Classical mechanics

  • Thread starter johnson12
  • Start date
  • #1
18
0
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.
 

Answers and Replies

  • #2
8
0
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
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,836
251
Hi johnson12! :smile:

Infinite length? … presumably only in one direction? :wink:

Use conservation of energy.
 
  • #4
18
0
are you referring to this equation:

[tex]\frac{\partial L}{\partial x_{i}}[/tex] - [tex]\frac{d}{dt}[/tex][tex]\frac{\partial L}{\partial \dot{x_{i}}}[/tex] = 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).
 
  • #5
8
0
are you referring to this equation:

[tex]\frac{\partial L}{\partial x_{i}}[/tex] - [tex]\frac{d}{dt}[/tex][tex]\frac{\partial L}{\partial \dot{x_{i}}}[/tex] = 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.
 
  • #6
18
0
I get that [tex]T(\dot{x}) = \frac{1}{2}m \dot{x}^{2}[/tex]
[tex] U(x) = mgx [/tex],

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

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?
 

Related Threads on Classical mechanics

  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
6
Views
755
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
4K
Top