A bead of mass m is constrained to move without friction along a straight fricti

Question

A bead of mass m is constrained to move without friction along a straight frictionless rigid wire (AB) of length L. The wire AB rotates about the vertical (OA) with constant angular velocity. The angle between wire and vertical is constant.

Set up the Largrangian for the system. (note: only a single generalized coordinate is needed. e.g., the distance along the wire from point A.) (Take the zero of potential energy to lie in the x-y plane, where the end of the wire traces out circular motion.)

Find the equation of motion for the bead. Note that the velocity can be expressed in terms of two orthagonal unit vectors.

Let me know if you have any questions or you'd like me to upload a diagram I drew. THANK YOU!

Explanation / Answer

let r be the distance along the wire

then x = r sin theta cos( wt + phi) where w is the angular velocity and phi is a phase angle

then y = r sin theta sin(wt + phi)

z = L cos theta - r cos theta

then x' = r' sin theta cos(wt + phi) - r w sin theta sin(wt + phi)

y' = r' sin theta sin(wt + phi) + r w sin theta cos(wt + phi)

z' = -r' cos theta

L = 1/2 mv^2 - m g z

L = 1/2 m (x'^2 +y'^2 + z'^2) - m g z

L = 1/2 m( r'^2 sintheta^2 cos(wt + phi)^2 + r^2 w^2 sintheta^2 sin(wt + phi)^2 - 2 r' r w sin theta^2 sin(wt + phi) cos(wt +phi)

+ r'^2 sin(theta)^2 sin(wt + phi)^2 + r^2 w^2 sintheta^2 cos(wt + phi)^2 + 2 r' r w sin theta^2 sin(wt + phi) cos(wt + phi)
+ r'^2 cos(theta)^2)
- m g (L cos theta - r cos theta)

L= 1/2 m ( r'^2 sin(theta)^2 + r^2 w^2 sin(theta)^2) - m g ( L cos theta - r cos theta)

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