Lecture 2. A Day of Principles. (with an example that applies ‘ em all at the end). The principle of virtual work. d’Alembert’s principle. Hamilton’s principle. Principle of virtual work says the work done by a virtual displacement from an equilibrium position must be zero .
(with an example that applies ‘em all at the end)
The principle of virtual work
the work done by a virtual displacement from an equilibrium position must be zero.
This can be used in statics to find forces
We need to move this up to dynamics
d’Alembert’s principle says that
can be treated as an
We eliminated internal forces last time, so the fs are external
We suppose we know them, and since the displacement is tiny
they don’t change during the virtual displacement
What I’d like to do is gather all of this up into what I’ll call configuration space
The dimension of q is N = 3K
We’ll see more as we go along.
The general rule, which we’ll investigate as we go along,
there are (at least) as many qs as there are degrees of freedom.
We’ll take up the matter of constraints — relations among the qs —later
Bottom line: we can write the rs in terms of the qs.
summation convention does not apply to i in this formula
Define the generalized force
This force goes with the kth generalized coordinate
dotted into something. If you like manipulations
you’ll LOVEwhat comes next
First a bit of what I call “integration by parts”
This is a useful trick in deriving things
We still aren’t done. The second term on the right is cool
but we need to play with the first term
Remember that there is a sum on k here!
I can split Qk into two pieces: the potential and nonpotential parts
From here on in Qk will be understood to be the nonpotential generalized forces
We’ll have an algorithm for the calculation later.
If the qk are independent, then we have the Euler-Lagrange equations
Principle of virtual work
d’Alembert’s principle (inertial forces)
Independence of the generalized coordinates
All the rest was clever manipulation
We will spend a lot of time dealing with cases where the qs are not independent. . . . . but not today . . .
This is a formalism that leads to the Euler-Lagrange equations
and it will help us when we need to consider constraints
(relations between the dqs).
This is generally posed in terms of the Lagrangian
but that eliminates the generalized forces
which I’d like to include
Let me define
where W denotes the work, potential and nonpotential
The action integral depends on the path between the two end points.
The actual path will be the path that minimizes the action integral.
where the zero denotes the desired path
The first piece is fine; we need to play with the second.
I can integrate the first part. All the paths hit the end points, and the integral is zero.
and this gives us the Euler-Lagrange equations
No matter how we look at it, we have a governing system
We get the Euler-Lagrange equations if the dqk are independent
We will have issues regarding independence
and solutions for them.
We will have issues regarding the generalized forces
and we’ll develop techniques for finding them.
A special kind of friction: “viscous” friction, damper/dashpot
the Rayleigh dissipation function
note double summation
the coefficients are the most general
they will usually be much simpler
The (unconstrained) Euler-Lagrange equations become
where Qk no longer includes the friction forces
We can look at some mechanical systems that can be viewed as
collections of point masses
OK. Away we go . . .
1. Find T and V as easily as you can
2. Apply geometric constraints to get to N coordinates
3. Assign generalized coordinates
4. Define the Lagrangian
derivative of the first generalized coordinate
6. Differentiate that result with respect to time
7. Differentiate the Lagrangian
with respect to the same generalized coordinate
8. Subtract that and set the result equal toQ1
Repeat until you have done all the coordinates
(start without the generalized forces)
Steps 1-4 lead us to
Put the physical variables back so it looks more familiar
This is all without forcing or damping — let’s add those
add a torque, t1
If y1 changes, f1 does work
Q1 = f1
If q changes, f1 does no work
Q2 = 0
If q changes, t1 does work
Q2 = t1
If y1 changes, t1 does no work
Q1 does not change
We added the generalized forces
Now we need the Rayleigh dissipation function
So, the Rayleigh dissipation function for this problem is
but let me say a few things that we will revisit.
a pair of coupled second order ordinary differential equations
It would be nice to have first order equations
There are lots of ways to do this, and we’ll look at many of them
but the simplest is to let
If you supply a force, a torque and initial conditions
you can solve this set numerically.
You can solve for the variables, which I won’t do because it is pretty messy,
and you’ll wind up with
Later on we’ll learn to call this a state vector