How Do You Convert Foucault Pendulum Equations to Polar Coordinates?

[beth92]

Homework Statement

For a Foucalt Pendulum:
Relative to horizontal Cartesian x and y axes fixed to the Earth (with x as East) the equations of motion for horizontal motion are:

x′′ + ω₀²x -2ωy′ = 0 and y′′ + ω₀²y + 2ωx′ = 0

[where x′, x′′, y′, y′′ are first and second time derivatives of x and y]

Convert into standard polar coordinates (ρ,φ) where x=ρcosφ and y=ρsinφ and show that:

ρ′′ + ρ(ω₀²-φ′²-2ωφ′) = 0

and

ρφ′′ + 2ρ′(φ′+ω) = 0

Homework Equations

The Attempt at a Solution

I'm just not sure how to convert the derivatives of x and y into polar coordinate form, eg., how to express x′ in terms of ρ′ and φ′ etc. There is no cos or sin term in the resulting equations and I'm not sure where they go...I'd appreciate some help here!

 

[rude man]

I guess it's just going thru the motions:

x' = ρ'cosψ - ρψ'sinψ
x'' = ρ''cosψ - ρ'ψ'sinψ - {ρ'ψ'sinψ + ρψ''sinψ + ρψ'²cosψ}

etc. for y' and y''

then equating your given equations to each other and to 0 and substituting the x', x'', y' and y'' expressions now in terms of ρ, ψ and their 1st and 2nd derivatives.