# Lagrangian, mass attached to spring on plane

1. May 15, 2010

### leonidas24

1. The problem statement, all variables and given/known data
A block of mass m moves on a horizontal, frictionless table. It is connected to the centre of the table by a massless spring, which exerts a restoring force F obeying a nonlinear version of Hooke's law,

$$F = -kr + ar^3$$

where r is the length of the spring. Show that the maximum energy for which the block remains bound to the centre of the table is approximately

$$E \approx \frac{k^2}{4a}$$

if a is small and positive. Draw a diagram to support your answer.

2. Relevant equations
Lagrangian:
$$L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2) - \frac{kr^2}{2} + \frac{ar^4}{4}$$

Energy for radial coordinate:
$$E = \frac{m\dot{r}^2}{2}+ \frac{J^2}{2mr^2}+ \frac{kr^2}{2} - \frac{ar^4}{4}$$

Energy and angular momentum $$J$$ are contants of motion.

3. The attempt at a solution
I'm really not sure how to tackle this one. Just a nudge in the right direction is all I'm after.

Last edited: May 15, 2010
2. May 16, 2010

### daschaich

Here's a nudge: since a is positive, at large distances where a|r|^3 > k|r|, the force pushes the block away from the center of the table, instead of dragging it back! (Weird spring!)

My question: does the problem say how the block is moving on the table? That is, do we know that the block is revolving around the center of the table, or could it be moving radially? I ask because if it's just moving radially, things are pretty simple.

3. May 16, 2010

### diazona

Well, my first thought upon seeing "small and positive" is perturbation theory...

...although on second thought, I'm not sure that's what the problem is getting at. Think about it like an orbit: when you express an orbiting body's energy in terms of angular momentum (as you did), you can make a plot of the effective potential as a function of radius and visually identify the characteristics of open and closed orbits. That should give you a clue as to where to go with the math.

4. May 16, 2010

### daschaich

I would recommend checking the simple J=0 case before worrying about orbits. After all, J doesn't seem to appear in the problem statement itself.

5. May 16, 2010

### diazona

No, but it does emerge as a conserved quantity corresponding to the φ-independence of the Lagrangian. That's part of the standard procedure for solving orbital motion problems. There isn't anything in this problem to indicate that the motion is purely radial. If it were, I'd expect it to say so explicitly, and then it would be a significantly easier problem (easier than you'd expect when studying Lagrangian mechancs).

6. May 16, 2010

### colmane

Does Emax even exist? E (total energy) should be a const .If Emax does exist is there a Emin?

7. May 16, 2010

### diazona

Total energy is a constant of the motion, sure, which means that whatever its initial value is, it doesn't change as the block moves. But the total energy of the.system can be set to various values (e.g. depending on how fast the block is moving when it starts out). If it starts with too much energy, it won't remain bound to the center of the table and will fly off. The objective of the problem is to find out just how much energy is required for that to happen, at least approximately.

8. May 16, 2010

### colmane

but how can the block flying out of the table? no critical point of spring provided.

9. May 16, 2010

### diazona

Sure there is, it comes from the force law for the spring.

10. May 16, 2010

### daschaich

Yes, the J=0 case is very easy. Since there's nothing in the problem statement to indicate that the use of the lagrangian is required, and given that the OP is a freshman, I wonder if we're making the question harder than it is meant to be.

Besides, the J=0 case immediately produces the requested result, starting from which shows that the J^2 term in the energy is suppressed by a, which is small.

Colmane, note what I said earlier: since a is positive, at large distances the spring pushes the block away from the center of the table, instead of dragging it back! (Weird spring!)

Last edited: May 16, 2010
11. May 16, 2010

### diazona

But not every problem which requires the use of the Lagrangian explicitly says so. Since leonidas24 started doing this with the Lagrangian, I assumed (justifiably, I think) that this is for a class in classical mechanics.

And sure, setting J=0 leads to showing that E=k²/4a exactly, but the problem did specify that you should find E≈k²/4a, i.e. inexactly. And besides, I don't see how the conclusion you draw at J=0 applies to arbitrary (large) J.

12. May 16, 2010

### daschaich

The J=0 result is an approximation to the full solution because it neglects possible angular motion. I would expect J to be bound from above by the finite restoring force, but I haven't worked through such corrections in detail, other than to note that they should be suppressed by a.

Hopefully it is illuminating (and apparently sufficient) to solve the simple case first. leonidas24, can you let us know if you've gotten that far, and whether further refinements are needed?

13. May 17, 2010

### colmane

can anyone explain why J=0 case is so important??? what happens in J=0 case?

14. May 17, 2010

### daschaich

The problem gets very, very easy.