Next semester I’m taking the graduate classical mechanics class, and so I’m trying to spend a little time this summer brushing up on my old undergraduate mechanics. It’s been about three years since I’ve had to do much of it, and I’ve dug out my old books and am working on some problems. My undergrad class text was Thornton and Marion
A particle of mass m rests on a smooth plane. The plane is raised to an inclination angle θ at a constant rate ω (θ = 0 at t = 0), causing the particle to move down the plane. Determine the motion of the particle.
The position of the particle is uniquely determined by two coordinates: the distance up the ramp r and the angle of the ramp θ. And θ(t) is fully determined by the fact that the particle is on the ramp* and the ramp’s tilting is not affected by the particle. So really we just have to find r(t). To construct the Lagrangian we need the potential and kinetic energy of the particle. The potential energy is
The kinetic energy is just K = (1/2)mv2, but v has components in the r and θ directions. If you’ve followed this far you probably know how to write the kinetic energy in polar form, but if not here’s a brief summary. First we need the velocity in polar coordinates. This is done in considerable detail in Marion and Thornton, and in a good if dense form here. The result is that
Where the dots denote differentiation with respect to time as usual, and the hat symbols denote unit vectors. Kinetic energy is proportional to v2 so we need to square the above expression:
And the dot products for the unit vectors are just 1 if the unit vectors are identical and 0 otherwise because the vectors are orthogonal. This means we finally have
Ok. Subtract the potential energy from the kinetic energy and we have the Lagrangian:
Now we need to find the partial derivatives with respect to r and r-dot so we have something to put in the differential equation. It’s pretty straightforward:
and
Which we can plug straight into the Euler-Lagrange equation for r.
leading to
So after taking the derivitive of r-dot, canceling the m’s, moving the rightmost term over, and casting θ in terms of ω as instructed in the problem statement, we have
Which is a fairly straightforward differential equation. You can find the homogeneous solutions and then get the particular solution using Green’s functions. The final result with an initial position r0 and initial velocity 0 is:
A lot more complicated than the stationary inclined plane, no? But it’s the right answer, and it was found by a reasonably straightforward procedure. Attempting this problem with forces would have been nightmarish. Just for fun, let’s graph this with g = 9.8, r = 10, and ω = 0.2 in the appropriate units.
Looks about like we expect. The particle starts off slow and picks up speed as the ramp angle is increased. Problem done!
*If this ceases to be true our assumptions about θ(t) are no longer valid and the solution method fails. This corresponds to the ball falling off as the ramp angle goes about 90 degrees. This need not happen if the angular rotation is fast enough, and we could use undetermined multipliers to actually find the minimum required angular velocity, the constraint forces (Lagrange’s method can be used to find the forces, just as the forces can be integrated through distance to find energy), etc. This is a somewhat more difficult problem.
3 responses so far ↓
1 Carl Brannen // Jun 20, 2008 at 11:16 pm
Here it is Friday and I feel like making trouble when I should do something useful. But I’m stuck in Moses Lake and there’s not a lot to do here. And I’ll see if you’ve got LaTeX implemented here…
Let the particle mass be m. At time t, it is in an environment where the slope that it is on has a steepness of $latex \omega t$, so the net force due to gravity along the ramp is just $latex - m g \sin(omega t)$.
The movement of the ramp also causes an acceleration of the mass. Think of a situation where there is no gravity. The mass would be accelerated outward with a centripetal acceleration. The formula for centripetal force is $latex m r \omega^2$. Therefore, the total force on the particle is:
$latex
F = m \ddot{r} = m r \omega^2 - m g \sin(\omega t),
$
which is what you get after quite some work. As far as solving the above, no thanks.
2 Carl Brannen // Jun 21, 2008 at 7:20 pm
Force is a great way of analyzing problems if you can avoid having to deal with constraints. This is a good example of that sort of thing.
To get an example where Lagrangian formalism really does help things, look for a situation where you can’t choose coordinates that split the motion of the particle into constrained and unconstrained motion. In this case, radius and angle split the problem beautifully. By the way, this is a secret that can be useful on the physics qualifying exams (or whatever they call them).
Physics is also taught as faith, and the professors have the duty to impart the faith as best describes the current belief. So, just like in church, the faithful have a tendency to pile up evidence even when there is no true evidence, just a lack of evidence against. This is one of those cases.
By and large, Lagrangian and Hamiltonian formalism is preferred because it works in great generality, and it is necessary for how the foundations of physics (especially particles and quantum mechanics which is my favorite branch) are written.
But it is not the case that the foundations can be written in only that way. In fact, elementary particles are the best example in the world of situations where it is not necessary to deal with forces of constraint. One can describe the elementary particles by writing down their Hamiltonian, and then derive the Feynman diagrams, or one can, alternatively, write down their Feynman diagrams directly. I prefer the second way, though I am pretty much alone in this.
When you build things up from the equations of motion rather than the Hamiltonian / Lagrangian formalism, it gives you a different sense of what “simple” and “elegant” means. I think this is a good reason to explore the methods. I should write a blog post on what this has to do with the masses of elementary particles.
My favorite classical mechanics problem has to do with drilling a hole near the edge of a coin. When you spin it, what positions of the hole are stable? Bottom? Side? Top? None? As I later owned a pinball business, I ended up counting enough quarters that I came upon some with holes drilled in them and could verify that theory was correct.
3 Chris Hertlein // Jun 23, 2008 at 5:30 pm
On our Classical Mechanics final, we had to solve the double pendulum (for small angles to save paper) using the Euler-Lagrange stuff. Not bad, but it took pages and pages of bookkeeping and was generally annoying, as I wanted to be sure I was correct and combed through my algebra making sure I didn’t accidentally drop something.
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