Find the angular velocity of the Absolute Motion System

[Northbysouth]

Homework Statement

Point A is given a constant acceleration a to the right starting from rest with x essentially zero. Determine the angular velocity w of link AB in terms of x and a.

I have attached an image of the question

Homework Equations

w = θ'

α = θ''

Pythagoras and trig

The Attempt at a Solution

x = 2bcos(θ)

x' = -2bθ'sin(θ)

x'' = -2bθ''sin(θ)-2bθ'²cos(θ)

a = -2bαsin(θ) - 2bw²cos(θ)

I know that bsin(θ) = √b²-x²/4

and I know that bcos(θ) = x/2

Substituting these is

a = -2α√b²-x²/4 - 2w²x/2

After here is where I get stuck. I'm not sure how to deal with the angular acceleration α. I can see that I have part of the final answer (shown in the image) but I'm not sure about the last few steps.

Any help would be appreciated. Thank you

 

[TSny]

x' = -2bθ'sin(θ)

Maybe you can get θ' from this. But you'll need to find x' in terms of x and a.

 

[Northbysouth]

Looking at the problem again, I had thought that maybe I could use

x' = V0 + at

where V0 is zero, but this introduces another unknown. Am I over thinking this?

 

[Northbysouth]

Ahh, wait. I got it. I was overthinking it. I only needed to take the first derivative. From there I can solve for θ'

θ' = -x'/2bsin(θ)

Knowing that a is constant I can use:

x' = v

v² = v₀² + 2ax

where v₀ = 0

plug in √2ax for x'

Then i can also see that 2bcos(θ) = √4b²-x²

Cheers for your help TSny

But I still have a negative in my final answer which shouldn't be there...

 

[TSny]

The signs are going to depend on your choice of positive directions. If you choose counterclockwise as positive $\omega$, then note that $\omega = -\dot{\theta}$.