Answer:
[tex]\theta = 49.81^0[/tex]
Explanation:
Given that:
[tex]\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m[/tex]
If we let the piece of the close lose contact at ∠θ;
Then ; from force balance;
we have:
[tex]\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})[/tex]
where;
[tex]v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77[/tex]
Again:
[tex]\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0[/tex]
