Max angular velocity when spring released from 360° twisted position

[physicsfun_12]

Homework Statement

I linear spring has one end attached to a ceiling, the other end hanging freely. It was noted that the spring was extended by 120mm when a mass of 2.5kg was attached at the bottom.

If the moment of inertia of the mass is 10kg.m² about its vertical axis, calculate the maximum angular velocity of the spring when the system is released from from the 360 degree twisted position. Assume no change in height and the system is conservative.

I am not to sure how to approach this. I think it may have something to so with the following formulas:

Homework Equations

K.E=1/2.K.theta² and K.E=1/2.I.w²

where k=torsional stiffness, I=moment of inertia and w=angular velocity

The Attempt at a Solution

Other than those equations, I'm really not sure.

Any help much appriciated with this, thanks in advance.

Mike

 

[Yitzach]

Wikipedia informs me that $$\kappa$$ is also called the spring constant. The reason for the "2.5 kg mass extends the spring 120 mm" is so you can find it using Hooke's law where k is preferred. You should be able to get much closer to the answer with that.