- #1
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?
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?