Question

A bead of mass m slides without friction along a rotating wire in the shape of a parabola with zar2, as shown below. The wire is rotating around the z-axis with constant angular velocity w z=ar2 (a) (0.5 point) Determine the Lagrangian for the system in terms of the coordinate r b) (1 point) Apply the Lagrange Equations to obtain the equation of motion. You (c) (0.5 points) Suppose that the bead is moving in a perfect circle of radius R, with (where r is the distance from the z-axis) may leave it as a differential equation and do not need to solve for r(t) r and z not changing. Given that you know a, for what value(s) of w, if any, is this motion possible (d) (0.5 points) Is the situation above stable or unstable?

Answer 1

A velocity vector for the given bead v = r r + r w emptyset + dz/dt z: r is radical distance of bead from z = ar^2: dz/dt = 2ra r so, Lagrangian for given system L = 1/2 m v v - mgz = 1/2 m(r^2 + r^2 w^2 + (dz/dt)^2) - mg ar^2 = 1/2 m(r^2 + r^2 w^2 + (2r ar)^2) - mg ar^2 L = 1/2 m[(1 + 4r^2 a^2)r^2 + r^2 w^2] - mg ar^2 b equation of Motion: - d/dt(partial differential L/partial differential r) = partial differential L/partial differential r partial differential L/partial differential r = partial differential/partial differential r(1/2 m(1 + 4r^2 a^2)r^2 + r^2 w^2] - mg ar^2 = 4ra^2 mr^2 + 2r w^2 m/2 - 2 ma gr partial differential L/partial differential r = 1/2 m(1 + 4r^2 a^2)2r = mr(1 + 4r^2 a^2) \r\n d/dt(partial differential L/partial differential r) = m(1 + 4r^2 a^2)r so, putting in eqn of motion m(1 + 4r^2 a^2)r = 4ra^2 mr^2 + rw^2 m - 2ma rg (1 + 4r^2 a^2) d^2 r/dt^2 = 4ra^2 (dr/dt)^2 + (w^2 - 2ag) r(t) c since,