Question

Consider the bead on a rotating circular hoop system that we saw in lecture. Recall that R is the radius of the hoop, ω is the angular velocity of rotation, and m is the mass of the bead that is constrained to be on the hoop as it rotates about the z-axis. We found that we could solve the constraints in terms of a single coordinate θ and that the Cartesian coordinates were given by

x = −R sin(ωt)sinθ, y = R cos(ωt)sinθ, z = −R cosθ,

and the Lagrangian was
L =1/2 m R^2 ( θ˙^2 + ω^2 sin^2 θ)+ m g R cosθ

Since L does not depend explicitly on time, there will be a corresponding constant of the motion h. Give the expression for h in terms of θ and ˙ θ. Does h correspond to the total (kinetic plus potential) energy of the bead? The Euler-Lagrange equation for this system gave
¨ θ = ω^2 sin θcosθ −g R sinθ
Using the trick that ¨ θ = d(1/2˙ θ^2)/dθ, integrate this equation to get a relation between ˙ θ^2 and θ that involves an integration constant. Show that this is equivalent to the conservation of h.

Answer 1

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