Answer
5.5 meters per second
Work Step by Step
We consider the maximum force:
$\omega = \sqrt{\frac{F}{ml^2}}\sqrt{\frac{300}{60\times12}}=.6455$
This corresponds to a maximum speed of 7.75, which is less than eight, so the speed requirement is met. Thus, we must find the initial velocity that corresponds to this value of omega: Using conservation of angular momentum, we find:
$v = 5.5 \ m/s$
