Solving a Collar's Angular Momentum Problem: Finding f(θ) and Max θ Reached

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harmyder
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Homework Statement


Initially collar is at rest with theta = 0. Find [tex]\dot\theta=f(\theta).[/tex] And find maximum theta reached.

Diagram of a bead A sliding on a curved wire of radius r inside a square frame accelerating to the right with acceleration a; the wire arcs from the upper left down to the lower right, and the angle theta is measured from the frame's upper-right corner to the bead.


Homework Equations


I don't know if i allowed to use angular momentum about top-right frame corner as it's accelerating.

But i definitely will need second Newton law:)

The Attempt at a Solution


I know that there mg, N forces. But i can't figure out another force due to acceleration of the frame.
 
haruspex said:
Are you comfortable using non-inertial frames of reference?

I think I'm not, but i believe there is no need in it here. Important thing is that i need to calculate θ' based on θ not on t.
 
harmyder said:
I think I'm not, but i believe there is no need in it here. Important thing is that i need to calculate θ' based on θ not on t.
It does help here, but it is never essential.
What force/acceleration equations can you write? It doesn't matter if you think they are wrong or incomplete, just post what you get.
 
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haruspex said:
It does help here, but it is never essential.
What force/acceleration equations can you write? It doesn't matter if you think they are wrong or incomplete, just post what you get.

The collar undergoes three forces - [itex]m\mathbf{g}, \mathbf{N}, \mathbf{F}_{ext}[/itex]. From which [itex]|m\mathbf{g}|=|N|[/itex], though i can't prove it.

With angle [itex]\theta[/itex] we have:
[itex]|m\mathbf{g}|[/itex] acting downward and [itex]\big(|\mathbf{N}| + |\mathbf{F}_{ext}|\big)\cos\theta[/itex] acting upward.

The maximum angle is reached with equilibrium.

And here i can't calculate the external force.
 
harmyder said:
The collar undergoes three forces - [itex]m\mathbf{g}, \mathbf{N}, \mathbf{F}_{ext}[/itex]. From which [itex]|m\mathbf{g}|=|N|[/itex], though i can't prove it.

With angle [itex]\theta[/itex] we have:
[itex]|m\mathbf{g}|[/itex] acting downward and [itex]\big(|\mathbf{N}| + |\mathbf{F}_{ext}|\big)\cos\theta[/itex] acting upward.

The maximum angle is reached with equilibrium.

And here i can't calculate the external force.
The external force acts on the frame, not on the collar.
Why do you think |mg|=|N|? Consider horizontal and vertical separately.
 
haruspex said:
It does help here, but it is never essential.
What force/acceleration equations can you write? It doesn't matter if you think they are wrong or incomplete, just post what you get.

I can't come up with solution in [itex]\hat{i}\times\hat{j}[/itex] frame.
Free-body diagram of a small bead on a curved wire, with normal force N, unit vectors e_n and e_t, weight mg pointing down, and inertial force ma pointing left in red.

Maybe i need to consider tangential parts of [itex]mg[/itex] and anther force, i think this another force is [itex]ma[/itex]? When they become equal, then no more change of angle.
 
The original post said "Initially collar is at rest with theta = 0" and according to the picture "theta= 0" is at the bottom! The collar won't move- [itex]\theta'= 0[/itex]. If you intended to the initial point to be at the top, with "theta" measured from the top, then, at each [itex]\theta[/itex], the downward acceleration vector, <0, -g>, can be written as the sum of two vectors, one parallel to the normal vector circular bar and one parallel to the tangent vector. Since the collar can't move normal to the bar, only the tangent vector is relevant.
 
HallsofIvy said:
The original post said "Initially collar is at rest with theta = 0" and according to the picture "theta= 0" is at the bottom! The collar won't move- [itex]\theta'= 0[/itex].
Have you noticed rightward horizontal acceleration of the frame?
 
harmyder said:
Maybe i need to consider tangential parts of [itex]mg[/itex] and anther force, i think this another force is [itex]ma[/itex]? When they become equal, then no more change of angle.
That is the non-inertial frame method I mentioned. You treat the acceleration as a 'fictitious' force ma acting the other way.
So what answer do you get?
 
harmyder said:
Have you noticed rightward horizontal acceleration of the frame?
NO! I hadn't. Thanks.