# Homework Help: Spring-Pendulum System

1. Sep 20, 2009

### Pengwuino

1. The problem statement, all variables and given/known data

Jackson 2.26. A particle of mass m is suspended by a massless spring of length L. It hangs, without initial motion, in a gravitational field of strength g. It is struck by an impulsive horizontal blow, which introduces an angular velocity w. If $$\omega$$ is sufficiently small, it is obvious that the mass moves as a simple pendulum. If $$\omega$$ is sufficiently large, the mass will rotate bout the support. Use a Lagrange multiplier to determine the conditions under which the string becomes slack at some point in the motion.

2. Relevant equations

In plane polar coordinates, the Lagrangian is

$$L = \frac{1}{2}m(\dot r^2 + r^2 \dot \theta ^2 ) + mgr\cos (\theta ) - \frac{1}{2}k(r - r_0 )^2$$

where $$r_0$$ is the unstretched length of the system.

There's a few things about this problem that I do not understand. When Jackson says that if $$\omega$$ is large enough, it will rotate about the support, he doesn't mean that the problem will become 3-D correct? Also, the constraint to this problem is something I can't figure out for the life of me. $$\theta$$ is not constrained and I can't imagine how r could be constrained so this problem has me stumped. What might the constraint be? I want to say $$r = r_0 + L$$ (L being the length of the spring) but that doesn't make any sense...

2. Sep 20, 2009

### kuruman

I don't have a copy of Jackson to check, but first you mention "spring" then in the end you say "string". If it is "spring", then "becomes slack" is meaningless (unless Jackson means "the spring is unstrained"). If it is "string", then "becomes slack" acquires meaning and the constraint should obviously be r = constant assuming that the string is inextensible.

3. Sep 20, 2009

### Pengwuino

Yes it's the string. We've kind of dismissed the idea of r = constant considering since the spring is in the system, r, the position of the mass, can't be constant. Given any angular velocity, the spring will stretch and the radial position of the mass will change. I know the constraint has to due with the radial length, however, considering $$\lambda \nabla \sigma$$ has to give us back the tension on the string for the external force. Figuring out what it is is stumping me right now. I'm really wondering why they gave us the length of the spring. What good does that do us considering, well, springs stretch.

4. Sep 27, 2009

### Pengwuino

Ok apparently it's a typo in the book and it's suppose to be a string, not a spring. Case closed and done :).

5. Oct 5, 2009

### nothingislost

so was the constraint that r is constant? How does the impulse force enter the equation?