I'm stuck in imaging the setup for this problem

Question

I have this problem, but I can't seem to figure out the setup of the bead and wire. How does this differ from a problem where the wire is rotating with the axis being the vertical diameter? What is the significance of the fixed point O in this case? Any hints on where to start? Thanks.

Answer 1

Not complicated with the simplification the given solution implies. The exact solution is, with $\tan\alpha=\dfrac{a\sin\theta}{b+a\cos\theta}$ this:

$$-m\omega^2\sqrt{b^2+a^2+2ab\cos \theta}\sin(\theta-\alpha)=m\ddot\theta a$$

With:

$\sqrt{b^2+a^2+2ab\cos \theta}$ being the distance from $O$ to $P$ (where the bead is, of course),

$m\omega^2\sqrt{b^2+a^2+2ab\cos \theta}$ being the centrifugal force acting on the bead,

$m\omega^2\sqrt{b^2+a^2+2ab\cos \theta}\sin(\theta-\alpha)$ being the projection of this force on the tangent to the circular wire the bead is and

$m\ddot\theta a$ is the bead's acceleration along the wire.

Explanation and simplification:

The centrifugal force acts in the direction of the line passing by $O$ and $P$ and outwards (the name is descriptive: centrifugal, latin for fleeing from the center). A simple diagram of forces will make it clear that the centrifugal force times the sinus of $\theta-\alpha$ accelerates the bead along the circular wire. The Coriolis force is of no influence here because it's always orthogonal to the velocity, so is always orthogonal to the path the bead is forced to follow.

You can get the given solution if it's the case that $b\gt\gt a$.

Being $b\gt\gt a$. We have that $\alpha\approx 0$ and $\sqrt{b^2+a^2+2ab\cos \theta}\approx\sqrt{b^2}=b$ Leading to $-m\omega^2b\sin\theta=m\ddot\theta a$ and finally to $\omega^2\dfrac ba\sin\theta+\ddot\theta=0$