5. A ball of mass m is suspended above the center of the rotating platform on the elastic
spring with stiffness k and initial (non-stretched) length lo as shown. The opposite end of
the spring is attached to the vertical pole fixed on axis of the platform. The platform starts
to rotate with angular velocity w. What is the angle a' that the spring makes with the
vertical? Consider all values of the angular velocity, from w=0 to w=inf and find the
conditions on w when the angle a' >0 and when a' =0.

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- maheshmeghwal9

can you show me the figure?

- anonymous

I have attached the diagram here. It is the one labelled diagram five.

##### 1 Attachment

- IrishBoy123

i've had a play with this and i get \(\alpha\) for a given \(\omega \) to be
\(tan \ \alpha = \frac{\omega^2 }{g}( l_{o} + \frac{mg}{k} + \epsilon )\)
where \(\epsilon\) is the extra extension in the spring for that given \(\omega\).
this makes sense as it implies that the thing rises but never gets to \(\alpha = \pi/2\) and at the same time \(\epsilon\) just gets bigger and bigger.
not that the question makes that much sense to me really.

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