2010 PDE2 Classical Mechanics

From TorontoMathWiki

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 $L:\mathbb{R}^n \times \mathbb{R} \times \overline{U} \to \mathbb{R}$ be smooth. $L$ is called a Lagrangian.

Let

I[w] =	∫	L(Dw(x),w(x),x)dx
	U

for smooth functions $w:\overline{U} \to \mathbb{R}$ satisfying $w = g$ on $\partial U$ . It is not hard to show that if $w$ is a minimizer of $I[\cdot]$ then it satisfies the Euler-Lagrange equation:

$-\sum_{i=1}^n (L_{p_i}(Du,u,x))_{x_i}+L_z(Du,u,x)=0$

in $U$ .

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 $F = - k x$ . 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 $x$ :

$U=\int_0^x -F dt=\int_0^x kt dt=\frac{kx^2}{2}.$

The kinetic energy $T$ of the system is the work needed to accelerate the system from rest to its present velocity. We have:

$T=\int Fdt = \int ma dt = \int m \frac{dv}{dt} dx = \int m \frac{dv}{dx} \frac{dx}{dt} dx= m \int v dv = m \frac{v^2}{2}.$

I'm confused why $U$ and $T$ are both defined as integrals of the force.

Thus the Lagrangian of the system is:

$L(x,\dot{x})=T-U=\frac{1}{2}\Big(m\dot{x}^2-kx^2\Big)$ .

The partial derivatives of $L$ are:

$\frac{\partial L}{\partial x}=-kx \quad \frac{\partial L}{\partial \dot{x}}=m\dot{x}$ .

The Euler-Lagrange equation is:

$\frac{\partial L}{\partial x}=\frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{x}}\Big)$ .

Hence $-kx=m\ddot{x}$ .

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 $ω 0$ . 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 $\varphi$ be measured from the positive x-axis. Since the circle is rotating with constant angular speed $ω 0$ , $\varphi=\omega_0 t$ . Then $x(\theta,t)=R\sin\theta \cos(\varphi)$ , $y(\theta,t)=R\sin\theta \sin(\varphi)$ , and $z (θ, t) = - R cosθ$ .

Taking the time derivatives we get $\dot{x}(\theta,\dot{\theta},t)=R(\dot{\theta}\cos\theta \cos\varphi-\omega_0 \sin\theta \sin\varphi)$ , $\dot{y}(\theta,\dot{\theta},t)=R(\dot{\theta}\cos\theta \sin\varphi+\omega_0\sin\theta \cos\varphi)$ , $\dot{z}(\theta,\dot{\theta},t)=R\dot{\theta}\sin\theta$ .

Adding the squares of the velocity components and multiplying by m/2, we get that the kinetic energy of the point is $K(\theta,\dot{\theta})=\frac{mR^2}{2}(\dot{\theta}^2+\omega_0^2\sin^2 \theta)$ .

Suppose that the gravitational potential energy is zero when $θ = 0$ . Then the gravitational potential energy is $U (θ) = m g R (1 - cosθ)$ . This was found by setting $\nabla U=mg$ . Abusing notation, $g = (0,0, g)$ . So $U (θ) = m g z + b = - m g R cosθ + b$ . We assumed that $U (0) = 0$ , hence $b = m g R$ .

Therefore the Lagrangian $L = K - U$ for the point is $L(\theta,\dot{\theta})=\frac{mR^2}{2}(\dot{\theta}^2+\omega_0^2\sin^2 \theta)-mgR(1-\cos\theta)$ .

Putting this into the Euler-Lagrange equation gives $\ddot{\theta}+\sin\theta(\frac{g}{R}-\omega_0^2\cos\theta)=0$ . This gives us a differential equation for $θ$ . Since we already know that the particle has distance R from the origin and azimuthal angle $\varphi=\omega_0t$ , we understand completely the particles motion.

2010 PDE2 Classical Mechanics

From TorontoMathWiki

Rolling bead on a rotating hoop

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