Bead sliding on a rigid straight wire

[Favicon]

Homework Statement

A bead is free to slide along a rigid, straight wire, whilst the wire is rotating at angular velocity ω about the z-axis and is tilted away from the z-axis at angle α. I have the equation of motion (EOM) and need to find an explicit solution for the distance of the bead along the wire (q) as a function of time.

Homework Equations

EOM: \ddot{q} - ω^2sin^2(α)q = -gcos(α)

The Attempt at a Solution

I assume the ansatz

q = Ae^{Bt}

which differentiates twice to give

\ddot{q} = B^2q

and substitute the second expression into the EOM, which gives (after a little rearrangement):

q = -\frac{gcos(α)}{B^2 - ω^2sin^2(α)}

I don't know what I've done wrong, but I'm sure I must have made a mistake somewhere since the result is an expression for q which has no dependence on time, despite my ansatz that q is a function of time. Can anyone explain what I should do differently here?

 

[tiny-tim]

Hi Favicon!

EOM: \ddot{q} - ω^2sin^2(α)q = -gcos(α)

I assume the ansatz

q = Ae^{Bt}

which differentiates twice to give

\ddot{q} = B^2q

and substitute the second expression into the EOM, which gives (after a little rearrangement):

q = -\frac{gcos(α)}{B^2 - ω^2sin^2(α)}

no, it's Ae^Bt = that

but wouldn't it be simpler to write \ddot{q} - ω^2sin^2(α)(q + gcos(α)/ω^2sin^2(α)), and make the obvious linear substitution?

 

[Favicon]

Hi Favicon!


no, it's Ae^Bt = that

but wouldn't it be simpler to write \ddot{q} - ω^2sin^2(α)(q + gcos(α)/ω^2sin^2(α)), and make the obvious linear substitution?

Please bear with me as I'm quite rusty on my differential equation solving. By "the obvious linear substitution" I take it you mean p(t) = q(t) + \frac{gcos(\alpha)}{\omega^2sin^2(\alpha)}

Then, since \alpha and \omega are both constant, we have \ddot{p} = \ddot{q}

Substituting those into the EOM (in the form you suggested) gives

\ddot{p} - \omega^2sin^2(\alpha)p = 0

No I take the ansatz p = Ae^{Bt} and substitute into the previous equation to get

B^2Ae^{Bt} - \omega^2sin^2(\alpha)Ae^{Bt} = 0

B = ±\omega sin(\alpha)

If I remember rightly, this means a general solution takes the form

p(t) = A_1e^{\omega sin(\alpha)t} + A_2e^{-\omega sin(\alpha)t}

Substituting for q then yields

q(t) = A_1e^{\omega sin(\alpha)t} + A_2e^{-\omega sin(\alpha)t} - \frac{gcos(\alpha)}{\omega^2sin^2(\alpha)}

Am I right so far? I can see that to go any further would require knowledge of q(t) at some specific time, e.g q(0). However, the textbook from which I'm working suggests that to determine a specific solution for q(t) would require knowledge of \dot{q}(0) at some specific time also. Where does this requirement come from?

 

[tiny-tim]

If I remember rightly, this means a general solution takes the form

p(t) = A_1e^{\omega sin(\alpha)t} + A_2e^{-\omega sin(\alpha)t}

Substituting for q then yields

q(t) = A_1e^{\omega sin(\alpha)t} + A_2e^{-\omega sin(\alpha)t} - \frac{gcos(\alpha)}{\omega^2sin^2(\alpha)}

yes

now you have two constants of integration, A₁ and A₂, so you need two initial conditions to find both (which will usually be a condition on q and a condition on q')

 

[Favicon]

Ah yes, that's a good point. Thanks very much for your help Tiny :)