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Lagrangian Mechanics

  • Thread starter Gogsey
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1). A bead is confined to moving on a wire in the shape of a porabola, given by y=bx^2. Write down the Lagrangian, with x as the generalized coordinate, and the equations of motion for this sytem.

We have L(x, bx^2)
For writing out the Lagrangian as a function of x, I get.:

L = m/2((xdot) + b(xdot0)^2 - mgbx^2

Then we get L = m/2((xdot^2) + 2b(xdot^2) +(b^2)(xdot^4)) - mgbx^2

But when I go to take tthe partial derivatives, everythin for the kinetic energy is in terms of xdot, and that leaves nothing for thetadot, so I'm a little confused.

2). Apply the Lagrangian method for a for a particle moving on a sphere using spherical coordinates.

so so x = rsin(theta)cos(phi), y = rsin(theta)sin(phi), z = rcos(theta)

so L = m/2(x^2 + y^2 + z^2) - U(r)

How do you get xdot, ydot, zdot? I know you just take the derivativebut with respect to what? Phi and Theta, since r is constant?
 

gabbagabbahey

Homework Helper
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the "dot" represents a time derivative; you take the derivative with respect to time and apply the chain rule
 
160
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Yeah thats what I thought, so I get

xdot = rcos(phi)cos(theta)(thetadot) - rsin(theta)sin(phi)(phidot)

ydot = rsin(phi)cos(theta)(thetadot) + rsin(theta)cos(phi)(phidot)

zdot = -rsin(theta)(thetadot)

Then you have to square then and put them into the Lagrangian expressio, which turns out to be a hug mess? It really gets nasty. Any trig identitites I can before squaring each term?
 
160
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Actually I think I found them.

Applying the dentities I got:

xdot = r(cos(theta + phi)(thetadot + phidot)

ydot = r(sin(theta + phi)(phidot + thetadot)

And z is the same as before.
 
160
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Ok, so I'm not sure if those time derivatives are correct or not? Do take the derivative of the x, y and z equations with respsct to r, theta and Phi?

Trouble is, when ypu do this then you get a god awful mess, then you have to square them, and now I'm lost.

Please Help
 
38
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Look at question no. 3 in the PDF.
 

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