Solving dynamics problems with matlab

Solving Dynamics Problems
in MATLAB
Brian D. Harper
Mechanical Engineering
The Ohio State University
A supplement to accompany
Engineering Mechanics: Dynamics, 6th
Edition
by J.L. Meriam and L.G. Kraige
JOHN WILEY & SONS, INC.
New York • Chichester • Brisbane • Toronto • Singapore

CONTENTS
Introduction 5
Chapter 1 An Introduction to MATLAB 7
Numerical Calculations 7
Writing Scripts (m-files) 10
Defining Functions 12
Graphics 13
Symbolic Calculations 21
Differentiation and Integration 24
Solving Equations 26
Chapter 2 Kinematics of Particles 37
2.1 Sample Problem 2/4 (Rectilinear Motion) 38
2.2 Problem 2/87 (Rectangular Coordinates) 41
2.3 Problem 2/126 (n-t Coordinates) 46
2.4 Sample Problem 2/9 (Polar Coordinates) 48
2.5 Sample Problem 2/10 (Polar Coordinates) 52
2.6 Problem 2/183 (Space Curvilinear Motion) 55
2.7 Sample Problem 2/16 (Constrained Motion
of Connected Particles) 57
Chapter 3 Kinetics of Particles 61
3.1 Sample Problem 3/3 (Rectilinear Motion) 62
3.2 Problem 3/98 (Curvilinear Motion) 65
3.3 Sample Problem 3/17 (Potential Energy) 67
3.4 Problem 3/218 (Linear Impulse/Momentum) 70
3.5 Problem 3/250 (Angular Impulse/Momentum) 72
3.6 Problem 3/365 (Curvilinear Motion) 73

Chapter 4 Kinetics of Systems of Particles 77
4.1 Problem 4/26 (Conservation of Momentum) 78
4.2 Problem 4/62 (Steady Mass Flow) 80
4.3 Problem 4/86 (Variable Mass) 83
Chapter 5 Plane Kinematics of Rigid Bodies 87
5.1 Problem 5/3 (Rotation) 88
5.2 Problem 5/44 (Absolute Motion) 93
5.3 Sample Problem 5/9 (Relative Velocity) 95
5.4 Problem 5/123 (Relative Acceleration) 99
5.5 Sample Problem 5/15 (Absolute Motion) 100
Chapter 6 Plane Kinetics of Rigid Bodies 107
6.1 Sample Problem 6/2 (Translation) 108
6.2 Sample Problem 6/4 (Fixed-Axis Rotation) 113
6.3 Problem 6/98 (General Plane Motion) 115
6.4 Problem 6/104 (General Plane Motion) 118
6.5 Sample Problem 6/10 (Work and Energy) 120
6.6 Problem 6/206 (Impulse/Momentum) 125
Chapter 7 Introduction to Three-Dimensional
Dynamics of Rigid Bodies 129
7.1 Sample Problem 7/3 (General Motion) 130
7.2 Sample Problem 7/6 (Kinetic Energy) 132
Chapter 8 Vibration and Time Response 137
8.1 Sample Problem 8/2 (Free Vibration of Particles) 138
8.2 Problem 8/139 (Damped Free Vibrations) 140
8.3 Sample Problem 8/6 (Forced Vibration of Particles) 143

INTRODUCTION
Computers and software have had a tremendous impact upon engineering
education over the past several years and most engineering schools now
incorporate computational software such as MATLAB in their curriculum. Since
you have this supplement the chances are pretty good that you are already aware
of this and will have to learn to use MATLAB as part of a Dynamics course. The
purpose of this supplement is to help you do just that.
There seems to be some disagreement among engineering educators regarding
how computers should be used in an engineering course such as Dynamics. I will
use this as an opportunity to give my own philosophy along with a little advice.
In trying to master the fundamentals of Dynamics there is no substitute for hard
work. The old fashioned taking of pencil to paper, drawing free body and mass
acceleration diagrams, struggling with equations of motion and kinematic
relations, etc. is still essential to grasping the fundamentals of Dynamics. A
sophisticated computational program is not going to help you to understand the
fundamentals. For this reason, my advice is to use the computer only when
required to do so. Most of your homework can and should be done without a
computer.
The problems in this booklet are based upon problems taken from your text. The
problems are slightly modified since most of the problems in your book do not
require a computer for the reasons discussed in the last paragraph. One of the
most important uses of the computer in studying Mechanics is the convenience
and relative simplicity of conducting parametric studies. A parametric study
seeks to understand the effect of one or more variables (parameters) upon a
general solution. This is in contrast to a typical homework problem where you
generally want to find one solution to a problem under some specified conditions.
For example, in a typical homework problem you might be asked something
about the trajectory of a particle launched at an angle of 30 degrees from the
horizontal with an initial speed of 30 ft/sec. In a parametric study of the same
problem you might typically find the trajectory as a function of two parameters,
the launch angle θ and initial speed v. You might then be asked to plot the
trajectory for different launch angles and speeds. A plot of this type is very
beneficial in visualizing the general solution to a problem over a broad range of
variables as opposed to a single case.

6 INTRODUCTION
As you will see, it is not uncommon to find Mechanics problems that yield
equations that cannot be solved exactly. These problems require a numerical
approach that is greatly simplified by computational software such as MATLAB.
Although numerical solutions are extremely easy to obtain in MATLAB this is
still the method of last resort. Chapter 1 will illustrate several methods for
obtaining symbolic (exact) solutions to problems. These methods should always
be tried first. Only when these fail should you generate a numerical
approximation.
Many students encounter some difficulties the first time they try to use a
computer as an aid to solving a problem. In many cases they are expecting that
they have to do something fundamentally different. It is very important to
understand that there is no fundamental difference in the way that you would
formulate computer problems as opposed to a regular homework problem. Each
problem in this booklet has a problem formulation section prior to the solution.
As you work through the problems be sure to note that there is nothing peculiar
about the way the problems are formulated. You will see free-body and mass
acceleration diagrams, kinematic equations etc. just like you would normally
write. The main difference is that most of the problems will be parametric studies
as discussed above. In a parametric study you will have at least one and possibly
more parameters or variables that are left undefined during the formulation. For
example, you might have a general angle θ as opposed to a specific angle of 20°.
If it helps, you can “pretend” that the variable is some specific number while you
are formulating a problem.
This supplement has eight chapters. The first chapter contains a brief introduction
to MATLAB. If you already have some familiarity with MATLAB you can skip
this chapter. Although the first chapter is relatively brief it does introduce all the
methods that will be used later in the book and assumes no prior knowledge of
MATLAB. Chapters 2 through 8 contain computer problems taken from chapters
2 through 8 of your textbook. Thus, if you would like to see some computer
problems involving the kinetics of particles you can look at the problems in
chapter 3 of this supplement. Each chapter will have a short introduction that
summarizes the types of problems and computational methods used. This would
be the ideal place to look if you are interested in finding examples of how to use
specific functions, operations etc.
This supplement uses the student edition of MATLAB version 7.1. MATLAB is
a registered trademark of The Mathworks, Inc., 24 Prime Park Way, Natick,
Massachusetts, 01760.

AN INTRODUCTION
TO MATLAB
This chapter provides an introduction to the MATLAB programming language.
Although the chapter is introductory in nature it will cover everything needed to
solve the computer problems in this booklet.
1.1 Numerical Calculations
When you open MATLAB you should see a prompt something like the
following.
EDU»
To the right of this prompt you will write some sort of command. The following
example assigns a value of 200 to the variable a. Simply type “a = 200” at the
prompt and then press enter.
EDU» a=200
a =
200
There are many situations where you may not want MATLAB to print the result
of a command. To suppress the output simply type a semi-colon “;” after the
command. For example, to suppress the output in the above case type “a = 200;”
and the press enter. The basic mathematical operations of addition, subtraction,
multiplication, division and raising to a power are accomplished exactly as you
would expect if you are familiar with other programming languages, i.e. with the
keys +, -, *, , and ^. Here are a couple of examples.
EDU» c=2*5-120/4
c =
-20
1

8 CH. 1 AN INTRODUCTION TO MATLAB
EDU» d=2^4
d =
16
MATLAB has many built in functions that you can use in calculations. If you
already know the name of the function you can simply type it in. If not, you can
find a list of functions in the “Help Desk (HTML)” file available under Help in
the main menu bar. Here are a few examples.
EDU» e=sin(.5)+tan(.2)
e =
0.6821
As with most mathematical software packages, the default unit for angles will be
radians.
EDU» f=sqrt(16)
f =
4
EDU» g=log(5)
g =
1.6094
EDU» h=log10(5)
h =
0.6990
In the last two examples be careful to note that log is the natural logarithm. For a
base 10 logarithm (commonly used in engineering) you need to use log10.
Range Variables
In the above examples we have seen many cases where a variable (or name) has
been assigned a numerical value. There are many instances where we would like
a single variable to take on a range of values rather than just a single value.
Variables of this type are often referred to as range variables. For example, to
have the variable x take on values between 0 and 3 with an increment of 0.25 we
would type “x=0:0.25:3”.
EDU» x=0:0.25:3
x =
Columns 1 through 7
0 0.2500 0.5000 0.7500 1.0000 1.2500 1.5000
Columns 8 through 13

INTRODUCTION TO MATLAB 9
1.7500 2.0000 2.2500 2.5000 2.7500 3.0000
Range variables are very convenient in that they allow us to evaluate an
expression over a range with a single command. This is essentially equivalent to
performing a loop without actually having to set up a loop structure. Here’s an
example.
EDU» x=0:0.25:3;
EDU» f=3*sin(x)-2*cos(x)
f =
Columns 1 through 7
-2.0000 -1.1956 -0.3169 0.5815 1.4438 2.2163 2.8510
Columns 8 through 13
3.3084 3.5602 3.5906 3.3977 2.9936 2.4033
Although range variables are very convenient and almost indispensable when
making plots, they can also lead to considerable confusion for those new to
MATLAB. To illustrate, suppose we wanted to compute a value y = 2*x-3*x^2
over a range of values of x.
EDU» x=0:0.5:3;
EDU» y = 2*x-3*x^2
??? Error using ==> mpower
Matrix must be square.
What went wrong? We have referred to x as a range variable because this term is
most descriptive of how we will actually use variables of this type in this manual.
As the name implies, MATLAB is a program for doing matrix calculations.
Thus, internally, x is really a row matrix and operations such as multiplication or
division are taken, by default, to be matrix operations. Thus, MATLAB takes the
operation x^2 to be matrix multiplication x*x. Since matrix multiplication of two
row matrices doesn’t make sense, we get an error message. In a certain sense,
getting this error message is very fortunate since we are not actually interested in
a matrix calculation here. What we want is to calculate a certain value y for each
x in the range specified. This type of operation is referred to as term-by-term. For
term-by-term operations we need to place a period “.” in front of the operator.
Thus, for term by term multiplication, division, and raising to a power we would
write “.*”, “./” and “.^” respectively. Since matrix and term-by-term addition or
subtraction are equivalent, you do not need a period before + or -. The reason that
our earlier calculations did not produce an error is that they all involved scalar
quantities. To correct the error above we write.
EDU» x=0:0.5:3;
EDU» y = 2*x-3*x.^2
y =

0 0.2500 -1.0000 -3.7500 -8.0000 -13.7500 -21.0000
Note that we wrote “2*x” instead of “2.*x” since 2 is a scalar. If in doubt, it
never hurts to include the period.
1.2 Writing Scripts (m-files)
If you have been working through the examples above you have probably noticed
by now that you cannot modify a line once it has been entered into the worksheet.
This makes it rather difficult to debug a program or run it again with a different
set of parameters. For this reason, most of your work in MATLAB will involve
writing short scripts or m-files. The term m-file derives from the fact that the
programs are saved with a “.m” extension. An m-file is essentially just a list of
commands that you would normally enter at a prompt in the worksheet. After
saving the file you just enter the name of the file at the prompt after which
MATLAB will execute all the lines in the file as if they had been entered. The
chief advantage is that you can alter the script and run it as many times as you
like.
To get started, select “File…New…M-File” from the main menu. A screen like
that below will then pop up.

Now enter the program line by line into the editor. Basically, you just enter the
commands you would normally enter at a prompt (do not, however, enter >> at
the beginning of a line). If you want to enter comments, begin the line with “%”.
These lines will not be executed. Once you have finished entering the script into
the editor you can save the file (be sure to remember where you saved it). The
file name can contain numbers but cannot begin with a number. It also cannot
contain any mathematical operators such as +, -, *, / etc. Also, the name of the
file should not be the same as a variable it computes.
To run the script all you have to do is type the name of the program at a prompt
(>>), but without the “.m” extension. A common error at this point is that
MATLAB will give some response indicating that it hasn’t a clue what you are
doing. The reason this happens is that the file you saved is not in the current path.
You can check (or change) the path by selecting File…Set Path… from the main
menu.
Following is a simple example of a script file that sets up a range variable x and
then calculates two functions over the range.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% This script calculates two
% functions over a specified range
x=0:0.2:1;
f=sin(x)./(1+x)
g=sin(x).*exp(x)
%%%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%
The program was saved as example1, so here is the output you should see when
the program is entered at a prompt.
EDU» example1
f =
0 0.1656 0.2782 0.3529 0.3985 0.4207
g =
0 0.2427 0.5809 1.0288 1.5965 2.2874
To save space we will, in the future, show the output of a script immediately after
the script is given without actually showing the file name being entered at a
prompt.

1.3 Defining Functions
Although MATLAB has many built in functions (see the “Help Desk (HTML)”
file available under Help in the main menu bar), it is sometimes advantageous to
define your own functions. While user defined functions are actually special
types of m-files (they will be saved with the .m extension) their use is really quite
different from the m-files discussed above. For example, suppose you were to
create a function that you call f1. You would then use this function as you would
any built in function like sin, cos, log etc.
Since functions are really special types of m-files you start by opening an editor
window with “File…New…M-File”. The structure of the file is, however, quite
different from a script file. Following is an m-file for a function called load1.
Note that you should save the file under the same name as the function. The file
would thus show up as load1.m in whatever directory you save it. As with script
files, you need to be sure the current path contains the directory in which the file
was stored.
%%%%%% a function m-file %%%%%%%%%%%%%%%%
function y = load1(x)
% this file defines the function load1(x)
y = 2*(x-1)+3*(x-2).^2;
%%%%%% end of function m-file %%%%%%%%%%%
It is important to understand the structure of a function definition before you try
to create some on your own. The name of this function is “load1” and will be
used as you would any other function. y and x are local variables which
MATLAB uses to calculate the function. There is nothing special about y and x,
other variable names will work as well. Now, suppose you were to type load1(3)
at a prompt or within a script file. MATLAB will calculate the expression for y
substituting x = 3. This value of y will then be returned as the value for load1(3).
Here are a few examples using our function load1.
EDU» load1(3)
ans =
7
EDU» f=load1(5)-load1(3)
f =
28

EDU» load1(load1(3))
ans =
87
EDU» x=1:0.2:2;
EDU» g=load1(x)
g =
3.0000 2.3200 1.8800 1.6800 1.7200 2.0000
1.4 Graphics
One of the most useful things about a computational software package such as
MATLAB is the ability to easily create graphs of functions. As we will see, these
graphs allow one to gain a lot of insight into a problem by observing how a
solution changes as some parameter (the magnitude of a load, an angle, a
dimension etc.) is varied. This is so important that practically every problem in
this supplement will contain at least one plot. By the time you have finished
reading this supplement you should be very proficient at plotting in MATLAB.
This section will introduce you to the basics of plotting in MATLAB.
MATLAB has the capability of creating a number of different types of graphs.
Here we will consider only the X-Y plot. The most common and easiest way to
generate a plot of a function is to use range variables. The following example
will guide you through the basic procedure.
EDU» x=-3:0.1:3;
EDU» f=x.*exp(-x.^2);
EDU» plot(x,f)
The procedure is very simple. First define a range variable covering the range of
the plot, then define the function to be plotted and issue the plot command. After
typing in the above you should see a graph window pop up which looks
something like the following figure.

Things like titles and axis labels can be added in one of two ways. The first is by
line commands such as those shown below.
EDU» xlabel('x')
EDU» ylabel('x*exp(-x^2)')
EDU» title('A Simple Plot')
Take a look at the graph window after you type each of the lines above. At this
point, the graph window should look something like that shown below.

There are many other things that you can add or change on the plot. The easiest
way to do this is with the second of the two approaches mentioned above. In the
graph menu select “Tools…Edit Plot”. Now you can edit the plot in a number of
different ways. To see some of the options try either double clicking or right
clicking on the graph. You should spend some time experimenting with a graph
window to see the range of possibilities. In particular, be sure to try clicking the
show plot tools button .
Once you have things like you want them you can save the graph to a file. You
can also export the graph to some format suitable for inserting in, say, a word
processor. Select “File…Export Setup” in the graph window to see the range of
possible formats.

Parametric Studies
One of the most important uses of the computer in studying Mechanics is the
convenience and relative simplicity of conducting parametric studies (not to be
confused with parametric plotting discussed below). A parametric study seeks to
understand the effect of one or more variables (parameters) upon a general
solution. This is in contrast to a typical homework problem where you generally
want to find one solution to a problem under some specified conditions. For
example, in a typical homework problem you might be asked to find the reactions
at the supports of a structure with a concentrated force of magnitude 200 lb that
is oriented at an angle of 30 degrees from the horizontal. In a parametric study of
the same problem you might typically find the reactions as a function of two
parameters, the magnitude of the force and its orientation. You might then be
asked to plot the reactions as a function of the magnitude of the force for several
different orientations. A plot of this type is very beneficial in visualizing the
general solution to a problem over a broad range of variables as opposed to a
single case.
Parametric studies generally require making multiple plots of the same function
with different values of a particular parameter in the function. As a simple
example, consider the following function.
32
55 axxxf +−+=
What we would like to do is gain some understanding of how f varies with both x
and a. It might be tempting to make a three dimensional plot in a case like this.
Such a plot can, in some cases, be very useful. Usually, however, it is too
difficult to interpret. This is illustrated by the following three dimensional plot of
f versus x and a.
% This script produces a 3-d plot of
% f versus x and a.
x=linspace(-10,10,30);
a=linspace(-3,3,30);
[X,A]=meshgrid(x,a);
F=5+X-5*X.^2+A.*X.^3;
B=0.*A+2;
mesh(X,A,F)
colormap gray
%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%

The plot above certainly is interesting but, as mentioned above, not very easy to
interpret. In most cases it is much better to plot the function several times (with
different values of the parameter of interest) on a single two-dimensional graph.
We will illustrate this by plotting f as a function of x for a = -1, 0, and 1. This is
accomplished in the following script.
% script for plotting f versus x for several
% values of a.
x=-5:0.1:5;
g=5+x-5*x.^2;
f1=g-x.^3; % a = -1
f2=g; % a = 0
f3=g+x.^3; % a = 1
plot(x,f1,x,f2,x,f3)
xlabel('x')
ylabel('f')
%%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%%%

Running this script results in the following plot.
Parametric Plots
It often happens that one needs to plot some function y versus x but y is not
known explicitly as a function of x. For example, suppose you know the x and y
coordinates of a particle as a function of time but want to plot the trajectory of
the particle, i.e. you want to plot the y coordinate of the particle versus the x
coordinate. A plot of this type is generally called a parametric plot. Parametric
plots are easy to obtain in MATLAB. You start by defining the two functions in
terms of the common parameter and then define the common parameter as a
range variable. Then issue a plot command using the two functions as arguments.
The following script illustrates this procedure.

% This script file illustrates parametric plotting.
% A function f is plotted versus another function g.
% The two functions are related by the
% common parameter a.
a=-1:0.05:3.5;
f=10*a.*(2-a);
g=sin(3*a);
plot(g,f)
xlabel('g')
ylabel('f')
%%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%%%%%%%%
The range selected for the parameter can have a big, and sometimes surprising
effect on the resulting graph. To illustrate, try increasing the upper limit on the
range on a a few times and see how the graph changes.
You can, of course, also plot g as a function of f.

% Another example of parametric plotting
a=-1:0.05:6;
f=10*a.*(2-a);
g=sin(3*a);
plot(f,g)
xlabel('f')
ylabel('g')
%%%%%%%%% end of script %%%%%%%%%%%%%%%%%

1.5 Symbolic Calculations
One of the most useful features of many modern mathematical software packages
is the capability of doing symbolic math. By symbolic math we mean
mathematical operations on expressions containing undefined variables rather
than numbers. Maple is, perhaps, the best known computer algebra program.
Many other programs (including MATLAB) have a Maple engine that is called
whenever symbolic operations are performed. The Maple engine is part of the
Symbolic Math Toolbox so, if you do not have this toolbox, you will not be able
to do symbolic math. If you have a student edition of MATLAB you should have
the Symbolic Math Toolbox. If you do not have the Symbolic Math Toolbox, you
will need to perform whatever symbolic calculations are done in this manual by
hand.
How does MATLAB decide whether to call Maple and attempt a symbolic
calculation? Perhaps the best way to learn this is by studying the examples below
and by practicing. Maple will always be used to manipulate any expression
represented as a character string (i.e. an expression enclosed in ‘quotation’
marks). In the remainder of this manual we will refer to character strings of this
type as Maple expressions. Here are a few examples of Maple expressions.
EDU» 'x+a*x^2-b*x^3'
ans =
x+a*x^2-b*x^3
EDU» f='A*sin(x)*exp(-x^2)'
f =
A*sin(x)*exp(-x^2)
Instead of placing the entire expression in quotations, you can also first declare
certain variables to be symbols and then write the expressions as usual. This is
more convenient in some situations but can get confusing. Here are some
examples of this type.
EDU» syms x a b A % Sets x, a, b and A as symbols
EDU» g=x+a*x^2-b*x^3
g =
x+a*x^2-b*x^3
EDU» f=A*sin(x)*exp(-x^2)
f =
A*sin(x)*exp(-x^2)
The above are just examples of Maple expressions and do not involve any
symbolic math. Two of the most important applications of symbolic math will be
discussed in the next two sections, namely symbolic calculus (integration and

differentiation) and symbolic solution of one or more equations. The purpose of
the present section is to introduce you to the basic procedures of symbolic math
as well as to give a few other useful applications. One particularly useful
symbolic operation is substitution and is illustrated by the following examples.
EDU» g='x+a*x^2-b*x^3'
g =
x+a*x^2-b*x^3
EDU» subs(g,'a',3) % substitutes 3 for a in g. Note the quotation marks about a.
ans =
x+3*x^2-b*x^3
EDU» g1=subs(g,'x','(y-c)')
g1 =
((y-c))+a*((y-c))^2-b*((y-c))^3
EDU» syms x y A B
EDU» f=A*sin(x)*exp(-B*x^2)
f =
A*sin(x)*exp(-B*x^2)
EDU» subs(f,B,4) % Note that B doesn't have to be in quotes since it
was declared to be a symbol.
ans =
A*sin(x)*exp(-4*x^2)
Another very convenient feature available in the Symbolic Toolbox provides the
capability of defining “inline” functions thus avoiding the necessity for creating
special m-files as described above. Here is an example.
EDU» f=inline('x^2*exp(-x)')
f =
Inline function:
f(x) = x^2*exp(-x)
EDU» f(3)
ans =
0.4481
In the previous section we plotted the function 32
55 axxxf +−+= for several
values of a. It turns out to be somewhat more convenient to do this with an inline
function. Let’s see how it works.

EDU» f=inline('5+x-5*x^2+a*x^3')
f =
Inline function:
f(a,x) = 5+x-5*x^2+a*x^3
At this point we might anticipate a minor problem. When we plot this as a
function of x we will have a range variable as one of the arguments for f. This
creates a problem as MATLAB will be performing the numerical calculations
and it will take the operations in f as matrix operations rather than term by term.
Fortunately we can use the “vectorize” operator to insert periods in the
appropriate places.
EDU» f=vectorize(f)
f =
Inline function:
f(a,x) = 5+x-5.*x.^2+a.*x.^3
EDU» x=-5:0.1:5;
EDU» plot(x,f(-1,x),x,f(0,x),x,f(1,x))

1.6 Differentiation and Integration
Here we will consider only symbolic differentiation and integration since these
are most useful in the mechanics applications in this manual. For differentiation
we will use the Maple diff command. Care should be exercised here since diff is
also a MATLAB command. To guarantee that you are using the Maple
command, be sure that you use diff only on Maple expressions.
EDU» f='a*sec(b*t)';
EDU» diff(f, 't') % Note that t must be placed in quotation marks since
it is a symbol.
ans =
a*sec(b*t)*tan(b*t)*b
We can avoid using the quotation marks by making a, b and t symbols.
EDU» syms a b t
EDU» f=a*sec(b*t);
EDU» diff(f, t)
ans =
a*sec(b*t)*tan(b*t)*b
To find higher order derivatives use the command diff(f, x, n) where n is the
order of the derivative. For example, to find the third derivative of alog(b+x) we
would write
EDU» diff('a*log(b+x)', 'x', 3)
ans =
2*a/(b+x)^3
Symbolic integration will be performed with the int command. The general
format of this command is int(f, x, a, b) where f is the integrand (a Maple
expression), x is the integration variable and a and b are the limits of integration.
If x, a and b have not been declared symbols with the syms command, they must
be placed in quotation marks. If the integration limits are omitted, the indefinite
integral will be evaluated. Here are several examples of definite and indefinite
integrals.
EDU» int('sin(b*x)','x')
ans =
-1/b*cos(b*x)

EDU» int('sin(b*x)','x','c','d')
ans =
-(cos(b*d)-cos(b*c))/b
EDU» syms x b c d
EDU» f=b*log(x);
EDU» int(f,x)
ans =
b*x*log(x)-b*x
EDU» int(f,x,c,d)
ans =
b*d*log(d)-b*d-b*c*log(c)+b*c
If a definite integral contains no unknown parameters either in the integrand or
the integration limits, the int command will provide numerical answers. Here are
a few examples.
EDU» int('x+3*x^3','x',0,3)
ans =
261/4
EDU» int('log(x)','x',2,5) % Don’t forget that log is the natural logarithm.
ans =
5*log(5)-3-2*log(2)
Note that Maple will always try to return an exact answer. This usually results in
answers containing fractions or functions as in the above examples. This is very
useful in some situations; however, one often wants to know the numerical
answer without having to evaluate a result such as the above with a calculator. To
obtain numerical answers use Maple’s eval function as in the following
examples.
EDU» eval(int('x+3*x^3','x',0,3))
ans =
65.2500
EDU» eval(int('log(x)','x',2,5))
ans =
3.6609

1.7 Solving Equations
Our preferred approach to solving equations is to use Maple’s solve command
available in the Symbolic Toolbox. We will also discuss briefly MATLAB’s built
in function fzero which can be used to solve single equations numerically.
Solving single equations
Here are a couple of examples illustrating how to use Maple’s solve command to
solve a single equation symbolically. The first example is the (hopefully) familiar
quadratic equation.
EDU» f='a*x^2+b*x+c=0';
EDU» solve(f, 'x')
ans =
1/2/a*(-b+(b^2-4*a*c)^(1/2))
1/2/a*(-b-(b^2-4*a*c)^(1/2))
EDU» g='a*sin(x)-cos(x)=1';
EDU» solve(g, 'x')
ans =
pi
atan2(2*a/(1+a^2),(a^2-1)/(1+a^2))
Note that both examples produced two solutions with each solution being placed
on separate lines. One of the solutions to the second example is the number π.
The second solution illustrates something you should always watch out for. In
many cases you may get a solution in terms of a function that you are not familiar
with, i.e. atan2 that appears in the second solution to the second example above.
Whenever this happens you should type “help atan2” at a prompt in order to learn
more about the function.
If the variable being solved for is the only unknown in the equation, solve will
return a number as the result. Here are a couple of examples.
EDU» f='2*x^2+4*x-12';
EDU» solve(f, 'x')
ans =
-1+7^(1/2)
-1-7^(1/2)
EDU» g='5*sin(x)-cos(x)=1';
EDU» solve(g, 'x')
ans =

pi
atan(5/12)
If you want the answer as a floating point number instead of an exact result, use
the Maple command eval. For example, with g defined as above we would write
EDU» eval(solve(g, 'x'))
ans =
3.1416
0.3948
Although Maple will always attempt to return an exact (symbolic) result, you
may occasionally have a case where a floating point number is returned without
using eval. Here is an example.
EDU» f='10*sin(z)=2*z+1';
EDU» solve(f, 'z')
ans =
.12541060097666150276238831440340
Whenever this occurs, Maple was unable to solve the equation exactly and
switched automatically to a numerical solution. In a certain sense this is very
convenient as the same command is being used for both symbolic and numerical
calculations. The problem is that nonlinear equations quite often have more than
one solution, as illustrated by the examples above. If you have the full version of
Maple then you will have various options available for finding multiple
numerical solutions. But since we are dealing now with numerical rather than
symbolic solutions it makes more sense just to use MATLAB.
The appropriate MATLAB command for the numerical solution of an equation is
fzero. This command numerically determines the point at which some defined
function has a value of zero. This means that we need to rewrite our equation in
terms of an expression whose zeros (roots) are solutions to the original
expression. This is easily accomplished by rearranging the original equation into
the form expression = 0. We will call this expression g in this example. Rewriting
the above equation f we have )sin(1012 zzg −+= . The general format for fzero
is fzero(‘function_name’, x0) where ‘function_name’ is the name of the m-file
function that you have created and x0 is an initial guess at the root (zero) of the
function. The initial guess is very important since, if there is more than one
solution, MATLAB will find the solution closest to the initial guess. Well, you
may be wondering how we can determine an approximate solution if we do not
yet know the solution. Actually, this is very easy to do. First we define a function
g(z) whose roots will be the solution to the equation of interest. Next, we plot this
function in order to estimate the location of points where g(z) = 0. Here’s how it
works for the present example.

%%%%% function m-file %%%%%%
function y = gz(z)
y = 2*z + 1 - 10*sin(z);
%%%%% end of m-file %%%%%%%%
Above is our user defined function (m-file) which we have named gz. Now we
can plot this function over some range as illustrated below.
EDU» z=-1:0.01:3;
EDU» plot(z,gz(z),z,0) % we also plot a line at g=0 to help find the roots.
EDU» xlabel('z')
EDU» ylabel('g(z)')
The plot shows that the function is zero at about z = 0 and 2. The first root is
clearly that found by Maple’s solve command above. Now we can easily find
both roots with the fzero command.
EDU» fzero('gz', 0)
ans =
0.1254
EDU» fzero('gz', 2)
ans =
2.4985

Thus, the two solutions are z = 0.1254 and 2.4985.
Finding Maxima and Minima of Functions
Here we will illustrate two methods for finding maxima and minima of a
function.
Method 1. Using the symbolic diff and solve functions.
The usual method for finding maxima or minima of a function f(x) is to first
determine the location(s) x at which maxima or minima occur by solving the
equation =
df
dx
0 for x. One then substitutes the value(s) of x thus determined
into f(x) to find the maximum or minimum. Consider finding the maximum and
minimum values of the following function:
3
21 xxf −+=
Before proceeding, it is a good idea to make a plot of the function f. This will
give us a rough idea where the minima and maxima are as a check on the results
we obtain below. The following script will plot f as a function of x.
% This script plots f as a function of x
x = -2:0.05:2;
f = 1 + 2*x - x.^3;
plot(x, f)
xlabel('x')
ylabel('f')
%%%%%%%%%% end of script %%%%%%%%%%%%%

This plot shows we have a minimum of about 0 at x of about −0.9 and a
maximum of about 2 at x of about 0.9. Now we can proceed to find the results
more precisely.
EDU» f = '1+2*x-x^3'
f =
1+2*x-x^3
EDU» dfdx = diff(f,'x')
dfdx =
2-3*x^2
EDU» solve(dfdx,'x')
ans =
1/3*6^(1/2)
-1/3*6^(1/2)
If an equals sign is omitted (as above) solve will find the root of the supplied
Maple expression. In other words, it will solve the equation dfdx = 0. It is more
convenient in the present case to have the solutions as floating point numbers.
Thus, we should have used eval as in the following.

EDU» eval(solve(dfdx,'x'))
ans =
0.8165
-0.8165
The above results are the locations (x) where the minima and maxima occur. To
find the maximum and minimum values of the function f we use subs to
substitute these results back into f.
EDU» subs(f,'x',0.8165)
ans =
2.0887
EDU» subs(f,'x',-0.8165)
ans =
-0.0887
Thus, fmax = 2.0887 at x = 0.8165 and fmin = − 0.0887 at x = − 0.8165.
Method 2. Using MATLAB’s min and max functions.
EDU» x = -2:0.05:2;
EDU» f = 1 + 2*x - x.^3;
These two lines are identical to those in the script used to plot f above. If you
have just run that script you do not need to execute these lines. Now we can find
the minimum and maximum values of f by typing min(f) and max(f).
EDU» min(f)
ans = -3
EDU» max(f)
ans = 5
These are clearly not the answers we obtained above. What went wrong? It is
important to understand that min and max do not find true minima and maxima in
the mathematical sense, i.e. they do not find locations where df/dx = 0. Instead,
they find the minimum or maximum value of all those calculated in the vector f.
To see this, go back and look at the plot of f above and you will see that the
minimum and maximum values of f are indeed − 3 and 5. This is another good
reason to plot the function first. To avoid the above difficulty we can change the
range of x to insure we pick up the true minima and maxima.
EDU» x = -1:0.05:1;

EDU» f = 1 + 2*x - x.^3;
In most cases, we want not only a minimum or maximum but also the location
where they occur. This being the case we should use the following command.
EDU» [f_min,i] = min(f)
f_min = -0.0880
i = 5
The command [f_min, i] = min(f) finds the minimum value of f and assigns it to
the variable f_min. The index of this value is assigned to the variable i. Now we
can find the location of the minimum by printing the 5th
value of the vector x.
EDU» x_min=x(i)
x_min = -0.8000
Now we can repeat the above for the maximum.
EDU» [f_max,i] = max(f)
f_max = 2.0880
i = 37
EDU» x_max=x(i)
x_max = 0.8000
You may have noticed at this point that the answers are somewhat different from
those found by the first method above. The reason is, once again, that min and
max do not find true minima and maxima. To obtain more accurate results you
can use a closer spacing for the range variable (x). To see this, repeat the above
for a spacing of 0.01 instead of 0.05.

Solving several equations simultaneously
EDU» eqn1='x^2+y^2=12'
eqn1 =
x^2+y^2=12
EDU» eqn2='x*y=4'
eqn2 =
x*y=4
In the above we have defined two equations which we will now solve for the two
unknowns, x and y.
EDU» [x,y] = solve(eqn1,eqn2)
x =
5^(1/2)-1
-1-5^(1/2)
5^(1/2)+1
1-5^(1/2)
y =
5^(1/2)+1
1-5^(1/2)
5^(1/2)-1
-1-5^(1/2)
Note that four solutions have been found, the first being 15 −=x , 15 +=y .
The following example illustrates a case where there are more unknowns than
there are equations. In this case, the specified variables will be solved for in
terms of the others.
EDU» f='x^2 + a*y^2 = 0'
f =
x^2 + a*y^2 = 0
EDU» g='x-y = b'
g =
x-y = b

EDU» [x,y] = solve(f,g)
x =
1/2/(a+1)*(-2+2*(-a)^(1/2))*b+b
1/2/(a+1)*(-2-2*(-a)^(1/2))*b+b
y =
1/2/(a+1)*(-2+2*(-a)^(1/2))*b
1/2/(a+1)*(-2-2*(-a)^(1/2))*b
In this case, there are two solutions. Now let’s consider an example with three
equations.
EDU» eqn1='x^2+y^2=12'
eqn1 =
x^2+y^2=12
EDU» eqn2='x*y=4'
eqn2 =
x*y=4
EDU» eqn3='x-y=z'
eqn3 =
x-y=z
EDU» [x,y,z] = solve(eqn1,eqn2,eqn3)
x =
5^(1/2)+1
1-5^(1/2)
5^(1/2)-1
-1-5^(1/2)

y =
5^(1/2)-1
-1-5^(1/2)
5^(1/2)+1
1-5^(1/2)
z =
2
2
-2
-2

KINEMATICS OF
PARTICLES
Kinematics involves the study of the motion of bodies irrespective of the forces
that may produce that motion. MATLAB can be very useful in solving particle
kinematics problems. Problem 2.1 is a rectilinear motion problem illustrating
integration with the int command. The formulation of this problem results in an
equation that cannot be solved exactly except with some rather sophisticated
mathematics. When this occurs it is generally easiest to obtain either a graphical
or numerical solution. This problem illustrates both approaches. Problem 2.2 is a
rectangular coordinates problem that illustrates diff and solve. Problem 2.3 is a
relatively straightforward problem where MATLAB is used to generate a plot
that might be useful in a parametric study. The r-θ components of velocity and
acceleration are plotted in problem 2.4 as well as the trajectory of a particle. In
problem 2.5, the r-θ components of the velocity are determined using symbolic
differentiation (diff). The problem also illustrates how computer algebra can
simplify what might normally be a rather tedious algebra problem. The diff
command is further illustrated in problems 2.6 and 2.7. Problem 2.7 is
particularly interesting in that it requires differentiation with respect to time of a
function whose explicit time dependence is unknown. This happens rather
frequently in Dynamics so it is useful to know how to accomplish this with
MATLAB.
2

38 CH. 2 KINEMATICS OF PARTICLES
2.1 Sample Problem 2/4 (Rectilinear Motion)
A freighter is moving at a speed of 8 knots when
its engines are suddenly stopped. From this time
forward, the deceleration of the ship is
proportional to the square of its speed, so that
2
kva −= . The sample problem in your text
shows that it is rather easy to determine the
constant k by measuring the speed of the boat at
some specified time. Show how k could be
found by (a) measuring the speed after some
specified distance and (b) measuring the time
required to travel some specified distance. In
both cases let the initial speed be v0.
Problem Formulation
(a) Since time is not involved, the easiest approach is to integrate the equation
vdv = ads.
dskvadsvdv 2
−== ∫∫ −=
sv
v
dsk
v
dv
00






=
v
v
ks 0
ln
With this result it is easy to find k given v at some specified s. To illustrate,
assume that v0 = 8 knots and that the speed of the boat is determined to be 3.9
knots after it has traveled one nautical mile.






=
9.3
8
ln)1(k k = 0.718 mi-1
(b) Here we follow the general approach in the sample problem. Integrating a =
dv/dt yields
∫∫ −=
tv
v
dtk
v
dv
0
2
0
0
0
vv
vv
kt
−
=−
0
0
1 ktv
v
v
+
=
To obtain the distance s as a function of time we integrate v = ds/dt
∫ ∫∫ +
===
t ts
dt
ktv
v
vdtsds
0 0 0
0
0
1
( )01ln
1
ktv
k
s +=

KINEMATICS OF PARTICLES 39
This equation turns out to be very difficult to solve for k. A good mathematician
or someone familiar with symbolic algebra software might be able to find the
general solution for k in terms of the so-called LambertW function (LambertW(x)
is the solution of the equation yey
= x). Even if this solution were found it would
be of little use in most practical situations. For example, you would have to spend
some time familiarizing yourself with the function. Once this is done you would
still have to use a program like Maple or a mathematical handbook to evaluate
the function.
For these reasons it is probably easiest to find k either graphically or numerically.
Obtaining a numerical solution with MATLAB is so easy that there is little
reason not to use this approach. It is generally advisable though to use a graphical
approach even when a numerical solution is being obtained. This is the best way
to identify whether there are multiple solutions to the problem and also serves as
a useful check on the numerical results. Thus, both approaches are illustrated
below.
The usual way to generate a graphical solution is to rearrange the equation so as
to give a function that is zero at points that are solutions to the original equation.
Rearranging the equation above in this manner yields,
( ) 01ln 0 =+−= ktvksf
Given values of s, t, and v0, f can be plotted versus k. The value of k at which f =
0 provides the solution to the original equation.
MATLAB Worksheet and Scripts
Although the integrations are simple in this problem, we'll go ahead and evaluate
them symbolically for purposes of illustration.
EDU» s_a='-1/k'*int('1/x','x','v0','v') % part (a)
s_a = -1/k*(log(v)-log(v0))
EDU» s_b=int('v0/(1+k*x*v0)','x',0,'t') % part (b)
s_b = log(1+t*k*v0)/k
The following script implements a graphical solution for part (b). To illustrate,
we take v0 = 8 knots and assume that the boat is found to move 1.1 nautical miles
after 10 minutes.

%%%%%%%%%%%% Script %%%%%%%%%%%%%%%%%%%%
% This script plots f as a function of k for
% given values of v0, s, and t
v0 = 8; % initial speed (knots)
s = 1.1; % distance (nautical miles)
t = 10/60; % time (hours)
k = 0:0.01:0.5;
f = s*k-log(1+k*v0*t);
line = k*0;
plot(k,f,k,line)
xlabel('k')
ylabel('f')
The above graph shows that k is about 0.34 mi-1
. Now let's try to find a solution
with the symbolic solve. First we write a Maple expression for f given our
assumed values for v0, s, and t.
EDU» f = '1.1*k - log(1+k*8*10/60) = 0'
f = 1.1*k - log(1+k*8*10/60)=0
EDU» solve(f, 'k')
ans =
0.
.33923053342470867736031513009887
Thus, k = 0.3392 mi-1
.

2.2 Problem 2/87 (Rectangular Coordinates)
A long-range rifle is fired at A with the projectile
hitting the mountain at B. (a) If the muzzle
velocity is u ٌ 400 m/s, determine the two
angles of elevation θ which will permit the
projectile to hit the mountain target B and plot
the two trajectories. (b) Determine the smallest
muzzle velocity that will allow the projectile to
strike at B and the angle at which it must be
fired. Repeat the plot for part (a) and include the
trajectory of the projectile for this minimum
initial velocity.
Problem Formulation
Place a coordinate system at A with x positive to the right and y positive up. The
initial components of the velocity are,
( ) θcos0 uvx = ( ) θsin
0
uvy =
The acceleration is constant with components 0=xa and gay −= . Integrating
these two accelerations twice and applying initial conditions yields (see page 44
of your text if you need additional details),
x = ucosθ t y = usinθ t -1/2gt2
Plotting y in terms of x for different times t will yield the trajectory of the
projectile. This type of plot is called a parametric plot since the items plotted (x
and y) are each known in terms of another parameter (t).
Anytime you have a projectile motion problem and you know the coordinates of
a point on the trajectory (our point B) you should solve for x and y (as we have
done above) and then obtain two equations by substituting the coordinates of the
points. These two equations can then be solved for two unknowns. Note that in
most cases one of the two unknowns will be the time of flight.
Part (a) Substituting x = 5,000 m, y = 1,500 m and u = 400 m/s gives
5000 = 400cosθ t 1500 = 400sinθ t -1/2(9.81)t2

We will let MATLAB solve these two equations simultaneously. MATLAB
actually finds four solutions, however two of the four can be discarded since they
involve negative times. The other two solutions correspond to the two solutions
shown in the illustration accompanying the problem statement. The results are,
θ1 = 26.6° and θ2 = 80.6°
Part (b) It should be intuitively obvious why there must be a minimum initial
velocity below which the projectile cannot reach B. How do we go about finding
it? We still have the two equations for the coordinates of point B,
5000 = ucosθ t 1500 = usinθ t -1/2(9.81)t2
however there are now three unknowns (u, θ, t). Suppose for the moment that the
launch angle θ were given and we were asked to calculate the required initial
speed u so that the projectile strikes B. In this case we would have two equations
and two unknowns. From this observation we see that u is a function of θ from
which we get our general solution strategy:
(a) Eliminate t from the above two equations and solve for u as a
function of θ.
(b) Differentiate this function with respect to θ to find the location of
the minimum.
Solving the first equation for u gives
θcos
5000
t
u =
Substituting into the second yields 2
2
1
tan50001500 gt−= θ . This equation is
now solved for ( ) gt /1500tan50002 −= θ which can be substituted back into
u to give
( ) g
u
/1500tan50002cos
5000
−
=
θθ
We will let MATLAB differentiate this equation and solve for the minimum
speed and the associated launch angle. The result is
umin = 256.8 m/s at θ = 53.3°

Part (a)
EDU>> eqn1='5000 = 400*cos(x)*y'
eqn1 =
5000 = 400*cos(x)*y
EDU>> eqn2='1500 = 400*sin(x)*y-1/2*9.81*y^2'
eqn2 =
1500 = 400*sin(x)*y-1/2*9.81*y^2
EDU>> [x,y] = solve(eqn1,eqn2)
x =
-1.7350349681568043127091034228443
-2.6858972177500184529892570306420
.45569543583977478547338635263748
1.4065576854329889257535399604352
y =
-76.452007728583426926316546391465
-13.920516408172829238156414017816
13.920516408172829238156414017816
76.452007728583426926316546391465
MATLAB finds four solutions but two can be excluded because they have time
(y) being negative. The two positive times and corresponding angles are
t B1 = 13.9205 secs θ1 = 0.4557 rads (26.6°)
and t B2 = 76.4520 secs θ2 = 1.4066 rads (80.6°)
%=========== script #1 ======================
% Note that we need to set up two time scales in order to
% insure that both trajectories end at point B.
t1 = 0:0.01:13.92;
t2 = 0:0.01:76.45;
x1 = 400*cos(.455695)*t1;
y1 = 400*sin(.455695)*t1-1/2*9.81*t1.^2;
x2 = 400*cos(1.4066)*t2;
y2 = 400*sin(1.4066)*t2-1/2*9.81*t2.^2;
plot(x1/1000,y1/1000,x2/1000,y2/1000)
xlabel('x (km)')
ylabel('y (km)')
title('plot for part (a)')
%======= end script #1 ======================

Part (b)
EDU>> u = '5000/cos(x)/sqrt(2*(5000*tan(x)-1500)/9.81)'
u =
5000/cos(x)/sqrt(2*(5000*tan(x)-1500)/9.81)
EDU>> dudx = diff(u,'x');
EDU>> solve(dudx,'x')
ans =
.93112656063638185561346315653632
-.63966976615851476361785853510343
Only one of the two solutions is between 0 and 90°.
EDU>> .9311*180/pi
ans =
53.3481

EDU>> subs(u,.9311)
ans =
256.7581
Thus, umin = 256.8 m/s at θ = 53.3°
Before plotting we need to find the time to reach point B. This can be done from
the equation x = 5000 = ucosθ.
EDU>> 5000/256.7581/cos(.9311)
ans =
32.6217
%=========== script #2 ======================
% this script should be run after script 1
t3 = 0:0.01:32.62;
x3 = 256.8*cos(.9311)*t3;
y3 = 256.8*sin(.9311)*t3-1/2*9.81*t3.^2;
plot(x1/1000,y1/1000,x2/1000,y2/1000,x3/1000,y3/1000)
title('plot for part (b)')
%======= end script #2 ======================

2.3 Problem 2/126 (n-t Coordinates)
A baseball player releases a ball with initial
conditions shown in the figure. Plot the radius of
curvature of the path just after release and at the
apex as a function of the release angle θ. Explain
the trends in both results as θ approaches 90°.
Problem Formulation
Just after release
ρ
θ
2
0
cos
v
gan ==
θ
ρ
cos
2
0
g
v
=
At the apex
At the apex, vy = 0 and v = vx = v0cosθ. Since v is horizontal, the
normal direction is vertically downward so that an = g.
( )
ρ
θ 2
0 cosv
gan ==
( )
g
v
2
0 cosθ
ρ =
MATLAB Script
%%%%%%%%%%%%%%%%% Script %%%%%%%%%%%%%%%%%%%%
v0 = 100;
g = 32.2;
theta = 0:0.01:pi/2;
rho_i = v0^2/g./cos(theta);
rho_a = (v0*cos(theta)).^2/g;
plot(theta*180/pi,rho_i, theta*180/pi,rho_a)
xlabel('theta (deg)')
title('radius of curvature (ft)')
axis([0 90 0 800])
% note that we need to limit the vertical axis since the
% initial rho approaches infinity as theta approaches
% 90 degrees.
%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%

Note that as θ approaches 90°, the initial ρ goes to infinity while ρ at the apex
approaches zero. When θ = 90°, the ball travels along a straight (vertical) path.
As you recall, straight paths have a radius of curvature of infinity. At the apex,
the velocity will be zero giving a radius of curvature of zero.

2.4 Sample Problem 2/9 (Polar Coordinates)
Rotation of the radially slotted arm is governed
by θ = 0.2t + 0.02t3
, where θ is in radians and t
is in seconds. Simultaneously, the power screw
in the arm engages the slider B and controls its
distance from O according to r = 0.2 + 0.04t2
,
where r is in meters and t is in seconds.
Calculate the magnitudes of the velocity and
acceleration of the slider as a function of time t.
(a) Plot v, vr and vθ for t between 0 and 5 sec. (b)
Plot a, ar and aθ for t between 0 and 5 sec. (c)
Plot the path of the slider B and compare with
the result in your book.
Problem Formulation
The first part of this problem solution will be identical to that in the Sample
Problem in your text except that everything will be left in terms of t. To
summarize,
2
04.02.0 tr += tr 08.0=& 08.0=r&&
3
02.02.0 tt +=θ 2
06.02.0 t+=θ& t12.0=θ&&
Now all we have to do is substitute these expressions into the definitions for the
velocity and acceleration. As usual, there is no need to make an explicit
substitution when using the computer.
trvr 08.0== & θθ
&rv = 22
θvvv r +=
2
θ&&& rrar −= θθθ
&&&& rra 2+= 22
θaaa r +=
The plot for part (c) can be obtained by using the suggestion in your book. First
write
θcosrx = θsinry =
and then plot y versus x. This type of plot is called a parametric plot since y is no
known explicitly as a function of x. Instead, both x and y are known in terms of
another parameter, namely t.

MATLAB Scripts
%======= script #1 ============================
% this script plots the velocities for part (a)
t = 0:0.01:5;
r = 0.2+0.04*t.^2;
rd = 0.08*t;
rdd = 0.08;
theta = 0.2*t+0.02*t.^3;
thetad = 0.2+0.06*t.^2;
thetadd = 0.12*t;
vr = rd;
vtheta = r.*thetad;
v=sqrt(vr.^2+vtheta.^2);
plot(t,v,t,vr,t,vtheta)
xlabel('time (seconds)')
ylabel('velocity (m/s)')
title('part (a) velocity')
%====== end of script#1=======================

%======= script #2 ============================
% this script plots the accelerations for part (b)
% run script # 1 first
ar = rdd - r.*thetad.^2;
atheta = r.*thetadd+2*rd.*thetad;
a=sqrt(ar.^2+atheta.^2);
plot(t,a,t,ar,t,atheta)
xlabel('time (seconds)')
ylabel('acceleration (m/s^2)')
title('part (b) acceleration')
%=========end of script 2======================
%======= script #3 ====================
% This script plots the path of the
% slider for part (b).
% Run script # 1 first.
x = r.*cos(theta);
y = r.*sin(theta);
plot(x, y)
xlabel('x (m)')
ylabel('y (m)')
title('part (c) Path of the Slider')
%=========end of script 3===============

2.5 Sample Problem 2/10 (Polar Coordinates)
A tracking radar lies in the vertical plane of the
path of a rocket which is coasting in unpowered
flight above the atmosphere. For the instant
when θ = 30°, the tracking data give r = 25(104
)
feet, r& = 4000 ft/s, and θ& = 0.8 deg/s. Let this
instant define the initial conditions at time t = 0
and plot vr and vθ as a function of time for the
next 150 seconds. You may assume that g
remains constant at 31.4 ft/s2
during this time
interval.
Problem Formulation
Place a Cartesian coordinate system at the radar with x positive to the right
and y positive up. Since the rocket is coasting in unpowered flight we can
use the equations for projectile motion.
( )tvxx βcos00 += ( ) 2
00
2
1
sin gttvyy −+= β
Where x0 = 25(104
)sin(30) ft, y0 = 25(104
)cos(30) ft,
v0 is the initial speed (5310 ft/sec, see the sample
problem) and β is the angle that v0 makes with the
horizontal. From the figure shown to the right we
can find the angle between v0 and the r axis as
o1
11.41)4000/3490(tan == −
φ . Since the r axis is
60° from the horizontal, β = 60 – 41.11 = 18.89°.
With r and θ defined as in the sample problem we have, at any time t
22
yxr += ( )yx /tan 1−
=θ
Now we find vr and vθ from their definitions.
rvr
&= θθ
&rv =
Substitution of x and y into the above equations and carrying out the derivatives
with respect to time gives vr and vθ as functions of time. The results are very

messy and will not be given here. Remember, though, that substitutions such as
this can be made automatically when using computer software such as
MATLAB.
MATLAB Scripts
%%%%%%%%%%%%%%%%%%%% Script # 1 %%%%%%%%%%%%%%%%%%%%%%%%%
% This script obtains symbolic results for the r and
% theta components of the velocity
syms x y t x0 y0 v0 beta g
x = x0+v0*cos(beta)*t
y = y0+v0*sin(beta)*t-1/2*g*t^2
r = sqrt(x^2+y^2)
theta = atan(x/y)
vr = diff(r,t);
vtheta = r*diff(theta,t);
% The results from these operations will be copied and
% pasted into script #2 for plotting. Before they can be
% used they have to be "vectorized", which places periods
% in front of the operators to insure term by term rather
% than matrix operations. We'll go ahead and do that here
% and then copy and paste the vectorized results.
vr = vectorize(vr)
vtheta = vectorize(vtheta)
Output from Script #1
x = x0+v0*cos(beta)*t
y = y0+v0*sin(beta)*t-1/2*g*t^2
r = ((x0+v0*cos(beta)*t)^2+(y0+v0*sin(beta)*t-1/2*g*t^2)^2)^(1/2)
theta = atan((x0+v0*cos(beta)*t)/(y0+v0*sin(beta)*t-1/2*g*t^2))
vr =
1./2./((x0+v0.*cos(beta).*t).^2+(y0+v0.*sin(beta).*t-
1./2.*g.*t.^2).^2).^(1./2).*(2.*(x0+v0.*cos(beta).*t).*v0.*cos(beta)+2.*(y0+v0.*
sin(beta).*t-1./2.*g.*t.^2).*(v0.*sin(beta)-g.*t))
vtheta =
((x0+v0.*cos(beta).*t).^2+(y0+v0.*sin(beta).*t-
1./2.*g.*t.^2).^2).^(1./2).*(v0.*cos(beta)./(y0+v0.*sin(beta).*t-1./2.*g.*t.^2)-
(x0+v0.*cos(beta).*t)./(y0+v0.*sin(beta).*t-1./2.*g.*t.^2).^2.*(v0.*sin(beta)-
g.*t))./(1+(x0+v0.*cos(beta).*t).^2./(y0+v0.*sin(beta).*t-1./2.*g.*t.^2).^2)

%%%%%%%%%%%%%%%%%%%% Script # 2 %%%%%%%%%%%%%%%%%%%%%
% This script plots the r and theta components of the
% velocity as functions of time.
x0 = 25*10^4*sin(30*pi/180);
y0 = 25*10^4*cos(30*pi/180);
v0 = 5310;
beta = 18.89*pi/180;
g = 31.4;
t = 0:0.1:150;
% The following expressions are copied and pasted from the
% results obtained with script #1.
vr = 1./2./((x0+v0.*cos(beta).*t).^2+(y0+v0.*sin(beta).*t-
1./2.*g.*t.^2).^2).^(1./2).*(2.*(x0+v0.*cos(beta).*t).*v0.*
cos(beta)+2.*(y0+v0.*sin(beta).*t-1./2.*g.*t.^2).*
(v0.*sin(beta)-g.*t));
vtheta = ((x0+v0.*cos(beta).*t).^2+(y0+v0.*sin(beta).*t-
1./2.*g.*t.^2).^2).^(1./2).*(v0.*cos(beta)./(y0+v0.*sin(bet
a).*t-1./2.*g.*t.^2)-(x0+v0.*cos(beta).*t)./
(y0+v0.*sin(beta).*t-1./2.*g.*t.^2).^2.*(v0.*sin(beta)-
g.*t))./(1+(x0+v0.*cos(beta).*t).^2./(y0+v0.*sin(beta).*t-
1./2.*g.*t.^2).^2);
plot(t, vr, t, vtheta)
xlabel('t (sec)')
ylabel('velocity (ft/sec)')

2.6 Problem 2/183 (Space Curvilinear Motion)
The base structure of the firetruck ladder rotates
about a vertical axis through O with a constant
angular velocity =θ& Ω. At the same time, the
ladder unit OB elevates at a constant rate =φ& Ψ,
and section AB of the ladder extends from within
section OA at the constant rate =R& Λ. Find
general expressions for the components of
acceleration of point B in spherical coordinates
if, at time t = 0, θ = 0, φ = 0, and AB = 0.
Express your answers in terms of Ω, Ψ, Λ, R0
and t, where R0 = OA and is constant. Plot the
components of acceleration of B as a function of
time for the case Ω =10 deg/s, Ψ = 7 deg/s, Λ =
0.5 m/s, and R0 = 9 m. Let t vary between 0 and
the time at which φ = 90°.
Problem Formulation
The components of acceleration in spherical coordinates are,
φθφ 222
cos&&&& RRRaR −−=
( ) φφθθ
φ
θ sin2
cos 2 &&& RR
dt
d
R
a −=
( ) φφθφφ cossin
1 22 && RR
dt
d
R
a +=
The components may be obtained as functions of time by substituting,
tRR Λ+= 0 , tΩ=θ and tΨ=φ
Differentiation and substitution will be performed in MATLAB. The results are,
( ) ( )( )ttRaR ΨΩ−ΨΛ+= 222
0 cos
( ) )sin(2)cos(2 0 ttRta ΨΛ+ΩΨ−ΨΩΛ=θ

( ) ( ) ( )tttRa ΨΨΩΛ++ΨΛ= cossin2 2
0φ
MATLAB Scripts
%%%%%%%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%%%%%%%%%%%%%%%
% This script calculates the components of the
% acceleration symbolically
% O = Omega; P = Phi; L = Lambda
syms O P L t R0
R = R0+L*t;
theta = O*t;
phi = P*t;
% Even though there are some obvious simplifications
% in this case, we still write the most general
% expressions for the spherical components of the
% acceleration. In this way we can consider other types
% of time dependence without modifying the script.
a_R = diff(R,t,2)-R*diff(phi,t)^2-R*diff(theta,t)^2
*cos(phi)^2
a_theta = cos(phi)/R*diff(R^2*diff(theta,t),t)-
2*R*diff(theta,t)*diff(phi,t)*sin(phi)
a_phi = 1/R*diff(R^2*diff(phi,t),t)+R*diff(theta,t)^2
*sin(phi)*cos(phi)
%%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%%%%%%%%%
Output of script #1
a_R =
-(R0+L*t)*P^2-(R0+L*t)*O^2*cos(P*t)^2
a_theta =
2*cos(P*t)*O*L-(2*R0+2*L*t)*O*P*sin(P*t)
a_phi =
2*P*L+(R0+L*t)*O^2*sin(P*t)*cos(P*t)
%%%%%%%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%%%%%%%%%%%
% This script plots the components of the acceleration
% as functions of time
% O = Omega; P = Phi; L = Lambda
L = 0.5; O = 10*pi/180;
P = 7*pi/180; R0 = 9;
tf = pi/2/P; % time at which phi=pi/2
t=0:0.05:tf;
a_R = -(R0+L*t)*P^2-(R0+L*t).*O^2.*cos(P*t).^2;
a_theta = 2*cos(P*t)*O*L-(2*R0+2*L*t)*O*P.*sin(P*t);
a_phi = 2*P*L+(R0+L*t)*O^2.*sin(P*t).*cos(P*t);
plot(t,a_R,t,a_theta,t,a_phi)

xlabel('time (sec)')
title('acceleration (m/s^2)')
%%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%%%%%%%%%
2.7 Sample Problem 2/16 (Constrained Motion of Connected Particles)
The tractor A is used to hoist the bale B with the
pulley arrangement shown. If A has a forward
velocity vA, determine an expression for the
upward velocity vB of the bale in terms of x. Put
the result in nondimensional form by
introducing the velocity ratio η = vB/vA and
nondimensional position χ = x/h. Plot η versus χ
for 0 ≤ χ ≤ 2.
Problem Formulation
The length L of the cable can be written
( ) ( ) tsconsxhyhtsconslyhL tan2tan2 22
+++−=++−=
Now, 0=L& will be used to obtain a relation between vA (= x& ) and vB (= y& ).
22
20
xh
xx
yL
+
+−==
&
&&
222
1
xh
xv
v A
B
+
=

The nondimensional result is now obtained by substituting AB vv η= and hx χ= .
2
12
1
χ
χ
η
+
=
Even though these operations are rather easily performed by hand, it is
instructive to have MATLAB do them. In particular, it will be instructive to see
how to evaluate L& even though x and y are not known explicitly as functions of
time.
MATLAB Worksheet and Script
EDU» syms t h vA vB chi eta
EDU» x = sym('x(t)'); y = sym('y(t)');
EDU» L = 2*(h-y)+sqrt(h^2+x^2)
L =
2*h-2*y(t)+(h^2+x(t)^2)^(1/2)
Note that we need to differentiate L with respect to time. Both x and y depend on
time, however, exactly how they depend on time is not known. It turns out that
this is not a problem. All we need to do is let MATLAB know x and y depend on
time by writing x = sym('x(t)') and y = sym('y(t)'). To illustrate, consider the
following.
EDU» diff(x, t)
ans =
diff(x(t),t)
This derivative would have been evaluated as zero had we not declared x to be a
function of t. Now let’s proceed by differentiating L with respect to t. We give
the result a name (eqn) to facilitate later substitutions.
EDU» eqn = diff(L,t)
eqn =
-2*diff(y(t),t)+1/(h^2+x(t)^2)^(1/2)*x(t)*diff(x(t),t)
EDU» eqn = subs(eqn,diff(y,t),eta*vA) % substitutes vB = ηvA for y& (diff(y,t))
eqn =
-2*eta*vA+1/(h^2+x(t)^2)^(1/2)*x(t)*diff(x(t),t)
EDU» eqn = subs(eqn,diff(x,t),vA)
eqn =

-2*eta*vA+1/(h^2+x(t)^2)^(1/2)*x(t)*vA
EDU» eqn = subs(eqn,x,chi*h)
eqn =
-2*eta*vA+1/(h^2+chi^2*h^2)^(1/2)*chi*h*vA
Now let’s remember that eqn is just a name for dL/dt which is zero. We now
solve this equation for η (eta). Also recall that solve(eqn, eta) actually solves the
equation eqn = 0 for eta.
EDU» eta = solve(eqn, eta)
eta =
1/2*chi*h/(h^2+chi^2*h^2)^(1/2)
We note finally that the h cancels in the above expression yielding the result
given in the problem formulation section above. Now we can produce the
required plot.
%%%%%%% Script for plotting chi versus eta %%%%%
chi = 0:0.01:2;
eta = 1/2*chi./sqrt(1+chi.^2);
plot(chi,eta)
xlabel('x/h')
title('v_B/v_A')
%%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%

KINETICS OF
PARTICLES
The kinetics of particles is concerned with the motion produced by unbalanced
forces acting on a particle. This chapter considers three approaches to the
solution of particle kinetics problems: (1) direct application of Newton’s second
law, (2) work and energy, and (3) impulse and momentum. Problem 3.1 is a
rectilinear motion problem where solve is used solve three equations
symbolically for three unknowns. In problem 3.2, dsolve is used to solve a
second order differential equation with initial conditions. The absolute path of a
particle is then plotted using polar. Problem 3.3 uses MATLAB to study the
effect of initial spring stretch upon the velocity of a slider. A physical
interpretation of the results is also required. Problem 3.4 is a typical ballistic
pendulum problem requiring both work/energy and conservation of momentum
to relate the velocity of a projectile to the maximum swing angle of a pendulum.
Problem 3.5 is a relatively straightforward conservation of angular momentum
problem where MATLAB is used to generate a plot that might be useful in a
parametric study. In problem 3.6, two equations are solved symbolically for two
unknowns using solve. The maximum value of a function is then determined
using diff and solve.
3

62 CH. 3 KINETICS OF PARTICLES
3.1 Sample Problem 3/3 (Rectilinear Motion)
The 250-lb concrete block A is released from
rest in the position shown and pulls the 400-lb
log up the 30° ramp. Plot the velocity of the
block as it hits the ground at B as a function of
the coefficient of kinetic friction µk between the
log and the ramp. Let µk vary between 0 and 1.
Why does the computer not plot results for the
entire range specified?
Problem Formulation
The constant length of the cable is L = 2sC + sA
(see figure). Differentiating this expression
twice yields a relation between the acceleration
of A and C (note that aC = aLOG).
AC aa += 20 (1)
From the free-body diagram for the log
[ ]0==Σ yy maF 0)30cos(400 =−N
[ ]xx maF =Σ Ck aTN
2.32
400
)30sin(4002 =+−µ
Substituting N yields,
Ck aT
2.32
400
)30sin(4002)30cos(400 =+−µ (2)
From the free-body diagram for block A
[ ]maF =Σ↓ AaT
2.32
250
250 =− (3)
MATLAB will be used to solve the three equations above for aA, aC and T in
terms of µk. Since the accelerations are constant, dav AA 22
= where d is the
vertical distance through which block A has fallen. Thus, the velocity of A when
it strikes the ground (d = 20 ft) is

KINETICS OF PARTICLES 63
BAAf vav == 40
MATLAB Scripts
%%%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%%%%%%%%%%%%%%
% This script solves the three equations developed
% in the problem formulation section for the two
% accelerations (aC and aA) and the tension T.
% When solving multiple equations, it is a good idea
% to let the variables solved for be single characters.
% Thus, we set x = aC, y = aA, z = T.
eqn1 = '2*x+y = 0';
eqn2 = '400*muk*cos(theta)-2*z+400*sin(theta)=400/32.2*x';
eqn3 = '250-z = 250/32.2*y';
[x,y,z]=solve(eqn1,eqn2,eqn3)
%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%%%%%%
Output from Script #1
x = 7.9674*muk - 6.9
y = -15.9349*muk + 13.8
z = 142.857+123.7179*muk
The only thing we are interested in here is aA (y in our solution above). Thus,
kAa µ9349.158.13 −=
Note that the accelerations may be either positive or negative depending on the
value of µk . The largest value of µk for which the block will move up can thus
be found by solving the equation aA = 0 for µk . This yields µk = 13.8/15.935 =
0.866.
%%%%%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%%%%%%%%%%%
% This script plots the vB (the velocity of the
% block as it hits the ground) versus the friction
% coeficient mu_k
muk = 0:0.01:1;
aA=-15.9349*muk+13.8;
vB = sqrt(40*aA)
plot(muk,vB)
xlabel('mu_k')
ylabel('v_B (ft/sec)')
%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%%%%%%%%%

At first sight, it seems that no results are plotted beyond the limiting value for µk
(0.866) that was determined above. If the plot were in color, you would see that
MATLAB has actually plotted zeros beyond this point. From a numerical point
of view this occurs because MATLAB will plot only the real part of complex
numbers. If you have MATLAB print the values of vB you will find that the real
parts of all the complex numbers generated are zero. Whenever imaginary or
complex values result there is usually some physical explanation. In this
problem, the physical explanation is that the log will not slide up the incline if the
coefficient of friction is too large.

3.2 Problem 3/98 (Curvilinear Motion)
The particle P is released at time t = 0 from the
position r = r0 inside the smooth tube with no
velocity relative to the tube, which is driven at
the constant angular velocity ω0 about the
vertical axis. Determine the radial velocity vr,
the radial position r, and the transverse velocity
vθ as functions of time t. Plot the absolute path
of the particle during the time that it is inside the
tube for r0 = 0.1 m, l = 1 m, and ω0 = 1 rad/s.
Problem Formulation
From the free-body diagram to the right,
( )2
0 θ&&& rrmmaF rr −===Σ
2
0
2
ωθ rrr == &&&
Any book on differential equations will have the solution
to this equation in terms of the hyperbolic sine and
cosine,
)cosh()sinh( 00 tBtAr ωω +=
The constants A and B are found from the initial conditions. These conditions are
that r = r0 and 0=r& at t = 0. The second condition comes from the fact that the
particle has no velocity (initially) relative to the tube. Before evaluating this
condition we must first differentiate r with respect to time.
)sinh()cosh( 0000 tBtAr ωωωω +=&
BBArtr =+=== )0cosh()0sinh()0( 0
000 )0sinh()0cosh(0)0( ωωω ABAtr =+===&
From the above we have B = r0 and A = 0. Thus,
)cosh( 00 trr ω=
From this we can obtain the radial and transverse velocities,

)sinh( 000 trrvr ωω== & )cosh( 000 trrv ωωθθ == &
The absolute path of the particle will be graphed using polar plotting. For this we
need r as a function of θ. Since θ = ω0t we have,
)cosh(0 θrr =
We want to plot this function only up to the point where the particle leaves the
tube. Substituting r = 1 we have 1 = 0.1cosh(θ), or θ = cosh-1
(10) = 2.993 rads.
Thus, the particle leaves the tube when θ = 2.993 rads (171.5°).
MATLAB Worksheet
EDU» theta = 0:0.01:2.993;
EDU» r = 0.1*cosh(theta);
EDU» polar(theta,r)

3.3 Sample Problem 3/17 (Potential Energy)
The 10-kg slider A moves with negligible
friction up the inclined guide. The attached
spring has a stiffness of 60 N/m and is stretched
δ m at position A, where the slider is released
from rest. The 250-N force is constant and the
pulley offers negligible resistance to the motion
of the cord. Plot the velocity of the slider as it
passes C as a function of the initial spring stretch
δ. Let δ vary between –0.4 and 0.8 m and
explain the results when δ exceeds a value of
about 0.65 m.
Problem Formulation
The change in the elastic potential energy is
( ) ( )( )222
1
2
2 2.1
2
1
2
1
δδ −+=−=∆ kxxkVe
The other results in the sample problem are unchanged,
150)6.0(25021 ==′−U J ( ) 22
0
2
)10(
2
1
2
1
vvvmT =−=∆
9.58)30sin2.1)(81.9(10 ==∆=∆ hmgVg J
( ) ( )( )( )222
21 2.160
2
1
9.5810
2
1
150 δδ −+++==′− vU
This equation can be solved for v either by hand or by using MATLAB. The
result is
δ1440958
10
1
−=v
First we will solve the work/energy equation symbolically for v.
EDU» eqn='150=5*v^2+58.9+30*((1.2+del)^2-del^2)'

eqn = 150=5*v^2+58.9+30*((1.2+del)^2-del^2)
EDU» solve(eqn,'v')
ans =
.1*(958.-1440.*del)^(1/2)
-.1*(958.-1440.*del)^(1/2)
%%%%%%%%%%%%%%%% Script %%%%%%%%%%%%%%%%%
% This script plots the velocity of the slider
% at C versus the initial spring stretch del
del = -0.4:0.01:0.8;
v = 1/10*sqrt(958-1440*del);
plot(del,v)
axis([-0.4 0.8 0 4])
xlabel('initial spring stretch (m)')
%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%

At first sight, it seems that no results are plotted beyond δ ≅ 0.65 m. If the plot
were in color, you would see that MATLAB has actually plotted zeros beyond
this point. Why? One reason is to observe from the above equation that v
becomes imaginary when δ > 958/1440 = 0.665 m. MATLAB interprets
imaginary numbers as complex numbers with a real part equal to zero. When
asked to plot a complex number it will plot only the real part. Thus, we have
zeros plotted after δ = 0.665 m. But this is a numerical reason instead of a
physical explanation. Usually, imaginary answers signify a situation that is
physically impossible for some reason. One way of understanding this is as
follows. If the spring is initially compressed it will, at least for some part of the
motion, be pushing up and thus be aiding the 250 N force in overcoming the
weight of the slider. If the spring is initially stretched, it will always be pulling
back on the slider. Thus the 250 N force will have to overcome not only the
weight but also the spring force. It stands to reason then that there will be some
value for the initial spring stretch beyond which the 250 N force will not be able
to pull the slider all the way to C. This value is found from the limiting case
where v = 0. Thus, the block never reaches C if δ > 0.665 m.

3.4 Problem 3/218 (Linear Impulse/Momentum)
The ballistic pendulum is a simple device to
measure the projectile velocity v by observing
the maximum angle θ to which the box of sand
with embedded particle swings. As an aid for a
laboratory technician, make a plot of the
velocity v in terms of the maximum angle θ.
Assume that the weight of the box is 50-lb while
the weight of the projectile is 2-oz.
Problem Formulation
(1) Impulse/Momentum
During impact, 0=∆G and 21 GG =
bvv 




 +
=





+





2.32
5016/2
)0(
2.32
50
2.32
16/2
bvv 401=
where v is the velocity of the projectile while vb is the velocity of the box of sand
immediately after impact.
(2) Work/Energy
Now we use the work/energy equation with our initial
position being the position where the pendulum is still
vertical (θ = 0) and the final position is that where the
pendulum has rotated through the maximum angle θ.
( ) hmgvmVTU bg ∆+−=∆+∆==′−
22
21 0
2
1
0
where m is the combined mass of the box and the
projectile.
( )θcos1)6)(2.32(22 −=∆= hgvb
θcos17882401 −== bvv

MATLAB Worksheet
EDU» theta = 0:0.01:pi/2;
EDU» v = 7882*sqrt(1-cos(theta));
EDU» plot(theta*180/pi,v)
EDU» xlabel('theta (degrees)')
EDU» title('velocity of projectile (ft/s)')

3.5 Problem 3/250 (Angular Impulse/Momentum)
The assembly of two 5-kg spheres is rotating
freely about the vertical axis at 40 rev/min with
θ = 90°. The force F that maintains the given
position is increased to raise the base collar and
reduce the angle from 90° to an arbitrary angle
θ. Determine the new angular velocity ω and
plot ω as a function of θ for 0 ≤ θ ≤ 90°. Assume
that the mass of the arms and collars is
negligible.
Problem Formulation
Since the summation of moments about the
vertical axis is zero we have conservation of
angular momentum about that axis. The spheres
are rotating in a circular path about the vertical
axis. The angular momentum of a particle
moving in a circular path of radius r with
angular velocity ω is ω2
mrH = . Thus, from the
conservation of angular momentum we have,
ωω 2
0
2
0 22 mrmr = 02
2
0
ωω
r
r
=
where )45cos(6.01.0 0
0 +=r and ( )2/cos6.01.0 θ+=r
MATLAB Script
%%%%%%%%%%% script %%%%%%%%%%%%%%%
theta = 0:0.05:pi/2;
r0 = 0.1+0.6*cos(pi/4);
r = 0.1+0.6*cos(theta/2);
w0 = 40*2*pi/60;
w = r0^2./r.^2*w0;
plot(theta*180/pi, w)
xlabel('theta (degrees)')
ylabel('omega (rad/s)')
%%%%%%%% end of script %%%%%%%%%%%

This diagram “begins” at θ = 90° where ω = ω0 = 40(2π)/60 = 4.19 rad/s.
3.6 Problem 3/365 (Curvilinear Motion)
The 26-in. drum rotates about a horizontal axis
with a constant angular velocity Ω = 7.5 rad/sec.
The small block A has no motion relative to the
drum surface as it passes the bottom position θ =
0. Determine the coefficient of static friction µs
that would result in block slippage at an angular
position θ; plot your expression for 0 ≤ θ ≤ 180°.
Determine the minimum required coefficient
value µmin that would allow the block to remain
fixed relative to the drum throughout a full
revolution. For a friction coefficient slightly less
than µmin, at what angular position θ would
slippage occur?
Problem Formulation
From the free body and mass acceleration diagrams,
[ ]nn maF =Σ 2
cos Ω=− mrmgN θ
[ ]tt maF =Σ 0sin =− θmgF

For impending slip we have NF sµ= . Substituting F into the above and solving
gives,
θ
θ
θ
θ
µ
cos8925.1
sin
cos
sin
2
+
=
Ω+
=
rg
g
s
The last two questions can be answered only after plotting µs as a function of θ.
%%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%%%%%%%%%
% This script solves our two equations symbolically
% for mu_s and N
% O = Omega
% x = mu_s
% y = N
syms theta O g m r x y
eqn1 = y-m*g*cos(theta)-m*r*O^2;
eqn2 = x*y-m*g*sin(theta);
% remember that we write our equations in the form
% expression = 0 and then omit the "=0"
[x,y] = solve(eqn1,eqn2)
%%%%%%%%%%%%%%end of script %%%%%%%%%%%%%%%%%%%%
Output of script #1
x =
g*sin(theta)/(g*cos(theta)+r*O^2)
y =
m*g*cos(theta)+m*r*O^2
%%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%%%%
% This script plots mu_s as a function of theta
theta = 0:0.02:pi;
mu_s = sin(theta)./(1.8925+cos(theta));
plot(theta*180/pi, mu_s)
title('coefficient of static friction')
%%%%%%%%%%%%%%end of script %%%%%%%%%%%%%%%%%%%%

If the block is not to slip at any angle θ, the coefficient of friction must be greater
than or equal to any value shown on the plot above. Thus, the minimum required
coefficient value µmin that would allow the block to remain fixed relative to the
drum throughout a full revolution is equal to the maximum value in the plot
above. The location where this maximum occurs can be found by solving the
equation 0/ =θµ dd s for θ. This θ can then be substituted into µs to yield the
required value for µmin. Here’s how we do this with MATLAB.
EDU» syms theta
EDU» mu_s = sin(theta)./(1.8925+cos(theta));
EDU» dmu = diff(mu_s,theta)
dmu = cos(theta)/(757/400+cos(theta))+sin(theta)^2/(757/400+cos(theta))^2
EDU» theta_m = solve(dmu,theta)
theta_m =
-atan(1/302800*236697316401^(1/2))+pi
atan(1/302800*236697316401^(1/2))-pi
EDU» eval(theta_m)
ans =
2.1275
-2.1275
EDU» mu_min = subs(mu_s,theta,2.1275)
mu_min = 0.6224
From the above we see that µmin = 0.622. If µs is slightly less than this value, the
block will slip when θ = 2.128 rads (121.9°).

KINETICS OF SYSTEMS
OF PARTICLES
This chapter concerns the extension of principles covered in chapters two and
three to the study of the motion of general systems of particles. The chapter first
considers the three approaches introduced in chapter 3 (direct application of
Newton’s second law, work/energy, and impulse/momentum) and then moves to
other applications such as steady mass flow and variable mass. Problem 4.1
considers an application of the conservation of momentum to a system comprised
of a small car and an attached rotating sphere. MATLAB is used to plot the
velocity of the car as a function of the angular position of the sphere. The
absolute position of the sphere is also plotted. Problem 4.2 uses the concept of
steady mass flow to study the effects of geometry upon the design of a sprinkler
system. One of the main purposes of this problem is to illustrate how a problem
can be greatly simplified using non-dimensional analysis. In particular, an
equation containing seven parameters is reduced to a non-dimensional equation
with only three parameters. Problem 4.3 is a variable mass problem in which
MATLAB is used to integrate the kinematic equation adxvdv = .
4

78 CH. 4 KINETICS OF SYSTEMS OF PARTICLES
4.1 Problem 4/26 (Conservation of Momentum)
The small car, which has a mass of 20 kg, rolls
freely on the horizontal track and carries the 5-
kg sphere mounted on the light rotating rod with
r = 0.4 m. A geared motor drive maintains a
constant angular speed θ& = 4 rad/s of the rod. If
the car has a velocity v = 0.6 m/s when θ = 0,
plot v as a function of θ for one revolution of the
rod. Also plot the absolute position of the sphere
for two revolutions of the rod. Neglect the mass
of the wheels and any friction.
Problem Formulation
Since 0=Σ xF we have conservation of momentum in the x
direction. The diagram to the right shows the system at θ = 0
and at an arbitrary angle θ. From the relative velocity
equation, the velocity of the sphere is the vector sum of the
velocity of the car (v) and the velocity of the sphere relative
to the car ( θ&r ).
( ) 15)6.0(5)6.0(200
=+==θxG N•s
( ) ( ) θθθθ sin825sin520 −=−+= vrvvGx
&
Setting ( ) ( )θθ xx GG ==0
and solving yields,
θsin32.06.0 +=v
Now let time t = 0 be the time when θ = 0 and place an x-y coordinate system at
the center of the car as shown in the diagram so that x(t) is the position of the
center of the car. Since v = dx/dt and θ = 4t we have,
( )( ) ( )( )∫ ∫ −+=+==
t t
ttdttvdtx
0 0
4cos108.06.04sin32.06.0
The x and y components of the sphere can now be determined as,
( )ttrxxs 4cos32.06.008.0cos ++=+= θ

KINETICS OF SYSTEMS OF PARTICLES 79
( )trys 4sin4.0sin == θ
The absolute position of the sphere can be obtained by plotting ys versus xs. The
time required for two revolutions of the arm is 4π/4 = π seconds.
MATLAB Scripts
%%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%%%%%%
% This script plots v as a function of theta
theta = 0:0.02:2*pi;
v = 0.6 + 0.32*sin(theta);
plot(theta*180/pi, v)
ylabel('v (m/s)')
%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%
%%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%
% This script plots the position of the sphere
t = 0:0.01:pi;
xs = 0.08+0.6*t+0.32*cos(4*t);
ys = 0.4*sin(4*t);
plot(xs, ys)
xlabel('x (m)')
ylabel('y (m)')
title('position of the sphere')
%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%

4.2 Problem 4/62 (Steady Mass Flow)
The sprinkler is made to rotate at the constant
angular velocity ω and distributes water at the
volume rate Q. Each of the four nozzles has an
exit area A. Write an expression for the torque M
on the shaft of the sprinkler necessary to
maintain the given motion. Here we would like
to study the effects of the geometry of the
sprinkler upon this torque. To this end, it is
helpful to introduce the non-dimensional
parameters M′ = M/4ρAru2
, Ω = ωr/u, and β =
b/r where u = Q/4A is the velocity of the water
relative to the nozzle and ρ is the density of the
water. Plot the non-dimensional torque M′
versus β for Ω = 0.5, 1, and 2. Let Ω0 be the
non-dimensional velocity Ω at which the
sprinkler will operate with no applied torque.
Plot Ω0 versus β. For both plots let β range
between 0 and 1.
Problem Formulation
The figure to the right shows the three components of the
absolute velocity of the water at the exit. u (= Q/4A) is the
velocity of the water relative to the nozzle. The mass flow
rate m′ = ρQ. Taking clockwise as positive, the application
of equation 4/19 of your text yields,
( )022
0 −−+=−=Σ urbrQMM ωωρ

( )( )22
brurQM +−= ωρ
Now we want to introduce the non-dimensional parameters defined in the
problem statement. For many undergraduate students, non-dimensional analysis
is a very confusing topic. It is important to realize that the difficulty is really that
of determining which non-dimensional parameters are appropriate for a particular
problem. If these parameters have already been defined, as in this problem, all
you have to do is substitute. In this case we merely substitute M = 4ρAru2
M′, ω
= Ωu/r, and b = rβ into the equation above. When this is done many terms will
cancel yielding,
( )2
11 β+Ω−=′M
Setting M′ = 0 we can solve for Ω0,
20
1
1
β+
=Ω
MATLAB Scripts
%%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%%%%%%
% This script plots the non-dimensional torque
% M' as a function of beta = b/r for Omega =
% 0.5, 1, and 2
beta = 0:0.01:1;
Mp = inline('1-Omega*(1+beta.^2)')
plot(beta, Mp(0.5,beta),beta, Mp(1,beta),beta, Mp(2,beta))
xlabel('beta = b/r')
title('non-dimensional torque')
%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%
% This script plots the non-dimensional angular
% velocity Omega0 as a function of beta = b/r
beta = 0:0.01:1;
Omega0 = 1./(1+beta.^2);
plot(beta, Omega0)
xlabel('beta = b/r')
%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%

4.3 Problem 4/86 (Variable Mass)
The open-link chain of length L and mass ρ per
unit length is released from rest in the position
shown, where the bottom link is almost touching
the platform and the horizontal section is
supported on a smooth surface. Friction at the
corner guide is negligible. Determine (a) the
velocity v1 of end A as it reaches the corner and
(b) its velocity v2 as it strikes the platform. Plot
v1 and v2 as functions of h for L = 5 m.
Problem Formulation
Let x be the displacement of the chain and T be
the tension in the chain at the corner as shown in
the diagram to the right. Note that the
acceleration of the horizontal and vertical
sections are both equal to x&& .
For the horizontal section,
[ ]xx maF =Σ ( )xxhLT &&−−= ρ
For the vertical section,
[ ]yy maF =Σ↓ xhTgh &&ρρ =−
Substituting T from the first equation into the second and simplifying gives,
xL
gh
x
−
=&&
Now we use the relation adxvdv = to write,
dx
xL
gh
vdv
hLv
∫∫
−
−
=
00
1






=
−
= ∫
−
h
L
ghdx
xL
gh
v
hL
ln22
0
2
1
( )hLghv /ln21 =

After end A has passed the corner it will be in free-fall until it hits the platform.
With y positive down we have vdv = gdy yielding,
( ) ghvv =− 2
1
2
2
2
1
Substituting for v1 and solving,
( ))/ln(122 hLghv +=
EDU» syms v v1 v2 h g x L
First we solve the equation ( ) 0/
00
1
=−− ∫∫
−hLv
dxxLghvdv for v1, omitting (as usual)
the “=0”.
EDU» eqn1 = int(v,v,0,v1)-int(g*h/(L-x),x,0,L-h)
eqn1 =
1/2*v1^2+log(h)*g*h-log(L)*g*h
EDU» solve(eqn1,v1)
ans =
(-2*log(h)*g*h+2*log(L)*g*h)^(1/2)
-(-2*log(h)*g*h+2*log(L)*g*h)^(1/2)
MATLAB has found two solutions. The first is the one we want since it is
positive. This solution can easily be simplified to the result given above in the
problem formulation section. Once v1 is known it is rather easy to find v2. We’ll
do it symbolically here for purposes of illustration.
First we copy and paste the first solution above to define v1. Then we solve the
equation 0
0
2
1
=− ∫∫
hv
v
gdyvdv for v2. Note how the result for v1 is automatically
substituted.
EDU» v1 = (-2*log(h)*g*h+2*log(L)*g*h)^(1/2);
EDU» eqn2 = int(v,v,v1,v2)-int(g,x,0,h)
eqn2 =
1/2*v2^2+log(h)*g*h-log(L)*g*h-g*h

EDU» solve(eqn2,v2)
ans =
(-2*log(h)*g*h+2*log(L)*g*h+2*g*h)^(1/2)
-(-2*log(h)*g*h+2*log(L)*g*h+2*g*h)^(1/2)
Once again, the first solution will simplify to that given in the problem
formulation section above.
%%%%%%%%%%%%%%%%%%% Script %%%%%%%%%%%%%%%%%%%%
% This script plots v1 and v2 as functions of
% h for L = 5 m
L = 5; g = 9.81;
h = 0:0.01:L;
v1 = sqrt(2*g*h.*log(L./h));
v2 = sqrt(2*g*h.*(1+log(L./h)));
plot(h, v1, h, v2)
xlabel('h (m)')
%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%

PLANE KINEMATICS OF
RIGID BODIES
This chapter extends the kinematic analysis of particles covered in chapter 2 to
rigid bodies by taking into account the rotational motion of the body. Thus, the
motion of rigid bodies involves both translation and rotation. Problem 5.1 is a
straightforward problem that compares the angular position of a rotating disk to
the total number of revolutions. Problem 5.2 is an interesting application of
absolute motion analysis. The problem illustrates the usefulness of non-
dimensional analysis in conducting a parametric study of the effects of geometry
upon the performance of a hydraulic lift. In problem 5.3 the velocity of the piston
in a reciprocating engine is plotted versus the angular orientation of the crank.
The maximum velocity and the corresponding orientation (of the crank) are
obtained using max after first trying diff and solve. The instantaneous center of
zero velocity is used in problem 5.4 to relate the velocity of a vertical control rod
to the angular velocity of a rotating bar in a switching device. Problem 5.4 is a
fairly straightforward relative acceleration problem. Problem 5.5 considers the
reciprocating engine of problem 5.3 but uses absolute rather than relative motion
analysis. The problem illustrates the use of computer algebra for carrying out
some tedious calculations including differentiation and substitution.
5

88 CH. 5 PLANE KINEMATICS OF RIGID BODIES
5.1 Problem 5/3 (Rotation)
The angular velocity of a gear is controlled
according ω = 12 – 3t2
where ω, in radians per
second, is positive in the clockwise sense and
where t is the time in seconds. Find the net
angular displacement ∆θ and the total number of
revolutions N through which the gear turns in
terms of the time t. Plot ∆θ and N for time t up
to 4 seconds.
Problem Formulation
You may want to have a look at sample problem
5/1 in your text before continuing with this
problem. In particular it is important to note the
difference between the angular displacement ∆θ
and total number of revolutions N. The angular
displacement is the integral of ω over time and
can be positive or negative depending on whether
the rotation is clockwise or counterclockwise.
Referring to the graph of ω to the right, ∆θ will
be the total area under the curve up to a particular
time. The area is negative when the curve dips
below the time axis. As the name implies, the
total number of revolutions simply counts the number of times the disk rotates
and is not concerned with whether the rotation is clockwise or counterclockwise.
Thus, N will be proportional to the magnitude of the area under the angular
velocity curve. For example, suppose the gear rotates two revolutions clockwise
followed by two revolutions counterclockwise. In this case ∆θ = 0 but N = 4.
Based upon the above it is essential to know beforehand the time when ω = 0 (i.e.
when the gear changes direction) in order to calculate N. Setting 12 – 3t2
= 0
gives t = 2 seconds. For time greater than 2 sec. we need to break the integral up
into two parts when calculating N. For ∆θ we do not need to keep track of when
the gear changes directions. Thus,




=
dt
dθ
ω ( ) 3
0 0
2
12312 ttduudt
t t
−=−==∆ ∫ ∫ωθ

PLANE KINEMATICS OF RIGID BODIES 89
for t < 2 sec ( ) ( )3
0 0
2
12
2
1
312
2
1
ttdttdtN
t t
−=−== ∫ ∫ π
ω
π
for t > 2 sec ( ) ( ) 







−−−=−= ∫∫∫ ∫ dttdttdtdtN
tt
2
2
2
0
2
2
0 2
312312
2
1
2
1
2
1
π
ω
π
ω
π
( )3
1232
2
1
ttN +−=
π
In summary,
3
12 tt −=∆θ all t
( )
( ) 









>+−
<−
=
sec21232
2
1
sec212
2
1
3
3
ttt
ttt
N
π
π
Although the expression for N is rather simple, plotting the result can be a little
tricky due to the change in the expression at time t = 2 sec. How do we tell
MATLAB to stop plotting one expression and start plotting the other?
Actually, this is considerably easier to do in MATLAB than in some other
applications (e.g. Maple) due to the fact that you are not limited to a single range
variable for the ordinate. Thus we can easily set up two time variables t1 and t2
and then write the two expressions in terms of those two variables.
MATLAB Scripts
%==== script #1 ==========================
% This script plots the angular velocity as
% a function of time. Note the way a horizontal
% line at zero is plotted. This is useful for
% visualizing where omega becomes negative
t = 0:0.05:4;
omega = 12 - 3*t.^2;
zero = 0*t;
plot(t,omega,t,zero)
title('angular velocity (rad/s)')
%======= end of script ===================

%==== script #2 ==========================
% This script plots the angular position
% (in degrees) as a function of time.
t = 0:0.05:4;
theta = (12*t - t.^3)*180/pi;
zero = 0*t;
plot(t,theta,t,zero)
title('angular position (degrees)')
%======= end of script ===================

%==== script #3 ==========================
% This script plots the number of revolutions
% as a function of time. Note that different
% time variables are used for the two different
% expressions for N
t1 = 0:0.05:2;
N1 = (12*t1 - t1.^3)/2/pi;
t2 = 2:0.05:4;
N2 = (32 - 12*t2 + t2.^3)/2/pi;
plot(t1,N1,t2,N2)
title('total number of revolutions')
%======= end of script ===================

5.2 Problem 5/44 (Absolute Motion)
Derive an expression for the upward velocity v
of the car hoist system in terms of θ. The piston
rod of the hydraulic cylinder is extending at the
rate s& . Plot the non-dimensional velocity v / s& as
a function of θ for b/L = 0.1, 0.5, 1, and 2.
Problem Formulation
From the diagram to the right,
θsin2by =
θθ cos2 && by =
If the angular velocity θ& were known as a
function of θ we would be finished. The motion
of the car hoist system is controlled by the
extension rate s& of the hydraulic cylinder rather
than the angular velocity. Thus, to complete the
problem we need to relate θ& and s& .
θcos2222
LbbLs −+=
θθ sin2002 && Lbss ++=
θ
θ
sinLb
ss&& =
Substituting,
θ
θ
θ
θ tan
cos22
cos
sin
2 22
L
LbbLs
Lb
sbs
v
−+
==
&&
( ) ( )
θ
θββ
θ
θ
tan
cos212
tan
cos/2/12
/
22
−+
=
−+
=
LbLb
sv &
where β = b/L.

MATLAB Script
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% v is the non-dimensional velocity
v = inline('2/tan(theta)*sqrt(1^2+beta^2-2*beta*cos(theta))')
v = vectorize(v);
% the above adds "." to insure term by term rather than
% matrix operations
th = 10*pi/180:0.01:pi/2;
deg = th*180/pi;
plot(deg,v(.1,th),deg,v(.5,th),deg,v(1,th),deg,v(2,th))
axis([10 90 0 5])
title('non-dimensional velocity')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

5.3 Sample Problem 5/9 (Relative Velocity)
The common configuration of a reciprocating
engine is that of the slider crank mechanism
shown. If crank OB has a clockwise rotational
speed of 1500 rev/min; (a) Plot vA versus θ for
one revolution of the crank. (b) Find the
maximum speed of the piston A and the
corresponding value of θ.
Problem Formulation
Let l be the length of connecting rod AB. Start with the relative velocity equation
vB = vA + vB/A
The crank pin velocity is vB = rω and is normal to OB. The velocity of A is
horizontal while the velocity of B/A has magnitude lωAB and is directed
perpendicular to AB. The angle β can be found in terms of θ by using the law of
sines,
θβ sinsin
l
r
= Also, θββ 2
2
2
2
sin1sin1cos
l
r
−=−=
From the vector diagram to the right,
βθ coscos / ABB vv = . Thus, ω
β
θ
rv AB
cos
cos
/ =
Also from the diagram, βθ sinsin / ABBA vvv += .
Substitution yields
( )














−
+=+=
θ
θ
θωβθθω
2
2
2
sin1
cos
1sintancossin
l
r
l
r
rrvA
Note that vA has been expressed explicitly in terms of θ by substituting for cosβ
and tanβ. This has been done only for sake of clarity. When working with a

computer such substitutions will be automatic once β has been defined in terms
of θ ( ( )lr /sinsin 1
θβ −
= ).
In part (b) we first try to find the angle θ at which the maximum value of vA
occurs by solving the equation dvA/dθ = 0 for θ. Although a solution is obtained it
is extremely messy and occupies many screen widths. For this reason we switch
to an approximate approach which utilizes Matlab’s max function. This method
gives the maximum speed of the piston as vA = 69.6 ft/sec at θ = 72.2°.
%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%
% This script solves for vA using symbolic
% algebra
syms r L theta omega
vB = r*omega;
beta = asin(r/L*sin(theta));
vBA = cos(theta)/cos(beta)*r*omega;
vA = vB*sin(theta)+vBA*sin(beta)
%%%%%%%%%%%%% end of script %%%%%%%%%%%
Output of script #1
vA =
r*omega*sin(theta)+cos(theta)/(1-r^2/L^2*sin(theta)^2)^(1/2)*
r^2*omega/L*sin(theta)
%%%%%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%%%
% This script plots the velocity of piston A
% as a function of theta
r = 5/12;
L = 14/12;
o = 1500*2*pi/60; % o=omega
x = 0:0.01:2*pi; % x=theta
vA = r*o*sin(x)+cos(x)./(1-r^2/L^2*sin(x).^2).^(1/2)*r^2*o/L.*sin(x);
plot(x*180/pi,vA)
title('velocity of piston A (ft/sec)')
%%%%%%%%%%%%% end of script %%%%%%%%%%%

Part (b)
%%%%%%%%%%%%%%%%%%% Script #3 %%%%%%%%%%%%%%
% This script solves for the values of theta which
% make vA a maximum
r = 5/12;
L = 14/12;
omega = 1500*2*pi/60;
syms theta
% The following is copied and pasted from the output of
% script #1
vA = r*omega*sin(theta)+cos(theta)/(1-
r^2/L^2*sin(theta)^2)^(1/2)*r^2*omega/L*sin(theta)
dvAdtheta = diff(vA,theta)
solve(dvAdtheta,theta)
%%%%%%%%%%%%%%%%%% end of script %%%%%%%%%%%
Output of script #3
vA =
125/6*pi*sin(theta)+625/84*cos(theta)/(1-25/196*sin(theta)^2)^(1/2)*
pi*sin(theta)
dvAdtheta =
125/6*pi*cos(theta)-625/84*sin(theta)^2/(1-25/196*sin(theta)^2)^(1/2)*pi+
15625/16464*cos(theta)^2/(1-25/196*sin(theta)^2)^(3/2)*pi*sin(theta)^2+
625/84*cos(theta)^2/(1-25/196*sin(theta)^2)^(1/2)*pi

ans = [output omitted]
The output was omitted because it was very messy and extended over several
screen widths making it practically unusable. For some reason, these problems do
not occur in Matlab 5. Due to these difficulties we try a different approach, using
Matlab’s max function.
After running script #1 type the following into the command window,
EDU>> [vA_max, i] = max(vA)
vA_max =
69.5514
i =
127
EDU>> x(i)
ans =
1.2600
EDU>> ans*180/pi
ans =
72.1927
Thus, the maximum speed of the piston is vA = 69.6 ft/sec at θ = 72.2°.

5.4 Problem 5/123 (Relative Acceleration)
The two rotor blades of radius r = 800-mm
rotate counterclockwise with a constant angular
velocity ω about the shaft at O mounted in the
sliding block. The acceleration of the block is
aO. Determine the magnitude of the acceleration
of the tip A of the blade in terms of r, ω, aO, and
θ. Plot the acceleration of A as a function of θ
for one revolution if aO = 3 m/s. Consider three
cases: ω = 2, 4, and 6 rad/s.
Problem Formulation
The acceleration of A relative to O is
( ) ( )tOAnOAOA aaaa //
rrrr
++=
The acceleration of O is to the right while the normal
relative acceleration must point from A towards O. Since
ω is constant, the tangential relative acceleration will be
zero. These considerations lead to the vector diagram
shown to the right. Using the law of cosines,
( ) ( ) θcos2 /
2
/
2
nOAOnOAOA aaaaa −+=
( ) ( ) θωω cos2 2222
raraa OOA −+=
MATLAB Script
%======== script ==============================
aA = inline('sqrt(3^2+0.8^2*omega^4-2*3*0.8*omega^2*cos(theta))')
th = 0:.05:2*pi;
plot(th,aA(2,th),th,aA(4,th),th,aA(6,th))
xlabel('theta (radians)')
ylabel('acceleration (m/s^2)')
%========= end of script =======================

5.5 Sample Problem 5/15 (Absolute Motion)
The common configuration of a reciprocating
engine is that of the slider crank mechanism
shown. If crank OB has a clockwise rotational
speed of 1500 rev/min; (a) Plot vA and vG versus
time for two revolutions of the crank. (b) Plot aA
and aG versus time for two revolutions of the
crank.
Problem Formulation
This problem appears in sample problems 5/9 and 5/15 in your text. Sample
problem 5/9 considers a relative velocity analysis while sample problem 5/15
uses a relative acceleration analysis. Generally speaking, the easiest approach to
use with a computer is an absolute motion analysis, provided you have software
capable of doing symbolic algebra and calculus such as MATLAB. We will use
the present problem to illustrate this approach.

We start by using the law of sines (sin(β)/r = sin(θ)/l) to express β as a function
of θ






= −
θβ sinsin 1
l
r
where l is the length of connecting rod AB and β is the angle between AB and the
horizontal. Now place an x-y coordinate system at O with x positive to the right
and y positive up and write expressions for the coordinates of A and G in terms of
θ and β
βθ coscos lrxA −−=
βθ coscos rrxG −−= ( ) βsinrlyG −=
where r is the distance from B to G (4 in. in the figure). All that is needed to find
the velocities vA, vGx, and vGy is to differentiate these expressions with respect to
time. The magnitude of the velocity of G is then found from
22
GyGxG vvv +=
The accelerations aA, aGx, and aGy are then found by differentiating vA, vGx, and
vGy with the magnitude of the acceleration of G being obtained from,
22
GyGxG aaa +=
Since we will be differentiating with respect to time, the first thing we will do in
the computer program is to define θ as a function of time. Then, when we write
the above expressions for β, xA, xG, and yG, the computer will automatically
substitute for θ rendering each of these as functions of time. Assuming that θ is
initially zero,
tttt 1.157
60
)2(1500
)( ===
π
ωθ
The problem statement asks us to plot versus time for two revolutions (θ = 4π
radians) of the crank. The time required for two revolutions is 4π/157.1 = 0.08
sec.

%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%%%%%%%
% This script uses symbolic algebra to find
% the velocity and acceleration of A and G
% w = constant angular velocity
% rb = r_bar (4 in. in the figure)
syms r rb L w t
theta = w*t;
beta = asin(r/L*sin(theta));
xA = -r*cos(theta)-L*cos(beta)
xG = -r*cos(theta)-rb*cos(beta)
yG = (L-rb)*sin(beta)
vA = diff(xA,t)
vGx = diff(xG,t)
vGy = diff(yG,t)
vG = sqrt(vGx^2+vGy^2)
aA = diff(vA,t)
aGx = diff(vGx,t)
aGy = diff(vGy,t)
aG = sqrt(aGx^2+aGy^2)
%%%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%
Output of Script #1
xA = -r*cos(w*t)-L*(1-r^2/L^2*sin(w*t)^2)^(1/2)
xG = -r*cos(w*t)-rb*(1-r^2/L^2*sin(w*t)^2)^(1/2)
yG = (L-rb)*r/L*sin(w*t)
vA =
r*sin(w*t)*w+1/L/(1-r^2/L^2*sin(w*t)^2)^(1/2)*r^2*sin(w*t)*cos(w*t)*w
vG =
((r*sin(w*t)*w+rb/(1-r^2/L^2*sin(w*t)^2)^(1/2)*r^2/L^2*
sin(w*t)*cos(w*t)*w)^2+(L-rb)^2*r^2/L^2*cos(w*t)^2*w^2)^(1/2)
aA =
r*cos(w*t)*w^2+1/L^3/(1-r^2/L^2*sin(w*t)^2)^(3/2)*r^4*sin(w*t)^2*
cos(w*t)^2*w^2+1/L/(1-r^2/L^2*sin(w*t)^2)^(1/2)*r^2*cos(w*t)^2*w^2-
1/L/(1-r^2/L^2*sin(w*t)^2)^(1/2)*r^2*sin(w*t)^2*w^2
aG =
((r*cos(w*t)*w^2+rb/(1-r^2/L^2*sin(w*t)^2)^(3/2)*r^4/L^4*sin(w*t)^2*
cos(w*t)^2*w^2+rb/(1-r^2/L^2*sin(w*t)^2)^(1/2)*r^2/L^2*cos(w*t)^2*w^2-

rb/(1-r^2/L^2*sin(w*t)^2)^(1/2)*r^2/L^2*sin(w*t)^2*w^2)^2+(L-rb)^2*
r^2/L^2*sin(w*t)^2*w^4)^(1/2)
The results for the velocities and accelerations will be used to produce the
required plots in the following two scripts. The easiest thing to do is to copy the
results from the worksheet into the scripts. Before doing this you should use the
vectorize command to place periods at appropriate places in order to insure term
by term rather than matrix operations. After running the script type
“vectorize(vA)” in the worksheet and then copy and paste the result into the
script. Repeat this process for vG, aA, and aG.
%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%%
% This script plots vA and vG versus time
r = 5/12;
rb = 4/12; % r_bar
L = 14/12;
w = 1500*2*pi/60; % w=omega
t = 0:0.001:0.08;
vA = r.*sin(w.*t).*w+1./L./(1-r.^2./L.^2.*sin(w.*t).^2).^
(1./2).*r.^2.*sin(w.*t).*cos(w.*t).*w;
vG = ((r.*sin(w.*t).*w+rb./(1-r.^2./L.^2.*sin(w.*t).^2).^
(1./2).*r.^2./L.^2.*sin(w.*t).*cos(w.*t).*w).^2+(L-
rb).^2.*r.^2./L.^2.*cos(w.*t).^2.*w.^2).^(1./2);
plot(t, vA, t, vG)
%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%% Script #3 %%%%%%%%%%%%%%%%%
% This script plots aA and aG versus time
r = 5/12;
rb = 4/12;
L = 14/12;
w = 1500*2*pi/60; % w=omega
t = 0:0.001:0.08;
aA = r.*cos(w.*t).*w.^2+1./L.^3./(1-r.^2./L.^2.*
sin(w.*t).^2).^(3./2).*r.^4.*sin(w.*t).^2.*cos(w.*t).^2.*w.
^2+1./L./(1-r.^2./L.^2.*sin(w.*t).^2).^(1./2).*r.^2.*
cos(w.*t).^2.*w.^2-1./L./(1-r.^2./L.^2.*sin(w.*t).^2).^
(1./2).*r.^2.*sin(w.*t).^2.*w.^2;
aG = ((r.*cos(w.*t).*w.^2+rb./(1-r.^2./L.^2.*
sin(w.*t).^2).^(3./2).*r.^4./L.^4.*sin(w.*t).^2.*cos(w.*t).
^2.*w.^2+rb./(1-r.^2./L.^2.*sin(w.*t).^2).^(1./2).
*r.^2./L.^2.*cos(w.*t).^2.*w.^2-rb./(1-r.^2./L.^2.*
sin(w.*t).^2).^(1./2).*r.^2./L.^2.*sin(w.*t).^2.*w.^2).^2+(
L-rb).^2.*r.^2./L.^2.*sin(w.*t).^2.*w.^4).^(1./2);
plot(t, aA, t, aG)
%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%

For further study
Suppose you wanted to solve this problem for the case where crank OB has a
constant angular acceleration α = 60 rad/s2
. It turns out that you can solve this
problem using exactly the same approach as above, changing only one line
defining the dependence of θ upon time. Assuming that the system starts from
rest at θ = 0, the appropriate expression for θ is θ (t) = ½αt2
= 30t2
. The
accelerations for this case are shown below in case you want to give it a try.

PLANE KINETICS OF
RIGID BODIES
This chapter concerns the motion (translation and rotation) of rigid bodies that
results from the action of unbalanced external forces and moments. In problem
6.1, five equations are solved for five unknowns. The problem illustrates an
alternative to the blunt (but straightforward) simultaneous solution of multiple
equations. Instead, the equations are solved in such a way that there is never
more than one unknown. In this way, the results are immediately obtained via
automatic substitution, thus avoiding some tedious algebra. The force at the
hinge of a pendulum is plotted versus the angular position of the pendulum in
problem 6.2. The algebra is rather simple in this case and MATLAB is used
primarily for purposes of plotting. Problems 6.3 and 6.4 consider rigid bodies in
general plane motion. solve is used in problem 6.3 to solve two equations
simultaneously for two unknowns. The maximum acceleration of a point on the
rigid body is then obtained by using diff and solve. In this problem, solve finds
five solutions to the equation and it is necessary to determine which of these is
physically correct. This problem is also interesting since a very natural “guess”
of the value for the maximum acceleration turns out to be incorrect. Problem 6.4
is an example of a kinetics problem that also requires some kinematics. The
angular acceleration of a bar is determined by summing moments. The kinematic
equation θαωω dd = is then integrated to obtain the angular velocity. Problem
6.5 is an interesting work and energy problem that is complicated considerably
by the fact that a spring is engaged for only part of the motion of a rotating bar.
Symbolic algebra simplifies this problem considerably, though it is still rather
tedious. It is common in Dynamics to find problems that require a combination of
methods for their solution. Problem 6.6 is a good example involving both
conservation of momentum and work/energy.
6

108 CH. 6 PLANE KINETICS OF RIGID BODIES
6.1 Sample Problem 6/2 (Translation)
The vertical bar AB has a mass of m =150 kg
with center of mass G midway between the ends.
The bar is elevated from rest at θ = 0 by means
of the parallel links of negligible mass, with a
couple M applied to the lower link at C. Plot the
force at A and at B as functions of θ (between 0
and 60°) for two cases, (a) a constant couple M
= 5 kN-m, and (b) a constant angular
acceleration α = 5 rad/sec2
.
Problem Formulation
The free-body diagram (FBD) and mass
acceleration diagram (MAD) are shown to the
right. Since the vertical bar undergoes curvilinear
translation, the acceleration of all points on the
bar will be identical. Thus, we can obtain the
acceleration of G immediately from that of point
A which moves in a circular path about C. Also
note that BD is a two force member since the
mass is negligible.
Here we take a somewhat different approach
than that in your textbook. The main reason for
this is that the case of constant moment and
constant angular acceleration require a slightly
different approach. The difference between the
two approaches is easiest seen by writing all the
equations first.
From the free-body diagram of the connecting link AC,
[ ]0=Σ CM 0=− tArM (1)
As pointed out in the sample problem, the force and moment equations are
identical to the equilibrium equations whenever the mass is negligible.
From the free-body diagram of the vertical bar,

PLANE KINETICS OF RIGID BODIES 109
[ ]damM A =Σ AGAGAB rrmrrmBr θαθωθ sincoscos 2
+= (2)
[ ]tt amF =Σ αθ rmmgAt =− cos (3)
[ ]nn amF =Σ 2
sin ωθ rmmgAB n =+− (4)
And, finally, the kinematics equation
∫ ∫=
ω θ
θαωω
0 0
dd (5)
At this point we have a total of 5 equations and 6 unknowns (M, α, ω, At, An, and
B). In part (a), M is specified while in part (b) α is given. In both cases we will
have 5 equations and 5 unknowns; however, it will be necessary to solve no more
than one equation at a time provided they are done in the right order. This is what
yields a different procedure for parts (a) and (b).
Part (a)
With M known we can find At from Equation (1) and then substitute the result
into Equation (3) to get α as a function of θ.
r
M
At =
θα cos
r
g
rm
At
−=
As usual, notice that there is no need to make an explicit substitution of At into α.
The computer will make such substitutions automatically. Now we substitute α
into Equation (5) and integrate,
θθω
θ
d
r
g
rm
At
∫ 





−=
0
2
cos
2
1






−= θθω sin22
r
g
rm
At
Finally, we can substitute into Equations (2) and (4) to find B and then An.

( )θαθω
θ
sincos
cos
2
+=
AB
AG
r
rrm
B
2
sin ωθ rmmgBAn −+=
22
tn AAA +=
Part (b)
With α known instead of M we have to take a different approach, starting with
Equations (5) and (3),
∫ ∫ ===
θ θ
αθθαθαω
0 0
2
222 dd
αθ rmmgAt += cos
Now we can substitute into Equations (1), (2) and (4) to find M, B and then An.
tArM =
( )θαθω
θ
sincos
cos
2
+=
AB
AG
r
rrm
B
2
sin ωθ rmmgBAn −+=
22
tn AAA +=

MATLAB Scripts
%========= script for part a ===========
m = 0.15; g = 9.81; r = 1.5;
rAB = 1.8; rAG = 1.2; rAC = r;
M = 5;
theta = 0:0.1:60;
thr = theta*pi/180;
At = M/r;
alpha = At/m/r - g/r*cos(thr);
omsq = 2*(At*thr/m/r - g/r*sin(thr));
B = m*r*rAG*(omsq.*cos(thr)+alpha.*sin(thr))/rAB./cos(thr);
An = B + m*g*sin(thr)-m*r*omsq;
A = sqrt(At.^2 + An.^2);
plot(theta,B,theta,A)
ylabel('force (kN)')
title('part (a): constant M')
======= end of script=====================

%========= script for part b ===========
m = 0.15; g = 9.81; r = 1.5;
rAB = 1.8; rAG = 1.2; rAC = r;
alpha = 5;
theta = 0:0.1:60;
thr = theta*pi/180;
omsq = 2*alpha*thr;
At = m*g*cos(thr) + m*r*alpha;
M = r*At;
B = m*r*rAG*(omsq.*cos(thr)+alpha.*sin(thr))/rAB./cos(thr);
An = B + m*g*sin(thr)-m*r*omsq;
A = sqrt(At.^2 + An.^2);
plot(theta,B,theta,A)
ylabel('force (kN)')
title('part (b): constant alpha')
======= end of script=====================

6.2 Sample Problem 6/4 (Fixed-Axis Rotation)
The pendulum has a mass of 7.5 kg with a mass
center at G and a radius of gyration about the
pivot O of 295 mm. If the pendulum is released
from rest when θ = 0, plot the total force
supported by the bearing at O along with its
normal and tangential components as a function
of θ. Let θ range between 0 and 180°.
Problem Formulation
The free body and mass acceleration diagrams are
identical to those in the sample problem. The main
difference in approach is that we will obtain results
at an arbitrary angle θ rather than at 60°.
[ ]αOO IM =Σ αθ 2
0cos mkrmg = θα cos2
0k
rg
=
[ ]θαωω dd = θθθωω
ω θ
sincos 2
0
0 02
0 k
rg
d
k
rg
d∫ ∫ ==
θω sin
2
2
0
2
k
rg
=
[ ]2
ωrmFn =Σ 2
sin ωθ rmmgOn =−
[ ]αrmFt =Σ αθ rmmgOt =+− cos
After substituting for α and ω we have,
θsin21 2
0
2








+=
k
r
mgOn θcos1 2
0
2








−=
k
r
mgOt
The magnitude of the force at O is,
( ) ( )22
tn OOO +=

After substituting r = 0.25 m, 0k = 0.295 m, m = 7.5 kg, and g = 9.81 m/s2
all
forces will be functions of θ only.
MATLAB Script
%%%%%%%%%%%%%%%% Script %%%%%%%%%%%%%%%%%%%
% This script plots the force at O and its
% components as functions of theta
rb = 0.25;
k0 = 0.295;
mg = 7.5*9.81;
theta = 0:0.01:pi;
On = mg*(1+2*(rb/k0)^2)*sin(theta);
Ot = mg*(1-(rb/k0)^2)*cos(theta);
O=sqrt(On.^2+Ot.^2);
td=180*theta/pi; % converts to degrees
plot(td,O,td,On,td,Ot)
xlabel('theta (degs)')
title('Force at O (Newtons)')

6.3 Problem 6/98 (General Plane Motion)
The slender rod of mass m and length l is
released from rest in the vertical position with
the small roller at end A resting on the incline.
(a) Determine the initial acceleration of A (aA)
and plot aA versus θ for 0 ≤ θ ≤ 90°. (b)
Determine the maximum value of aA over this
range and the angle θ at which it occurs.
Problem Formulation
The free-body and mass acceleration
diagrams are shown to the right. Shown on
the mass acceleration diagram are the two
components of the acceleration of the center
of mass G obtained from the following
kinematic relation,
aG = aA + (aG/A)n + (aG/A)t
where (aG/A)n = 2
2
ω
l
=0, (aG/A)t = α
2
l
.
[ ]damIM A +=Σ α θαα cos
22212
1
0 2 l
ma
ll
mml A−+=
[ ]xx amF =Σ 





−= θαθ cos
2
sin
l
ammg A
These two equations can be solved simultaneously to give,
( )
θ
θθ
α 2
cos34
cossin/6
−
=
lg
θ
θ
2
cos34
sin4
−
=
g
aA
At first, the answer to part (b) seems obvious. Intuitively, we would like to say
that the maximum acceleration is aA = g and occurs at θ = 90°. But this intuition
neglects the effects of the bar’s rotation upon the acceleration. As we will see
below, the maximum acceleration is somewhat larger than g.
The maximum acceleration is obtained in the usual manner. The orientation
where the maximum occurs is first found by solving the equation 0/ =θddaA

for θ. This angle is then substituted back into the expression for aA to yield the
maximum acceleration.
When using the solve command to solve simultaneous equations, be sure that the
variables solved for are single characters. Thus, in the following we let x = aA
and y = α. Also remember that the equations solved must be in the form where
the right hand side is zero. The equals sign is omitted.
EDU» syms x y L theta g m
EDU» eqn1 = 1/12*m*L^2*y+m*L/2*y*L/2-m*x*L/2*cos(theta)
eqn1 = 1/3*m*L^2*y-1/2*m*x*L*cos(theta)
EDU» eqn2 = m*g*sin(theta)-m*(x-L/2*y*cos(theta))
eqn2 = m*g*sin(theta)-m*(x-1/2*L*y*cos(theta))
EDU» [x,y]=solve(eqn1,eqn2)
x = -4*g*sin(theta)/(-4+3*cos(theta)^2)
y = -6*cos(theta)*g*sin(theta)/L/(-4+3*cos(theta)^2)
The result for x will be used in the script below to plot aA versus θ. It is
convenient to go ahead and do part (b) first as we have everything already set up.
EDU» aA = subs(x,g,9.81)
aA = -981/25*sin(theta)/(-4+3*cos(theta)^2)
EDU» daA = diff(aA,theta)
daA =
-981/25*cos(theta)/(-4+3*cos(theta)^2)-5886/25*sin(theta)^2/
(-4+3*cos(theta)^2)^2*cos(theta)
EDU» solve(daA,theta)
ans =
1/2*pi]
atan(1/6*3^(1/2)*6^(1/2))
-atan(1/6*3^(1/2)*6^(1/2))
-atan(1/6*3^(1/2)*6^(1/2))+pi
atan(1/6*3^(1/2)*6^(1/2))-pi
MATLAB has found five solutions. Now we use the eval command to put these
in a form easier to understand.

EDU» eval(ans)
ans =
1.5708
0.6155
-0.6155
2.5261
-2.5261
Only two of the five solutions are in the range from 0 to 90°. These two solutions
are π/2 (90°) and 0.6155 rad (35.3°). Substitution will reveal which is the
maximum. Of course, we could also look at the plot below to see that the second
solution corresponds to a maximum.
EDU» subs(aA,theta,.6155)
ans = 11.3276
EDU» subs(aA,theta,pi/2)
ans = 9.8100 % this result shouldn’t be surprising
Thus, (aA)max = 11.33 m/s2
when θ = 35.3°.
%%%%%%%%%%%%%%%%%% Script %%%%%%%%%%%%%%%%%
% This script plots aA versus theta
g = 9.81;
theta = 0:0.01:pi/2;
aA = -4*g*sin(theta)./(-4+3*cos(theta).^2);
plot(theta*180/pi, aA)
ylabel('a_A (m/s^2)')
%%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%

6.4 Problem 6/104 (General Plane Motion)
The uniform 12-ft pole is hinged to the truck bed
and released from the vertical position as the
truck starts from rest with an acceleration a. If
the acceleration remains constant during the
motion of the pole, derive an expression for the
angular velocity ω in terms of a, g, and L where
L is the length of the pole. Plot ω versus θ for a
= 3 ft/s2
.
Problem Formulation
The free-body and mass acceleration diagrams
are shown to the right. Shown on the mass
acceleration diagram are the three components
of the acceleration of the center of mass G
obtained from the following kinematic relation,
aG = aO + (aG/O)n + (aG/O)t
where aO = a, (aG/O)n = 2
ωr , (aG/O)t = αr , and
2/Lr = = 6 feet.
[ ]damIMO +=Σ α θααθ cos
22212
1
sin
2
2 L
ma
LL
mmL
L
mg −+=
( )θθα cossin
2
3
ag
L
+=
Now we integrate the relation θαωω dd = to obtain,
( ) ( )( )θθθθθω
θ
sincos1
2
3
cossin
2
3
2
1
0
2
ag
L
dag
L
+−=+= ∫
( )( )θθω sincos1
3
ag
L
+−=

MATLAB Script
%%%%%%%%%%%%%%%%%%%% Script %%%%%%%%%%%%%%%%%%%%
L = 12; g = 32.2; a = 3;
theta = 0:0.01:pi/2;
omega = sqrt(3/L*(g*(1-cos(theta))+a*sin(theta)));
plot(theta*180/pi,omega)
ylabel('omega (rad/s)')
%%%%%%%%%%%%%%%%%end of script %%%%%%%%%%%%%%%%%

6.5 Sample Problem 6/10 (Work and Energy)
The 4-ft slender bar weighs 40 lb with a mass
center at B and is released from rest in the
position for which θ is essentially zero. Point B
is confined to move in the smooth vertical guide,
while end A moves in the smooth horizontal
guide and compresses the spring as the bar falls.
Plot the angular velocity of the bar and the
velocities of A and B as a function of θ from 0 to
90°. The stiffness of the spring is 30 lb/in.
Problem Formulation
From the figure to the right, the lengths CB and CA are
2sinθ and 2cosθ respectively. Using the instantaneous
center C we can write the two velocities in terms of the
angular velocity ω.
θωω cos2== CAvA θωω sin2== CBvB
Now we need to divide the range for θ into two distinct
intervals depending upon whether or not the spring has
been engaged. Since the velocities for A and B are known
in terms of ω and θ, we need to find only the angular
velocity in these two intervals. From the diagram we see
that A will first contact the spring at an angle
)24/18(sin 1−
=θ = 0.8481 rads (48.6°).
(a) Before the spring is engaged (θ ≤ 48.6°).
]
2
1
2
1
[ 22
ωIvmT += ( ) 222
4
2.32
40
12
1
2
1
cos2
2.32
40
2
1
ωθω 





+=∆T
( ) 22
cos348282.0 ωθ−=∆T
][ hWVg ∆=∆ )1(cos80)2cos2(40 −=−=∆ θθgV
We now substitute into the energy equation 021 =∆+∆=′− gVTU ,
( ) ( )1cos80cos348282.00 22
−+−= θωθ , from which we find

θ
θ
ω 2
cos34
cos1
829.9
−
−
=
(b) After the spring is engaged (48.6 ≤ θ ≤ 90°). The kinetic and potential
energies are the same as in part (a). At any angle θ, point A has moved 2sinθ feet
to the left. Thus, the spring is compressed by 2sinθ − 18/12 feet.
]
2
1
[ 2
kxVe = 0
12
18
sin21230
2
1
2
−





−











=∆ θ
ft
in
in
lb
Ve
2
2
3
sin2180 





−=∆ θeV
Again, we substitute into the energy equation 021 =∆+∆+∆=′− eg VVTU ,
( ) ( )
2
22
2
3
sin21801cos80cos348282.00 





−+−+−= θθωθ
θ
θθθ
ω 2
2
cos34
65sin144cos16sin216
457.2
−
−−−
=
MATLAB Scripts
In the following, terms ending with “_a” are for part (a) where θ is between 0 and
48.6° while terms ending with “_b” are for part (b) where θ is greater than 48.6°.
%%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%%%%%%
% This script solves the energy equation for omega
% for cases a and b
syms theta omega
DT = 0.8282*(4-3*cos(theta)^2)*omega^2;
DVg = 80*(cos(theta)-1);
DVe = 180*(2*sin(theta)-3/2)^2;
U12_a = DT + DVg
U12_b = DT + DVg + DVe
omega_a = solve(U12_a,omega)
omega_b = solve(U12_b,omega)
%%%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%

Output of script #1
U12_a =
(4141/1250-12423/5000*cos(theta)^2)*omega^2+80*cos(theta)-80
U12_b =
(4141/1250-12423/5000*cos(theta)^2)*omega^2+80*cos(theta)-
80+180*(2*sin(theta)-3/2)^2
omega_a =
200/(-16564+12423*cos(theta)^2)*41410^(1/2)*((-4+3*cos(theta)^2)*
(cos(theta)-1))^(1/2)
-200/(-16564+12423*cos(theta)^2)*41410^(1/2)*((-4+3*cos(theta)^2)*
(cos(theta)-1))^(1/2)
It shouldn’t be surprising that MATLAB has found two solutions to the equation.
As you can see, they have the same magnitude with one being positive and the
other negative. The solution of interest is the positive one which turns out to be
the second solution. This solution is copied and pasted into script #2.
omega_b =
50/(4141+12423*sin(theta)^2)*(-(41410+124230*sin(theta)^2)*
(16*cos(theta)+65+144*sin(theta)^2-216*sin(theta)))^(1/2)
-50/(4141+12423*sin(theta)^2)*(-(41410+124230*sin(theta)^2)*
(16* cos(theta)+65+144*sin(theta)^2-216*sin(theta)))^(1/2)
In this case, the first solution is positive.
%%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%
% This script plots omega versus theta. Note that
% two range variables are used to divide the range
% for theta into the two regions.
% a = theta for theta < 48.6 deg. (.8481 rads)
% b = theta for theta > 48.6 deg.
% The results for omega are copied and pasted from
% the output of script #1
a = 0:0.001:0.8481;
b = 0.8481:0.001:pi/2;
omega_a = -200./(-16564+12423*cos(a).^2)*41410^(1/2).*((-
4+3*cos(a).^2).*(cos(a)-1)).^(1/2);
omega_b = 50./(4141+12423*sin(b).^2).*(-
(41410+124230*sin(b).^2).*(16*cos(b)+65+144*sin(b).^2-
216*sin(b))).^(1/2);
plot(a*180/pi, omega_a, b*180/pi, omega_b)

ylabel('omega (rad/sec)')
%%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%% Script #3 %%%%%%%%%%%%%%%%%
% This script plots vA and vB versus theta.
a = 0:0.001:0.8481;
b = 0.8481:0.001:pi/2;
omega_a = -200./(-16564+12423*cos(a).^2)*41410^(1/2).*((-
4+3*cos(a).^2).*(cos(a)-1)).^(1/2);
omega_b = 50./(4141+12423*sin(b).^2).*(-
(41410+124230*sin(b).^2).*(16*cos(b)+65+144*sin(b).^2-
216*sin(b))).^(1/2);
vA_a = 2*omega_a.*cos(a);
vA_b = 2*omega_b.*cos(b);
vB_a = 2*omega_a.*sin(a);
vB_b = 2*omega_b.*sin(b);
plot(a*180/pi, vA_a, b*180/pi, vA_b, a*180/pi, vB_a, b*180/pi, vB_b)
ylabel('velocity (ft/sec)')
%%%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%

For Further Study
A striking feature of the velocity curves above is the sudden change in shape
when the spring is engaged at about θ = 49°. The details depend very much upon
the relative magnitudes of the weight and spring constant. To illustrate, the figure
below shows the angular velocity for several different values of the spring
constant k.
Note that for stiff springs, the angular velocity goes to zero before reaching θ =
90°. The physical explanation for this is that, for a stiff spring, the bar will
rebound before it reaches the horizontal position.

6.6 Problem 6/206 (Impulse/Momentum)
Determine the minimum velocity v that the
wheel may have and just roll over the
obstruction. The centroidal radius of gyration of
the wheel is k, and it is assumed that the wheel
does not slip. Plot v versus h for three cases: k =
½, ¾, and 1 m. For each case take r = 1 m.
Problem Formulation
During Impact: Conservation of Angular Momentum
As usual, we neglect the angular impulse of the
weight during the short interval of impact. With this
assumption we have conservation of angular
momentum about point A. Immediately before
impact, the center of the wheel is not moving in a
circular path about A and we need to use the formula
for general plane motion. Note that 2
mkI = .
( ) ( )hrmv
r
v
mkhrmvmkdvmIH A −+=−+=+= 22
ωω
We will use primes to denote the state immediately after impact. Since the wheel
now rotates about A we can use the simpler formula ωAA IH = . Note that, by the
parallel axis theorem, 2
mrII A += ( )22
rkm += .
( )r
v
rkIH AA
′
+=′=
′ 22
ω
Setting ′
= AA HH and solving yields,






+
−=′ 22
1
rk
rh
vv
After Impact: Work-Energy
( ) mghIVT Ag +′−==∆+∆ 22
0
2
1
0 ω

( ) mgh
r
v
rkm =




 ′
+
2
22
2
1
Substituting the result for v′ into the above equation followed by simplification
yields,
( )
rhrk
rkghr
v
−+
+
= 22
22
2
MATLAB Scripts
%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%
% symbolic solution for v
syms m g k r h v vp
Ib = m*k^2; IA = m*(k^2+r^2);
HA = Ib*v/r+m*v*(r-h);
HAp = IA*vp/r;
vp = solve(HA-HAp,vp) % conservation of angular momentum
eqn = 1/2*IA*vp^2/r^2-m*g*h; % work/energy
solve(eqn,v)
%%%%%%%%%%% end of script %%%%%%%%%%
Output of script #1
vp =
v*(k^2+r^2-r*h)/(k^2+r^2)
ans =
(2*g*h*r^2+2*g*h*k^2)^(1/2)*r/(k^2+r^2-r*h)
-(2*g*h*r^2+2*g*h*k^2)^(1/2)*r/(k^2+r^2-r*h)
The first solution is the one we want.
%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%
% This script plots v versus h for the specified values of k.
% The result for v below was obtained by substituting g = 9.81
% and r = 1 in to the symbolic result found in script #1
v = inline('4.429*sqrt(h+h*k^2)/(1+k^2-h)');
v = vectorize(v); % puts “.”’s in expression
h = 0:0.01:1;
plot(h,v(h,1/2),h,v(h,3/4),h,v(h,1))
xlabel('h (m)')
ylabel('v (m/s)')
%%%%%%%%%%% end of script %%%%%%%%%%

INTRODUCTION TO THREE-
DIMENSIONAL DYNAMICS
OF RIGID BODIES
This chapter presents a brief introduction to rigid body dynamics in three
dimensions. In problem 7.1, the general 3-D motion of three connected bars is
investigated. In particular, the angular velocities of two of the bars are plotted
versus the length of the third bar. solve is used to solve four equations
symbolically for four unknowns. In problem 7.2 we consider a bent plate rotating
about a fixed axis. The problem illustrates a simplified version of what engineers
might do in a real design situation. Two dimensions of the bent plate are left as
variables and the objective of the problem is to find all suitable values of those
dimensions which satisfy several constraints simultaneously. Here we illustrate a
graphical approach to this type of design problem.
7

130 CH. 7 INTRODUCTION TO 3-D DYNAMICS OF RIGID BODIES
7.1 Sample Problem 7/3 (General Motion)
Crank CB rotates about the horizontal axis with
an angular velocity ω1 = 6 rad/s, which is
constant for a short interval of motion that
includes the position shown. Link AB has a ball-
and-socket fitting on each end and connects
crank DA with CB. Let the length of crank CB
be d mm (instead of 100 mm as in the sample
problem in your text) and plot ω2 and ωn as a
function of d for 0 ≤ d ≤ 200 mm. ω2 is the
angular velocity of crank DA while ωn is the
angular velocity of link AB.
Problem Formulation
Our analysis will follow closely that in the sample problem in your text.
vA = vB + ωn×rA/B
where vA = 50ω2j vB = 6di rA/B = 50i + 100j + dk
Substitution into the velocity equation gives
d
d nznynx
10050
650 2 ωωωω
kji
ij +=
Expanding the determinant and equating the i, j, and k components yields the
following three equations
( ) 01006 =−+ nznyd ωω ( ) 050 2 =+− nxnz dωωω 02 =− nynx ωω
At this point we have three equations with four unknowns. As explained in the
sample problem in your text, the fourth equation comes by requiring ωn to be
normal to vA/B
ωn• rA/B = 50ωnx + 100ωny + dωnz = 0
These four equations will be solved simultaneously for ω2, ωnx, ωny, and ωnz.
Once this is done,

3-D DYNAMICS OF RIGID BODIES 131
222
nznynxn ωωωω ++=
MATLAB Scripts
%%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%%%%%%%
% This script solves four equations for
% omega_2 (w) and the three components of
% omega_n (x, y, and z). The magnitude of
% omega_n is then found from its components.
syms w x y z d
eqn1 = d*(6+y)-100*z;
eqn2 = 50*(w-z)+d*x;
eqn3 = 2*x-y;
eqn4 = 50*x+100*y+d*z;
[w,x,y,z]=solve(eqn1,eqn2,eqn3,eqn4)
omega_n = sqrt(x^2+y^2+z^2);
omega_n = simplify(omega_n)
%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%
Output of script #1
w = 3/50*d (ω2)
x = -3*d^2/(d^2+12500)
y = -6*d^2/(d^2+12500)
z = 750*d/(d^2+12500)
omega_n = 3*5^(1/2)*(d^2/(d^2+12500))^(1/2) (ωn)
%%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%%
% This script plots omega_2 and the magnitude of
% omega_n as functions of d
d = 0:0.5:200;
omega_2 = 3/50*d;
omega_n = 3*5^(1/2)*(d.^2./(d.^2+12500)).^(1/2);
plot(d, omega_2, d, omega_n)
xlabel('d (mm)')
title('angular velocity (rad/s)')
%%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%

7.2 Sample Problem 7/6 (Kinetic Energy)
The bent plate has a mass of 70 kg per square
meter of surface area and revolves around the z-
axis at the rate ω = 30 rad/s. Let the dimensions
of part B be a and b where a is the dimension
parallel to the x-axis and b is the dimension
parallel to the z-axis. Part A remains unchanged.
(a) Find all suitable values for a and b which
satisfy the following conditions: a ≤ 0.2 m, b ≤
0.6 m, and 15 ≤ T ≤ 30 J where T is the kinetic
energy of the plate. (b) Find a and b for the case
where T = 40 J and H0 = 5 N•m•s where H0 is
the magnitude of the angular momentum about
O.
Problem Formulation
Substitution of ωx = 0, ωy = 0, and ωz = ω into equations 7/11 and 7/18 of your
text yields.
HO = ω(−Ixzi −Iyzj +Izzk)
222222
zzyzxzOzOyOxO IIIHHHH ++=++= ω

2
2
1
ωzzIT =
The moments and products of inertia for part A remain unchanged. For part B, mB
= 70ab where a and b are in meters. The moments and products of inertia for part
B are
( ) 













++=+=
2
222
2
125.
12
a
ma
m
mdII B
B
zzzz












+=+=
22
0
ba
mdmdII Bzxxzxz
( ) 





+=+=
2
125.00
b
mdmdII Bzyyzyz
The total moment and products of inertia are found by adding the above to those
found for part A (see the sample problem in your text). After substituting for mB
and simplifying we have,
baabIzz
3
33.23094.100456.0 ++=
22
5.17 baIxz = 2
375.400273.0 abI yz +=
Substitution of these results will give HO and T as functions of a and b.
Part (a)
The most efficient way to show the acceptable ranges for a and b is to find the
required relationship between these two dimensions in order to satisfy the upper
and lower bounds on T. This is accomplished by substituting these bounds for T
in the equation above and then solving that equation for b as a function of a. To
illustrate, consider the lower limit on T (15 J). Substituting T = 15 into the
equation above gives
( ) baabIT zz
32
500,102.492052.230
2
1
15 ++===
Solving for b,
( )2
643
078921.0
aa
b
+
= (for T = 15 J)
A similar result can be obtained for the upper limit,

( )2
643
17035.0
aa
b
+
= (for T = 30 J)
Plotting these two functions defines the acceptable regions for a and b.
Part (b)
This is similar to (a) except that we solve two equations (T = 40 and H0 = 5)
simultaneously for two unknowns, a and b. The result is a = 0.1211 m and b =
0.4852 m.
MATLAB Scripts
%%%%%%%%%%%%%%% Script #1 %%%%%%%%%%%%%%%%%%%%%%
% The first part of this script solves the equations
% T = 15 and T = 30 symbolically for b in terms of a.
% These two results are named b15 and b30 respectively.
% When using the symbolic solve it is best to rewrite
% the equations in the form expression = 0. The "=0"
% is automatically understood by MATLAB and doesn't
% have to be typed in. In the following, our two
% expressions are named eqn1 and eqn2.
syms a b
mB = 70*a*b;
omega = 30;
Izz = mB/12*a^2+mB*(0.125^2+(a/2)^2)+0.00456;
Ixz = mB*a/2*b/2;
Iyz = mB*0.125*b/2+0.00273;
H0 = omega*sqrt(Ixz^2+Iyz^2+Izz^2)
T = 1/2*Izz*omega^2
eqn1 = T - 15;
eqn2 = T - 30;
b15 = solve(eqn1,b)
b30 = solve(eqn2,b)
% This part of the script solves the two equations
% for part b.
eqn3 = T - 40;
eqn4 = H0 - 5;
[a,b]=solve(eqn3,eqn4)
%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%%%
Output of script #1
H0 =
1/10000*(27562500000000*a^4*b^4+1722656250000*a^2*b^4+2149875000*a*b^2+25
42185+49000000000000*a^6*b^2+4593750000000*a^4*b^2+19152000000*a^3*b+107
666015625*a^2*b^2+897750000*a*b)^(1/2)

T = 2625*a^3*b+31500*a*b*(1/64+1/4*a^2)+513/250
b15 = 8632/109375/a/(64*a^2+3)
b30 = 18632/109375/a/(64*a^2+3)
a =
-.118672955347671873813795187
-.924961825359232183759243317e-1-.20405912363720769438094726*i
-.924961825359232183759243317e-1+.20405912363720769438094726*i
-.257514497287494286960690298e-2-.25491731461066579787563883*i
-.257514497287494286960690298e-2+.25491731461066579787563883*i
.349640543257908784471004656e-2-.265040090095767094717337313*i
.349640543257908784471004656e-2+.26504009009576709471733731*i
.903798369106820783468910094e-1-.204312206452426767071840559*i
.9037983691068207834689100945e-1+.2043122064524267670718405*i
.121063125678745863921655544
b =
-.499591682557401531621796988
.286519678237340239067367533+.281089760611403745879347469*i
.286519678237340239067367533-.281089760611403745879347469*i
.64375775885513160849563080e-1-.778487038145858201006376746*i
.64375775885513160849563080e-1+.77848703814585820100637674*i
-.5367082781855869430616764e-1-.579398660286892007965618348*i
-.5367082781855869430616764e-1+.57939866028689200796561834*i
-.29810587868680635995547322+.28455901589818267092944534*i
-.29810587868680635995547322-.284559015898182670929445341*i
.485167563521434010754982656
We see from the above that MATLAB has found ten solutions to our equations.
Of these only the first and last are real. The first solution has negative values for
a and b and thus can be excluded. This leaves only the last solution,
a = 0.1211 m; b = 0.4852 m
%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%%%%%%
% This script plots the two expressions for b
% obtained with script #1
a = 0.01:0.001:0.2;
b15 = 8632/109375./a./(64*a.^2+3);
b30 = 18632/109375./a./(64*a.^2+3);
plot(a, b15, a, b30)
xlabel('a (m)')

ylabel('b (m)')
axis([0 0.2 0 0.6])
%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%%%%%%
The two curves above represent the values of a and b for which T is exactly 15 or
30 J. Thus, the acceptable values of a and b satisfying the condition 15 ≤ T ≤ 30
J are all those combinations lying on or between the two curves.

VIBRATION AND TIME
RESPONSE
This chapter considers an important class of Dynamics problems that involve
linear or angular oscillations of a body or structure about some equilibrium
position or configuration. Very few of the homework problems in your text
require you to actually plot the oscillations of a body versus time. For this reason,
all of the problems in this chapter will involve such plots. This is very useful in
visualizing the time response of a vibrating system, especially for the case of
damped or forced vibrations. Problem 8.1 looks at the effects of damping
coefficient upon time response while Problem 8.3 considers the effects of initial
conditions.
8

138 CH. 8 VIBRATION AND TIME RESPONSE
8.1 Sample Problem 8/2 (Free Vibration of Particles)
The 8-kg body is moved 0.2 m to the right of the
equilibrium position and released from rest at
time t = 0. Plot the displacement as a function of
time for three cases, c = 8, 32, and 56 N•s/m.
The spring stiffness k is 32 N/m.
Problem Formulation
As in the sample problem, the natural circular frequency is
28/32/ === mknω rad/sec. Now we find the damping ratio for each case
(from ζ = c/2mωn) with the result ζ = 0.25, 1, and 1.75 for c = 8, 32, and 56
N•s/m respectively.
(a) ζ = 0.25. Since ζ < 1, the system is underdamped. The damped natural
frequency is 937.11 2
=−= ζωω nd rad/sec. The displacement and velocity are
)937.1sin()sin( 2/
ψψωζω
+=+= −−
tCetCex t
d
tn
)937.1cos(937.1)937.1sin(5.0 2/2/
ψψ +++−= −−
tCetCex tt
&
From the initial conditions 2.00 =x and 00 =x& we find C = 0.207 m and ψ =
1.318 rad.
(b) ζ = 1. For ζ = 1, the system is critically damped. The displacement and
velocity are
( ) ( ) tt
etAAetAAx n 2
2121
−−
+=+= ω
( ) tt
etAAeAx 2
21
2
2 2 −−
+−=&
From the initial conditions 2.00 =x and 00 =x& we find A1 = 0.2 m and A2 = 0.4
m/s.
(c) ζ = 1.75. Since ζ > 1, the system is overdamped. The displacement and
velocity are
tt
tt
eBeBeBeBx
nn 372.6
2
628.0
1
1
2
1
1
22
−−



 −−−



 −+−
+=+=
ωζζωζζ

VIBRATION AND TIME RESPONSE 139
tt
eBeBx 372.6
2
628.0
1 372.6628.0 −−
−−=&
From the initial conditions 2.00 =x and 00 =x& we find B1 = 0.222 m and B2 =
−0.0219 m/s.
MATLAB Script
%%%%%%%%%%%%%%% Script %%%%%%%%%%%%%%%%%%%%%%
t=0:0.01:5;
C = 0.207; psi = 1.318;
xa = C*sin(1.937*t+psi).*exp(-t/2);
A1 = 0.2; A2 = 0.4;
xb = (A1+A2*t).*exp(-2*t);
B1 = 0.222; B2 = -0.0219;
xc = B1*exp(-0.628*t)+B2*exp(-6.372*t);
x0 = 0.*t;
plot(t,xa,t,xb,t,xc,t,x0)
ylabel('x (m)')
%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%%

8.2 Problem 8/139 (Damped Free Vibration)
The mass of a critically damped system having a
natural frequency nω is released from rest at an
initial displacement x0. (a) Determine the time t
required for the mass to reach the position
01.0 xx = if 4=nω rad/s. (b) Plot the non-
dimensional displacement x/x0 for nω = 2, 4, and
8 rad/s.
Problem Formulation
Start with the equation for the critically damped case on page 606 of your text.
( ) tnetAAx ω−
+= 21
First we have to evaluate the constants in terms of x0 using the initial conditions.
This will require the derivative of x.
( ) t
n
t nn etAAeAx ωω
ω −−
+−= 212&
Now, substituting the initial conditions,
( ) 100 Axtx ===
( ) nAAtx ω1200 −===&
These two equations give
01 xA = and nxA ω02 =
( ) t
n
netxxx ω
ω −
+= 00
or ( ) t
n
net ω
ωη −
+= 1
where η is the non-dimensional displacement x/x0.
Part (a) will require the solution to the equation

( ) t
et 4
411.0 −
+=
For part (b) we will simply plot η for three different natural frequencies nω .
Part (a)
EDU>> f = '(1 + 4*t)*exp(-4*t) = 0.1'
f =
(1 + 4*t)*exp(-4*t) = 0.1
EDU>> solve(f,'t')
ans =
-.24044468956330014201486061769676
.97243004246685726447599505623170
Thus, the time t required for the mass to reach the position 01.0 xx = when
4=nω rad/s is t = 0.972 sec.
Part (b)
%========= script for part b ========
eta = inline('(1 + wn*t).*exp(-wn*t)')
t = 0:0.05:3;
plot(t,eta(t,2),t,eta(t,4),t,eta(t,8))
title('non-dimensional displacement')
%========= end of script ============
eta =
Inline function:
eta(t,wn) = (1 + wn*t).*exp(-wn*t)

8.3 Sample Problem 8/6 (Forced Vibration of Particles)
The 100-lb piston is supported by a spring of
modulus k = 200 lb/in. A dashpot of damping
coefficient c = 85 lb-sec/ft acts in parallel with
the spring. A fluctuating pressure p = 0.625
sin(30t) (psi) acts on the piston, whose top
surface area is 80 in2
. Plot the response of the
system for initial conditions x0 = 0.05 ft and
=0x& 5, 0, and −5 ft/sec.
Problem Formulation
The particular (steady state) solution was found in the sample problem in your
text,
( )φω −= tXxp sin
where X = 0.01938 m, φ = 1.724 rad and ω = 30 rad/sec. Also from the sample
problem, 8.27/ == mknω rad/sec and ζ = c/2mωn = 0.492.
The complete solution is found by adding the complementary (transient) and
particular solutions. Since the system is underdamped (ζ < 1), the complementary
solution is,
)sin( ψωζω
+= −
tCex d
t
c
n
where 2.241 2
=−= ζωω nd rad/sec. The displacement of the system is
( )φωψωζω
−++=+= −
tXtCexxx d
t
pc
n
sin)sin(
)724.130sin(01938.0)2.24sin(68.13
−++= −
ttCex t
ψ
The velocity is found by differentiating x,
( )φωωψωζωψωω ζωζω
−++−+= −−
tXteCteCx d
t
nd
t
d
nn
cos)sin()cos(&
( ) )724.130cos(5814.0)2.24sin(68.13)2.24cos(2.2468.13
−++−+= −
tttCex t
ψψ&
The constants C and ψ are found from the initial conditions,

0192.0sinsinsin05.00 −=+== ψφψ CXCx
0887.0sin7.13cos2.24cossincos0 −−=+−= ψψφωψζωψω CCXCCx nd
&
The first equation can be solved for C = 0.0692/sinψ . Substitution into the
second equation gives ψ .








+
= −
035.1
675.1
tan
0
1
x&
ψ
ψsin
0692.0
=C
This yields the following values for C and ψ .
x0 = 0.05 ft, =0x& 5 ft/s, C = 0.259 ft, ψ = 0.271 rad
x0 = 0.05 ft, =0x& 0 ft/s, C = 0.081 ft, ψ = 1.017 rad
x0 = 0.05 ft, =0x& −5 ft/s, C = −0.178 ft, ψ = −0.400 rad
MATLAB Script
%%%%%%%%%%%%%%%% Script %%%%%%%%%%%%%%%%%%%%%
% This script plots x versus t for the three
% sets of initial conditions
x = inline('C*sin(24.2*t+psi)*exp(-13.68*t)+.01938*sin(30*t-1.724)')
x = vectorize(x);
t = 0:0.001:0.5;
plot(t,x(.259,.271,t),t,x(.081,1.017,t),t,x(-.178,-.4,t))
xlabel('t (sec)')
ylabel('x (ft)')
%%%%%%%%%%%%% end of script %%%%%%%%%%%%%%%%%%

Notice how quickly the three cases converge to the steady state solution.

Solving dynamics problems with matlab

More Related Content

What's hot

Viewers also liked

Similar to Solving dynamics problems with matlab

Recently uploaded

Solving dynamics problems with matlab