Solving the Geneva Wheel Problem Homework

[Rellek]

Homework Statement

I'll attach the screenshot, it has everything.

Homework Equations

This is in the section of absolute motion analysis.

The Attempt at a Solution

I actually was able to solve the problem correctly, but it was annoyingly complicated and I'm curious as to whether there's a better way.

So, first off, I wanted to find a way to write $\varphi$ in terms of $\theta$.

So, recognize that tan$\varphi$ = Dsin$\varphi$/Dcos$\varphi$

The D*sin$\varphi$ is always going to be equal to 100sinθ.

For cos$\varphi$ it was a bit more tricky, but it's easy to recognize the relationship because the distance stays constant.

100√2 - 100cos(θ) = D*cos$\varphi$

So now we have:

tan$\varphi$ = 100sinθ/(100√2 - 100cosθ)

And now I take the derivative. This is a quotient rule, and the derivative is going to be with respect to time, so through the chain rule we will have our angular velocities pulled out. Note that this derivative has been simplified from the raw form just a little.

sec²$\varphi$*ω_phi = ω_θ (√2*cosθ - 1)/(√2 - cosθ)²

So now the next conundrum is how to get rid of the damn sec²$\varphi$

Well, we recongnize that sec = Hypotenuse / Adjacent side,

And we've solved for the adjacent side already, it is equal to the cos$\varphi$ equation, which is:

100√2 - 100cosθ

But what is the hypotenuse? Well, by the pythagorean theorem:

H² = (100sinθ)² + (100√2 - 100cosθ)²

So, now we have our values, and since it is sec²$\varphi$, we will use:

H² / B²

Which gives (after some more excessive simplifying):

ω_phi(3 - 2√2 cosθ)/(√2 - cosθ)² = ω_θ (√2*cosθ - 1)/(√2 - cosθ)²

Now notice the denominators are the same, they cancel out. Then just divide to solve for the angular velocity with respect to $\varphi$

You GET:

ω_phi = ω_θ (√2 cosθ - 1)/(3 - 2√2 cosθ)

And that IS the correct answer, it's just that it took a really long time and a ton of failed attempts before I finally solved it. I'm just wondering if there's a better way! Thanks!

 

[NascentOxygen]

Your approach looks about as straight-forward as I could make it. I don't quite get your answer, but I've probably tripped on the trig and have't gone back to check.

You know your answer is correct, so well done!