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Lecture 9 – Appendix A:
To ensure that we are all on the same page let’s do the single pendulum problem in
some detail with various coordinate choices and compare the
Newtonian and Lagrangian approaches. A mass m is suspended in a
gravitational field on a rigid, massless bar of length l. A deflection
angle is defined with respect to the downward direction. Define x ˆ as
the horizontal direction and y ˆ as the vertical (up) direction. The
acceleration of gravity is then in the y ˆ direction. The tricky part of
analyzing this problem in rectangular coordinates is the a priori
unknown constraint force due to the bar that ensures that the motion of
the mass is along a circle of radius l. The simplest way to think about the problem is
in terms of cylindrical (or 2-D polar) coordinates defined by the angle . The
rectangular coordinates of the mass with respect to the relaxed, down position ( 0 )
are then defined as
^1 cos^ ,
sin.
y l
x l
(A. 1 )
We obtain the equation of motion most easily by considering the angular
acceleration, moment of inertia and torques with respect to the point where the
pendulum is suspended. Since the constraint force due to the suspending bar acts
through the suspension point, it provides no torque. The moment of inertia of the
mass with to the suspension point is
2 I ml , while the torque due to gravity is
mgl sin. Thus the angular version of Newton’s equation is (moment of inertia
times angular acceleration = torque)
2
sin sin 0.
g
ml mgl
l
(A. 2 )
In the small angle limit ( 1 ) this becomes the familiar harmonic oscillator equation
0,
g
l
(A. 3 )
with natural frequency 0
g l.
To obtain Eq. (A. 3 ) directly in rectangular coordinates we must go through a free-
body analysis for the mass m. There is a vertical force on the mass due to gravity,
ˆ g
F mgy
. The component of this gravity force along the direction of the bar
(denoted by
ˆ b x ˆ^ sin y ˆcos, directed away from the point of suspension) is
ˆ cos g
F b mg
. Since the mass cannot accelerate in the direction of the bar (because
the bar is rigid with length l ), the bar must supply a constraint force such that the
radial component of acceleration vanishes. Until we have solved the problem, the
only thing we know about this constraint force is its direction, i.e ., along the bar. We
can write the force applied to the mass by the bar as
ˆ bar bar
F F b
, where we
allow the magnitude of the force to vary with the orientation,^ ^ , and Fbar ^ ^0
corresponds to the bar being under tension. With this notation the rectangular version
of Newton’s equations become
(^)
^
(^)
^
2 2 2 2 2 2
ˆ : sin cos sin
ˆ sin
ˆ : 1 cos sin cos
ˆ cos.
g bar bar
g bar bar
d
x mx m l ml
dt
F F x F
d
y my m l ml
dt
F F y mg F
(A. 4 )
The complexity of this result should serve as an indication of how the correct choice
of coordinates above simplifies the problem. To simplify this system of rectangular
equations we multiply the first equation by cos and the second by sin and sum.
This procedure eliminates the contribution of the constraint force and we have
(^)
2 2
2
cos cos sin sin sin cos
cos cos sin sin cos sin
sin 0.
ml ml ml
mg mg mg mg
g
l
(A. 5 )
This component of the vector Newton’s equation, which is just the component
orthogonal to the direction of the support bar, reproduces Eq. (A. 2 ). If we take the
small angle limit we get back to Eq. (A. 3 ).
2 2
2
2
1 cos
cos.
TOT
TOT
E T V ml mgl
E mgl mgl
ml
(A. 8 )
Thus the constraint force is given by (note that it depends on the total energy)
2
cos cos
2
3cos 2.
bar TOT
TOT
F mg E mgl mgl
l
E
mg
l
(A. 9 )
This discussion is meant to suggest that as smart but lazy physicists we would really
prefer to analyze mechanics in terms of scalar quantities rather vector quantities.
This is precisely what Mr. Lagrange has done for us. The analysis due to Lagrange,
which was introduced in the Lecture and will be developed more completely in
Lecture 20, is presented entirely in terms of scalar quantities and is based on the
calculus of variations (see Chapter 9 in Boas). The underlying physics principle is
the “Principle of Least Action” (or Hamilton’s Principle). The “Action” S in physics
is defined by the time integral of the Lagrangian L ,
S dt L dt T V ,
(A. 10 )
Where (as above) T is the kinetic energy describing a system and V is the potential
energy. Typically T and V are functions of coordinates ( e.g ., of point particles) k
x
and the corresponding velocities k
x
. Application of the calculus of variations ( i.e .,
the Euler-Lagrange equation) to find the extreme value of the action leads to
Lagrange’s Equation (for the 1-D case)
0
d L L
dt x x
^
^ ^
(A. 11 )
as a “simpler” replacement for Newton’s vector equation. We have replaced having
to worry about vector components with having to worry about partial derivatives. We
can apply this formalism to the pendulum by treating ,
as the relevant coordinates
and writing
^ ^ ^
^
2 2
2
, 1 cos 1 cos
sin 0
sin 0.
m m
T l V mgl L l mgl
d
ml mgl
dt
g
l
(A. 12 )
As advertised, Lagrange does lead to the same equation of motion as Newton. Note
that we have explicitly used the constraint to make this a 1-dimensional problem (no
motion in the radial direction), just as we did in the Newton analysis at the beginning
of this Appendix. We could also keep the problem 2-D and use the method of
Lagrange multipliers to explicitly solve for the constraint force.
