2010 PDE2 Classical Mechanics
These are notes on classical mechanics as discussed in 2010S MAT1061HS PDE2.
The following gives further details to pp. 34-36 of File:Colliander279Notes.pdf.
Let be smooth. L is called a Lagrangian.
for smooth functions satisfying w = g on . It is not hard to show that if w is a minimizer of then it satisfies the Euler-Lagrange equation:
This is shown on pp. 431-433 of Evans. The only part of the proof I had trouble with was the use of the chain rule to determine i'(τ) on p. 433.
- What problem did you have? Colliand 15:46, 23 February 2010 (UTC)
If we have a point at the end of a spring where x = 0 is the rest position of the particle, for the spring to obey Hooke's law means that there is a constant k such that the force F at the point is F = − kx. The potential energy U of the system is the amount of work that must be done to bring the system from the equilibrium position to its present position x:
The kinetic energy T of the system is the work needed to accelerate the system from rest to its present velocity. We have:
I'm confused why U and T are both defined as integrals of the force.
Thus the Lagrangian of the system is:
The partial derivatives of L are:
The Euler-Lagrange equation is:
File:Pde1.pdf is the presentation on the Legendre transform for the March 2 class.
Rolling bead on a rotating hoop
Take a circle of radius R in three space such that one diameter is a subset of the z-axis and another diameter is a subset of the x-axis. Suppose that the circle is rotating about the z-axis with constant angular speed ω0. We consider the motion of a point of mass m that is sliding without friction on the circle. There is a constant gravitational field with gravitational acceleration g.
As the circle rotates and the point moves along the circle, the point is constrained to lie on the surface of a sphere of radius R. Therefore we can describe the position of the point with spherical coordinates. Let θ be measured from the negative z-axis and be measured from the positive x-axis. Since the circle is rotating with constant angular speed ω0, . Then , , and z(θ,t) = − Rcosθ.
Taking the time derivatives we get , , .
Adding the squares of the velocity components and multiplying by m/2, we get that the kinetic energy of the point is .
Suppose that the gravitational potential energy is zero when θ = 0. Then the gravitational potential energy is U(θ) = mgR(1 − cosθ). This was found by setting . Abusing notation, g = (0,0,g). So U(θ) = mgz + b = − mgRcosθ + b. We assumed that U(0) = 0, hence b = mgR.
Therefore the Lagrangian L = K − U for the point is .
Putting this into the Euler-Lagrange equation gives . This gives us a differential equation for θ. Since we already know that the particle has distance R from the origin and azimuthal angle , we understand completely the particles motion.