# Classical mechanics

Hello all, I'm having trouble with the following problem:

Pb: A chain with constant density and infinite length is slipping down from the table without friction. Determine the position of the tip of the chain at time t.

I know there are a few ways to approaching this problem, namely from newtons equations, or lagranges equations, but I am quite rusty with this, so any suggestions would help a lot, thanks.

Hi,

do you know the Euler–Lagrange equations ?
You have to find an expression for the potential and kinetic energy,
the difference is the Lagrange-function. Put this function in the Langrange equations
and you get a second order diff.-equation.

kind regards

tiny-tim
Homework Helper
Hi johnson12! Infinite length? … presumably only in one direction? Use conservation of energy.

are you referring to this equation:

$$\frac{\partial L}{\partial x_{i}}$$ - $$\frac{d}{dt}$$$$\frac{\partial L}{\partial \dot{x_{i}}}$$ = 0, i=1,2,3

where L = T - U is the lagrange function,

how can I use this to model my problem?
(ps. im apologize if my physics is wrong, unfortunately im a math major).

are you referring to this equation:

$$\frac{\partial L}{\partial x_{i}}$$ - $$\frac{d}{dt}$$$$\frac{\partial L}{\partial \dot{x_{i}}}$$ = 0, i=1,2,3

where L = T - U is the lagrange function,

how can I use this to model my problem?
(ps. im apologize if my physics is wrong, unfortunately im a math major).

Yes, but we need only one variable x_1 = y for example. Now you have to find an expression
for the kinetic energy T which is simple and for the potential energy U which is simple
as well. U depends of course at your point of reference. Make a sketch.

I get that $$T(\dot{x}) = \frac{1}{2}m \dot{x}^{2}$$
$$U(x) = mgx$$,

$$\frac{\partial L}{\partial x}= - m g$$
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{x}}= m\ddot{x}$$
Lagranges equation implies $$m\ddot{x} + mg = 0$$

but I'm a little confused, if the chain is of infinite length, would it then have infinite mass?

so how can the chain move?