Download mb ρ F δr F

Lagrangian Dynamics 2008/09
Constraint forces and work
Lecture 4: Dealing with Constraints; Generalised Coordinates & Velocities
Consider the above system subject to an external force F ext acting along r ba .
ρ
Degrees of freedom
A general rigid body is specified fully by R (the centre of mass position) and three angles
of rotation, giving the inclination of a set of axes in the body with respect to those in the
lab frame (we return to this in depth later on). Hence there are 6 degrees of freedom (below
abbreviated as dof), rather than 3N. We deduce that of the constraint equations, 3N-6 are
independent (not obvious).
z0
z
Axes fixed
in LAB
R
ma
F ext
mb
(held fixed)
In equilibrium
F constraint (r) = − k δr = − F ext
1
1
F constraint (r) · dr = − k(δr)2 = − (F ext )2 /k
2
2
which vanishes in the limit of large k. In general:
δU =
Z
Constraint forces do no work in any small instantaneous displacement of the system consistent with the constraints themselves.
x0
O
x
Fconstraint
with δr = (r − ρ ba ) r̂ ba . The work done by the constraint force in reaching this state is
y0
y
δr
Axes fixed on body
Technical Note: The limit of a continuous material can be taken by sending N → ∞; for a
rigid body, the number of degrees of freedom is still 6.
Even given that we have a rigid body, there may be further constraints. For example, a disc
confined to a plane has 2 centre of mass coordinates and one rotation angle giving a total of
3 dof. For our previous example of a hoop rolling inside a cylinder, the hoop rolls without
slipping so that the tangential velocity of the hoop and the cylinder at the point of contact
must be equal. Moreover the centre of the hoop is confined to a circle. This reduces the
problem to one dof. In problem solving, the first step is very often to identify the number
and type of degrees of freedom.
A small instantaneous displacement of this kind is called a virtual displacement. The point
is that constraint forces are there to prevent certain types of virtual displacement; only if
they fail to do this can work be done. And if they fail, they are no longer constraints.
Technical Note: The above result does not mean that the constraint forces can do no work
during the actual motion of a system. Consider, for example, a particle constrained to lie
on a surface which is itself moving: there may then be a component of the actual particle
velocity in the direction of the constraint force, so that work is done. We return to this later.
Holonomic Constraints
|r ab | = ρ ab
Rigid body:
What is a constraint? For all the examples we will meet in this course, a constraint is a
limiting approximation valid for the problem at hand.
This is an example of a holonomic constrant, ie there is a relation
Example: Two particles a, b are connected by a light inextensible rod of length ρba . In
reality the rod has a large stiffness k – there is a harmonic restoring force (Hooke’s Law)
when the rod is stretched to length rba = |r ba | :
which is an algebraic equation between coordinates, ie not a differential relation; not an
inequality. Explicit time dependence is permitted, however.
F ba (r ba ) = −k(r ba − ρ ba ) r̂ ba
1
U(r ba ) = k (rba − ρba )2 + constant
2
⇔
U
f (r a , r b , . . . , rN ; t) = 0
Holonomic constraints reduce the number of degrees of freedom.
They are important because each holonomic constraint reduces by one the number of independent differential equations we have to solve!
y’
^r
ma
ba
ρba
mb
ρba
rba
a
Example: Consider a cylindrical
log rolling down an inclined plane as
shown.
x’
In the limit k → ∞, rba = ρ ba always, and the corresponding degree of freedom is eliminated.
1
2
θ
This system has 3 dofs: (x′ , y ′, θ). If we impose rolling conditions (no-slide and no-bounce)
on the centre of mass of the log,
x′ = aθ + constant
and
y′ = a ,
ie two holonomic constraints, then only one dof remains. We can choose either θ or x′ as
the independent variable. The hoop rolling in the cylinder is similar.
Algebraic constraints mean that we can eliminate variables before solving the equations of
motion. This wouldn’t be true if the log started to slide, or to bounce.
Technical Note: The rolling constraint for a log (see above) can also be viewed as a differential
constraint (ẋ′ = aθ̇), however this can be integrated immediately (x′ = aθ + constant) to
give an algebraic relation, which is holonomic. Almost all the rolling constraints met later
in the course are holonomic in the same way.
Section III: From Newton to Lagrange
In this section we shall derive Lagrange’s equations in an important but relatively difficult way. Things will get easier when we actually start using the Lagrangian formalism.
Furthermore, we shall present an easier derivation later using the calculus of variations.
Nonholonomic constraints
Common examples of these are: (i) inequalities (ii) differential equations of constraint:
Generalised coordinates and velocities
As usual we begin our discussion with Newton’s Second Law:
Example of an inequality constraint: a particle constrained to move inside a rectangular box:
The constraints:
(
ma r̈ a = F a
a = 1, 2, · · · , N
There are 3N Cartesian coordinates, r a = (xa , ya , za ). As a shorthand notation, we write
the entire list (x1 , y1, z1 , x2 , y2 , z2 , · · · , xN , yN , zN ) as
0 ≤ x ≤ a
0 ≤ y ≤ b
b
are enforced by collisions with the walls of the box,
but they obviously don’t reduce the number of dofs.
xi
where i = 1, 2, · · · , 3N
with each corresponding component of the force Fi . The list of masses is
a
mi = (m1 , m1 , m1 , m2 , m2 , m2 , · · · , mN , mN , mN )
Example of differential constraint: A vertical disc is free to roll without slipping on a
horizontal plane. This is basically a unicycle! There are 4 degrees of freedom, (x, y, θ, φ).
The angle θ gives the orientation of the
vertical disc wrt to the coordinate axes.
(think about it).
If holonomic constraints are present, not all the xi are independent. For M constraints,
there must be a set of independent variables
qi = qi (x1 , x2 , · · · , x3N ; t)
Rolling constraints: as the disc rolls such
that φ → φ + dφ, then x → x + dx, and
y → y + dy with
where the index i now only runs from 1 to 3N − M. Our aim will be to generate 3N − M
second order DE’s between the qi ’s; these are the Equations of Motion (EoMs) of the system.
(Actually solving them is a different problem!)
a
φ
θ
y
dx = a cos θ dφ
dy = a sin θ dφ
x
There is no reason for the qi ’s to be cartesian coordinates; they are very often angles, etc. In
general the qi ’s represent a set of generalised coordinates; for shorthand we write the full set
as {q}. The transformation to our generalised coordinates is invertible: given {q} we can use
the M constraint equations to recover the 3N original variables if the 3N − M generalised
coordinates are known, ie we may write
These constraint equations cannot be integrated to eliminate a dof. The physical interpretation is that in the 4-D coordinate space, any point (x, y, θ, φ), can be reached from any other
(try it on your unicycle). Hence there can be no real reduction in the number of degrees of
freedom and variables cannot be eliminated. Differential constraint equations like these can
only be solved after a full general solution of the unconstrained problem has been found.
The important point is that the original variables {x} cannot be varied independently without violating the constraints, where as {q} (by construction) can be so varied.
From now on, we consider only holonomic systems but keep an eye out for places where our
constraints might break down or cease to be holonomic.
The time derivatives of the generalised coordinates (ie, the quantities {q̇}) are called the
generalised velocities.
3
4
xi = xi ({q}, t)