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1-D Lagrange and Hamilton equation gives different results.

  • #1
44
5

Homework Statement


This was supposed to be an easy question. I have a question here that wants you to describe a yoyo's acceleration (in one dimension) using Lagrangian mechanics. I did and got the right answer. Now I want to use Hamilton's equations of motion but I get a wrong number. Here is the excerpt

"A yoyo attached to a mass less string is suspended vertically from a fixed point and the other end is wrapped several times around a uniform cylinder of mass, m and radius R. When the cylinder is released it moves vertically down, rotating as the string unwinds. Write down the Lagrangian equation using the distance x as your generalized coordinate. Find the Lagrange equation of motion and show that the cylinder accelerates downward with #\ddot{x}# = 2g/3"

Homework Equations


I used Hamilton equation xdot = partial H/partial P
To help,
L = (3/4) m xdot^2 + mgx
and
H = (3p^2)/(4m) - mgx

The Attempt at a Solution


[/B]
I used Hamilton's equation and took the derivative
d/dt (xdot) = xddot = d/dt(3p/4m)
xddot = (3 xddot)/2
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,405
2,841
The momentum, ##p##, that enters into the Hamiltonian should be the "canonical momentum" which is determined from the Lagrangian. You should find that ## p \neq m \dot{x}##.
 
  • #3
44
5
The momentum, ##p##, that enters into the Hamiltonian should be the "canonical momentum" which is determined from the Lagrangian. You should find that ## p \neq m \dot{x}##.
I get ##p = \partial L / \partial \dot{q} = (6/4) m \dot{x}## but I dont know how/where to replace this ##p## into. Back into L and then re equate the Hamiltonian then use the Hamiltonian equation?
 
  • #4
TSny
Homework Helper
Gold Member
12,405
2,841
OK for your result for ##p##.

I don't understand what you mean when you say "re equate the Hamiltonian". There is a specific prescription for constructing the Hamiltonian from the Lagrangian and the canonical momentum.
 
  • #5
44
5
OK for your result for ##p##.

I don't understand what you mean when you say "re equate the Hamiltonian". There is a specific prescription for constructing the Hamiltonian from the Lagrangian and the canonical momentum.
We im stuck at ##p = (6/4)m \dot{x}## do I take this result and somehow put ##p## back into the lagrange?
 

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