# Homework Help: Legrangian Eqn. of Motion

1. Sep 1, 2007

### Mindscrape

Two blocks, each of mass M, are connected by an extensionless, uniform string of length l. One block is placed on a smooth horizontal surface, and the other block hangs over the side, the string passing over a frictionless pulley. Describe the motion of the system when the string has a mass m.

By Hamilton's principle
$$L = T - U$$

the kinetic energies will be
$$T = 1/2 m \dot{x}^2 + 1/2 m \dot{y}^2$$

and if the potential is defined to be zero at the horizontal, the potential will be
$$U = -Mgy + U_{string}$$

This is the part I need a quick help on. The x block has a zero potential because it stays along the horizontal where the zero potential is defined, and the hanging block will have a potential of -Mgy, and I know that the mass of the string contributing to the potential will increase until finally it reaches as the string moves down. So I was thinking that

$$U_{string} = -\frac{m}{t}*g*y$$

That gives the mass per unit time for a given length y, which would also be

$$U_{string} = -m g \dot{y}$$

But units don't work out correctly unless I divide U_string by t, which would create a discontinuity and not make any sense. I don't know why I am having so much trouble with such a simple prospect.

2. Sep 2, 2007

### Irid

Why did you use (m/t)? You should find the center of mass of the part of the string which hangs below the table. If the total mass is m, then the hanging part is
$$m'=m\frac{y}{l}$$
and mass center is in the middle (y/2). This information should give you the potential energy of the string.