In Taylor, read sections 7.2 to 7.3 for today, 7.4 to 7.5 for Friday.
Lagrange's equations:
where
We looked at an example applying Lagrange's equation to a simple, unconstrained, 1-`dimensional problem. Now we will investigate the use of Lagrange's equation with constrained systems.
One of the strengths of Lagrangian mechanics is the ease with which constrained systems are handled. By constrained systems, I mean systems whose motion is restricted, not simply restricted to a line or a plane, but to any arbitrary one dimensional path or two dimensional surface. For instance, a bead that is free to slide along a wire bent to an arbitrary shape, or a mass confined to move on an arbitrary surface. One need simply express the kinetic and potential energy in terms of appropriate variables; one variable in the case of one-dimensional motion, and two variables in the case of two-dimensional motion.
As a simple example, write down the Lagrangian for a simple pendulum of mass m and length l and solve for the motion.
The position of the pendulum bob can be described by it's (x,y) position, but since the length l is constant, the bob's motion is restricted to a one-dimensional path, and we can express its location in terms of the angle φ with x = l sinφ and y = l ( 1 - cosφ ). In terms of φ the kinetic energy is T = ½ m l²φdot², and the gravitational potential energy is U = mgl ( 1 - cosφ ). The tension doesn't appear in the Lagrangian approach; it is a constraint force that keeps the bob on its proper path, but doesn't contribute to the kinetic or potential energies. The Lagrangian for the pendulum is
This Lagrangian has only one generalized coordinate, φ, so there is one equation:
This evaluates to
This is the familiar result for a pendulum. If we make the small angle approximation, sinφ ≈ φ, the equation predicts simple harmonic motion with angular frequency ω = √(g/l).
One of the great advantages of Lagrangian mechanics is the ability to use virtually anything as a coordinate for describing the motion. In the example of the simple pendulum, we used the angle φ of the pendulum to the vertical as the coordinate. Let's look at a more complex example, a double pendulum.
A double pendulum is made by suspending a mass m1 from a fixed point by a massless rod of length l1, and suspending a mass m2 from m1 by a massless rod of length l2. Both rods are free to pivot in the x--y plane.
We can express the positions of the two masses in terms of just two coordinates, φ1 and φ2, the angles of the rods l1 and l2 with respect to the vertical. Measuring y vertically downward from the point of attachement, the position of m1 is (x1, y1) = l1(sinφ1, cosφ1) and the position of m2 is (x2, y2) = l1(sinφ1, cosφ1) + l2(sinφ2, cosφ2). The kinetic energy of m1 is T1 = ½ m1 v² = ½ m1 (x1dot² + y1dot²) = ½ m1l1² φ1dot². The kinetic energy of m2 is T2 = ½m2(x2dot² + y2dot²) = ½m2l1²φ1dot² + ½m2l2²φ2dot².
The potential energy of m1 is U1 = -m1gy1 = -m1gl1cosφ1, and the potential energy of m2 is U2 = -m2gy2 = -m2g(l1cosφ1 + l2cosφ2).
The Lagrangian is
or, after rearranging terms
The number of degrees of freedom for a system is the minimum number of coordinates that must be specified to describe the state of the system. Or, stated slightly differently, the number of coordinates that can be independently varied. Notice that these aren't necessarily the same as the number of generalized coordinates.
Systems where the number of degrees of freedom is equal to the number of generalized coordinates are called holonomic. The problems we will encounter in this course are holonomic.
An example of a non-holonomic system is a billiard ball, where the orientation of the ball is important. The billiard ball can be placed initially with a mark positioned at the top of the ball, and after being rolled across the table and returned to its original position, the mark may not be at the top. To completely describe the position of the ball requires 5 coordinates, 2 positions and 3 rotation angles. But there are only 2 degrees of freedom -- it can be rolled in the E-W or N-S directions, independently.
The derivatives of the Lagrangian are known as a generalized force (∂L/∂qi = Fi) and a generalized momentum (∂L/∂qidot = pi). When thought of this way, Lagrange's equations can be expressed in the form
an expression that looks deceivingly like Newton's second law. But notice that the generalized momentum and force can be completely different from a traditional linear momentum and force in Newtonian mechanics. They are just the appropriate derivatives of the Lagrangian. For example the Lagrange equation in φ yielded a torque for the generalized force, and an angular momentum for the generalized momentum.