Dot Physics

Mechanics an Example of the Pendulum

This post has been sitting in my mind for quite some time. Really, it is about mechanics – not about pendulums. What is the goal in mechanics (classical mechanics, if you like)? Generally, it is to find out how something changes over time. If you could get an equation of motion, that would do it.

As Matt (Built on Facts) did a while ago, it can be shown that you can get the equation of motion for a mass on a spring with normal Newtonian mechanics or with Lagrangian mechanics. Let me summarize two different ways of looking at the motion of an object.

The Newtonian Way

Maybe that isn’t the best name for it, but here is the basic idea. Find all the forces acting on an object and then use the momentum principle.

i-03125cd383771a8a9fde8d61a08b6584-2009-12-03_la_te_xi_t_1_12.jpg

So, if you know how the momentum changes you can find some way to find the position of the thing. In this method, you could break forces into two kinds:

  • Forces that you can calculate right away.
  • Forces that do whatever they can to constrain an object.

Let me show two examples. First – a planet orbiting a star. Here is a diagram (simplified)

i-e356567cbc649c613b3706533558a103-2009-12-03_untitled_3.jpg

This is an example of forces you can calculate right away. The gravitational force depends on the position of the two objects, so there is no problem. What about another seemingly simple case, a block sliding down an inclined plane.

i-700a1a95ecd074c51322719dd8c35200-2009-12-03_untitled_4.jpg

Again, the gravitational force is not a problem. It is the Fsurface that is the problem. How do you calculate this force? You have to use some tricks. Basically, Fsurface is whatever it needs to be in order to keep the block from going into the inclined plane. One way to do this is to say that the acceleration of the block perpendicular to the plane is zero. This would give a magnitude of the surface force as:

i-eff7dd9cf462abb78fccd9d0e9cad890-2009-12-03_la_te_xi_t_1_13.jpg

Where theta is the incline of the plane. In the Newtonian way, it is these forces of constraint that can be the real problem. The above example is simple, but what about a block sliding down a circular path (like a skate boarder in a half-track)? In this case, this force of constraint is not constant. You can do such a problem the Newtonian way, but it can get messy.

Lagrangian – the constraint way

In the Lagrangian way, you can pick some variables that describe the object – really these variables can be anything. The Lagrangian is then:

i-6193cc4c0411e63694d323dbe15ea864-2009-12-03_la_te_xi_t_1_14.jpg

Where T is the ‘kinetic energy’ and V is the ‘potential’. Those are in quotes because it is possible to choose variables that describe the system such that the T is not actually the kinetic energy. Anyway, the point is that the path of motion is such that the Lagrangian is a minimum along this path. I know that is complicated – but if you want to explore this more, check out Edwin Taylor site www.eftaylor.com/software/ActionApplets/LeastAction.html.

In the end, really the Lagrangian way gets you essentially the same equation of motion that you would get from the Newtonian way.

Pendulum Example – Newtonian

Here I will briefly show how to use these two methods for a pendulum. I am skipping over a lot of the Lagrangian details because it can get tricky – and anyway, it is not my main point (as you will soon see). So, suppose I have a mass m at the end of a string of length a. Finally, suppose I release it from rest at some initial angle. Here is a diagram.

i-d0f0d537efcbeb4e802f2e67755a1b75-2009-12-03_untitled_5.jpg

In the Newtonian way, the goal is to get a relationship between acceleration and position – or something close. If you approach this from the typical starting point of finding the forces, it gets complicated. What is an expression for the tension in the string? The difficult thing is that this force is not just whatever it needs to be to make the acceleration in that direction zero (like it was for the inclined plane) because it is accelerating that way (circular motion).

Here is the trick. Think polar coordinates. In polar coordinates, the mass can only accelerate in the direction of theta. This means that I only need to worry about forces in the theta direction. Here is a diagram of the pendulum at a certain instant. I have also drawn my axes (that move):

i-1dd9e3e305b0d337367d97fa823eb716-2009-12-04_untitled_6.jpg

Since the mass can only move in the theta direction, here is the newtonian equation in the theta direction:

i-ea7f89cabbe40e90255451769cf143d3-2009-12-04_la_te_xi_t_1_1.jpg

Here I have used the usual convention of double-dots to represent the second derivative with respect to time. Theta-double-dot is the angular acceleration. Needless to say, this is the answer. If you want you could do some more tricks – like only consider small theta.

Pendulum Example – Lagrangian

The first step in using the Lagrangian is to choose a coordinate that can represent the situation. In this case, it can only move one way, so theta will work. Now I need the kinetic energy and the potential in terms of theta and its time derivatives.

i-abdef7dad98e5f5ce0b68f6f577dd4aa-2009-12-04_la_te_xi_t_1_3.jpg

I just realized that I have been using different things to represent the length of the pendulum. Oh well – I am going to keep going. If you put this into Lagrange’s equation, you will see that you get the exact same equation as with the Newtonian way.

Ok, this was way longer than I wanted it to be. I am going to put the rest in Part II. Just as a hint, in part II I am going to do this one more way.

Update:

There was a typo – as pointed out by Paul (see comments). I fixed it.

Comments

  1. #1 Paul Orwin
    December 4, 2009

    It has been a long time since I did this type of physics, but how come in your example of the Lagrangian method T=1/2 m (theta double dot)^2 instead of (theta dot)^2? shouldn’t kinetic energy relate to velocity, not acceleration? Or am i confused?

  2. #2 Rhett Allain
    December 4, 2009

    @Paul,

    Because it is a typo – you are correct. I will fix that.

  3. #3 Paul Orwin
    December 5, 2009

    Yay! My frosh physics profs would be so proud :)

  4. #4 sikiƟ
    December 5, 2009

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