Circular Accelerometer

Referring to the figure in the text of the problem, let $\theta$ be the angle formed between the directions 1) bead-centre of the wire and 2) centre of the wire-down along the wire's vertical axis of symmetry (sorry, I am too lazy to draw).

The bead's weight generates a tangential acceleration $g\sin\theta$ in the anti-clockwise sense. The acceleration of the wire generates a tangential acceleration $a\cos\theta$ in the clockwise sense. Since the bead is in equilibrium, we have:

$a\cos\theta = g\sin\theta$

that is

$\tan\theta = 2$

Since the radius of the wire is 1, the distance of the bead equilibrium position to the vertical axis is $\sin\theta$ , which expressed through the tangent gives:

$\sin\theta = \frac{\tan\theta}{\sqrt{1 + \tan^2\theta}} = \frac{2}{\sqrt{5}} = 0.894427...$

This can be very easily done. First of all , the bead is expriencing , two accelartions, 1 horizotal, 1 verital. Now By Vector Law, the resultant of the two perpendicular acceleration vectors are $\sqrt{(2g)^+(g)^2}$ = $a_r$ = $g$ $\sqrt{5}$ .

Hence now, this acceleration is equal to the centrepetal accelaration $a_c$ = $v^2/r$ where r=1m

Hence $a_r$ = $a_c$ or $g$ $\sqrt{5}$ = $v^2$

Hence 1/2mv^2 = mgh, now h = $\sqrt{5}$ /2 from the uppermost part of the axis.

Now it will look something like this

Now by pythagoras, we get Distance = 0.89.. (approx)