Time scale for the rapid transient

[glebovg]

Homework Statement

Time scale for the rapid transient (overdamped bead on a rotating hoop).

The governing equation m*r*φ'' = -b*φ' - m*g*sinφ + m*r*(ω^2)*sinφ*cosφ, can be reduced to ε(d^2φ/dτ^2) + dφ/dτ = f(φ).

Using phase plane analysis it can be shown that the equation ε(d^2φ/dτ^2) + dφ/dτ = f(φ) has solutions that rapidly relax to the curve where dφ/dτ = f(φ). Here f(φ) = sinφ(γcosφ - 1), ε = (m^2*g*r)/b^2, τ = t/T, and T is chosen to be T = b/(m*g).

a) Estimate the time scale T_fast for this rapid transient in terms of ε, and then express T_fast in terms of the original dimensional quantities m, g, r, ω, and b.
b) Rescale the original differential equation, using T_fast as the characteristic time scale, instead of T_slow = b/mg. Which terms in the equation are negligible on this time scale?
c) Show that T_fast ≪ T_slow if ε≪1. (In this sense, the time scales T_fast and T_slow are widely separated.)

Relevant equations

γ = (r*ω^2)/g

 

[lippyka]

Homework Equations See statement.The Attempt at a Solution a) T_fast = εT_slow = (m^2*g*r*b^(-2))*(b*m^(-1)*g^(-1))= m^2*g^(-1)*r*b^(-3).b) The equation can be rescaled as: ε(d^2φ/dτ_fast^2) + dφ/dτ_fast = f(φ), where τ_fast = t/T_fast. The terms that are negligible on this time scale are those that have a higher order in ε, such as ε(d^2φ/dτ_fast^2).c) T_fast ≪ T_slow if ε ≪ 1, where ε = (m^2*g*r)/b^2. Since b is always positive, the only way for ε ≪ 1 is for m, g, and r to be small.