# Question #18121

Oct 16, 2015

I found: $\omega = 28.9 \frac{\text{rad}}{\sec}$

#### Explanation:

Maximum tension is $50 , 000 N$ so using Newton's Second Law, $\Sigma \vec{F} = m \vec{a}$, we get that:
$T = m {a}_{c} = m {v}^{2} / 4$ where ${a}_{c}$ is centripetal acceleration;
$50 , 000 = 15 \cdot {v}^{2} / 4$
and:
$v = 115.5 \frac{m}{s}$
This linear velocity can be converted into rotational, $\omega$, by remembering that:
$v = \omega r$
so:
$\omega = \frac{v}{r} = \frac{115.5}{4} = 28.9 \frac{\text{rad}}{\sec}$