moment, momentum, impulse, energy, work
ix
Contents Preface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ����������������������������xi Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ������������������������� xiii 1
Topic AK1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 1 1.1 Application of the Linear Momentum Principle to Determine the Velocity of Particles 1 1.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2
Topic AK2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . �����������������������������9 2.1 Application of the the Linear Momentum Principle and Motion of the Mass Center of a System of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3
Topic AK3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ������������������������� 37 3.1 Application of the Angular Momentum Principle to Determine the Angular Velocity of the System of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 3.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
4
Topic AK4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ������������������������� 65 4.1 Determination of Velocity and Acceleration of Particles by Given Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 4.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 4.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
5
Topic AK5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ������������������������� 69 5.1 Differential Equations of Motion of a Rigid Body: Rotational Motion of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 5.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 5.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
6
Topic AK6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ������������������������� 91 6.1 Study of Impact Dynamics between Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 6.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 6.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
7
Topic AK7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . �������������������������129 7.1 Application of the Virtual Displacements Principle to Problems on Balanced Forces Applied to a Mechanical System with One Degree of Freedom �������������������������������������������� 129 7.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 7.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
8
Topic AK8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . �������������������������149 8.1 Projectile Motion Analyses with Quadratic Air Resistance ���������������������������������������������������� 149 8.2 Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
Author Biography. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . �������������������������161
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Preface This series of books contains practical problems in Engineering Mechanics, which were mainly translated and revised from selected problems prepared by A. A. Yablonski (Theoretical Mechanics, 1972, in Russian). It is intended to introduce this unique work to Western academia. These books are the product of material covered in many classes over a period of four decades in several universities. The series comprises of four books: Statics, Kinematics, Dynamics, and Advanced Kinetics. This fourth book, Advanced Kinetics, consists of eight sets of problems in which students are introduced to various advanced topics of Engineering Mechanics. Each set involves 30 problems, which can be assigned as individual homework as well as test/exam problems. Solutions to similar sample problem for each set are provided.
xiii
Acknowledgments The author acknowledges that this work is essentially a translation and a revision of selected problems provided by Professor A. A. Yablonski (Collection of Problems for Course Projects in Theoretical Mechanics, 2nd ed., Vischaya Shkola Publishers, 1972, in Russian). The author intended to introduce this unique work to Western academia, which is the product of material covered by him in many classes over a period of four decades of his career at a number of universities and colleges. The author would like to acknowledge Michelle Pederson who prepared graphics for this book.
1
CHAPTER 1
Topic AK1 1.1
APPLICATION OF THE LINEAR MOMENTUM PRINCIPLE TO DETERMINE THE VELOCITY OF PARTICLES
The initial velocity of the object of mass m is v0 upward along the inclined surface with angle of inclination α with the horizon. In the same direction, the force P⃗ is acting on the object (Figure 1.1). Knowing a variation of the force P⃗ in time P = P(t) and the friction coefficient f, define the velocity of the object at time instants t1, t2, t3 and verify the answer obtained for the time instant t1 using a differential equation of motion (Newton’s second law). The parameters necessary to solve the problems are given in the table below. Prob. No. 1 2 3 4 5 6 7 8 9 10 11 12 13 `14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
α deg
m kg
ʋ0 m/s
t1
25 37 21 32 24 40 25 23 20 27 35 42 30 23 18 39 15 26 30 15 28 31 21 18 33 45 38 26
35 20 25 10 16 40 20 16 12 50 10 12 10 20 14 20 24 15 15 22 13 11 12 10 18 8 17 9
5.4 0 0 4.5 9 4 8 7.6 0 12 5 3 8 8.5 9 3 10 13 7.2 8.2 10 6 4.5 7 0 9 5 7.5
4 6 4 5 4 4 5 6 6 2 6 3 4 5 7 5 6 8 3 2 8 7 5 4 8 4 7 4
t2 s
t3
10 10 10 10 8 8 8 11 10 6 10 8 12 8 12 9 10 16 10 9 12 9 10 14 10 8 10 12
18 15 16 16 16 12 11 13 14 12 16 14 16 15 18 17 15 20 16 11 16 14 16 29 16 12 18 17
P0
P1
P2
P3
150 160 120 40 0 300 0 0 0 200 100 120 150 150 0 100 60 0 90 110 0/200 0/100 40 50 0/140 0 100 100/70
250 180 0 100 160 0 0 0 0 200 200 120 100 0 100 100 140 90 120 50 100 120 80 0 180 120 280 70
N 100 200 200 0 120 400 0 75 100 0 50 60 0 40 0 300 0 110 150 70 0 160 50 100 180 0 190 0
200 160 200/120 180 120/0 300 300 200 140 300 100 180 150 100 140 150 180 150 110 110 200/0 0 120 50 160/0 150 170/0 140/70
f 0.1 0.25 0.1 0.12 0.08 0.06 0.2 0.12 0.2 0.08 0.24 0.15 0.18 0.07 0.15 0.12 0.2 0.22 0.3 0.15 0.14 0.11 0.22 0.09 0.17 0.1 0.21 0.2
2
1. TOPIC AK1
29 30
60 33
20 10
9.5 10.2
5 6
6 7
11 10
0 120
400 70
400/100 0
200 120
0.25 0.15
To plot the variation of the force P⃗ with its given values P0, P1, P2, P3 corresponding to the time instants t0, t1, t2, t3, the relationship P = P(t) between those time instants will be considered as linear. The values of the force P⃗ provided in the table above as fractions are indicating that the magnitudes of the force “jumps” at that given time instant. A nominator of the fraction shows a magnitude of the force at the end of the time interval, a denominator—at the beginning of the succeeding time interval.
1.2
SAMPLE PROBLEM
Given parameters: m 300 m = 40 kg; v0 = 10 s ; t0 = 0; t1 = 3 s; t2 = 8 s; t3 = 12 s; P0 = 0; P1 = 250 N; P2 = 200 N; P3 = 150 N; α = 30°; f = 0.1. Define: v1, v2, and v3 at time instants t1, t2, t3.
1.3
SOLUTION
First, we show the forces acting on the given object (Figure 1.1): the gravity force G⃗, the normal reaction force N⃗, the given force P,⃗ and the friction force F⃗ (along the inclined surface and opposite to the initial velocity). Next, we will plot the P = P(t) graph according to the given values of P0, P1, P2, and P3 (Figure 1.2).
x
N
P L
F
G O Figure 1.1.
a
1.3 SOLUTION
P (N) C
300 B
250
D
200
E
150
I
K
t2=8
t3=12
M O
t1=3
t (s)
Figure 1.2.
We construct the equation of the liner momentum—impulse principle on the axis x in the time interval from “0” to t1 for the given object which is considered as a particle: mv1x – mv10x = � Six , (1.1)
where
� Six = – Gt1 sin α – Ft1 + SPx.
The “x” component of the impulse of the variable force P from “0” to t1 seconds: t1
SPx = � Pdt . 0
This integral is defined as an area of the triangle OBM in the graph for P = P(t): 3 ∙ 250 SPx = = 375 N ∙ s . 2 Considering the friction force as:
F = f N = f G cos α ,
Equation (1.1) can be rewritten as: mv1x – mv0x = – mgt1 sin α – f mg cos α t1 + 375 . From here:
375 v1x = v0x – gt1 sin α – fg cos α t1 + m
3
4
1. TOPIC AK1
or
375 m = 10 – 14.72 – 2.56 + 9.38 = 2.1 s . v1x = 10 – 9.81 ∙ 3 ∙ 0.5 – 0.1 ∙ 9.81 ∙ 0.87 ∙ 3 + 40 Hence:
m v1 = v1x = 2.1 s .
We need to note that the friction force F⃗ is directed opposite to the velocity. Therefore, before the calculations it is necessary to find out if during the time t1 the velocity changes from the initial upward direction along the inclined surface (v0x > 0), and, consequently, the friction force is directed downward. Thus, it is necessary to find out if there is any time instant t* < t1 when the velocity of the object will be equal to zero under the action of the constant forces G⃗, N⃗, F⃗ and the force P⃗ linearly varying by line OB: 250 P= t. 3 We will write an equation of the linear momentum–impulse principle for the assumed time interval from “0” to t*: mvx – mv0x = – mgt* sin α – fmgt* cos α + SPx, where
250 1 125 2 t* ∙ = t* . vx = 0, SPx t* ∙ 3 2 3 Hence, we will obtain the following equation to define t*:
or
125 2 t* – mg(sin α + f cos α ) t* + mv0x = 0 , 3 t*2 –
or
(196.2 + 34.1) ∙ 3 400 ∙ 3 t* + =0, 125 125
t*2 – 5.52t* + 9.6 = 0 . The solution of this equation shows that there is no time instant when the velocity of the object under the acting forces will be equal to zero. To determine the velocity of the object at the time instant t2 we will use the linear momentum–impulse principle for the time interval t2 – t1: (1.2) mv2x – mv1x = � Six ,
where
� Six = – G(t2 – t1) sin α – F(t2 – t1) + SPx.
The x component of the impulse of the variable force P⃗ during the time interval t2 – t1 will be expressed by the area of the trapezoid MBCI of the P = P(t) graph:
SPx =
5(250 + 300) = 1,375 N ∙ s . 2
1.3 SOLUTION
Hence, Equation (1.2) can be rewritten as: mv2x – mv1x = – mg(t2 – t1) sin α – fmg cos α (t2 – t1) + 1,375 , from where: 1,375 1,375 = 2.1 – 9.81 ∙ 5 ∙ 0.5 – 0.1 ∙ 9.81 ∙ 0.87 ∙ 5 + v2x = v1x – g(t2 – t1) sin α – fg cos α (t2 – t1) + m 40 m = 2.1 – 24.52 – 4.27 + 34.38 = 7.68 . s Hence:
m v2 = v2x = 7.68 . s
Before constructing Equation (1.2), we need to check if during the time interval t2 – t1 the initial velocity ʋ1(ʋ1x > 0) doesn’t change its direction. In fact, at the origin of the motion in this segment the force P > (G sin α + f G cos α) and it continues to increase. This means that the velocity of the object can’t change its initial direction. To define the velocity v3 at the time instant t3 we will apply the linear momentum–impulse principle for the time interval t3 – t2: (1.3) mv3x – mv2x = � Six ,
where
� Six = – G(t3 – t2) sin α – fG cos α (t3 – t2) + SPx.
The x component of the impulse of the variable force P⃗ during the time interval t3  t2 will be expressed by the area of trapezoid IDEK:
S = Px
4(200 + 150) = 700 N ∙ s . 2
Then:
700 = 7.68 – 9.81 ∙ 4 ∙ 0.5 – 0.1 ∙ 9.81 ∙ 0.87 ∙ 4 + 17.5 v3x = v2x – g(t3 – t2) sin α – fg cos α (t3 – t2) + 40
m = 7.68 – 19.62 – 3.41 + 17.5 = 2.15 . s
Hence:
m v3 = v3x = 2.15 . s
Before constructing Equation (1.3) we need to verify if during the time interval t3  t the velocity v2 (v2x > 0) doesn’t change its initial direction and the friction force is directed downward along the inclined surface. Thus, it is necessary to find out if there exists any time instant τ*< t3 when the velocity of the object will be equal to zero under the actions of the constant forces G⃗, N⃗, F⃗ and the force P⃗ linearly varying by line DE presented in the third segment of the graph (Figure 1.2). The equation of the line DE is defined as: 200 – 150 P = 200 – τ = 200 – 12.5τ , 4 where τ is the time counted since the time instant t2 = 8 s .
5
6
1. TOPIC AK1
Next, we will apply the linear momentum–impulse principle for the time interval between τ0 = 0 and τ = τ* : mvx – mv2x = – mgτ^* sin α – fmg cos α τ* + SPx , where for the considered case: vx = 0 and
SPx =
200 + (200 – 12.5 τ*) τ* = 200 τ* – 6.25 τ*2 . 2
Hence, the following equation is obtained to define τ* : 6.25 τ*2 – 200 τ* + mg(sin α + f cos α) τ* – mv2x = 0 ,
or
τ*2 +
230.3 – 200 40 ∙ 7.68 τ* – =0, 6.25 6.25 τ*2 + 4.85 τ* – 49.1 = 0 .
From here:
τ* = – 2.42 ± √5.86+49.1 = – 2.42 ± 7.41 (s) .
Hence,
τ* = 4.99 s .
The velocity will be equal to zero at τ = 4.99 s, but (t3 – t2) < τ*, therefore the velocity will not change its initial direction during the time interval t3 – t2 . Finally, we will verify the obtained velocity v1 at time instant t1 using the differential equation of motion of the object: mẍ =�Fxi . For the given system of forces: or
mẍ = – G sin α – F + P
mẍ = – mg sin α – fmg cos α + P ,
P ẍ = – g sin α – fg cos α + , m
250 where P = t is the equation of the line OB. m Therefore:
25 ẍ = – g sin α – fg cos α + t. 12
Integrating the obtained differential equation, we will obtain:
25 2 ẋ = – g(sin α + f cos α) t + t +C. 24
m To determine the integral constant C we will use the initial condition of the problem: at t = 0: v0x = x0̇ = 10 . s
1.3 SOLUTION
m Substituting the initial conditions in the obtained differential equation, we will find that C = 10 s . Thus, the equation of the velocity changes in the time interval from zero to t1 : 25 ẋ = – g(sin α + f cos α) t + s t2 + 10 . At time instant t = 3 s:
25 m v1 = x1̇ = – 9.81(0.5 + 0.1 ∙ 0.87) ∙ 3 + ∙ 9 + 10 = 2.1 s . 24
7
9
CHAPTER 2
Topic AK2 2.1
APPLICATION OF THE LINEAR MOMENTUM PRINCIPLE AND MOTION OF THE MASS CENTER OF A SYSTEM OF PARTICLES
The object 2 (Figures 2.1–2.30) is driven via the motor mounted on the object 1 (Figures 2.2, 2.3, 2.8, 2.11, 2.16, 2.17, 2.19, 2.21–2.23, 2.25, 2.27), or it is a part of the selfmoving device (Figures 2.1, 2.4–2.7, 2.9, 2.10, 2.12–2.15, 2.18, 2.20, 2.24, 2.26, 2.28–2.30). The angular velocity of this object varies according to the following equation: ω2 = ω0 [1 – exp(–st)]. Define the equation of the motion and the velocity of the object 1 if the resistance force on this object is given as: ⃗ . R⃗ = bv⃗ + Ffr
The following parameters are given: mi – the mass of the ith object; ri, Ri – the radii of the small and large circles, respectively; b – a constant coefficient; f – the friction coefficient between the object 1 and the stationary surface (Ffr = fN); and α, β – the angles as shown in Figures 2.1–2.30. In all problems the ropes are weightless and nonextendable. The slips of the ropes and the rolling wheels are ignored.
10
2. TOPIC AK2
3
4
ω2 2
1 β
α
Figure 2.1.
ω2
2
1
3
β α
Figure 2.2.
2.1 APPLICATION OF THE LINEAR MOMENTUM PARINCIPLE 11
ω2
3 4
2
1
α Figure 2.3.
1
ω2
5
3 4
2 β α
Figure 2.4.
12
2. TOPIC AK2
3
2 ω2
1 4
β α
Figure 2.5.
2 4
ω2
β
α Figure 2.6.
3
1
2.1 APPLICATION OF THE LINEAR MOMENTUM PARINCIPLE
3 4
1
5
2 ω2
α
β
Figure 2.7.
3
5
ω2 2
4
1 α Figure 2.8.
13
14
2. TOPIC AK2
3
4
ω2 2
α
1
5 Figure 2.9.
2 3
ω2
5 1
4 β α Figure 2.10.
2.1 APPLICATION OF THE LINEAR MOMENTUM PARINCIPLE
3
2 ω2
5
4
α
α
1
Figure 2.11.
3
4
5 α Figure 2.12.
2
ω2 1
15
16
2. TOPIC AK2
4
2
3 ω2
5
2 α
ω2
1
Figure 2.13.
5
3
2
α
4 Figure 2.14.
ω2
2
1
2.1 APPLICATION OF THE LINEAR MOMENTUM PARINCIPLE
4
3
2
1
ω2 α Figure 2.15.
ω2
4
2
3 3 1
α Figure 2.16.
17
18
2. TOPIC AK2
2 ω2 4
3 1
α Figure 2.17.
4 3
4
ω2
2 α
Figure 2.18.
ω2
2 1
2.1 APPLICATION OF THE LINEAR MOMENTUM PARINCIPLE
2 ω2
4
3
1 5
α
Figure 2.19.
4
ω2 2
3
5
α Figure 2.20.
1
19
20
2. TOPIC AK2
3 4
α 1 ω2
2 Figure 2.21.
3
ω2
4
2
1 5
α Figure 2.22.
2.1 APPLICATION OF THE LINEAR MOMENTUM PARINCIPLE
3 4
1 2 ω2 α Figure 2.23.
3
4
5
3
2 ω2
3 3 α Figure 2.24.
1
21
22
2. TOPIC AK2
ω2
2 3
4 1 α Figure 2.25.
3
2
4
2
60°
2
60°
ω2
Figure 2.26.
α
1
2.1 APPLICATION OF THE LINEAR MOMENTUM PARINCIPLE
ω2
3 2 1 4
α Figure 2.27.
2
5 3
ω2
4 1
α Figure 2.28.
23
24
2. TOPIC AK2
3
5
2
ω2
4 α
1
Figure 2.29.
2
3 ω2 4
1
5 α Figure 2.30.
2.2
SAMPLE PROBLEM
The object 2 is a part of the selfmoving device (Figure 2.31). The angular velocity of this object varies according to the following equation: ω2 = ω0 [1 – exp(–st)].
2.3 SOLUTION
25
Define the equation of the motion and velocity of the object 1 if the resistance force on this object is given as: R⃗ = bv⃗ + F⃗fr .
2.3
SOLUTION
For the mechanical system consisted of the objects 1–6 and given in Figure 2.31 we will apply the linear momentum– impulse principle: (2.1) K⃗ – K⃗0 = �S i⃗ext , where K⃗0, K⃗ – the linear momentums of the mechanical system at the initial time instant t = 0 and at the time instant t, respectively; �S ie⃗ xt – a geometrical sum of impulses of all external forces acting on the given mechanical system during the time interval t.
Since at t = 0 the mechanical system was at the rest, then K0 = 0. Therefore, Equation (2.1) can be rewritten as:
K⃗ = �S i⃗ext .
K⃗ = �S ix⃗ext ,
(2.3a)
(2.3b)
(2.2)
Equation (2.2) in the algebraic form on axes x and y can be written as:
K⃗ = �S iy⃗ext .
The linear momentum of the given mechanical system is determined as a geometrical sum of the linear momentum of all objects of this system: K⃗ = K⃗1 + 2K2 + K3⃗ + K4⃗ + K5⃗ + K6⃗ .
The linear momentum of each object is determined as a product of the mass of this object and the absolute velocity its mass center (Figure 2.32). Let’s determine the linear momentums of the objects 1–6. The linear momentum of the object 1:
K⃗1 = m1 v1⃗ .
The object 2 (two wheels) and the object 3 (a conveyor belt) have the common mass center at point C2–3. The velocity ⃗ 2. Then, the linear momentum of these objects of this point is equal to the velocity of the point C2, e.g., ʋ⃗C2–3 = ʋC will be determined as: ⃗ = 2K2⃗ + K3⃗ = (2m2 + m3 ) vC ⃗ 2. K2–3
The linear momentums of the objects 4–6 will be determined as: ⃗ 4, K4⃗ = m4 vC
⃗ 5, K5⃗ = m5 vC ⃗ 6, K6⃗ = m6 vC
26
2. TOPIC AK2
where v⃗C2, v⃗C4, v⃗C5, v⃗C6 – the absolute velocities of the mass centers of the corresponding objects. The nonmoving reference system xOy we attach to the nonmoving surface along which the object 1 is moving. The moving reference system x1O1y1 we attach to the object 1. The absolute velocity of the mass center of each object is equal to the geometrical sum of the transfer and the relative velocities. The transfer velocity of the object 1 will be the same for the transfer velocities of the other objects: v⃗1 = ve⃗ .
The absolute velocity of the mass center C2–3 :
⃗ 2r = v1⃗ + vC ⃗ 2r , v⃗C2 = v⃗e + vC
where v⃗C2r – is the relative velocity of the mass centers C2 of the wheels during the rolling without a slip. The magnitude of this velocity: vC2r = ω2 R2 . The absolute velocity of the mass center C4 :
v⃗C4 = ve⃗ = v1⃗ .
The absolute velocity of the mass center C5 :
⃗ 5r = v1⃗ + vC ⃗ 5r , v⃗C5 = ve⃗ + vC
where v⃗C5r – is the relative velocity of the mass centers C5 . The magnitude of this velocity: vC5r = 1 (vLr – vMr), 2
where vLr,vMr – are relative velocities of the points L and M of the object 5. These velocities can be determined via angular velocity ω4 of the object 4: vLr = ω4R4; The angular velocity ω4 can be defined as: ω4 =
vMr = ω4r4.
vC5r ω2 R2 = . R4 R4
r Then: vLr = ω2R2 and vMr = ω2R2 4 , R4 r r vC5r = 1 �ω2R2 – 4 ω2R2� = 1 ω2R2 �1 – 4 �. 2 R4 2 R4 The absolute velocity of the mass center C6 : ⃗ 6 = vC ⃗ 5. vC Next, we will find the x components of the absolute velocities of the mass centers of the objects: v1x = ve = ẋ ,
2.3 SOLUTION
27
where x – is the coordinate of the mass center C1;
vC2x = ve – vC2r cos α = ẋ – ω2R2 cos α ;
(2.4a)
vC4x = ve = ẋ ;
(2.4b)
r 1 vC5x = ve – vC5r cos β = ẋ – ω2R2 ∙ �1 – 4 � cos β ; 2 R4
(2.4c)
vC6x = vC5x.
(2.4d)
Hence, the x component of the linear momentum of the entire mechanical system:
r 1 Kx = m1ẋ + (2m2 + m3)(ẋ – ω2R2 cos α) + m4ẋ + (m5 + m6)�ẋ – ω2R2 �1 – 4 � cos β � 2 R4 or r 1 Kx = mẋ – ω2R2 �(2m2 + m3) cos α + (m5 + m6) ∙ �1 – 4 � cos β �, (2.5) 2 R4 where m = m1 + 2m2 + m3 + m4 + m5 + m6 .
28
2. TOPIC AK2
2
4
ω2
3
2 ω2
5
1
6
β
α Figure 2.31.
2.3 SOLUTION
2
3
2
ω2
4
C2 C4
VC2r
C23 VC2r
G2
C2
G3
VC2r
ω2
G2
G23 G4
L
5
Ve C5
O1
O
G1
C6 G5
1
C1
6 M
29
G6 β
α
Figure 2.32.
The sum of the x components of the impulses of all external forces:
�Six ext = Gt sin α – � bẋdt – � Ffr dt , 0 0 where G – is a weight of entire system. t
t
If to consider that:
t x � bxḋ t = b� dx = bx 0 0 where N – is the normal pressure force, we will obtain:
and
Ffr = fN ,
t �S ext ix = Gt sin α – bx – f� Ndt . 0 Substituting the expressions (2.5) and (2.6) into Equation (2.3a) we will obtain:
(2.6)
r t 1 mẋ – ω2R2 �(2m2 + m3) cos α + (m5 + m6) ∙ (1 – 4 ) cos β� = Gt sin α – bx – f� Ndt. (2.7) 2 R4 0 The y components of the absolute velocities of the mass centers of the objects: v1y = 0 ,
(2.8a)
30
2. TOPIC AK2
vC2y = – vC2r sin α = – ω2R2 sin α ,
(2.8b)
vC4y = 0 ,
(2.8c)
r 1 vC5y = vC5r sin β = ω2R2 �1 – 4 � sinβ, (2.8d) 2 R4 vC6y = vC5y . (2.8e) Hence, the y component of the momentum of the entire system: r 1 Ky = – (2m2 + m3) ω2R2 sin α + (m5 + m6) ω2R2 �1 – 4 � sin β 2 R4 or r 1 Ky = ω2R2 � (m5 + m6) �1 – 4 � sin β – (2m2 + m3) sin α �. 2 R4 The sum of the y components of the impulses of all external forces: ext = – Gt cos α + � Ndt . �S iy 0 Substituting the expressions (2.9) and (2.10) into Equation (2.3b) we will obtain: t
r t 1 ω2R2 � (m5 + m6) �1 – 4 � sin β – (2m2 + m3) sin α� = – Gt cos α + � Ndt. 0 2 R4 From here: r t 1 � Ndt = Gt cos α + ω2R2 � (m5 + m6) �1 – 4 � sin β – (2m2 + m 3) sin α�. 0 2 R4 Substituting the expression (2.12) in Equation (2.7) we obtain:
(2.9)
(2.10)
(2.11)
(2.12)
r 1 mẋ + bx = ω2R2 �(2m2 + m3) (cos α + f sin α ) + (m5 + m6) �1 – 4 � (cos β – f sin β)� + Gt(sin α – f cos α) , 2 R4 (2.13)
where
ω2 = ω0 [1 – exp(–st)].
Dividing both sides of Equation (2.13) into m we obtain:
ẋ + ηx = h[1 – exp(–st)] – qt , where
η=
(2.14)
b , m
r 1 ω0 R2 �(2m2 + m3)(cos α + f sin α ) + 1 (m5 + m6)�1 – 4 �(cos β – f sin β)� , 2 R4 h= m q = g(f cos α – sin α) . The solution of Equation (2.14) consists of the general solution of the corresponding firstorder homogeneous equation ẋ + ηx = 0 and the particular solution of this equation: x = x1 + x2 . (2.15)
2.3 SOLUTION
31
A general solution of the firstorder homogeneous equation is defined as: x1 = Cexp(–ηt). A particular solution of Equation (2.14): x2 = Q1exp(–st) + Q2t + Q3. Substituting the particular solution x2 into Equation (2.14), we obtain: –Q1s exp(–st) + Q2 + η[Q1exp(–st) + Q2t + Q3] = h[1 – exp(–st)]–qt . Equating the coefficients of the variable parameters exp(–st) and t, as well as the other terms, we will obtain the following equations: –Q1s + ηQ1 = –h , ηQ2 = –q , Q2 + ηQ3 = h , from where:
h Q1 = s – η , q Q2 = – η , h q Q3 = η –+ η2 .
Hence,
h h q q x = x1 + x2 = C exp(–ηt) + s–η exp(–st) + η+ η2 – η t . (2.16) The constant C we will find from the initial condition: at t = 0
x0 = 0 .
Using this condition we can find the coefficient C :
h 1 q C = – s–η– η � η + h� .
Thus, the equation of motion of the object 1 will be expressed as:
h 1 q q x = s–η [exp(–st) – exp(–ηt)] + η � η + h� [1 – exp(–ηt)] – η t . (2.17)
The velocity of the object 1 :
h q q hs q v=ẋ = s–η [ηexp(–st) – sexp(–st)] + �η + h� exp(–ηt) – η = s–η [exp(–ηt) – η [1 – exp(–ηt)] . – exp(–st)]
(2.18)
32
2. TOPIC AK2
The expression (2.18) reveals that the velocity of the object 1 after certain time interval will be equal to zero and hence the object will stop. Therefore, Equation (2.17) is valid only for this time of interval. To study the motion of the given mechanical system we will apply the equation of the center of mass motion: ma⃗C = �F⃗iext ,
(2.19)
where �F⃗iext – is a geometrical sum of all external forces acting on the given system. The differential equations of the center of mass motion in algebraic form:
ext mẍC = �F xi = F ext x , ext mÿC = �F yi = F ext y . The coordinates of the center of mass:
(2.20a) (2.20b)
m1xC1 + (2m2 + m3) xC2–3 + m4 xC4 + m5 xC5 + m6 xC6 , xC = m yC =
(2.21a)
m1 yC1 + (2m2 + m3) yC2–3 + m4 yC4 + m5 yC5 + m6 yC6 .
(2.21b)
m
Differentiating the expressions (2.21a) and (2.21b) twice by time, and considering the following: ̈ 1 = ẍ, xC
̈ 2–3 = xC ̈ 2, xC
̈ 6 = xC ̈ 5, xC
̈ 1 = y1̈ = 0, yC
̈ 2–3 = yC ̈ 2, yC
̈ 6 = yC ̈ 5, yC
we will obtain:
̈ 2 + m4xC ̈ 4 +(m5 + m6)xC ̈ 5 , m1ẍ + (2m2 + m3)xC ẍC = m
(2.22a)
̈ 2 + m4yC ̈ 4 +(m5 + m6)yC ̈ 5 . (2m2 + m3)yC ÿC = (2.22b) m Using the x and y components of the velocities of the mass centers provided in Equations 2.25a and 2.25b and 2.8a – 2.8d, we can determine the x and y components of the mass centers of the objects: dω2 dvC2x = ẍ – R ẍ = cos α , C 2 2 dt dt
(2.23a)
dvC4x = ẍ , = ẍ C 4 dt
(2.23b)
dvC5x r 1 dω2 = ẍ – R �1 – 4� cos α , ẍC5 = dt 2 dt 2 R4
(2.23c)
2.3 SOLUTION
dω2 dvC5y ÿC2 = = ẍ – R sin α , dt dt 2
(2.23d)
̈ 4 = 0 , yC
dvC5Y r dω2 1 = – R2 �1 – 4 � cos α , ÿC5 = dt 2 dt R4
33
(2.23e)
(2.23f )
Substituting the expressions (2.23a – 2.23f ) in Equations (2.22a) and (2.22b):
dω 1 dω2 r R2 �1 – R 4 � cos β � , (2.24a) (m1ẍ + 2m2 + m3)�ẍ – dt 2 R2 sin α� + m4ẍ + (m5 + m6)�ẍ – 2 dt 4 ẍC = m
dω 1 dω r –(2m2 + m3) dt 2 R2 sin α + (m5 + m6) ∙ 2 R2 dt 2 �1 – R4 � sin β , 4 y ̈C = m or
(2.24b)
dω 1 r mẍ – R2 dt 2 �(2m2 + m3) cos α + 2 (m5 + m6)�1 – R 4 � cos β� , 4 ẍC = m
yC ̈ =
R2
dω2 1 r4 dt �2 (m5 + m6) (1 – R4 � sin β – (2m2 + m3) sin α . m
The x and y components of the resultant vector of all external forces acting on the system: ext = G sin α – bẋ – Ffr , F xext = �F xi
(2.25a)
ext = –G cos α + N . (2.25b) F yext = �F yi Substituting the expressions (2.24a), (2.24b), (2.25a), and (2.25b) into the differential Equations (2.20a) and (2.20b) of the motion of the mass center of the system, we will obtain:
or and
mẍ – R2
dω2 1 r �(2m2 + m3) cos α + (m5 + m6)�1 – 4 � cos β � = G sin α – bẋ – fN dt 2 R4
mẍ + bẋ = R2
dω2 1 r �(2m2 + m3) cos α + (m5 + m6)�1 – 4 � cos β � + G sin α – fN dt 2 R4
(2.26a)
dω2 1 r R � (m5 + m6)�1 – 4 � sin β – (2m2 + m3) sin α = – G cos α + N ] . (2.26b) dt 2 2 R4 Considering that the angular acceleration of the wheels 2 are equal:
ε2 =
dω2 = ω0s exp(–st) , dt
34
2. TOPIC AK2
from Equation (2.26b) we can find the normal reaction force N : 1 r N = ω0s exp(–st) R2 � ( m5 + m6)�1 – 4 � sin β – (2m2 + m3) sin α� + G cos α. 2 R4
(2.27)
Substituting the expression (2.27) into Equation (2.26a) we will obtain the differential equation of the motion of the object 1: ẍ + ηẋ = hs exp(–st) – q ,
(2.28)
b where η = m , 1 r ω0 R2 �(2m2 + m3)(cos α + f sin α) + (m5 + m6)�1 – 4 �(cos β – f sin β)� , 2 R 4 h= m q = g(f cos α – sin α) . The general solution of the differential equation (2.28) will be: x = x1 + x2 ,
(2.29)
where x1 = C1 + C2 exp(–ηt) – is the general solution of the homogeneous equation: ẍ + ηẋ = 0 .
The particular solution x2 will be defined as: x2 = Q1 exp(–st) + Q2t . Substituting the particular solution x2 into Equation (2.28) we can find Q1 and Q2 : Q1s2 exp(–st) – ηsQ1 exp(–st) + ηQ2 = hs exp(–st) – q . Equating the coefficients of the exp(st) and the constant terms: Q1(s – η) = h , ηQ2 = –q , we will obtain:
h Q1 = s – η , q Q2 = – η .
Then the general solution of the differential equation (2.28) will be:
q h x = x1 + x2 = C1 + C2 exp(–ηt) + s – η exp(–st) – η t . hs q dx ẋ = = –ηC2 exp(–ηt) s – η exp(–st) – η . dt
(2.30) (2.31)
2.3 SOLUTION
The integral constants C1 and C2 can be determined using the initial conditions: at t = 0
x0 = 0 ,
35
x ̇0 = 0 .
At t = 0, Equations (2.30) and (2.31) can be written as:
h 0 = C1 + C2 + s – n , hs q 0 = –ηC2 – s – η – η , from where:
q h h hs q h C 1 = –C2 – s – η = η(s – η) + η2 – s – η = η + η2
.
Hence the equation of motion of the object 1 will be:
1 q q h h) ∙ �1 – exp(–ηt) – η �. (2.32) x – exp(–ηt)] + η ( η + = s – η [exp(–st)
The same equation has been obtained during the study of the motion of the mechanical system using the momentum– impulse principle for the system (2.17).
37
CHAPTER 3
Topic AK3 3.1
APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY OF THE SYSTEM OF PARTICLES
The object H of mass m1 rotates around a vertical axis z with angular velocity ω0. At point O of the channel AB of the object H at the distance AO from the point A measured along the channel, there is a particle K of mass m2. At certain time instants (t = 0) a pair of forces with the moment Mz = f1(t) starts acting on the system. At t = τ this pair of forces stops acting on the system. At the same time the particle K starts a relative motion from the point O along the channel AB (toward to the B) According to the equation of motion OK = s = f2 (t–τ) at t > τ. Determine the angular velocity of the object H at both t = τ and t = T ignoring the resistance to the rotation of the object H. The object H consider as a flat plate in the shapes, as shown in Figures 3.1–3.30. The required information is given in Tables 3.1 and 3.2. In the problems where the plate H is located on the vertical plane, a relative motion of the particle K will be created by the force acting on the same plane. For other problems it is assumed that the particle K is a selfmoving cart.
z
H
Figure 3.1.
A
2b
2a
B
38
3. TOPIC AK3
z
H
B α
A
Figure 3.2.
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z
H
A
Figure 3.3.
B
39
40
3. TOPIC AK3
z
B
α
H A
Figure 3.4.
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z .
b
B
H A
Figure 3.5.
a
z a
A B
Figure 3.6.
H
41
42
3. TOPIC AK3
z
H
B
a
a
Figure 3.7.
z
A
H
Figure 3.8.
A
B
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z
B
A H α
a/2 Figure 3.9.
α
a/2
43
44
3. TOPIC AK3
z H
a/2
a/4 B
A a
Figure 3.10.
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z
B
α a
Figure 3.11.
A
H
45
46
3. TOPIC AK3
z
A
Figure 3.12.
B
H
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z
H
A
B
a
Figure 3.13.
a
47
48
3. TOPIC AK3
z
H
B A
Figure 3.14.
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z
H
A
B
a
a
2a
Figure 3.15.
z b
b
a
B
H
α α
A
Figure 3.16.
49
50
3. TOPIC AK3
z
B H
α A
Figure 3.17.
z
H
B
A
Figure 3.18.
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z
A
H a
B
a
Figure 3.19.
z
H B
A
Figure 3.20.
51
52
3. TOPIC AK3
z
B
H
Figure 3.21.
A
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z
H
B
A
a
a
Figure 3.22.
z
a
a
B
A
Figure 3.23.
H
53
54
3. TOPIC AK3
z
B
H
α
A
a
Figure 3.24.
z
H
A
Figure 3.25.
B
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z
A
B
H
Figure 3.26.
z
a
a α α
A
Figure 3.27.
B α
55
56
3. TOPIC AK3
z
H
B
A
a
Figure 3.28.
z
a
a
a
B
H
Figure 3.29.
A
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
z
H B
A a
a
Figure 3.30.
Table 3.1
Mz = f 1 (t), N·m
τ, s
T, s
–29.6 t2
3
4
101
5
6
√3 (t – τ)2
0
–120t
4
6
√2 2 4 (t – τ)
30
0.4
21t
2
6
0.6√t – τ


15√t
4
6.5

2.5

0 πα 6
700t
√3
2√3
1.6
1
0.8

968
1
2
0
1.2

2

240√t
4
8
5
1.2

0.4
45
29.2t
3
4
0.5(t – τ) 5πa 2 18 (t – τ) πR 2 2 (t – τ) πa 4 √t – τ 3πR 2 4 (t – τ)
No.
m1, kg
m2, kg
ω0, s1
a, m
1
32
10
1
1
1.5
1.2

2
200
60
2


2
120
3
120
40
0
2



4
16
5
3


1
5
66
10
1.5
2
1.5
6
160
80
1.25
1.5
7
300
50
2
8
80
20
9
20
5
b, m R, m
α, deg
AO, m πR 6 √3 2
0 πα 2 πR 4
OK = s = f2(t – τ), m 5πR 12 (t –τ)
57
58
3. TOPIC AK3
10
100
40
2
2
√2


11
60
20
1
2


15
√2 2 0
12
40
10
3
1

2

13
24
4
4
1


14
40
10
2


15
120
50
4
1
16 17
60 50
10 10
5 2
18
120
50
19
90
20
√2 4 (t – τ)
90√t
4
5
40t
2
4
0
50t2
3
5

0.5
27√t
1
3
0.3(t – τ)
1

0
120t
1
4

2

0
330t2
2
3
1 
1.2 
1.6
30 30
74 69t
2 4
6 6
3
2
3
0.8

0.4 0.6 πR 2
0.5(t – τ) πa 2 2 (t – τ) 0.3√(t – τ)
324
3
5
30
1
1.5



0
135t
2
3
50
12
3
1

1.2

14t2
3
5
21
40
10
6


1

75√t
1
3
22
150
50
1
1.6
1.2
0.6

πa 6 √2 2 πR 2
163
4
5
23
90
20
2
√2
1


√3 2
210
2
3
24
50
12
3
0.6


60
0.2
27t2
2
6
25
36
8
5


0.5

0
20t
2
4
26
150
40
4
1.5

2

1170√t
1
2
27 28
120 15
30 4
0 2
1 0.6


60 
πa 6 0 0.1
25t 5.6t
2 3
3 4
29
20
5
5
0.6

0.6

0
6.3√t
4
5
30
150
50
0
1.6
1.2


1.6
652t
2
4
0.4(t – τ)2 πa 3 (t – τ)
0.6(t – τ) πR 2 8 (t – τ) πa 4 (t – τ) πR 2 12 (t – τ) √2 2 6 (t – τ) πR 2 (t – τ) √3 2 (t – τ) 0.4√(t – τ) πR 2 6 (t – τ) πa 2 2 (t – τ) (t – τ)2 0.4√(t – τ) 5πR 6 (t – τ) 0.2(t – τ)2
Note: a negative sign of Mz and ω corresponds to the clockwise direction of the rotation if we look from the positive side of the axis z.
3.1 APPLICATION OF THE ANGULAR MOMENTUM PRINCIPLE TO DETERMINE THE ANGULAR VELOCITY
Table 3.2: Axes moments of inertia of homogeneous plates
𝐽x mR2 2
z
𝐽y mR2 4
𝐽z mR2 2
z
R
R
O
y
O
y
O O
y
x
x
m(R2 + r2) 2
z
m(R2 + r2) 4
m(R2 + r2) 4
z
R r
r O
y
R
x
x
m(a2 + b2) 2
mb2 3
ma2 3
z
b
z
b
O
b
y
a
a
x z
O
a
y
b
x
O
y
a
ma2 3
0
a
ma2 3
z
y
O
a
x
x
m(3a2 + b2) 18
z
mb2 18
ma2 6
z
a
y
b/3
x a
O OO
b b/3
b
O O
x a
y
59
60
3. TOPIC AK3
3.2
SAMPLE PROBLEM
The given parameters: m1 = 200 kg; m2 = 80 kg; Mz = 592 t N ∙ m; ω0 = –2 s1; AO = 0.8 m; R = 2.4 m; a = 1.2 m; τ = 4 s; T = 6 s; OK = 0.5 (t – τ)2 m. Determine ωτ and ωT assuming the object H as a homogeneous round plate.
3.3
SOLUTION
To solve the problem we will apply an angular momentum principle for the mechanical system: dLz E dt = �Miz ,
where Lz – is an angular momentum of the mechanical system consisting of the object H and the particle K about axis z; E
E
�Miz = Mz – is the total moment of external forces applied to the mechanical system about axis z.
The following external forces are acting on the given mechanical system during the time interval from t = 0 to t = τ: the weight G⃗1of the object H; the weight G⃗2 of the particle K; the pair of forces with the moment Mz; and the reaction forces of two supports (Figure 3.31).
Assuming the rotation of the object H in the counter clockwise direction, we can determine an angular momentum of the given system, which consists of the angular momentum 𝐽zω of the object H and the angular momentum of the particle K located at the point O of the object H with the velocity v = ω ∙ O1O: m2v ∙ O1O = m2ω ∙ O1O . Then: Lz = 𝐽zω + m2ω ∙ (O1O)2 = [𝐽z + m2 (O1O)2]ω.
The total moment of the external forces is equal to the rotating moment Mz, as the other forces don’t create moments about axis z. Then, the equation of the angular momentum principle can be written as: d{[𝐽z + m2 ∙ (O1O)2]ω} = Mz, dt where Mz = ct (c = 592 N ∙ m/s).
(3.1)
Separating the variables in Equation (3.1) and integrating both sides, we will obtain: ωτ
[𝐽z + m2 ∙ (O1O)2] �
ω0
Then:
τ
dω =� ctdt. 0
[𝐽z + m2 ∙ (O1O)2] ∙ (ωτ – ω0) = cτ . 2 2
(3.2)
3.3 SOLUTION
61
Let’s find the numerical values of the parameters in Equation (3.2). The moment of the inertia of the object H about axis z can be determined from the principle of the relationship between the moment of inertias about parallel axis: 𝐽z = 𝐽zG + m1 a2,
where 𝐽zG – is the moment of the inertia of the homogeneous round plate object H about the vertical axis crossing the center of the gravity C of this object and parallel to the axis z: 𝐽zG =
Then: From Figure 3.31:
𝐽z =
m1R2 . 2
m1R2 200 ∙ 2.42 + m1 a2 = + 200 ∙ 1.22 = 864 kg ∙ m2. 2 2
O1O2 = OC2 + O1C2 = 1.62 + 1.22 = 4 m2. Therefore:
Then from Equation (3.2):
𝐽z + m2 ∙ O1O2 = 864 + 80 ∙ 4 = 1,184 kg ∙ m2. 2
1,184[ωτ – (–2)] = 592 ∙ 4 , 2
or
ωτ = 2 s–1.
During the time period from t = τ to t = T the forces acting on the given mechanical system are gravity forces G1 and G2, and the reaction forces (Figure 3.32). E
As the rotating moment Mz is removed, e.g., �Miz = 0, then:
dLz = 0, dt
Lz = const.
Next, we will determine the angular moments Lzτ at t = τ and Lzτ at t = T, and then we will equate their values. At t = τ:
2
Lzτ = (𝐽z + m2 ∙ O1O2)ωτ = 1,184 ∙ 2 = 2,368 kg ∙ m . s
At t > τ the velocity of the particle K is defined as a sum of the relative velocity v⃗r relative to the object H and the transfer velocity ve⃗ of its motion with the object H together. Therefore, at t = T we will show (Figure 3.32) two vectors of the particle momentums: m2vr⃗ and m2ve⃗ ․ At t = T:
LzT = 𝐽zωT + m2ωT ∙ O1KT2 – m2vr ∙ O1C.
62
3. TOPIC AK3
We can define: where
O1KT2 = O1C2 + CKT2 , CKT = OKT – OC, OKT = st = T = 0.5(T – τ)2 = 0.5(6 – 4)2 = 2 m,
e.g.,
CKT = 2 – 1.6 = 0.4 m, O1KT2 = 1.22 + 0.42 = 1.6 m2.
The relative velocity:
vr = ds = 2 ∙ 0.5(t – τ). dt At t = T: vr = 2 ∙ 0.5(6 – 4) = 2 m. s Therefore: Equating Lzτ and LzT:
LzT = 864ωT + 80T ∙ 1.6 – 80 ∙ 2 ∙ 1.2 = 992ωT – 192. 2,368 = 992ωT – 192,
we can obtain: ωT = 2.59s–1.
3.3 SOLUTION
Z D
YD
mv XD
v
MZ
A
90° H
O
O1 C
B
a
G2 G1
ZE
E
XE x Figure 3.31.
YE
y
63
64
3. TOPIC AK3
z D
YD
XD
A m2ve
O H
O1
ve
C 90° KT B vr
ZE
E
XE x Figure 3.32.
G2
m2vr
G1
YE
y
65
CHAPTER 4
Topic AK4 4.1
DETERMINATION OF VELOCITY AND ACCELERATION OF PARTICLES BY GIVEN EQUATIONS OF MOTION
From the given equations of motion of particle M in the table below, define an equation of path, position of particle M on path at t = t1 (sec), velocity, normal, tangent, and full accelerations of the particle M, and radius of curvature at t = t1. Show the defined parameters on the graph. Problem # 1 2
Equations of Motion of Particle M x = x(t), cm y = y(t), cm z = z(t) 2 –5t 3t –2t + 3 π 4 cos2 � t�+ 2 3
π 4 sin2 � t� 3
π – cos � t2�+ 3 3
π sin � t2�– 1 3
6
π 2 sin � t� 3 2 3t + 2
π – 3 cos � t�+ 4 3
7
3t2 – t + 1
3 4 5
8 9 10 11 12 13 14 15 16 17
4t + 4
–4 t+1
–4t
5t2 –
5 t–2 3
π 7 sin � t2�+ 3 6
π 2 – 7 cos � t2� 6
π – 4 cos � t� 3 –4t2 + 1
π – 2 sin � t�– 3 3
π 5 sin2 � t� 6
π – 5 cos2 � t�– 3 6
–2t – 2
–2 t+1
–3 t+2
π 5 cos � t2� 6
3t + 6
–3t
π – 5 sin � t2� 3
π 4 cos � t� 3 3t
π – 3 sin � t� 3
π 7 sin2 � t� – 5 6
π – 7 cos2 � t� 6
4t2 + 1
t1, sec ½
2t
1
1.5t
1
4t + 4
2
T
1
3t
½
2.5t
1
5t
1
4t + 8
2
T
1
2t
½
3t
1
1.5t
1
2t + 2
2
3t
1
1.5t
½
5t
1
66
4. TOPIC AK4
19
π 1 + 3 cos � t2� 3
20
2 – 3t – 6t2
18
22
π 6 sin � t2� – 2 6
23
3 – 3t + t
21
24 25 26 27 28 29 30
4.2
– 5t2 – 4
7t2 – 3
π – 4 cos � t� – 1 3
π 3 sin � t2� + 3 3
3.5t
1
6t
1
2t
0
4t
1
t
½
1.5t
1
2t
1
5t
1
6t
1
3.5t
1
– 3t
4t
1
3t2 + t + 3
5t
1
1.5t
1
3t
3–
3 t – 3t2 2
π 6 cos � t2� + 3 6 5t
4 – 5t2 +
5 t 3
– 6t
π – 4 sin � t� 3
π 8 cos2 � t� + 2 6
π – 8 sin2 � t� – 7 6
π – 3 – 9 sin � t2� 6 – 4t2 + 1
5t2 +
5 t–3 3
π 2 cos � t2� – 2 3
– 2t2 – 4
π – 9 cos � t2� + 5 6
π – 2 sin � t2� + 3 3
SAMPLE PROBLEM
From the given equations of motion (4.1) of particle M define an equation of path, position of particle M on path at t = t1 (sec), velocity, normal, tangent, and full accelerations of the particle M, and radius of curvature at t = t1. Show the defined parameters on the graph. 4 x= (cm) t+1 � y = –4t – 4 (cm) (4.1) z = 2t + 2 (cm) t1 = 0.
4.3
SOLUTION
Equations of motion (4.1) are parametric equations of particle M trajectory. To obtain an equation of trajectory on a usual coordinate form we will exclude time t from the first and second equations of the system, as well as from the second and third equations. Then: xy = –16,
(4.2)
y = –2z.
(4.3)
4.3 SOLUTION
67
Equation (4.2) expresses in plane xOy an equalsided hyperbola, where the coordinate axes are its asymptotes. In space this equation describes hyperbolic cylinder with the surface tracer line parallel to the axis Oz. Equation (4.3) expresses a straight line on the plane yOz, which crosses the origin of the coordinate system. In space it is a plane which contains the axis Ox. To find the velocity of the particle we need to find the x, y, and z components of velocity: 4 cm , vx = ẋ = – (t + 1)2 s
cm vy = ẏ = – 4 , s cm vz = ż = 2 . s Then the magnitude of the velocity:
5(t + 1)4cm . 2 �4 + v= + = s (t + 1)2 √vx2
vy2 + vz2
Similarly, we can find the x, y, and z components of the acceleration ax = ẍ =
8 ; a = ÿ = 0; az = z̈ = 0. (t + 1)3 y
A magnitude of the full acceleration of the particle M:
a = √ax2 + avy2 + az2 =
8 cm . (t + 1)3 s2
The coordinates, velocity, acceleration, and their x, y, and z components at time instant t = 0 are shown in Table 4.2. Table 4.2
Coordinates, cm x y z 4 4 2
Velocity, cm/s vx vy vz v 4 4 2 6
Accereration, cm/s2 Radius of Curvature, cm ay az a aτ an ρ 0 0 8 5.33 5.96 6.04
ax 8
Tangent acceleration we find by differentiating module of velocity (4.3): dv aτ = � � dt
dv 2 vx v̇x + 2 vy v̇y+ 2 vz v̇z vxax + vyay + vzaz –4 ∙ 8 cm = = = = – 5.33 2 . 2 2 2 dt + + v 6 s 2 √vx vy vz dv A negative sign for dt shows the particle motion is decelerated, and, hence, aτ⃗ and v⃗ are collinear but opposite to each other. Normal acceleration of particle at given time instant:
cm an = √a2 – aτ2 = √82 – 5.332 = 5.96 s2 .
68
4. TOPIC AK4
The radius of curvature of the trajectory at point where the particle M is located at t = 0 v2 62 ρ = a = 5.96 = 6.04 cm. n
The calculated values of aτ, an, and ρ are also shown in Table 4.2.
Using equation (4.2) we can draw a trajectory (Figure 4.1) and show a position of the particle M at the given time instant. The vector v⃗ is constructed by the components vx⃗, vy⃗ , and vz⃗. This vector v⃗ must be tangent to the particle trajectory. The vector a⃗ we find by components aτ⃗ and an⃗ as well, which double checks the correctness of the calculations.
z
y+2z=0
vz v M vx y
vy
16
4 4
an 16
x Figure 4.1.
0 1
at
a=ax xy=16
2
69
CHAPTER 5
Topic AK5 5.1
DIFFERENTIAL EQUATIONS OF MOTION OF A RIGID BODY: ROTATIONAL MOTION OF A RIGID BODY
At some time instant (t = 0) the force couple with the moment M (driving moment) or driving force P are applied to the link 1 of the given mechanism with angular velocity ω10. The mechanisms are schematically shown in Table 5.1. The required information is given. Determine the equation of the rotational motion of the link of the mechanism shown on the last column of the Table 5.1. Also determine the tension in the ropes at the given time instant. In the problems where the links 1 and 2 are in touch, determine the circular force at the contact point. The links 1 and 2 must be considered as homogeneous and solid if their radii of inertia ix1 and ix2 are not given. Nomenclature: m1, m2, m3 – are masses of the links 1 and 2, and the lifted load 3, respectively; MC – moment of the resistance forces applied to the driven link 2; and R1, r1, R2, r2 – the radii of large and small circles of the links 1 and 2, respectively.
Mc M
1
Figure 5.1.
2
3
70
5. TOPIC AK5
P
1
90° Mc 2
3
Figure 5.2.
1
M Figure 5.3.
Mc
2
3
5.1 DIFFERENTIAL EQUATIONS OF MOTION OF RIGID BODY
2
Mc
1
Figure 5.4.
M
3
90°
Mc
P
2
Figure 5.5.
1
3
71
72
5. TOPIC AK5
2
Mc
1
M 3 Figure 5.6.
1
Mc 2 90°
Figure 5.7.
P
3
5.1 DIFFERENTIAL EQUATIONS OF MOTION OF RIGID BODY
P
90°
Mc 2
1 3
Figure 5.8.
Mc
M
2
3 Figure 5.9.
1
73
74
5. TOPIC AK5
M
1
Mc
2
3
Figure 5.10.
90°
P
Mc
1
2
Figure 5.11.
3
5.1 DIFFERENTIAL EQUATIONS OF MOTION OF RIGID BODY
Mc
2
1
3
P
Figure 5.12.
90°
1 Mc 90°
Figure 5.13
P
2 3
75
76
5. TOPIC AK5
M
1
2
Mc
Figure 5.14.
3
2
Figure 5.15.
Mc
M
1
3
5.1 DIFFERENTIAL EQUATIONS OF MOTION OF RIGID BODY
1
Mc 90°
2
Figure 5.16.
P 3
2
1
Mc
M
3 Figure 5.17.
77
78
5. TOPIC AK5
1
2 M
Mc
3
Figure 5.18.
P
Mc
90°
2
1
Figure 5.19.
3
5.1 DIFFERENTIAL EQUATIONS OF MOTION OF RIGID BODY
M
Mc
2
1 3
Figure 5.20.
90°
P
1
Mc
Figure 5.21.
3
2
79
80
5. TOPIC AK5
2
Mc
1
90°
P 3
Figure 5.22.
Mc
1
2
Figure 5.23.
3
M
5.1 DIFFERENTIAL EQUATIONS OF MOTION OF RIGID BODY
2
Mc
P
1
90°
3
Figure 5.24.
2 Mc
1
M Figure 5.25.
3
81
82
5. TOPIC AK5
Mc
2
1
3
90° P Figure 5.26.
5.1 DIFFERENTIAL EQUATIONS OF MOTION OF RIGID BODY
P
90°
1
Mc 2 3
Figure 5.27.
M 1 Mc Figure 5.28:
3
2
83
84
5. TOPIC AK5
1
M
Mc
2
3
Figure 5.29.
P Mc
90°
1
2 3 Figure 5.30.
5.2 SAMPLE PROBLEM
Table 5.1 m1 m2 m3 R1 r1 R2 r2 No.
kg
ix1
cm
ix2 cm
M Nm
P N
MC Nm
ω10
t1
s1
s
85
Define Equation of Rotational Motion of this Link
1
100 300 500
20

60
40

50
2,100 + 20t

1000
2
2
1
2
300
500
70
50
20

60


10,200 + 100t
600
1
0.5
2
3
200 100 400
60

30
20
60
20√2
6,100 + 20t

800
0.5
2.5
1
4
100 250 300
20

50
30

40
1,000 + 40t

1400
1.5
2
1
5
150 300 600
30

50
20

30

5,500 + 200t
1500
2
1
2
6
400 250 600
70

30
20
70
20√2
4,800 + 10t

800
3
4
1
80
7
300 200 400
60
40
30
20
50
20

3,000 + 100t
500
0
3
2
8
300 250 700
50
30
40
20
40
30

9,700 + 50t
500
1
2
1
9
200 100 500
80
60
20

50√2

5,900 + 30t

600
2
3
2
10
250 100 400
40
20
30

30

2,500+50t

1200
0
1.5
2
11
150 300 700
40
30
60
30
30
40

3,900 + 50t
1000
1
2
1
12
100 200 600
30
20
60

20√2
60

5,700 + 50t
1500
2
2
1
13
180 100 300
50
40
30
20
30√2
20

2,700 + 200t
400
0.5
1
2
14
150
400
40
20
30

30

1,800 + 20t

700
1.5
2.5
1
15
300 180 500
20
10
50

10√2
50
700 + 40t

300
0
1.5
1
16
300 250 400
60
40
50
30
50
40

7,300 + 100t
1200
1
2
1
17
250 100 800
50
30
20

40

5,400 + 50t

900
2
2
1
18
200 100 600
20

50


50
1,900 + 20t

1500
0.5
1
2
19
250 150 400
50
30
30
20
40
20√2

14,200 + 200t
500
0.5
2
1
20
400 100 800
50
20
30

40

3,700 + 50t

1200
2
1
2
21
200 150 300
50
40
30
20
30√2
20

3,800 + 100t
800
1
1.5
2
80
22
250 100 800
60
20
10

50


9,700 + 200t
700
2
0.5
1
23
200
400
40
20
30

30

2,300 + 20t

900
0.5
1
2
24
100 200 500
30

40
20

30

12,600 + 100t
500
1.5
1
1
25
150
400
60

20

60

4,900 + 40t

800
0
1.5
2
26
250 200 500
50
20
40
30
40
30

3,500 + 150t
600
2
2
1
27
250 100 500
50
30
40
30
30√2
30

15,200 + 100t
700
1.5
1
1
28
60
200 900
20

60
10

50
900 + 10t

1500
0
2
2
29
50
200 500
20

40
30

25√2
2,100 + 20t

1000
2
1.5
1
30
300
60
50
30
20

40


7,200 + 50t
700
1.5
1
2
5.2
80 80
600
SAMPLE PROBLEM
The given parameters: m1 = 100 kg; m2 = 150 kg; m2 = 400 kg; M = 4,200 + 200t N ∙ m; MC = 2,000 N ∙ m; ω10 = 2 s–1; R1 = 60 cm; R2 = 40 cm; r2 = 20 cm; ix1 = 20√2 cm; ix2 = 30 cm.
86
5. TOPIC AK5
Determine equation φ2 = f(t) of the rotational motion of the link 2 of the mechanism, the circular force S at the contact point between the links 1 and 2, and the tension T of the rope at the time instant t1 = 1 s (Figure 5.31).
M
1
A
2
B
Figure 5.31.
5.3
3
Mc
SOLUTION
The forces applied to the link 1 of the mechanism are the gravity force G1⃗ , the driving moment M, the components Y⃗A and ZA⃗ of the reaction force at the pin support A, the circular force S⃗1, and the normal reaction force N1⃗ of the link 2 (Figure 5.32). The forces applied to the link 2 of the mechanism are the gravity force G2⃗ , the moment of the resistance forces MC, the components Y⃗B and ZB⃗ of the reaction force at the pin support B, the tension T⃗ of the rope holding the load 3, the circular force S2⃗ , and the normal reaction force N⃗2 of the link 1 (Figure 5.32). The forces applied to the load 3 are the gravity force G⃗3 and the tension T⃗ '.
It is obvious that:
S⃗2 = –S⃗1, N⃗1 = –N2⃗ , and T⃗ ' = –T⃗ .
5.3 SOLUTION
87
The differential equation of the rotational motion of the link 1 about fixed axis x1: 𝐽x1 φ̈1 = MxE1.
The total moment of external forces MxE1 applied to the link 1 of the given mechanical system (Figure 5.32) about axis x1: MxE1 = M – S1R1. The moment M brings to the motion the whole system and therefore it is considered with a positive sign, but the moment created by the force S1⃗ resists to the rotation of the link 1 and therefore it has a negative sign. Then a differential equation of the rotational motion of the link 1:
𝐽x1 φ̈1 = M – S1R1. The angular velocity φ̈1 of the link 1 can be expressed through the angular velocity φ̈2 of the link 2.
(5.1)
φ1̈ R2 R As: then φ̈1 = φ̈2 ∙ 2 . = , ̈ φ R1 R1 2 Then, Equation (5.1) can be written as:
R 𝐽x1 φ̈1 2 = M – S1R1. (5.2) R1 An angular momentum principle for the mechanical system can be applied to write a differential equation of the rotational motion of the link 2 (which holds the load 3) about axis x2:
dL x2 = MxE2. R1 The angular momentum of the system 2 – 3 about axis x2:
(5.3)
Lx2 = 𝐽x2 ω2 + m3vr2,
where 𝐽x2 ω2 – is the angular momentum of the link 2 rotating with the angular velocity ω2 around fixed axis x2; m3vr2 – is the angular momentum of the load 3 in translational motion with the velocity v. Since = ω2 r2:
Lx2 = (𝐽x2 + m3r22 ) ω2 = 𝐽'x2 φ̇2 ,
where 𝐽'x2 = 𝐽x2 + m3r22 – the moment of inertia of the system 2 – 3 about axis x2.
The total moment MxE2 of the external forces applied to the system 2 – 3 (Figure 5.32) about axis x2: MxE2 = S2R2 – G3r2 – MC.
The moment of the force S⃗2 brings to the motion the system 2 – 3 and therefore it is considered with a positive sign, but the moment created by the load G⃗3 and the moment of the resistance forces M⃗C resist to the motion of the system, and therefore they have negative signs.
88
5. TOPIC AK5
Hence, from Equation (5.3):
d 𝐽'x φ̇2 = S2R2 – G3r2 – MC. dt 2
We obtain the following differential equation of the rotation of the link 2:
𝐽'x2φ̈2 = S2R2 – G3r2 – MC.
(5.4)
In the system of Equations (5.2) and (5.4):
R 𝐽x1 φ̈2 2 =M – S1R1, R1
𝐽'x2 φ̈2 = S2R2 – G3r2 – MC
the forces S1 = S2 = S and the angular acceleration φ2̈ are unknown. To exclude S, the first equation of this system we multiply to R2 and the second equation to R1, then add the corresponding sides of these equations: R2 (𝐽x1 2 + 𝐽'x2 ∙ R1) φ̈2 = MR2 – (G3r2 + MC) R1, R1 and from here:
MR1R2 – (G3r2 + MC) R12 . (5.5) φ̈2 = 𝐽x1 R22 + 𝐽'x2 R12. Equation (5.5) defines the angular acceleration of the link 2 of the mechanism. Using the given values:
𝐽x1 = m1ix21 = 100(0.2√2)2 = 8 kg ∙ m2,
𝐽'x2 = 𝐽x2 + m3r22 = m2ix22 + m3r22 = 150 ∙ 0.32 + 400 ∙ 0.22 = 29.5 kg ∙ m2.
From Equation (5.5): φ̈ 2 =
(4,200 + 200t)0.6 ∙ 0.4 – (400 ∙ 9.81 ∙ 0.2 + 2,000) 0.62 = 4.034t + 0.4597 (s–2). 8 ∙ 0.42 + 29.5 ∙0.62
Twice integrating this equation:
φ2̇ = 2.017t2 + 0.4597t + C1,
φ̇2 = 0.672t3 + 0.23t2 + C1t + C2. To determine the integral constants we will use the initial conditions: at R 60 t = 0, φ20 = 0, φ̇20 = ω20 = ω10 ∙ 1 = 2 ∙ = 3 s–1. R2 40 –1 ̇ = C1, φ20 = C2, or C1 = 3 s , C2 = 0. Hence: φ20
5.3 SOLUTION
The equation of the angular velocity of the link 2:
φ̇2 = 2.017t2 + 0.4597t + 3 (s–1).
Then the equation of the rotational motion of the link 2: φ2 = 0.672t3 + 0.23t2 + 3t (rad). The circular force S can be determined from Equation (5.4): 𝐽'x2 φ2 + G3r2 + MC , ̈ S = S2 = R2 29.5(4.034 ∙ 1 + 0.4597) + 400 ∙ 9.81 ∙ 0.2 + 2,000 at t1 = 1 s S= = 7295 N. 0.4
M
1 A x1 2
x2 Figure 5.32.
3
G1
YA
N1
s1
2
ZA
N2
T
T’
B ZB Y B v
G2
ω1
s2
1
y1
R1
r2
ω2
R2
y2
Mc
G3
To determine the tension T of the rope, we will write a differential equation of the rotation of the link 2 (Figure 5.33): 𝐽x2 φ̈2 = S2R2 – Tr2 – MC,
89
90
5. TOPIC AK5
from where: T=
S2R2 – MC – 𝐽x2 φ̈2, r2
7,295 ∙ 0.4 – 2000 – 13.5(4.0334 ∙ 1 + 0.4597) = 4,285 N at t1 = 1 s T= . 0.2
N2
z
s2 ZB
x2 Figure 5.33.
G2
T
ω2 YB
2
Mc
y2
91
CHAPTER 6
Topic AK6 6.1
STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES
Problem 6.1. The cart 1 with total mass m1 = 6,000 kg is moving horizontally with velocity v1 = 2.5 m/s and hits stationary cart 2 with total mass including the container m2 = 4,000 kg. After the impact the cart gains the horizontal velocity u1 = 2 m/s, but the container of mass m0 = 500 kg flips about point A supported by the stopper. The container is assumed as a homogeneous rectangular parallelepiped (a = 0.8 m,h = 1.5 m). The vertical impact planes of the carts are assumed smooth. Determine the velocity of the cart 1 after the impact with the cart 2, and the impact impulse on the support stopper at point A.
1
Figure 6.1.
A
h a
2
92
6. TOPIC AK6
Problem 6.2. A load with mass m0 = 500 kg falls from the height h = 1 m to the point D of the rigid bar supported by the stationary pin support A and the spring B with the coefficient of the rigidity c = 20,000 N/cm . The impact of the load on the bar is assumed as an inelastic. The mass of the bar m = 6,000 kg and its length l = 4 m. The horizontal position of the bar shown in Figure 6.2 corresponds to the static deformation of the spring support under the weight of the bar only. Consider the bar as a slender homogeneous cylinder, and the weight as a particle. Determine the impact impulse on the bar at the point D and maximum deformation of the spring support assuming that the point B moves linearly.
h
Figure 6.2.
A
D
l/4
B
3l/4
Problem 6.3. The string holding the load with the mass m0 = 500 kg was broken and the load falls from the height h = 1 m on the platform rested on two identical and symmetrically positioned spring supports. The load strikes at the point A which is on the vertical symmetrical plane of the platform and at the distance d = 0.6 m from its center of gravity C. The load impacts the platform not elastically. The mass of the platform is m0 = 5,000 kg, and its radius of inertia horizontal axis of symmetry is iC = 0.5 m. Define the velocity of the center of gravity and the angular velocity of the platform at the end of the impact. Also, determine an impact impulse at the point A. Consider the platform as a rigid body and the load as a particle.
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES
d C
93
h
A
Figure 6.3.
Problem 6.4. The homogeneous cylindrical load of mass m = 200 kg and radius r = 0.2 m is moving via a horizontal conveyor belt with velocity v = 0.6 m/s without skidding on the discs 1 and 2. At some instant the conveyor suddenly stops. As the surface of the conveyor belt is not smooth, due to the sudden stop of the conveyor the cylinder will roll on the belt without skidding. The rolling friction is negligibly small. Determine an impact impulse on the absolute rough surface of the belt due to the sudden stop of the conveyor. Using Carnot’s theorem verify the obtained velocity of the center of gravity (or an angular velocity) of the cylinder. The cylinder travels a certain distance and hits the step of the height h = 0.03 m. Determine an impact impulse on this step.
2
Figure 6.4.
O2
h
C A
O1
1
94
6. TOPIC AK6
Problem 6.5. The homogeneous cylinder of mass m = 500 kg and radius r = 0.5 m rolls without initial velocity along an incline (α = 30°) at a distance s1 = 3 m and continues to roll on the horizontal surface without skidding. The coefficient of the rolling friction is δ = 0.2 cm. At what distance s2 should a support step of the height h = 0.1 m, be placed so the cylinder after hitting the corner F just climbs on it without further traveling and skidding? Also, determine both horizontal and vertical components of the impact impulse on the cylinder from the step.
C h Figure 6.5.
E
F
D
S1
B
S2
A
Problem 6.6. The pendulum consists of the weight (homogeneous circular disk of radius r = 0.1 m) suspended from the pivot of l = 1.2 m long. The mass of the weight m0 = 5 kg, and the mass of the pivot is negligibly small. The pendulum is displaced sideway from its resting, equilibrium position and it swings about fixed point O with angular velocity ω = 3 s–1. The pendulum hits a homogeneous rectangular parallelepiped D of mass m = 6m0 (a = 0.8 m, b = 0.4 m, h = 0.2 m) at point B. The coefficient of restitution during an impact is k = 0.5. The both surfaces of the pendulum and the object D are absolutely rough (no skidding during the impact). Determine the angular velocity of the object D about point A at the end of the impact and the impact impulse on the rough surface at the point A.
O
l b r
Figure 6.6.
D
B
a A
h
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES
95
Problem 6.7. The lever arm consists of two solid bars AB and AD connected under a right angle. The lever arm has a fixed horizontal axis of rotation A and it is held at point B via spring. AD = a = 1.5 m. The load with mass m0 = 100 kg falls from the height h = 0.5 m and hits the horizontal bar of the lever arm at point D. The mass of the bar is m = 1,000 kg, and its radius of inertia about axis of rotation is iA = 0.5 m. The position of the center of gravity C of the lever arm is defined by the coordinates xC = 0.4 m and yC = 0.3 m. Consider the load as a particle and the impact as inelastic. Determine an impact reaction on the load at point D, as well as both horizontal and vertical components of the impact impulse on the support A.
y
B
yc
Figure 6.7.
A
C
xc
a
D
h
x
96
6. TOPIC AK6
Problem 6.8. A homogeneous, hollow, thinwalled cylindrical load with the mass m0 = 500 kg and the radius r = 0.4 m is rested on the truck 1. To prevent its possible displacement relative to the cart, the load is held by the step and the inclined plane (the angle of inclination with the horizon α = 60°). The truck 1 with total mass (including the load) m1 = 3,000 kg moves linearly and horizontally, and with the velocity v1 = 3 m/s hits nonmoving trolley 2 of total mass m2 = 6,000 kg. At the end of the impact the truck 1 stops and the cylinder hits the inclined plane and rolls on it. During the impact the cylinder stays in contact with the inclined plane and it is not gliding due to the absolute roughness of the inclined plane. The impact surfaces of the truck and the trolley are smooth. Define an angular velocity of the cylinder at the end of its impact with the inclined plane. Using Carnot’s theorem, verify the obtained angular velocity of the cylinder. Also, define the velocity of the trolley 2 at the end of its impact with the truck 1.
2
B
Figure 6.8.
C
A
1
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES
97
Problem 6.9. An object D of mass m0 moves translationally along horizontal surface and with velocity v0 = 3 m/s hits the knot C of the nonmoving truss structure. The surfaces of the object D and the knot C at the impact point are smooth. A coefficient of restitution at the impact k = 0.5. The solid truss structure has a fixed pin support O and an elastic support A. BC = a = 2 m. The mass of the truss structure m = 20m0 and its radius of inertia about horizontal axis of rotation O is i0 = 1 m. Define the angular velocity of the truss structure at the end of the impact and using Carnot’s theorem verify the obtained result. Also, define the distance s traveled by the object D translationally after the impact if the friction coefficient f = 0.1.
B
O
a
A
D
Figure 6.9.
C
98
6. TOPIC AK6
Problem 6.10. A pendulum is displaced sideway from its resting, equilibrium position under angle α = 60° and it swings without initial velocity about fixed axis O due to the gravity. At its vertical position the pendulum hits with point F the rested object at position A. The distances from the axis O to the gravity center of the pendulum C and to the hit point F are given: OC = d = 0.9 m and OF = l = 1.1 m. The mass of the pendulum is m = 18 kg, and its radius of inertia about axis of rotation is i0 = 1 m. The object is considered as a particle and its mass m0 = 6 kg. A coefficient of restitution at the impact between the pendulum and the object k = 0.2. Due to the impact, the object falls from point A of the plane AB to the point D of the smooth horizontal plane DE. The plane DE is located h = 1 m below the plane AB. The impact of the object at point D is inelastic (k1 = 0). Define the impact impulse at point D and the equation of motion of the object after the impact in the coordinate system xDy. Also, define the angle of swing β after the pendulum hits the objects at point A.
O
d C
x Figure 6.10.
A D
y
E
h
l
F
B
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES
99
Problem 6.11. During the impact testing the pendulum copra of mass m0 = 500 kg is displaced sideways from its resting equilibrium position under angle α = 60° and it swings without initial velocity about fixed axis O due to the gravity. At its vertical position the pendulum hits with point A the center D of the rested vertical beam BF of mass m = 2,000 kg. The beam BF has a fixed cylindrical pin support B and an elastic support F (BF = 2a = 3.2 m). The beam is considered as a homogeneous slim rod. A coefficient of restitution at the impact k = 0.4. The distances: OC = d = 1.5 m, OA = l = m. Here, C is a center of gravity of the pendulum. The radius of inertia of the pendulum about axis of rotation i0 = 1.8 m. After the impact the pendulum swings under the angle β and it is held in that position by the special catching device. Define the impact impulse at point D and the distance from point B where during the impact the pin support B didn’t experience an impact impulse.
O l
A
d
F
C D B
Figure 6.11.
a a
100
6. TOPIC AK6
Problem 6.12. The truck 1 with total mass (including the container) m1 = 2,000 kg moves linearly and horizontally, and with the velocity v1 = 2.5 m/s hits the trolley 2 of total mass m2 = 8,000 kg moving with velocity v2 = 0.5 m/s in the same direction. After the mutual impact the truck 1 stops and the container rolls about the side A held by the support bar. Consider the container as a homogeneous right parallelepiped (a = 0.9 m, h = 1.2 m) of mass m0 = 500 kg. The vertical impact surfaces of the truck and the trolley are smooth. Define an angular velocity of the container at the end of the impact of the truck and the trolley. Using Carnot’s theorem, verify the obtained angular velocity of the container. Also, define the coefficient of restitution during the impact between the truck and the trolley.
h
a
Figure 6.12.
A
1
2
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES 101
Problem 6.13. The support axis O of the pivot of the pendulum moves translationally on horizontal plane with constant velocity. During this motion the pendulum maintains its vertical equilibrium position. The radius of inertia of the pendulum about axis O is i0 = 0.8 m. After the sudden stop of the support axis of the pivot, the pendulum swings and hits with the point A the resting object D of mass m0 = 2.5 m, where m is a mass of the pendulum. The impact surfaces of the pendulum and the object D are smooth at the impact point. A coefficient of restitution at the impact between the pendulum and the object is k = 0.6. The distances: OA = l = 1 m and OC = d = 0.7 m. Here, C is a center of gravity of the pendulum. After the impact the objects D moves translationally along the horizontal plane the distance s = 0.1 m. The friction coefficient between the object D and the surface is f = 0.1. Define the velocity of the axis O of the pivot of the pendulum just before it stops and the angle of swing β of the pendulum after its impact with object D.
O
d C A
Figure 6.13.
l D
s
102
6. TOPIC AK6
Problem 6.14. The solid construction of the right triangle shape with the sides AB = a=1 m and BD = b = 2 m is supported by the fixed pin support A and at point B by the spring. A load of mass m0 = 200 kg is dropped from the height h = 0.5 m and it hits the point D of the horizontal side BD of the resting construction. The mass of the construction is m = 2,000 kg and it is considered as a homogeneous triangle. The load is considered as a particle. Define the angular velocity of the system after the impact and using Carnot’s theorem verify the obtained expression for the angular velocity. Also, define the impact impulse on the support A.
h
b D
B
a A Figure 6.14.
Problem 6.15. During the transportation the loads from the position A slip without initial velocity along the plane inclined under the angle α = 30° with horizon. The load travel distance s1 = 3 m and continue a motion along the horizontal plane. Define the shortest distance BD must be installed a support D to stop the loads without flipping over. The calculation must be made for the homogeneous right parallelepiped load of mass m = 500 kg and sides b = 2a = 1 m. The friction coefficient f = 0.2. Also, define an impact impulse on the support D at the same conditions.
A
Figure 6.15.
a
b S1
B
S2
D
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES 103
Problem 6.16. The trolley 1 of total mass m1 = 6,500 kg moves linearly along a horizontal plane and hits the resting truck 2 of total mass (including the load) m2 = 4,000 kg. The load is a homogeneous hollow thinwalled cylinder of mass m0 = 500 kg and the radius r = 0.5 m. It is held by two step supports to prevent any possible displacement. After the impact the trolley 1 and the truck 2 gain the same speeds along the horizontal linear path. But the cylinder gains an angular velocity of the rotation about the side E of the step support DE. The surface of the step support is rough enough to prevent a slip of the cylinder during an impact. The cylinder doesn’t detach from the planes when it hits the side E. After an impact the cylinder climbs on the step DE of height h = 0.1 m. During this climb the velocity of the truck 2 gained after the impact is constant. The vertical planes of the impact surfaces between the trolley and the truck are smooth. Define an impact impulse on the cylinder from the step DE and he velocity of the trolley 1 before the impact with the truck 2.
1 C
B
Figure 6.16.
E
D
h
2
104
6. TOPIC AK6
Problem 6.17. The axis O of the pivot of the pendulum moves translationally along the horizontal plane with velocity v = 2 m/s. During this motion the pendulum maintains its vertical equilibrium position. The pendulum is a homogeneous slim rod of length l = 1 m and mass m0 = 20 kg. When the axis O suddenly stops, the pendulum gains an angular velocity and it hits at point E the resting homogeneous hollow thin walled cylinder of radius r = 0.2 m and mass m = 2m0. A coefficient of restitution at the impact between the pendulum and the object is k = 1/3. The impact surfaces between the pendulum and the cylinder are smooth. The horizontal plane is rough enough to prevent the slippage of the cylinder during an impact. Define the impact impulse on the axis O of the pivot when it stops suddenly. Also, define the angular velocity of the cylinder after its impact with the pendulum.
O
l
E Figure 6.17.
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES 105
Problem 6.18. The rigid bar of mass m = 8,000 kg and length l = 4 m is supported by the elastic spring at point A and the fixed pin support B. At static deformation of the spring A the bar is in the horizontal position. The coefficient of rigidity of the spring is c = 10,000 N/cm. The radius of inertia of the bar about horizontal axis of rotation B is iB = 2.2 m. A steam hammer hits the bar. The total mass of the hammerstockpiston system is m0 = 800 kg. Under
the steam pressure the entire system falls from the height h = 0.8 m and hits the bat in the middle. At the impact the velocity of the system is twice higher than the velocity at the free fall. A coefficient of restitution at the impact between the hammer and the bar is k = 0.2. The hammer with other related components is considered as a particle. The motion of all points of the bar is linear. Define a maximum deformation of the elastic support A assuming that the hammer after bouncing off from the bar is held by the back pressure of steam and it doesn’t fall again. Also, define an impact impulse on the support B.
h
A
l/2 Figure 6.18.
B
l/2
106
6. TOPIC AK6
Problem 6.19. The buffer bars are tested by the copra pendulum of mass m = 500 kg and the radius of inertia about fixed horizontal axis of rotation O is i0 = 1.2 m. The pendulum is swing under the angle α = 90° from its vertical equilibrium position and it is released without initial velocity. The pendulum hits at point A the buffer bar of mass m0 = 1,000 kg. The coefficient of the rigidity of the spring system is c = 10,000 N/cm . The coefficient of restitution
at the impact between the pendulum and the bar is k = 0.5. After the impact the pendulum is bouncing off at the
angle β and it is held in this position via a special catching device. The distances: OC = d = 0.9 m and OA = l = 1.5 m. Here, C is a center of gravity of the pendulum. The friction between the bar and the horizontal plane is ignored. Define a maximum compression of the buffer springs, an impact impulse at point A and the distance from point A to the center of the impact.
O
Figure 6.19.
l d
C
A
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES 107
Problem 6.20. In an epicycle mechanism the crank OC of a mass m = 2 kg and the gear 1 of a radius r1 = 30 cm are rotated with the angular velocities ω0 = 1.5 s–1 and ω1 = 1 s–1, respectively. The gear 2 has a mass m2 = 8 kg and the radius r2 = 10 cm. At some instant the gear 1 suddenly stops. Define the angular velocity of the crank after the impact and the impact impulses at points A and C. Consider the crank as a homogeneous slim rod and the gear 2 as a homogeneous solid disc.
ω1 1
O ωO
C A Figure 6.20.
2
108
6. TOPIC AK6
Problem 6.21. The conveyor belt is inclined under the angle α = 15° with the horizon. The radii of the pulleys are r = 0.2 m. The conveyor belt doesn’t slip over the pulleys 1 and 2. The homogeneous cubic load of mass m0 = 200 kg and size a = 0.5 m is placed on the conveyor belt. At some instant the conveyor suddenly stops and the load gains an angular velocity of the rotation about the side A fixed by the support bar. Define the smallest angular velocity of the pulleys when the conveyor stops and the load flips over. Also, define an impact impulse on the supporting bar when the pulleys rotating with the same angular velocity suddenly stop.
2 O2 1
O1
Figure 6.21.
A
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES 109
Problem 6.22. The homogeneous hollow thinwalled cylindrical load of mass m = 800 kg and radius r = 0.4 m rests between two step supports on the moving platform. When the platform suddenly stops, the step AB can’t hold the load. Hence, the cylinder climbs on the step and it rolls along the horizontal plate BE the distance BD = s = 1 m. Then it hits the side F of another step support EF of the height h = 0.1 m and turns around the side F. As a result, the center of the gravity of the cylinder vertically rises to the distance h1 = 0.07 m. The cylinder rolls from B to F without slip. The rolling friction coefficient of the cylinder is δ = 0.1 cm. The cylinder was not detached from the planes when it hits the step. The absolute roughness of the step surface prevents the slip of the cylinder during the impact. Define the initial velocity of the center of mass of the cylinder along the section BD. Also, define an impact impulse on the side F of the step EF.
B
Figure 6.22.
C A
D
F
h1 E
h
110
6. TOPIC AK6
Problem 6.23. The pendulum consists of the homogeneous slim rod AB of mass m = 4 kg and l = 0.7 m long, and the homogeneous solid sphere of mass m0 = 2 kg and radius r = 0.1 m. The axis A of the pendulum moves translationally over the horizontal plane with the velocity v = 1.4 m/s and the pendulum stays in the vertical equilibrium position. Due to the sudden stop of the axis of the pivot, the pendulum swings about this axis and in the same vertical position hits the point D of the resting vertical plate. The impact surfaces of the pendulum and the plane are smooth. The coefficient of restitution at the impact between the pendulum and the plane is k = 0.4. Define the angular velocity of the pendulum when the axis of the pivot suddenly stops. Using Carnot’s theorem, verify the obtained expression for the angular velocity. Also, define a swing angle β of the pendulum after its impact with the vertical plane and an impact impulse on the axis A of the pendulum.
A
l
B Figure 6.23
D
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES 111
Problem 6.24. During the impact load tests of the foundation, the copra pendulum rotates about fixed axis without initial velocity and due to its weight falls from the vertical position. In its horizontal position the pendulum hits at point A the middle of the upper face of the resting foundation. The mass of the pendulum is m0 = 500 kg, and its radius of inertia about the axis of rotation is i0 = 1.8 m. The mass of the foundation m = 10,000 kg. The coefficient of restitution at the impact is k = 0.2. The swung pendulum after the impact is held with the special catching device at that position. The distances: OC = d = 1.5 m and OA = l = 2 m. Here, C is a center of gravity of the pendulum. Define a maximum elastic draft of the foundation of the coefficient of the rigidity c = 16 ∙ 105 N/cm and an impact impulse at point A. Also, define a distance from the point O to the impact center.
A C
O
Figure 6.24.
d
l
112
6. TOPIC AK6
Problem 6.25. The truck 1 of total mass m1 = 3,000 kg moves linearly in horizontal direction with velocity v1 = 5 m/s and hits the truck 2 of total mass (including the load) of m2 = 2,000 kg moving in the same direction with velocity v2 = 1 m/s. The load is a homogeneous hollow thinwalled cylinder of mass m0 = 500 kg and the radius r = 0.5 m, which is held by two inclined plane supports to prevent its possible displacement along the truck. After the impact the velocities of the trucks are changed in the same linear and horizontal direction. The cylinder hits the plane AB inclined under the angle α = 60°, and rolls along this plane without slip and travels the distance s = 0.1 m. A rolling friction is negligibly small. The cylinder was not detached from the plane during the impact. The inclined plane is rough enough to prevent a slip of the cylinder during the impact. Assume after the impact that the cylinder moves along the inclined plane with constant velocity and the impact planes of the trucks are smooth. Define the velocity of the truck 1 after the impact with the truck 2. Also, define an impact impulse of the inclined plane obtained from the cylinder.
1
B
C A
Figure 6.25.
s
2
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES 113
Problem 6.26. In the epicyclic mechanism the crank OC of mass m = 3 kg and length l = 30 cm rotates with angular velocity ω0 = 2 s–1. The gear 1 of mass 2m2 = 10 kg and radius r2 = 10 cm rotates with angular velocity ω1 = 3 s–1. At certain time instant the gear 1 is suddenly stopped. Assuming the gear 2 as a homogeneous disc, and the crank as a homogeneous slim rod, define an angular velocity of the gear 2 after the impact. Also, define impact impulses at points A and C.
2
C
A
ω1 ωO O
1
Figure 6.26.
114
6. TOPIC AK6
Figure 6.27. The load of mass m = 400 kg is dropped from the height h = 1.2 m and it hits the point D of the rigid beam of mass m = 5,000 kg and l = 3 m long. The beam is supported by the fixed pin support A and elastic spring support B. At the rest the beam is in the horizontal position. An impact between the load and the beam is inelastic. Consider the beam as a homogeneous slim rod and the load as a particle. Define an angular velocity of the beam after the impact and using Carnot’s theorem, and verify the obtained expression for the angular velocity. Also, define an impact impulse at the support A.
B
A
Figure 6.27.
2l/3
D
l/3
h
6.1 STUDY OF IMPACT DYNAMICS BETWEEN RIGID BODIES 115
Problem 6.28. The mechanism of the gear with cams rotates about fixed horizontal axis O. The mass of the mechanism is m = 50 kg, and its radius of inertia about axis of rotation is i0 = 0.2 m. The mechanism dumps metallic ingots of mass m0 = 2 kg from the point A of the horizontal plane AB to the horizontal plane ED to horizontal distance from point A d = 1.5 m. The plane ED is located below the plane AB the height h = 1 m. An inelastic impact between the cam and the ingot (k1 = 0) takes place at the distance l = 0.4 m from the axis of rotation of the mechanism. The coefficient of restitution of the impact between the ingot and the horizontal plane at point E is k2 = 0.2. The ingot is assumed as a particle. Define an angular velocity of the gear before an impact. Also, define impact impulses at points A and E of the ingot.
O
l A E Figure 6.28.
d
D
h
116
6. TOPIC AK6
Problem 6.29. The pendulum of mass m = 100 kg is displaced sideway from its steady equilibrium position under angle α and it swings without initial velocity about fixed axis O due to the gravity. At its vertical position the pendulum hits with point A the rested homogeneous hollow thinwalled cylinder of mass m0 = 200 kg and radius r = 0.2 m. The radius of the inertia of the pendulum about the axis of rotation O is iO = 1 m. The distances: OC = d = 0.8 m and OA = l = 1.2 m. Here, C is a center of gravity of the pendulum. The coefficient of restitution of the impact between the pendulum and the cylinder is k = 0.6. After the impact the cylinder slides without rotation along the smooth plane. Then it hits the step DE of height h = 0.05 m and climbs on this step not further than point E. The cylinder was not detached from the point E during the impact with the step. The plane is rough enough to prevent a slip of the cylinder during the impact. Define the angle α of the initial sideway displacement of the pendulum. Also, define an impact impulse gained by the cylinder from the impact with the pendulum.
d
l
A
Figure 6.29.
O
C
E
D
h
6.2 SAMPLE PROBLEM 117
Problem 6.30. The lever arm consists of two rigid bars OD and OF connected under right angle. The sizes: OD = a = 1 m, OF = b = 1.5 m. The lever arm is supported by the fixed pin support O and the spring at point E. The mass of the lever arm is m = 400 kg, and its radius of inertia about axis of rotation O is iO = 0.4 m. At static deformation of the spring the lever arm is at rest and its rod OD is in the horizontal position. The load A of mass mA = 20 kg falls from the height h = 0.5 m and hits the horizontal bar of the lever arm at point D. An impact between the load and the rod OD is inelastic (k1 = 0). The lever arm gains an angular velocity and at point F hits a resting object B of mass mB = 120 kg. The coefficient of restitution of this impact is k2 = 0.2. Consider the load A and the object B as particles. Define the velocity of the object B after the impact with the rod OF. Also, define an impact impulse at the object B.
h
D
A O
d
b
E F
B
Figure 6.30.
6.2
SAMPLE PROBLEM
A homogeneous hollow thinwalled cylindrical load of mass m = 500 kg and the radius r = 0.5 m is rested on the moving platform and it is held by two step supports to prevent any possible displacement along the platform (Figure 6.31). Due to the sudden stop of the platform, the cylinder hits the side D of the step BD of height h = 0.1 m and climbs on this step. Then, the cylinder rolls the distance DE along the horizontal surface DK and hits another detent (the plane KN inclined under the angle α = 60° with horizon). The cylinder travels FN = s = 0.1 m distance along the surface DK. The cylinder rolls without slipping and the rolling friction is negligibly small. The cylinder doesn’t detach from the planes when it hits the side step and the inclined plane. The surfaces of both the step and the inclined plane are rough enough to prevent a slip of the cylinder during the impacts. Define the velocity of the
118
6. TOPIC AK6
platform before it stops. Also, define impact impulses on the cylinder from both the step and the inclined plane. Using Carnot’s theorem, verify the obtained expressions for the angular velocities of the cylinder after the impacts with both the step and the inclined plane.
s N
F
KE
B
D
C
h
A
Figure 6.31.
6.3
SOLUTION
Due to the sudden stop of the platform, a translational motion of the cylinder instantly converts to the rotational motion about the side D of the step BD, e.g. the cylinder experiences the hit. We will apply a change of the angular momentum (kinetic moment) of the mechanical system during the impact. As an axis of moments we will consider the fixed horizontal axis provided along the side D of the step BD (Figure 6.32): LIID – LID = �MD (S⃗iE ).
In Figure 6.32 the positions I and II corresponding to the beginning and the end of the impact on the side D of the step BD coincide. The sum of the moments of the external impact impulses applied to the cylinder about axis D: �MD (S⃗iE ) = 0.
This means that the impact impulse S⃗D crosses the axis D. Therefore: LIID = LID. The kinetic moment of the cylinder about the axis D at the beginning of the impact: LID = mvCI (r – h), where vCI = v is the velocity of the center of the gravity of the cylinder at the beginning of the impact, which equals to the velocity of the platform before a sudden stop.
6.3 SOLUTION 119
The kinetic moment of the cylinder about the axis D at the end of the impact: LIID = 𝐽D ωII,
where 𝐽D is the moment of the inertia of the cylinder about axis D and ωII is the angular velocity of the cylinder at the end of the impact. Then:
mr2 3 + mr2� ωII = mr2ωII. LIID = (𝐽C + mr2)ωII = � 2 2 As: LIID = LID, then:
3 2 mr ωII = mv(r – h). 2
From here:
2(r – h)v . (6.1) ωII = 3r2 Let’s verify the obtained expression (6.1) for the angular velocity of the cylinder using the Carnot’s theorem for the case when an ideal elastic support is applied to the mechanical system: TI – TII = T*, where TI is the kinetic energy of the particles system at the beginning of the impact; TII is the kinetic energy of the particles system at the end of the impact; and T* is the kinetic energy corresponding to the lost velocities of the system’s particles. The kinetic energy of the cylinder before it hits the step: TI = The kinetic energy of the cylinder after it hits the step:
y
vCI
sDy
vCII
90°
D ωII B
Figure 6.32.
1 2 𝐽 ω . 2 D II
TII =
1 2 mv . 2
II C
A
sDx
I
x
Next, we determine the kinetic energy of the cylinder which corresponds to the lost velocities of its particles Δvi⃗ I = vi⃗ I – vi⃗ II (Figure 6.33):
120
6. TOPIC AK6
T* =
1 �mi (vi⃗ I – vi⃗ II)2 2
=
=
=
Here
1 �mi (vi⃗ I – vi⃗ II) (vi⃗ I – vi⃗ II) 2
1 �mi (vi⃗ I vi⃗ I – 2vi⃗ I vi⃗ II + vi⃗ II vi⃗ II) 2
1 2 2 �mi (vi⃗ I – 2vi⃗ I vi⃗ II cos γi + vi⃗ II). 2
viI = v is the velocity of the particle Mi before the impact; and viII = ωI DMi is the velocity of the particle Mi after the impact.
Then: T* =
1 �miv2 2
– �mivωIIDMi cos γi
1 �mi ωII2 (DMi)2 2 1 = v2 �mi 2 +
– vωII�miDMi cos γi
1 1 1 ωII �mi(DMi)2 = mv2 – vωII �mi yi + 𝐽D ωII2 . 2 2 2
+
However,
�mi yi = myC = m(r – h).
Therefore, T* =
Hence,
or
1 2 1 mv – mvωII (r – h) + 𝐽D ωII2 . 2 2
1 1 1 2 1 mv – 𝐽D ωII2 = mv2 – mvωII (r – h) + 𝐽D ωII2 . 2 2 2 2 𝐽D ω2II = mvωII (r – h),
e.g.,
and from here:
3 2 2 mr ωII = mvωII (r – h), 2 ωII 2(r – 2h)v. 3r
6.3 SOLUTION 121
Δvi viI
viII y
II
γi
90°
D
B
C
γi
ωII
Figure 6.33.
yc
I
x
A
Next, we apply the kinetic energy work principle for the mechanical system, which corresponds to the lift of the cylinder on the step BD from the position II to the position III (Figure 6.34):
Because the cylinder rotates about axis D, then:
1 𝐽 ω 2 – 1 𝐽 ω2 = – Gh D III 2 D II 2
or
TIII – TII = �AEi.
3 3 2 2 mr ωIII – mr2 ωII2 = –mgh, 2 2
and from here: 4gh ωIII = �ωII2 – 2 . 3r
vCIII
C
90°
D
B Figure 6.34.
III
vCII
90°
ωIII
ωII
II C
A
h
(6.2)
x
Next, we apply the kinetic energy work principle for the mechanical system, which corresponds to the rolling of the cylinder on the segment DE from the position III to the position IV (Figure 6.35):
As in this displacement �AEi = 0, then
TIV – TIII = �AEi . TIV = TIII ,
122
6. TOPIC AK6
which means: vCIV = vCIII
ωIV = ωIII.
and
(6.3)
When the cylinder touches the inclined plane, the instantaneous axis of rotation of the cylinder (the instantaneous center of velocities) suddenly moves from point E to the point F, e.g., the cylinder experiences the hit.
IV
vCIV C vCIII
ωIV
Figure 6.35.
E
ωIII B
C
III
G
D
Next, we apply the kinetic energy work principle for the mechanical system at the impact. The axis of the moments will be a fixed horizontal axis, which corresponds to the generatrix F of the cylinder (IV and V positions of the cylinder corresponding to prior and after the impact, coincide) (Figure 6.36): LVF – LIVF = �MF (SE⃗i ).
As the impact impulse S⃗F applied to the cylinder crosses the axis F: �MF (SE⃗i ) = 0. Therefore,
LVF = LIVF .
To calculate the kinetic moments of the cylinder LIVF and LVF we will apply the theorem of the kinetic moment of the system for the general case of its motion. The kinetic moment of the cylinder about axis F prior to the impact: LIVF = mvCIV r cos α + 𝐽C ωIV,
where vCIV = ωIVEC, as the point E will be the instantaneous center of velocities. Then,
2 LIVF = mωIV r2 cos α + mr ωIV = mωIV r2 �cos α + 1�. 2 2 The kinetic moment of the cylinder about axis F after the impact:
LVF = mvCV r + 𝐽C ωV,
where vCV = ωV ∙ CF, because the point F is the instantaneous center of velocities.
6.3 SOLUTION 123
Then, 2 LVF = mωV r2 + mr ωV = 3 mωV r2. 2 2 Because LVF = LIVF , then: 3 mω r2 = mω r2 �cos α + 1�, V IV 2 2 and from here ω (2 cos α + 1)) ωV = IV . (6.4) 3 Using Carnot’s theorem we will verify the obtained expression for the angular velocity of the cylinder (for the case when an ideally inelastic support is applied to the system of particles):
TIV – TV = T*. The kinetic energy of the cylinder prior of its impact with the inclined plane: 2 2 2 2 1 mr2 2 2 TIV = 1 mvC2IV + 1 𝐽C ωIV = 1 mωIV r + ∙ ωIV = 1 ∙ 3mr ω2IV = 1 𝐽F ωIV . 2 2 2 2 2 2 2 2 The kinetic energy of the cylinder after the impact with the inclined plane:
2 2 TV = 1 mvCV + 1 𝐽C ωV2 = 1 mωV2 r2 + 1 ∙ mr ωV2 = 1 ∙ 3mr ωV2 = 1 𝐽F ωV2 . 2 2 2 2 2 2 2 2
sF
Figure 6.36.
V
vCIV
vCIV90° C ωV 90° s F ωIV F K
ε
IV
E
The kinetic energy of the cylinder corresponding to its lost velocities ∆vi⃗ IV = vi⃗ IV – vi⃗ V (Figure 6.37):
T* = 1 �mi (vi⃗ IV – vi⃗ V)2 2 = 1 �mi �v⃗iIVvi⃗ IV – 2vi⃗ IV vi⃗ V + vi⃗ V vi⃗ V� 2 = 1 �mi �vi2IV – 2viV cos φi + vi2V�. 2 Here: viIV = ωIV EMi – is the velocity of the particle Mi before the impact; viV = ωV EMi – is the velocity of the particle Mi after the impact. Then, T* = 1 �miω2iIV (EMi)2 2
124
6. TOPIC AK6
– 1 �2mi ωIV EMiωV FMi cos φi 2 2 + 1 �2miωV (FMi)2 2 2 = 1 ωIV �mi (EMi)2 2 2 – 1 ωIV ωV�2miEMi FMi cos φi + 1 ωV �mi (FMi)2. 2 From the triangle ∆Mi EF:
(EF)2 = (FMi)2 + (EMi)2 – 2FMi EMi cos φi.
Therefore, –�2miEMi FMi cos φi = �mi[(EF)2 – (FMi)2 – (EMi)2]
= �mi(EF)2 –
�mi(FMi)2 – �mi(EMi)2 = (EF)2 �mi – 𝐽F – 𝐽E.
Then, 2 T* = 1 ωIV �mi (EMi)2 + 1 ωIVωI [m(EF)2 – 𝐽F – 𝐽E] 2 2 2 + 1 ωV �mi (FMi)2 2 = 1 𝐽 ω 2 + 1 mω ω (EF)2 – 1 𝐽 ω ω – 1 𝐽 ω ω + 1 𝐽 ω 2 . IV V 2 E IV 2 2 F IV V 2 E IV V 2 F V However, 𝐽F = 𝐽E = 𝐽F(E) .
From triangle ∆CEF:
Therefore, Hence, or
EF = 2r sin α. 2
2 2 T* = 1 𝐽E ωIV + 1 mωIVωV 4r2sin2 α – 𝐽F(E) mωIVωV + 1 𝐽F ωV . 2 2 2 2 2 1 1 𝐽 ω2 – 1 𝐽 ω 2 = 1 𝐽 ω2 + 2mω ω 4r2sin2 α – 𝐽 IV V F(E) ωIVωV + 𝐽F ωV , 2 E IV 2 F V 2 E IV 2 2
–𝐽FωV2 = 2mωIVωV r2 sin2 α – 𝐽F(E)ωIVωV, 2 e.g., – 3 mr2ωV2 = 2mωIVωVr2 sin2 α – 3 mr2ωIVωV. 2 2 2 From here:
6.3 SOLUTION 125
or
α ωIV �4sin2 2 + 3� ωIV [–2(1 – cos α) + 3] ωV = = 3 3
ωV =
V viV
Δvi
F Figure 6.37.
ωIV �2cos α + 1� . 3
γi
90° v 90° iIV
K
ωV
C
ωIV
IV
E
Next, we apply the kinetic energy work principle for the mechanical system, which corresponds to the rolling of the cylinder along the inclined plane from the position V to the position VI at the distance s (Figure 6.38): TVI – TV =�AEi. Because TVI = 0, or
e.g.,
From here:
mv 2 J ω2 – � CV + C V � = Gs sin α 2 2 mr2 2 mωV2 r2 ω – – 2 V = – mgs sin α, 2 2 3 ω 2 r2 = gs sin α. 4 V
ωV = 2 �gs sin α. 4 3 From Equations (6.2)–(6.4) we can find: Considering Equation (6.5):
9ωV2 4gh ωII = � (2 cos α + 1)2 + 3r2 .
3gs sin α 2 gh ωII = 4 � (2 cos α + 1)2 + 3 . Thus, according to Equation (6.1), the velocity of the platform: 3gs sin α 2 gh v = 3r ∙ 2 � (2 cos α + 1)2 + 2 2(r – h) r
(6.5)
126
6. TOPIC AK6
3 ∙ 0.5 3 ∙ 9.81 ∙ 0.1 ∙ 0.866 4 ∙ 9.81 ∙ 0.1 m + = = 3.68 s . 0.5 – 0.1� (2 ∙ 0.5 + 1)2 3
VI
N
F
C
vCV C G
ωV
K
Figure 6.38.
s V
To find the impact impulse on the cylinder from the step, we will apply the linear momentum impulse principle for the mechanical system during the impact in algebraic form on x and y axes (Figure 6.32): E
mvCII – mvCI = �Six , x
x
mvCII – mvCI = �SiEx , y
or
y
–mvCII cos β – (–mvCI) = SDx, mvCII sin β = SDy.
From here: 2 2 SDx = mvCI – mvCII cos β = mv – mωII r r r– h = mv �1 – 2(r – h) �= 500 ∙ 3.68 �1 – 2(0.5 – 0.1) � = 1,055 N ∙ s, 2 2 3r 3· 0.5 SDy = mvCII sin β = mωII r√1 – cos2 β =
2(r – h)mv (r – h)2 2(0.5 – 0.1)500 ∙ 3.68 �1 – = ∙ 0.6 = 589 N ∙ s. 3∙0.5 3r r2
Then, the impact impulse on the cylinder from the step: 2
2
SD = �SDx + SDy = √1,0552 + 5892 = 1,208 N ∙ s.
To find the impact impulse on the cylinder from the inclined plane, we will apply the linear momentum impulse principle for the mechanical system during the impact in algebraic form on ξ and η axes (Figure 6.36): E
mvCV – mvCIV = �Siξ , ξ
ξ
E
mvCV – mvCIV = �Siη , η
or
η
–(–mvCIV) sin α = SFξ ,
6.3 SOLUTION 127
mvCV – mvCIV cos α = SFη . From here:
SFξ = mvCIV sin α = mωIV r sin α = m
3ωV r sin α, 2 cos α + 1
3ωV r cos α. SFη = mvCV – mvCIV cos α = mωV r – mωIV r cos α = mωV r – m 2 cos α + 1 Using the equation (6.5) we will find:
gs sin α 9.81 ∙ 0.1 ∙ 0.866 6 ∙ 500� 3 3 SFξ = sin α = ∙ 0.866 = 691 N ∙ s, 2 cos α +1 2 · 0.5 +1 6m�
SFη = 2m�
gs sin α 3 cos α 9.81 ∙ 0.1 ∙ 0.866 3 ∙ 0.5 �1 – � = 2 ∙ 500� �1 – � =133 N ∙ s. 3 2 cos α + 1 3 2 · 0.5 + 1
The impact impulse on the cylinder from the inclined plane: 2
2
SF = �SFξ + SFη = √6912 + 1332 = 704 N ∙ s.
129
CHAPTER 7
Topic AK7 7.1
APPLICATION OF THE VIRTUAL DISPLACEMENTS PRINCIPLE TO PROBLEMS ON BALANCED FORCES APPLIED TO A MECHANICAL SYSTEM WITH ONE DEGREE OF FREEDOM
Figures 7.1–7.30 schematically show the mechanism under balanced forces. The parameters required to solve the problems are given in Table 7.1. Applying the virtual displacements principle and ignoring the resistance forces, determine the parameters shown in the last column of Table 7.1. Note: In Figures 7.3, 7.6, 7.10, 7.14, 7.16, 7.18, 7.19, 7.25, and 7.30 the mechanisms are located on vertical planes. In other problems the mechanisms are located on horizontal planes. Nomenclature: Q and P – forces applied to mechanism, N; M – moment of the couple of forces, Nm; c – coefficient of rigidity of spring, N/cm; and h – deformation of spring, cm.
y
M
O
Figure 7.1.
A
60°
30°
B
P
x
130
7. TOPIC AK7
y
P
90°
M
x
O1
Figure 7.2.
z
M
r3
Q Figure 7.3.
O2
30°
A
r1
B
90°
r2
y
7.1 APPLICATION OF VIRTUAL DISPLACEMENTS PRINCIPLE TO PROBLEMS ON BALANCE FORCES 131
C D
90°
y
A
O
30° P
Figure 7.4.
y
M
A
O
Figure 7.5.
B
90°
O1
x B
P
x
132
7. TOPIC AK7
P
z
r1
90°
r2
r3
O
y
O1
Q
Figure 7.6.
y D
C
O
Figure 7.7.
A
x
45° 30°
B
P
7.1 APPLICATION OF VIRTUAL DISPLACEMENTS PRINCIPLE TO PROBLEMS ON BALANCE FORCES 133
y
C 60°
O
Q
90°
A
x
30°
P
Figure 7.8.
B
y
A
M Figure 7.9
30°
O
120°
30°
B
Q
x
134
7. TOPIC AK7
z r2 O1
Figure 7.10.
r3
r1
O
90°
A
y
Q
y A
M
O
Figure 7.11.
30°
B
45°
90°
30°
P x O1
7.1 APPLICATION OF VIRTUAL DISPLACEMENTS PRINCIPLE TO PROBLEMS ON BALANCE FORCES 135
D
y
M
O
60°
A
30°
P
30°
Figure 7.12.
y
O
x
O1
M
A
x C 45° B
Figure 7.13
90°
P
45°
O1
136
7. TOPIC AK7
z
P
B
120°
A
90°
Q 30°
O
Figure 7.14.
y
y B
A
Figure 7.15.
Q
P
30°
90°
C
O
D
x
7.1 APPLICATION OF VIRTUAL DISPLACEMENTS PRINCIPLE TO PROBLEMS ON BALANCE FORCES 137
z
O1
l1
d1
B
A
d2
Q
Figure 7.16.
P
y
A
90° M Figure 7.17.
l2
O
P
30°
D
C 30°
O1
B
x
O2
y
138
7. TOPIC AK7
z r1
O
y
Q r2 Figure 7.18.
z
r1 M
O1
Figure 7.19.
r2
O
A
90°
30°
P
B
y
7.1 APPLICATION OF VIRTUAL DISPLACEMENTS PRINCIPLE TO PROBLEMS ON BALANCE FORCES 139
y
P
B A
M Figure 7.20.
C
30°
O
y
C
Figure 7.21.
45°
M
O
A B
D
90°
Q
x
P
x
140
7. TOPIC AK7
y
C
45°
90°
P
M
O
A B
D
x
Figure 7.21.
y
P
Figure 7.22.
A
30°
45°
O
O1
M
B
x
7.1 APPLICATION OF VIRTUAL DISPLACEMENTS PRINCIPLE TO PROBLEMS ON BALANCE FORCES 141
y
D
C
90°
O
M
A 45°
P
45°
Figure 7.23.
x
B
y
A
C 30°
Figure 7.24.
D
90° P
90°
O
B Q
x
142
7. TOPIC AK7
z
O
15°
A
y
30° D
Q
90° 30°
P
B
Figure 7.25.
y
45° O Figure 7.26.
A
90° M
O1 30°
45°
60° B
P
x
7.1 APPLICATION OF VIRTUAL DISPLACEMENTS PRINCIPLE TO PROBLEMS ON BALANCE FORCES 143
P D
y
A
Figure 7.27.
30°
B
90°
O1
y
P 90°
Q
O
Figure 7.28.
x
30°
O
D
60°
30°
x B
144
7. TOPIC AK7
O
y B
90°
45° D
90°
Figure 7.29.
r1
Figure 7.30.
Q
A x
O1
z
M
r2
r3
O
P
r4
y
7.1 APPLICATION OF VIRTUAL DISPLACEMENTS PRINCIPLE TO PROBLEMS ON BALANCE FORCES 145
Table 7.1
No.
Q
P

100 200 
M Nm 20 100 10
h cm 4 
200



200
200 

2,000


11 12 13
cm OA = 10 O1A = 20 r1 = 20, r2 = 30, r3 = 40 OC/OA = 4/5 OA = 100 r1 = 15, r2 = 50, r3 = 20, O1A=80 OC = OA OC= = AC OA = 20 r1 = 20, r2= = 40, r3 = 20, O1A = 100 OA = 20 O1D = 60, AO = 20 OA = 40
14
OB = 2 OA
20


25
3
15
AC = OC = OD d1 = 80, d2 = 25, l1 = 100, l2 = 75 OA = 20
3,000


250
3
5,000


100
4
200 100 200 200 250
200 100 400 50 
100 50 150
2.5
1 2 3 4 5 6 7 8 9 10
N
c N/cm 
Notes 
P M Q c P

Weight of O1A ignored
P
10 10 
3 2 
Spring is compressed Weight of O1A ignored
P Q M


4

c

300 100 200


P P P
c P Q h Q
17 18 19 20 21 22 23 24
r1 = 20, r2 = 30, OA = 25 OA = AB = AC= = 50 OA = AB = AC = DC = 25 OA = 40 OC = 2 OA = 100 AD = OD = OB
200 50 
25
OD = DB = 0.8 AO
400


120
3
26 27 28 29
OA = 25 OB = AB OB = 5/4 (OA) AO = 30, BD = O1D r1 = 10, r2 = 36, r3 = 10, r4 = 20

500 450 
120 120
180 100
2 2 
Weights of OA and OB are ignored; spring is stretched Spring is compressed Weights of O1B and O2B are ignored; spring is stretched P – weight of pulley of radius r2 Weight of AB is ignored Spring is compressed Weights of AO and BO are ignored; spring is stretched Spring is stretched 

600




16
30
?
P P P P h P M M P h Q P
146
7. TOPIC AK7
7.2
SAMPLE PROBLEM
Given parameters: Q = 100 N; c = 5 N/cm; r1 = 20 cm; r2 = 40 cm; r3 = 10 cm; OA = l = 50 cm; α =30°; β = 90° (Figure 7.31). Ignoring the weights of the links OA and AB, define the spring deformation h.
P
y
1
sQ
r2
r1
s
1
2 O1
G1
A s1
K O
3
G3
sA
r3 3
sB
F
B
Q
Figure 7.31.
7.3
SOLUTION
The following balanced forces are acting on the given mechanism: F⃗ – spring elasticity force; G⃗1 – gravity force of the disc 1 with the gear 2; G⃗3 – gravity force of the gear 3; Q⃗ – gravity force of the load; and reaction forces of supports. The supports applied to the mechanism allow the following virtual displacements of its links: – an angle of turn δφ1 of the disc 1 along with the gear 2;
x
7.3 SOLUTION 147
– an angle of turn δφ3 of the gear 3; – a translational vertical displacement δsQ of the load Q; – a horizontal displacement δsB of the slider B; – a displacement δsA of the particle A perpendicular to OA.
The sum of work performed in the virtual displacements: QδsQ – FδsB = 0.
(7.1)
Next, we will determine a relationship between the virtual displacements of the particles of the system. The displacements of the load Q and the point of the rim of the disc 1 are equal as a thread holding the load Q is inelastic and the slipping between the thread and the disc 1 is ignored. Therefore, the virtual angle of turn of the disc 1 along with the gear 2: δφ1 =
δsQ r1
.
A virtual displacement of the point K of the rim of the gear 2: δs1 = r2 δφ1 =
r2 δs . r1 Q
The virtual displacements of the contact points of the gears 2 and 3 are equal as the slipping between them is ignored. Then, the virtual angle of turn of the gear 3: δφ3 =
δs1
=
r2 δsQ . r1 r3
r3 The virtual displacement of the point A of the crank rigidly attached to the gear 3:
r2 l δs . δsA = OA δφ3 = r 1 r3 Q To define a relationship between the virtual displacements δsA and δsB we find a position of the instantaneous center of rotation P of the link AB. Then: δsB PB . = δsA PA From here:
δsB =
PB δs . PA A
From the triangle ∆APB: PB 1 . = PA cos 30° Hence, r2 l δsB = δs . r1 r3 cos 30° Q The force of the elasticity of the spring is proportional to its deformation: F = ch.
148
7. TOPIC AK7
Substituting the expressions for the force of the elasticity and the virtual displacements of the system particles into Equation (7.1), we will obtain: r2l δs = 0. QδsQ – ch r1 r3 cos 30° Q From here: Q r r cos 30° 100 ∙ 20 ∙ 10 ∙ 0.87 h= ∙ 1 3 = = 1.74 cm. c r2l 5 ∙ 40 ∙ 50 Hence, the spring is compressed 1.74 cm.
149
CHAPTER 8
Topic AK8 8.1 PROJECTILE MOTION ANALYSES WITH QUADRATIC AIR RESISTANCE A projectile of mass m is launched with an initial velocity v0 at an initial launch angle of θ to the horizontal in the uniform gravitational field g (Figure 8.1). Ignoring the lift force determine variations of simulated projectile trajectories with drag coefficient, initial velocity, initial launch angle, and mass of the projectile both analytically (with approximation) and numerically. The following parameters are given in the table below: projectile’s mass m, density ρ, launch angle θ, reference area (usually the face area) A, drag coefficient CD, and initial launch velocity v0. No.
m g
ρ g cm3
θ deg
A cm2
CD 
v0 m/s
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
10 5; 10; 20 15 20 25 20 18 15; 20; 30 22 34 29 38 45 15; 25; 40 35 18 25 48 15 25; 45; 60 45 40 35 24 18 15; 35; 50
7.86 7.35 8.31 8.56 7; 8; 8.5 7.32 6.87 7.98 7.23 7.65 7; 7.5; 8 8.3 7.9 8.25 7.99 8.22 7.5; 8; 9 7.4 7.76 8.32 7.54 6.99 6.8; 7; 8 7.76 7.68 7.46
15; 30; 45 30 45 15 30 60 10; 25; 45 35 45 25 35 40 20; 30; 45 25 38 30 45 55 10; 25; 40 35 32 25 42 45 20; 35; 60 60
2.5 3.0 2; 3; 5 3.5 2.7 3.4 2.8 4.0 3; 5; 8 5.5 4.7 5.2 4.7 4.6 4; 5; 7 6 7 4.5 4.8 6.5 3; 5; 7 4 5.8 5 6.2 4.3
1.5 1.4 1.55 1.5; 2; 2.5 2.0 1.8 1.75 1.43 1.85 1.3; 1.5; 2 1.55 1.46 1.57 1.56 1.45 1.4; 1.8; 2.1 1.53 1.41 1.43 1.65 1.60 1.3; 1.6; 2.1 1.44 1.53 1.46 1.48
900 800 850 870 780 700; 800; 900 850 740 880 815 820 600; 750; 800 950 880 650 875 970 700; 800; 960 880 920 870 950 875 750; 850; 950 780 900
Verying Parameter θ m A CD ρ v0 θ m A CD ρ v0 θ m A CD ρ v0 θ m A CD ρ v0 θ m
150
8. TOPIC AK8
27 28 29 30
28 50 44 63
7.55 7.49 6.8; 7; 8 7.08
38 40 22 27
5.2 6 4.9 7.1
A CD ρ v0
1.38 890 1.5; 1.6; 1.7 915 1.58 895 1.57 650; 700; 840
y
v0
H
q O
x
Figure 8.1.
8.2 SAMPLE PROBLEM Consider a projectile of mass m that is launched with an initial (detonation) velocity v0 at an initial launch angle of θ to the horizontal in the uniform gravitational field g (Figure 8.2).
y FL
FA v0
FD
v G
H
q O
x
Figure 8.2: Projectile motion.The equation of projectile motion is:
m d v⃗ = G⃗ + F⃗D + FL⃗ , dt
(8.1)
⃗ is here m is the mass of the projectile; v⃗ is the projectile velocity; G⃗ is the force of gravity (projectile weight); FD d v⃗ = a⃗ is an acceleration of the projectile; and t is a time. Drag and lift are two the drag force; F⃗L is the lift force; dt
8.2 SAMPLE PROBLEM 151
components of the aerodynamic force FA. A drag force is the component of the surface force parallel to the flow direction. If we assume that the drag force is approximately proportional to the square of the projectile’s speed relative to the air, then we can write: C A ρ ⃗ = – D F vv⃗ , (8.2) FD 2 here CD is the drag coefficient; ρ is the mass density of the fluid; and A is the reference area (usually the face area AF). A lift force is the component of the aerodynamic force that is perpendicular to the approaching flow direction. It can be estimated using the following formula: CD AF ρ vv⃗ , (8.3) FL⃗ = 2 here A is the area normal to the lift force; and CL is the lift coefficient which is a function of the angle of attack, which is an angle between the fragment's reference line (chord) and the oncoming flow, Mach number, and Reynolds number. The analyses showed that a maximum lift coefficient will be at 15° angle of attack. Figure 8.3 shows a variation of the lift coefficient with angle of attack (https://en.wikipedia.org/wiki/Lift_coefficient#/media/File:Lift_curve.svg). Küchemann (2012) proposed an empirical relationship for predicting a lift force to drag force ratio: (8.4) FL 4(M + 3) = , FD M here M is the Mach number. Equation (8.4) was verified by wind tunnel tests and it is considered a quite accurate. Equation (8.1) can be written in algebraic form in 2D Cartesian coordinate system as: dvx = –FDx + FLx, m dt
(8.5)
d vy m = –G – FDy FLy . dt
(8.6)
1.8 1.6
Lift Coefficient
1.4 1.2
1 0.8 0.6 0.4 0.2 0 5
0
5
10
15
Angle of Attack, deg
Figure 8.3.
20
25
30
152
8. TOPIC AK8
Substituting Equations (8.2) and (8.3) in Equations (8.5) and (8.6), we obtain: (CD AF – CL A)ρ dvx =– vvx, dt 2mg
(8.7)
dvy (CD AF – CL A)ρ (8.8) =–g– vvy . dt 2mg Analytical Analysis If we consider only gravity and drag forces acting on projectile, as recommended by many researchers (e.g., Cooper, 1996), the equations of the projectile motion (8.7) and (8.8) in the first approximations: vvx ≈ vx2 and vvy ≈ vy2 can be rewritten as: CD Aρ 2 (8.9) dvx =– v , dt 2mg x dvy CD Aρ 2 (8.10) =–g– v . dt 2mg y Integrating Equations (8.9) and (8.10), and using the initial conditions: x = 0, y = 0, vx = v0 cos θ, vy = vy sin θ when t = 0, we can obtain: k2 vx0 , (8.11) vx = 2 k + gvx0 t tg vy0 – k tan � k � vy = k . k + vy0 tan �tg� k
(8.12)
Substituting vx = dx and vy = dy in Equations (8.11) and (8.12), and integrating one more time, we obtain: dy dt k2 + gvx0t k2 x = ln � � , (8.13) g k2 v 2 + k2 k2 ln � y 20 + 2 � , y = 2g vy k here k = � 2mg . CDAρ
(8.14)
Substituting vy = 0 in Equation (8.14), we can obtain a maximum height (H) of the projectile and time (T) required for reaching a maximum height v 2 + k2 k2 ln � y 0 2 � , (8.15) H = 2g k v k (8.16) T = tan1� y0�. g k Figure 8.4 shows the calculated trajectories of the steel (ρ = 7.86 g cm3) projectile of mass m = 11 g, face area A = 2.21 cm2, drag coefficient CD = 1.6, initial launch velocity v0 = 1,000 m/s, k = 21.8 m s1, for different initial launch angles: θ = 15°, 30°, and 45°. In Figures 8.5 and 8.6 the trajectories are shown for different k parameters (at v0 = 1,000 m/s and θ = 45°) and initial launch velocities v0 (at k = 21.8 m s1 and θ = 45°), respectively.
8.2 SAMPLE PROBLEM 153 350
Analytical Simulations m=11 g; A=2.21 cm2; v0=1,000 m/s; CD=0.7; k=33 m/s
300
250
45 45° 30 30° 15 15°
y, m
200
150
100
50
0
0
100
200
300
400
500
x, m
Figure 8.4. Projectile trajectories for different initial launch angles (k = 33 m s1). 500 Analytical Simulations m=11 g; A=2.21 cm2; V0=1,000 m/s; q=450 400 k=29 m/s k=33 m/s
300
y, m
k=39 m/s
200
100
0
0
100
200
300
400
x, m
Figure 8.5. Projectile trajectories for different k parameters (θ = 45°).
500
600
700
154
8. TOPIC AK8 350 Analytical Simulations m=11 g; A=2.21 cm2; k= 33 m/s; q=45°
300
Vo=600 m/s
250
Vo=800 m/s Vo=1,000 m/s
y, m
200
150
100
50
0
0
100
200
300
400
500
x, m
Figure 8.6. Projectile trajectories for different initial launch velocities (k = 33 m s1; θ = 45°).
Numerical Analysis An exact solution of Equations (8.7) and (8.8) can’t be found analytically. Therefore, we will solve them numerically. As we know, the acceleration components ax and ay are continuously changing as the velocity components vx and vy change. A basic concept of the numerical simulation is that over an adequately short time interval Δt we can consider the acceleration as constant. If we know the initial coordinates and velocity components of the projectile at some time t = t0 (usually t0 = 0), we can find their values at a time t + Δt using the formulas for constant acceleration. During a time interval Δt the average changes of the x and y components of the acceleration and the velocity of the projectile will be: ∆v , ∆v , (8.17) ax = x ay = y ∆vx = ax∆t, ∆vy = ay ∆t . ∆t ∆t The values of x and y components of the velocity at a time t + Δt will be: vx + ∆vx = vx + ax ∆t and vy + ∆vy = vy + ay ∆t.
(8.18)
The average of the velocity components during the time interval ∆t will be: ∆v ∆vy vx + x and vy + . 2 2 Then, during a time interval Δt the coordinates x and y change as:
∆x = �vx +
∆vx a ∆t2 ∆v a ∆t2 � ∆t = vx∆t + x and ∆y = �vy + y�∆t = vy∆t + y . (8.19) 2 2 2 2
The initial conditions can be stated: At t = 0:
x = 0, y = 0, vx = v0 cos θ, vy = v0 sin θ.
(8.20)
8.2 SAMPLE PROBLEM 155
Using Equations (8.18) and (8.19) we can find the position and velocity at the end of each interval in terms of their values at the beginning. A general algorithm of the simulations is presented in Figure 8.7. Identify the parameters, m, A, CD, CL , and r Choose the time interval Δt and the initial values of x, y, vx, vy , and t
Find flow function for arbitrary cross section area Transform function to Laplace equation with the boundary condition on the stream contours
Choose the maximum number of intervals N. The maximum time will be tmax = NΔt Using a Schwarz's integral calculate function Calculate the acceleration components ax and ay
Plot x, y, vx, vy, ax , and ay
Describe turbulent flow of air by 2layered Prandtle Taylor structure model
Find u(x, y) by LavrentyevShabat method Find equivalent radius by PolyaSzego method
Iterate these steps while n < N or t < tmax Calculate the new velocity components vx and vy using Equation 13
Calculate the new coordinates x and y using Equation 14 Increment the time by Δt Stop
Figure 8.7. General algorithm of numerical analysis.
Projectile trajectories were numerically simulated according to the algorithm provided in Figure 8.7 with time interval ∆t = 1 s for the fragment with mass m = 11g, face area A = 2.21 cm2, detonation velocity v0 = 1,000 m/s, and initial launch angle θ = 45°. A variation of numerically simulated projectile trajectories with drag coefficient, detonation velocity, initial launch angle, and mass of the fragment are shown in Figures 8.8–8.11, respectively.
156
8. TOPIC AK8
350
Numerical Simulations m=11 g; A=2.21 cm2; V0=1,000 m/s; q=45
300
Cd = 0.5; k=39 m/s Cd = 0.7; k=33 m/s Cd = 0.9; k=29 m/s
250
y, m
200
150
100
50
0
0
50
100
150
200
250
300
350
x, m Figure 8.8. Variation of numerically simulated projectile trajectories with drag coefficient.
400
450
500
8.2 SAMPLE PROBLEM 157
250 Numerical Simulations m=11 g; A=2.21 cm2; CD=0.7; k=33 m/s; q=45 600 m/s 200
800 m/s 1000 m/s
y, m
150
100
50
0
0
50
100
150
200
250
x, m
Figure 8.9. Variation of numerically simulated projectile trajectories with initial launch velocities.
300
350
158
8. TOPIC AK8
250
Numerical Simulations m=11 g; A=2.21 cm2; v0=1,000 m/s; CD=0.7; k=33 m/s 200
15° 15 30° 30 45° 45
y, m
150
100
50
0
0
50
100
150
200
250
300
x, m
Figure 8.10. Variation of numerically simulated projectile trajectories with initial launch angles.
350
400
REFERENCES 159
250
Numerical Simulations q=45 ; A=2.21 cm2; v0=1,000 m/s; CD=0.7; k=33 m/s
200
m=11 g m=50 g m=100 g
y, m
150
100
50
0
0
50
100
150
200
250
300
350
x, m Figure 8.11. Variation of numerically simulated projectile trajectories with fragment mass.
REFERENCES
Cooper, P. W. (1996). Explosives Engineering. New York:WileyVCH. Küchemann, D. (2012). The Aerodynamic Design of Aircraft. AIAA Education Series. Reston, VA: American Institute of Aeronautics & Astronautics.
161
Author Biography Dr. Sayavur I. Bakhtiyarov is a Professor at the New Mexico Institute of Mining and Technology (Socorro, NM, U.S.) and a Fulbright Fellow. Dr. Bakhtiyarov obtained his Ph.D. from the Russian Academy of Sciences in 1978, and a D.Sc. from the Azerbaijan National Academy of Sciences in 1992. His areas of expertise are: multiphase flows, nanotechnology, nonlinear fluid mechanics, tribology, rheology, and selfhealing composites. Dr. Bakhtiyarov has taught engineering courses for over four decades in several countries (Azerbaijan, Russia, China, UK, Turkey, U.S.). Dr. Bakhtiyarov has authored 350+ scientific publications in refereed scholarly journals, books, international conferences and symposia proceedings, and holds 14 patents. Furthermore, Dr. Bakhtiyarov was elected as a foreign member of the Russian Academy of Natural Sciences and International Ecoenergy Academy. He served as a Program Director of U.S. DOE and NASA research projects, and INSRP U.S. DOD coordinator for NASA’s Mars Science Lab mission. Dr. Bakhtiyarov is a lead organizer of the ASME annual symposia and forums, EditorinChief of two international journals, Mechanics and Solids (IJM&S) and Manufacturing Science and Technology (IJMS&T), and an Editorial Board Member of imanager's Journal on Engineering and Technology (IJET), Mathematics Applied in Science and Technology (MAST), International Journal of Applied Engineering Research (IJAER), International Journal of Dynamics of Fluids (IJDF), and FarEast Journal of Mathematics (FEJM).