An Example of Lagrange's Method 91see also Exercise 2.3.4 below. A bead of mass m placed on the wire will slide
down under the force of gravity. Note that the two equations in (2.3.32),
describing the helix, define two constraints and leave the bead with only
one degree of freedom. We assume that the friction force is negligible, so
that the restraining force N = (N\,N2,Ns), exerted by the wire on the
bead, is orthogonal to the velocity vector at every point of the motion. Our
objective is to derive the equations of the motion of the bead along the
wire.
First, let us look at what the Newton-Euler method will produce. By
Newton's Second Law, we get three equations
mx = Nx, my — Ny, m'z = Nz — mg, (2.3.33)with the six unknowns x,y,z,Ni,N2,Ns. Equations (2.3.32) are the two
constraint equations on x, y, z. The sixth equation comes from the orthog-
onality condition for frictionless motion:
r-N = xN 1 +yN 2 + zN 3 = 0. (2.3.34)To solve this system of six equations, it is natural to parameterize the helix
by setting
x = acosq, y = bsinq, z = cq, (2.3.35)where q is the generalized coordinate for this problem.
EXERCISE 2.3.4-5 Verify that equations (2.3.32) and (2.3.35) define the same
set of points. How should one interpret the inverse cosine in (2.3.32)?
The position vector of the bead is r(q) = x(q) i + y(q) j + z(q) k, and
the motion of the bead is determined by the function q = q(t). Note
that the tangent vector to the helix is dr/dq. Using the six equation, the
unknown reaction force N can be eliminated and a single second-order
ordinary differential equation for q = q(t) can be obtained; the details of
this approach are the subject of Problem 2.9, page 421.
We now look more closely at the alternative approach using the La-
grange method. Let p = m r be the momentum of the sliding bead. Then
equation (2.3.33) in vector form becomes
p = N -mgk. (2.3.36)Denote by ptan = P • (dr/dq) and Q = —mgk • (dr/dq) the tangential
components of the momentum p and the gravitational force —mg k. This