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Finding the Lagrangian of a bead sliding along a wire

  • Thread starter Rumor
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  • #1
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Homework Statement


"A bead with mass m slides without friction on a wire which lies in a vertical plane near the earth. The wire lies in the x-z plane and is bent into a shape conforming to the parabola az = x2, where a is a positive known constant. (X is horizontal and z is vertical) The particle moves in the x-z plane but its motion is constrained by the requirement that it remains on the wire.

Find the kinetic energy in terms of m, a, x, x-dot. (Eliminate z and z-dot using the constraint)"

Homework Equations


z = x2/a and z-dot = (x-dot)2/a
(I think these are the restraint equations, unless I did these wrong.)


The Attempt at a Solution



I know that K = (1/2)mv2, so in this case wouldn't that translate to K = (1/2)*m*((x-dot)2 + (z-dot)2)?

If so, then continuing on, K = (1/2)*m*(x-dot)2+((x-dot)2/a)2) or K = (1/2)*m*(x-dot)2+[(x-dot)4/a2]

Am I doing this right?
 

Answers and Replies

  • #2
rude man
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1.

Homework Equations


z = x2/a and z-dot = (x-dot)2/a
(I think these are the restraint equations, unless I did these wrong.)


z-dot = what now?
 
  • #3
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It's not specified within the problem given. I just rearranged the given equation ( az = x^2 ) and applied it to z-dot as well, since part of the problem later requires eliminating z and z-dot.
 
  • #4
rude man
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Well, I would have written z-dot = 2x(x-dot)/a.

Would you like to see how I got that? Might be a chance for you to make a fool out of me. It happens often enough! :-)
 

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