Spring and centrifugal force

1. Jun 6, 2012

Gavroy

hi

i found the following equation that a spring(with constant k) , a mass m rotating with angular velocity ω, will have a constant angle $\alpha$ to the vertical axis that is given by
$cos(\alpha)=\frac{g}{\omega^2 l}(1-\frac{m \omega^2}{k})$
this looks similar to something like $cos(\alpha)=F_{gravitation}/F_{centripetal}-F_{gravitation}/F_{spring}$ but i do not see the geometrical idea behind this.

i am not sure whether this equation is correct at all, I found it in my physics schoolbook

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2. Jun 6, 2012

nasu

The equation seems to be OK.
In order to find it you could apply Newton's second law for the vertical and horizontal directions. Use the equation for elastic force and centripetal acceleration, too.

3. Jun 6, 2012

Gavroy

thank you for this hint. now, i found it by myself.