Kleppner/Kolenkow: Two Points Around a Circle

[Cosmophile]

Hey, all. I've decided to go back and work on some old K&K problems that I didn't finish last time. Here's a neat one that's been giving me trouble.

I hadn't attempted problem a yet (I admittedly completely overlooked it by accident!). For b, I had a difficult time finding a good starting point. I initially took $B$'s frame of reference to be stationary and decided that would mean $A$ would be the only object moving.

For this case, I centered the circle at the origin $(0,0)$. Let $A$ and $B$ be two points on the circle. From $B$'s frame, $A$ travels around the circle. $A$'s position is given by the vector $\vec{R}$, drawn from $B$. $\vec{R}$ is a chord on the circle. $\theta$ is the angle made by $\vec{R}$ and the line $BO = L = AO$, where $O$ is the origin $(0,0)$.

Because I am finding the velocity of $A$ from $B$'s point of view, the velocity vector will not be purely tangential to the circle. So, in polar coordinates, I get:
$$\vec{v_A} = -v \sin(\theta) \hat{r} + v \cos(\theta) \hat{\theta}$$

After discussing this with a friend who has completed his undergrad, he informed me that I was wrong. From $B$'s point of view, both $A$ AND the circle are moving. If this is the case, I am having a difficult time finding a starting point. So, help is welcome as always!

 

[A.T.]

I hadn't attempted problem a yet

You should. It asks for the general solution, which you can then apply in b).

 

[Cosmophile]

You should. It asks for the general solution, which you can then apply in b).

Thanks, I'll take a look at it.

 

[Cosmophile]

For part a, I've found

$$B = A - R$$
$$v_B = v_A - \frac{dR}{dt} \qquad \to \qquad v_A = v_B + \frac{dR}{dt}$$