In Taylor, read sections 7.3 to 7.5 for today, and 7.5 to 7.6 for Friday.

S = ∫_{1}^{2} L(q_{i}, q_{i}dot, t) dt.

∂L/∂q_{i} - (d/dt)(∂L/∂q_{i}dot) = 0

L = T - U

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, cos&phi1;) + l2(sinφ2, cosφ2).
The kinetic energy of m1 is T1 = ½m1v² = ½ m1(x1dot² + y1dot²) = ½m1l1²φ1dot².
The kinetic energy of m2 is T2 = ½m2(x2dot² + y2dot²) = ½m2l1²φ1dot² + ½m_{2}l_{2}²φ2dot².

The potential energy of m1 is U1 = -m1gy1 = -m1gl1cosφ1, and the potential energy of m2 is U2 = -m2gy2 = -m2g(l1cosφ1 + l2cosφ2).

The Lagrangian is

L = T1 + T2 - U1 - U2 = ½m_{1}l_{1}²φ_{1}dot² + ½m_{2}l_{1}²φ_{1}dot² + ½m_{2}l_{2}²φ_{2}dot² + m_{1}gl_{1}cosφ_{1} + m_{2}g(l_{1}cosφ_{1} + l_{2}cosφ_{2})

or, after rearranging terms

L = ½(m_{1} + m_{2})l_{1}²φ_{1}dot² + (m_{1} + m_{2})gl_{1}cosφ_{1} + ½m_{2}l_{2}²φ_{2}dot² + m_{2}gl_{2}cosφ_{2}

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.

Please read this section. I will not cover this in class.

An Atwood machine consists of two masses, m1 and m2, connected by a string of fixed length l strung over a frictionless pulley of radius R and moment of inertia I. Write down the Lagrangian for the system in terms of position of m1, x, find the Lagrange equation for x, and solve for the acceleration. Compare with the Newtonian solution.

If m1 is at x, then m2 is at y = l - x - πR. The angular velocity of the wheel is related to xdot by ω = xdot/R Therefore the kinetic energy is

T = ½m1x1dot² + ½m2x2dot² + ½Iω² = ½(m1 + m2 + I/R²)xdot²

The potential energy is

U = -m1gx -m2gy = -(m1 - m2)gx + const

where we can ignore the terms independent of x, xdot and t. The Lagrangian is

L = ½(m_{1} + m_{2} + I/R²)xdot² + (m_{1} - m_{2})gx

The Lagrange equation for x is

∂L/∂x = (d/dt)(∂L/∂xdot

which becomes

(m_{1} - m_{2})g = (d/dt)(m_{1} + m_{2} + I/R²)xdot = (m_{1} + m_{2} + I/R²)xddot.

This yields an acceleration of

xddot = g (m_{1} - m_{2})/(m_{1} + m_{2} + I/R²).

To solve this problem with Newton's laws we need three equations:

m1g - F1 = m1xddot

F2 - m2g = m2xddot

(F1 - F2)R = Iωdot = (I/R)xddot

Adding the first two equations and using the third to substitute for F1 - F2, yields the same equation as above. The fact that you must realize that the tension on the two sides of the pulley are different is completely avoided in the Lagrangian approach.