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Dynamics
Ok, so I can do q1 and 3 this week but not 2 :P
A smooth straight wire rotates with constant angular speed ω about the vertical axis through a fixed point O on the wire, and the angle between the wire and the upward vertical is constant and equal to θ, where 0<θ<½pi. A bead is free to slide on the wire. Show that z(t), the height of the bead above O, satisfies the equation
z:-(ω² sin² θ ) z=-gcos² θ
where z:=d²z/dt²
and find z(t) if z(0)=h and z.(0)=0. Deduce that if h>gcot²θ/ω², then
z(t)->∞ as t->∞ but if h<gcot²θ/ω², then z(t)->-∞ as t->∞.
This is a big question it really is but I cant do any of it. I would be really grateful for help
....even if you can only do one bit of it 
Cheers
A smooth straight wire rotates with constant angular speed ω about the vertical axis through a fixed point O on the wire, and the angle between the wire and the upward vertical is constant and equal to θ, where 0<θ<½pi. A bead is free to slide on the wire. Show that z(t), the height of the bead above O, satisfies the equation
z:-(ω² sin² θ ) z=-gcos² θ
where z:=d²z/dt²
and find z(t) if z(0)=h and z.(0)=0. Deduce that if h>gcot²θ/ω², then
z(t)->∞ as t->∞ but if h<gcot²θ/ω², then z(t)->-∞ as t->∞.
This is a big question it really is but I cant do any of it. I would be really grateful for help
....even if you can only do one bit of it Cheers
lgs98jonee
Ok, so I can do q1 and 3 this week but not 2 :P
A smooth straight wire rotates with constant angular speed ? about the vertical axis through a fixed point O on the wire, and the angle between the wire and the upward vertical is constant and equal to ?, where 0<?<½pi. A bead is free to slide on the wire. Show that z(t), the height of the bead above O, satisfies the equation
z:-(?²sin²?)z=-gcos²?
where z:=d²z/dt²
and find z(t) if z(0)=h and z.(0)=0. Deduce that if h>gcot²?/?², then
z(t)->? as t->? but if h<gcot²?/?², then z(t)->-? as t->?.
This is a big question it really is but I cant do any of it. I would be really grateful for help....even if you can only do one bit of it
Cheers
A smooth straight wire rotates with constant angular speed ? about the vertical axis through a fixed point O on the wire, and the angle between the wire and the upward vertical is constant and equal to ?, where 0<?<½pi. A bead is free to slide on the wire. Show that z(t), the height of the bead above O, satisfies the equation
z:-(?²sin²?)z=-gcos²?
where z:=d²z/dt²
and find z(t) if z(0)=h and z.(0)=0. Deduce that if h>gcot²?/?², then
z(t)->? as t->? but if h<gcot²?/?², then z(t)->-? as t->?.
This is a big question it really is but I cant do any of it. I would be really grateful for help
....even if you can only do one bit of it Cheers
Smilie in wrong place
steve2005
Smilie in wrong place
sorted
Surely someone can help me 
I've done the first bit (havent looked at the later bits yet):
The tricky bit was coming up force on the bead that would make it fly up the wire. The point of wire that the bead is sat on is accelerating towards the axis of rotation at rω² m/s/s. Resolving this along the line of the wire gives rω²sinθ. The bead isnt fixed to the wire so it is as if the wire is sliding out from under it (providing the bead is light enough) with the same acceleration. Part of the acceleration of the bead up the wire is therefore rω²sinθ. The acceleration due to gravity (down the wire) resolves to gcosθ. The resultant acceleration up the wire is therefore given by
rω²sinθ-gcosθ
This is an acceleration in the direction of the wire, and we want z:, the acceleration upwards, so we make a small modification by resolving into a vertical component (we're not interested in the horizontal):
z: = cosθ(rω²sinθ-gcosθ )
All that remains is to replace that r, and simple trigonometry says that r=ztanθ (or r=zsinθ/cosθ ). Plugging that in:
z: = cosθ((zsinθ/cosθ )ω²sinθ-gcosθ )
and multiplying out:
z: = zω²sin²θ-gcos²θ
which rearranges to the equation you have:
z: - zω²sin²θ = -gcos²θ
I'll have a go at the other parts in a moment, my pizza just arrived. Sorry for the wordyness, that's just how I work when I do a mechanics question.
The tricky bit was coming up force on the bead that would make it fly up the wire. The point of wire that the bead is sat on is accelerating towards the axis of rotation at rω² m/s/s. Resolving this along the line of the wire gives rω²sinθ. The bead isnt fixed to the wire so it is as if the wire is sliding out from under it (providing the bead is light enough) with the same acceleration. Part of the acceleration of the bead up the wire is therefore rω²sinθ. The acceleration due to gravity (down the wire) resolves to gcosθ. The resultant acceleration up the wire is therefore given by
rω²sinθ-gcosθ
This is an acceleration in the direction of the wire, and we want z:, the acceleration upwards, so we make a small modification by resolving into a vertical component (we're not interested in the horizontal):
z: = cosθ(rω²sinθ-gcosθ )
All that remains is to replace that r, and simple trigonometry says that r=ztanθ (or r=zsinθ/cosθ ). Plugging that in:
z: = cosθ((zsinθ/cosθ )ω²sinθ-gcosθ )
and multiplying out:
z: = zω²sin²θ-gcos²θ
which rearranges to the equation you have:
z: - zω²sin²θ = -gcos²θ
I'll have a go at the other parts in a moment, my pizza just arrived. Sorry for the wordyness, that's just how I work when I do a mechanics question.
Is it something along the lines of....
if h>gcot²θ, then z:>0 so accelerating upwards
and if h<gcot²θ, then z:<0, so accelertating downwards
??
if h>gcot²θ, then z:>0 so accelerating upwards
and if h<gcot²θ, then z:<0, so accelertating downwards
??
A difficult question (and quite refreshing, I'm on a gap year and trying to get back into this for upcoming interviews). Finding z(t) involves solving the second order differential equation you end up with in part one. The general solution for an equation of this type (as you probably know) is:
z(t)=Aeλ1 t+Beλ2 t+C
I'll call this (1)
Where λ1 and λ2 are the solutions of the auxilliary equation:
λ²-ω²sin²θ=0
(so λ1=ωsinθ and λ2=-ωsinθ )
The constants take a bit of work to determine but it's all fairly straightforward algebra: Differentiate (1) to get z.(t), and use z.(0)=0 to arrive at A=B. Then use z(0)=h in the (1) to get 2A+C=h. Thirdly, use the equation arrived at in the first part, put z(0)=h into it and conclude that z: (0)=hω²sin²θ-gcos²θ. Differentiate z.(t) to get another form z: (t) and put z: (0)=hω²sin²θ-gcos²θ to hopefully get your three constants A, B and C. Let me know if you need any of this expanding on.
You should arrive at A=B=½h-(gcos²θ/2ω²sin²θ ) or ½h-(gcot²θ/2ω²) and C=gcos²θ/ω²sin²θ
So your final z(t) is defined as
z=(½h-(gcot²θ/2ω²))(eωtsinθ+e-ωtsinθ)+gcos²θ/ω²sin²θ
I'll call this (2)
I've checked that and I think it all works out, but I am known to slip up occasionally. I'll know for sure when I've done the rest of it, bear with me.
Edit: In fact, I think I have made a mistake. z.(t) needs to be a function involving h also and according to the above, it isn't. I'll try and find my error.
Edit2: I put z(0)=0 instead of z(0)=h, it's right now I think.
Edit3: Made another stupid error (left out a squared sign), I've corrected this post and now the very last part is fairly trivial:
Looking at (2) in pieces:
(eωtsinθ+e-ωtsinθ)
This part gets more positive with increasing time, since the first half just increases exponentially and the other half gets increasingly tiny (and remains positive).
+gcos²θ/ω²sin²θ
This is a constant, negative because g is negative and the rest is squared. Although it is negative it becomes less significant as t increases since the magnitude of the first half of (2) increases rapidly.
(½h-(gcot²θ/2ω²)) or ½(h-(gcot²θ/ω²))
This is the crux of the matter, the sign of z as t becomes large rests here because it is constant and it multiplies the significant (and positive) part of (2). Obviously, if h>gcot²θ/ω² then this constant is positive and the whole thing will get increasingly positive (z tends to ∞ ), similarly if h<gcot²θ/ω² then z tends to -∞.
z(t)=Aeλ1 t+Beλ2 t+C
I'll call this (1)
Where λ1 and λ2 are the solutions of the auxilliary equation:
λ²-ω²sin²θ=0
(so λ1=ωsinθ and λ2=-ωsinθ )
The constants take a bit of work to determine but it's all fairly straightforward algebra: Differentiate (1) to get z.(t), and use z.(0)=0 to arrive at A=B. Then use z(0)=h in the (1) to get 2A+C=h. Thirdly, use the equation arrived at in the first part, put z(0)=h into it and conclude that z: (0)=hω²sin²θ-gcos²θ. Differentiate z.(t) to get another form z: (t) and put z: (0)=hω²sin²θ-gcos²θ to hopefully get your three constants A, B and C. Let me know if you need any of this expanding on.
You should arrive at A=B=½h-(gcos²θ/2ω²sin²θ ) or ½h-(gcot²θ/2ω²) and C=gcos²θ/ω²sin²θ
So your final z(t) is defined as
z=(½h-(gcot²θ/2ω²))(eωtsinθ+e-ωtsinθ)+gcos²θ/ω²sin²θ
I'll call this (2)
I've checked that and I think it all works out, but I am known to slip up occasionally. I'll know for sure when I've done the rest of it, bear with me.
Edit: In fact, I think I have made a mistake. z.(t) needs to be a function involving h also and according to the above, it isn't. I'll try and find my error.
Edit2: I put z(0)=0 instead of z(0)=h, it's right now I think.
Edit3: Made another stupid error (left out a squared sign), I've corrected this post and now the very last part is fairly trivial:
Looking at (2) in pieces:
(eωtsinθ+e-ωtsinθ)
This part gets more positive with increasing time, since the first half just increases exponentially and the other half gets increasingly tiny (and remains positive).
+gcos²θ/ω²sin²θ
This is a constant, negative because g is negative and the rest is squared. Although it is negative it becomes less significant as t increases since the magnitude of the first half of (2) increases rapidly.
(½h-(gcot²θ/2ω²)) or ½(h-(gcot²θ/ω²))
This is the crux of the matter, the sign of z as t becomes large rests here because it is constant and it multiplies the significant (and positive) part of (2). Obviously, if h>gcot²θ/ω² then this constant is positive and the whole thing will get increasingly positive (z tends to ∞ ), similarly if h<gcot²θ/ω² then z tends to -∞.
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