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ME451
Kinematics and Dynamics
of Machine Systems
Dynamics of Planar Systems
Thursday, November 18, 2010
Initial Conditions - 6.3.4
© Dan Negrut, 2010
ME451, UW-Madison
Before we get started… [1]

Last Time


Derived the equations of motion (EOM) for any mechanical system of rigid
bodies undergoing 2D motion
The approach is boiler plate: get the generalized mass M, the constraints 
present in the model, and the generalized forces QA



Based on these quantities you can write the constrained equations of motion,
which constitute a set of differential and algebraic equations
Last lecture contained the most important slide of ME451
Today:

Example: formulating the equations of motion (EOM) for slider-crank
Discuss why and how to specify Initial Conditions (ICs)

HW due Tuesday, Nov 30:


6.3.3, 6.4.1 + MATLAB component


NOTE: Problem 6.4.1 references in its text Prob. 6.3.1. That’s a typo, it should be 3.6.1
(page 116). The dimensions are in meters (in “m”).
MATLAB component will be posted online
2
Summary of the Lagrange form of the
Constrained Equations of Motion

Equations of Motion:

Position Constraint Equations:
The Most
Important Slide
of ME451

Velocity Constraint Equations:

Acceleration Constraint Equations:
3
Example 6.3.5: Slider Crank Mechanism

Derive the equations of motion (EOM) for the slider
crank model in the figure
4
[New Topic]
The Need for Initial Conditions (ICs)

When dealing with a differential
equation, one needs a set of
initial conditions to uniquely find
the solution of the “problem”

In layman’s words, the “problem”
here can be formulated as
follows:


I give you at each point in plane
the direction (slope) of an
unknown function
Can you find me the unknown
function if additionally I also give
you the value of this function at
time t=0?
Picture shows the slopes
of the unknown function
everywhere in plane.
You simply have to start
somewhere and follow
the direction of the slope
5
ODE: Infinite Number of Solutions


6
0.1t
ODE Problem: y  0.1y 100e
Initial Condition: y0 =[-1000:50:1000]
The Need for Initial Conditions (ICs)

You just computed the accelerations, and are ready to apply
some numerical formula to integrate the acceleration

What you need though is a set of initial conditions (just like in
ME340, if you recall)

How many ICs can/should you specify at time t0?

Recall that you have a set of n generalized coordinates, but
they are not totally arbitrary, in that they need to satisfy the
set of m constraints present in the system:
7
simEngine2D Comment
² Recall t hat one of your MAT LAB assignment s was t o read in a body from an adm ¯le:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Body: 7
% i d of t hi s body
Mass: 3
% mass of body
Jbar : 0. 4
% mass moment of i ner t i a
xZer o: 0. 23
% i ni t i al X posi t i on
yZer o: - 0. 3
% i ni t i al Y posi t i on
phi Zer o: 1. 5707963267948
% i ni t i al or i ent at i on ( t hi s i s pi / 2)
xDot Zer o: 1. 0
% i ni t i al vel X
yDot Zer o: 0. 0
% i ni t i al vel Y
phi Dot Zer o: 0. 1
% i ni t i al vel Phi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
² In ot her words, you already have an est imat ion for q_0 and q_0
² You are good t o go if ©(q 0 ; 0) = 0m and and © q q_0 = º .
² If your init ial posit ion and init ial velocity are not consist ent wit h your set of kinemat ic const raint , you
will have t o come up wit h a set of consist ent init ial condit ions
8
Specifying Initial Conditions (ICs)

So you have m equations that must be satisfied by n generalized coordinates.
You have m constraints
(kinematic and driving, that is)

You need to know the initial configuration of the mechanism (positions and
velocities) to be able to start the numerical integration procedure

What does it take to uniquely define the initial configuration of the mechanism?


You can specify an additional set of ndof conditions that your generalized
coordinates must satisfy
Recall how we chose the driving constraints…
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Specifying Position ICs

Important: since you have ndof generalized coordinates that you can choose, it
will be up to you to decide the configuration in which the mechanism starts

For a simple pendulum: should it start in a vertical position or in a horizontal position?


You decide based on the problem you solve…
How do you exactly specify the conditions


You do exactly what you did for the kinematic analysis of a mechanism: you had
some excess DOFs and prescribed motions (drivers) D(q,t) to “occupy” them
For IC in dynamics you’ll specify ndof initial conditions that will implicitly determine
the configuration of the mechanism
Only at the beginning of the simulation,
you specify more constraints to
determine configuration of mechanism
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Specifying Position ICs



How do you know whether the set of conditions you specified to determine the
initial conditions are “healthy”?
Healthy is going to be that IC(q,t0) that leads to a nonsingular constraint
Jacobian:
If this is the case, one can apply Newton’s method to solve the system of
nonlinear equations below to get the initial configuration of the mechanism q0
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Specifying Velocity ICs

Specifying a set of *position* ICs is not enough

We are dealing with a second order differential problem: We need two sets of ICs



Position ICs (we’ve just done this)
Velocity ICs
Specifying Velocity ICs:

You can play it safe and take the same generalized coordinates that you assigned
initial positions and also specify for them some initial speeds (done most often)

You can decide to use a completely new set of generalized coordinates for which you
prescribe the value of the velocity at time t0 (strange, but not inconceivable)
12
Specifying Velocity ICs


Playing it safe, you find the velocities at time t0 as the
solution of the linear system
If you want to be fancy, you replace the matrix qIC with a
different matrix D, chosen such that the coefficient matrix
continues to be nonsingular (not going to elaborate on this)
13
ICs: Concluding Remarks

This IC issue is actually simple if you remember what we
did back when we dealt with the Kinematics problem

Note that there is no “dynamics” specifics here (although
the need for ICs is rooted in dynamics analysis)

We are in fact using concepts associated with the first part of the
course, namely “kinematics”
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Example 6.3.6: Simple Pendulum


How do you go about
specifying ICs?
I’d like the pendulum to
start from a vertical
configuration, and angular
velocity to be 2 rad/s.
Pin joint at point P
(between pendulum and ground)
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End ICs
Beginning Constraint Reaction Forces (6.6)
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Reaction Forces: The Framework

Remember that we jumped through some hoops to get rid
of the reaction forces that would show up in joints

I’d like to go back and recover them, since they are
important





Durability analysis
Stress/Strain analysis
Selecting bearings in a mechanism
Etc.
It turns out that the key ingredient needed to compute the
reaction forces in all joints is the set of Lagrange multipliers 
17
Reaction Forces: The Basic Idea

Recall the partitioning of the total force acting on our mechanical system

Applying a variational approach (principle of virtual work) we ended up
with this equation of motion

After jumping through hoops, we ended up with this:

It’s easy to see that
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The Important Observation

IMPORTANT OBSERVATION:

Actually, you don’t care for the “generalized” QC flavor of the reaction force,
but rather you want the actual force represented in the Cartesian global
reference frame



You’d like to have Fx, Fy, and a torque T that is due to the constraint
You report these quantities as they would act at a point P
The strategy:

Look for a force (the classical, non-generalized flavor), that when acting on
the body would lead to a generalized force equal to QC
19
The Nuts and Bolts



There is a joint acting
between Pi and Pj and we
are after finding the reaction
forces/torques Fi and Ti, as
well as Fj and Tj
Figure is similar to Figure
6.6.1 out of the textbook
Textbook covers topic well (pp. 234), I’m only modifying one thing:


The book expresses the reaction force/torque Fi and Ti in a body-fixed
reference frame
attached at point Pi
I didn’t see a good reason to do it that way

Instead, start by deriving in global reference frame OXY and then
multiply by AT to take it to the body-fixed reference frame
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The Main Result
(Expression of reaction force/torque in a joint)

Suppose that two bodies i and j
are connected by a joint, and that
the equation that describes that
joint, which depends on the
position and orientation of the two
bodies, is

Suppose that the Lagrange multiplier associated with this joint is

Then, the presence of this joint in the mechanism will lead at point P on
body i to the presence of the following reaction force and torque:
21
Comments
(Expression of reaction force/torque in a joint)

Note that there is a Lagrange multiplier associated with each constraint equation

Number of Lagrange multipliers in mechanism is equal to number of constraints

Each Lagrange multiplier produces (leads to) a reaction force/torque combo

Therefore, to each constraint equation corresponds a reaction force/torque
combo that throughout the time evolution of the mechanism “enforces” the
satisfaction of the constraint that it is associated with


Since each constraint equation acts between two bodies i and j, there will also
be a Fj/Tj combo associated with each constraint, acting on body j


Example: the revolute joint brings along a set of two kinematic constraints and
therefore there will be two Lagrange multipliers associated with this joint
According to Newton’s third law, they oppose Fi and Ti, respectively
Note that you apply the same approach when you are dealing with driving
constraints (instead of kinematic constraints)

You will get the force and/or torque required to impose that driving constraint
22
Reaction Forces
~ Remember This ~


As soon as you have a joint
(constraint), you have a Lagrange
multiplier 

As soon as you have a Lagrange
multiplier you have a reaction
force/torque:
The expression of  for all the usual
joints is known, so a boiler plate
approach renders the value of the
reaction force in all these joints
Just in case you want another form for the torque T above, note that
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Example 6.6.1: Reaction force in
Revolute Joint of a Simple Pendulum
Pendulum driven by motion:
1) Find the reaction force in the
revolute joint that connects
pendulum to ground at point O
2) Express the reaction force in
the
reference frame
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End Constraint Reaction Forces
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