The state of a system of n particles & subject to m constraints connecting some of the 3n rectangular coordinates is completely specified by s = 3n – m generalized coordinates.  Sometimes its convenient to represent the state of such a system by a point in an abstract s-dimensional space called CONFIGURATION SPACE. Each dimension in this space corresponds to one of the coordinates q j. This point specifies the CONFIGURATION of the system at a particular time. As the q j change in time (governed by the eqtns of motion) this point traces out a curve in configuration space. The exact curve depends on the initial conditions. Often we speak of “the path” of the system as it “moves” in configuration space. Obviously, this is NOT the same as the particle path as it moves in ordinary 3d space! Configuration Space

Lagrange’s Equations in Generalized Coordinates Section 7.4 In light of the preceding discussion we can now restate Hamilton’s Principle in slightly different language: Of all of the possible paths along which a dynamical system may move from one point to another in configuration space within a specified time interval, the ACTUAL PATH followed is that which minimizes the time integral of the Lagrangian for the system.

Repeat Hamilton’s Principle & Lagrange Equations derivation in terms of generalized coordinates: Use the fact that the Lagrangian L is a SCALAR.  L is invariant under coordinate transformation (Ch. 1). –Also allowed: Transformations which change L but leave the equations of motion unchanged. For example it can easily be shown that if L  L + (d[f(q j,t)]/dt), this obviously changes L but leaves Lagrange Equations of motion unchanged. (Valid for ANY f(q j,t)!). Always use the definition L = T - U. This is valid in a proper set of generalized coordinates. L can also contain an arbitrary constant. This comes from the arbitrary constant in PE U.

 It doesn’t matter if we express L in terms of rectangular coordinates & velocities x α,i & x α,i or generalized coordinates & velocities q j & q j. L = T(x α,i ) - U(x α,i ) (1) The transformation between rectangular & generalized coordinates & velocities is of the form: x α,i = x α,i (q j,t); x α,i = x α,i (q j,q j,t) So, we also have L = T(q j,q j,t) - U(q j,t) (2) The Lagrangians in (1) & (2) are the same! From (2), in generalized coordinates we have L  L(q j,q j,t)  Note!

Hamilton’s Principle in generalized coordinates:  δ∫L(q j,q j,t) dt = 0 (limits t 1 < t < t 2 ) This is again identical to the abstract calculus of variations problem of Ch. 6 with the replacements: δJ  δ∫Ldt, x  t, y i (x)  q j (t) y i (x)  (dq j (t)/dt) = q j (t) f[y i (x),y i (x);x]  L(q j,q j,t)  The Lagrangian L satisfies Euler’s equations with these replacements!

These become Lagrange’s Equations in generalized coordinates: (  L/  q j ) - (d/dt)[(  L/  q j )] = 0 (A) For a system with s degrees of freedom and m constraints, there are j = 1,2,..,s coupled equations like (A) + m constraint equations.  Together, these s + m equations give a complete description of the system motion!

Note that the validity of Lagrange’s Equations requires that: –All forces (other than constraint forces) must be conservative (that is a PE, U must exist). However, in graduate mechanics, it is shown how to generalize the Lagrangian formalism to dissipative (non-conservative) forces! –The equations of constraint must be relations that connect the coordinates of particles & they may be functions of time: f k (x α,i,t) = 0, k = 1,2,.. m (B) Constraints which may be expressed in the form (B)  Holonomic Constraints. If the constraint equations, (B), do not explicitly contain the time  Fixed or Scleronomic Constraints. Time dependent constraints  Rhenomic Constraints. Here, we only consider conservative forces & holonomic constraints.

Projectile motion (with no air resistance). Find the equations of motion in both Cartesian & polar coordinates. Cartesian: x = horizontal, y = vertical T = (½)mx 2 + (½)my 2, U = mgy  L = T - U = (½)m(x 2 + y 2 ) - mgy Lagrange’s Equations: (  L/  x) - (d/dt)[(  L/  x)] = 0, (1) (  L/  x) = 0, (  L/  x) = mx, (d/dt)[(  L/  x)] = mx, (1)  x = 0 (  L/  y) - (d/dt)[(  L/  y)] = 0, (2). (  L/  y) = -mg, (  L/  y) = my (d/dt)[(  L/  y)] = my. (2)  mg + my = 0; y = -g Simple Example 7.3

Polar: x = r cosθ, y = r sinθ r 2 = x 2 + y 2 ; tanθ = (y/x) T = (½)mr 2 + (½)mr 2 θ 2 U = mgr sinθ  L = T- U = (½)m(r 2 + r 2 θ 2 ) - mgr sinθ Lagrange’s Eqtns: (  L/  r) - (d/dt)[(  L/  r)] = 0, (3) (  L/  r) = mrθ 2 – mg sinθ, (  L/  r) = mr, (d/dt)[(  L/  r)] = mr (3)  rθ 2 – g sinθ - r = 0 (  L/  θ) - (d/dt)[(  L/  θ)] = 0, (4) (  L/  θ) = - mgr cosθ, (  L/  θ) = mr 2 θ (d/dt)[(  L/  θ)] = m(2rr θ + r 2 θ) (4)  gr cosθ + 2rr θ + r 2 θ = 0 Clearly, its simpler using Cartesian coordinates as the generalized coordinates!

A particle, mass m, is constrained to move on the inside surface of a smooth cone of half angle α (figure). It is subject to gravity. Determine a set of generalized coordinates & determine the constraints. Find Lagrange’s Equations of motion. Worked on the board!! Example Non-trivial! Constraint: z = r cotα  m m

See figure. The point of support of a simple pendulum of length b moves on a massless rim of radius a, rotating with a constant angular velocity ω. Obtain the expressions for the Cartesian components of the velocity and the acceleration of the mass m. Obtain also the angular acceleration for the angle θ shown in the figure. Worked on the board! Example 7.5 m (x,y) 

See figure. Find the frequency of small oscillations of a simple pendulum placed in a railroad car that has a constant acceleration a in the x-direction. Worked on the board!! Note: the railroad car is not an inertial frame! Example 7.6 gg

A bead slides along a smooth wire bent in the shape of a parabola, z = cr 2 as in the figure. The bead rotates in a circle or radius R when the wire is rotating about its vertical symmetry axis with angular velocity ω. Find the value of c. Worked on the board! Example 7.7

Consider the double pulley system shown in the figure. Use the coordinates indicated and determine the equations of motion. Worked on the board! Example 7.8