Solving a Vector Triangle Differential Equation

[phantomvommand]

Homework Statement

See picture below

Relevant Equations

Cosine rule,
Speed vector equation

By considering a vector triangle at any point on its circular path, at angle theta from the x -axis,

We can obtain that:
(rw)^2 + (kV)^2 - 2(rw)(kV)cos(90 + theta) = V^2

This can be rearranged to get:
(r thetadot)^2 + (kV)^2 + 2 (r* thetadot)(kV)sin theta = V^2.

I know that I must somehow solve this differential equation in theta, and integrate from theta = 0 to theta = 2pi.

How do I solve this equation?

Thank you!

 

[haruspex]

How do I solve this equation?

You are told k is very small, so you need to make an approximation.
And you know the extra is of order k², so you know which terms to keep.
I would start by solving the quadratic.

 

[phantomvommand]

You are told k is very small, so you need to make an approximation.
And you know the extra is of order k², so you know which terms to keep.
I would start by solving the quadratic.

Thank you! I was thinking that the quadratic equation formula couldn’t be applied to solve for thetadot because there is a sintheta*thetadot term. This doesn’t pose a problem I suppose?

 

[haruspex]

Thank you! I was thinking that the quadratic equation formula couldn’t be applied to solve for thetadot because there is a sintheta*thetadot term. This doesn’t pose a problem I suppose?

You can solve the quadratic for $\dot\theta$, creating a trig term inside a square root. But then you can use the small k approximation to get rid of the square root. You may need to use it again. Just make sure to keep terms up to k².