Answer:
The rotation frequency required is 23.78 RPM
Explanation:
Given;
radial acceleration, a = 32.7 m/s²
length of the beam, r = 5.29 m
The linear velocity is calculated as;
[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{ar}[/tex]
where;
v is linear velocity
The angular velocity is calculated as;
[tex]\omega = \frac{v}{r} \\\\Recall, v = \sqrt{ar} \\\\Then, \omega = \frac{\sqrt{ar}}{r}} \\\\ \omega = \frac{\sqrt{32.7 \times5.29}}{5.29}\\\\\omega = 2.49 \ rad/s\\\\Angular \ frequency \ is \ calculated \ as;\\\\\omega = 2\pi f\\\\f = \frac{\omega}{2\pi} \\\\f = \frac{2.49}{2\pi} \\\\f = 0.396 \ rev/s\\\\f = 23.78 \ rev/min[/tex]
Therefore, the rotation frequency required is 23.78 RPM