# Question #c6f97

Nov 14, 2016

$3.83$[m/s]

#### Explanation:

The giration radius is given by

$r = l \cos \theta$

The acceleration induced by the rotational movement is

${\vec{a}}_{r} = \left({\omega}^{2} r\right) \hat{i} = {v}^{2} / r \hat{i}$

Considering a noninertial rotating reference frame. In equilibrium, the weigth $- m g \hat{j}$ and the centrifugal force $m \left({v}^{2} / r\right) \hat{i}$ have null momentum sum concerning the hanging cable fulcrum.

${M}_{0} = \left(l \sin \theta\right) \left(m {v}^{2} / r\right) - \left(l \cos \theta\right) \left(m g\right) = 0$

then

$v = \sqrt{{\cos}^{2} \frac{\theta}{\sin} \theta l g} \approx 3.83$[m/s]