Lagrangian problems, constrained point masses
Problem:
A bead of mass m slides without friction on a circular loop
of radius r. The loop lies in a vertical plane and rotates about a vertical
diameter with constant angular velocity ω.
(a) For angular velocity
ω larger than some critical value ωc, the bead can undergo small
oscillations about an equilibrium point θ0 ≠ 0. Find ωc
and θ0(ω).
(b) Obtain the equations of motion for the small oscillations about θ0 as a function of
ω
and find the period of the oscillations.
Solution:
- Concepts:
Lagrange's equations
- Reasoning:
Lagrange's equations are the equations of motion.
- Details of the
calculation:
(a) L = T - U.
T = ½m(r2(dθ/dt)2 + r2sin2θ ω2),
U = -mgr cosθ.
L = ½m(r2(dθ/dt)2 + r2sin2θ ω2)
+ mgr cosθ.
Here r and ω are constants.
We have only one generalized coordinate.
d/dt(∂L/∂(dqi/dt)) -
∂L/∂qi = 0.
Lagrange's equation yields
d2θ/dt2 = sinθ cosθ ω2 - (g/r) sinθ.
d2θ/dt2|θequ
= 0 for a stationary point.
The angles θ = 0, θ = π, or cosθ = g/(rω2) define stationary
points.
We need g/(rω2) < 1 for the point θ = cos-1(g/(rω2)
to exist.
The critical frequency ωc
= (g/r)½.
(b) Consider the equilibrium point cosθ0 = g/(rω2).
For a stable point we need a restoring force. Let θ = θ0 + δ.
We need d2δ/dt2 = -cδ, with c a positive number.
d2δ/dt2 = sin(θ0 + δ)cos(θ0
+ δ) ω2 - (g/r) sin(θ0 + δ)
= d(sinθ cosθ ω2 - (g/r) sinθ)/dθ|θ0*δ (Taylor
series expansion)
= (cos2θ0 ω2 - sin2θ0
ω2 - (g/r) cosθ0)δ.
For a stable point we need cos2θ0 ω2 -
sin2θ0 ω2 - (g/r) cosθ0 to
be negative.
But
cos2θ0 ω2 - sin2θ0
ω2 - (g/r)cosθ0 = g2/(r2ω2)
- ω2 < 0 since g/(rω2) < 1 for the point to exist.
This point is always stable if it exists.
The equation of motion for small oscillations about θ0 is
d2δ/dt2 = -(ω2 - g2/(r2ω2))δ.
The frequency of small oscillations about this point is
Ω = (ω2 - g2/(r2ω2))½.
The period T = 2π/Ω.