Calculate Max Angular Velocity of Spring: Moment of Inertia Homework

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SUMMARY

The discussion focuses on calculating the maximum angular velocity of a linear spring system when released from a twisted position. The spring extends 120mm with a 2.5kg mass, and the moment of inertia is given as 10kg.m². Key equations include kinetic energy equations K.E=1/2.Kθ² and K.E=1/2.Iω², where K is the torsional stiffness, I is the moment of inertia, and ω is the angular velocity. The solution involves applying Hooke's Law to determine the spring constant (k) for accurate calculations.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with rotational dynamics and moment of inertia
  • Knowledge of kinetic energy equations in rotational motion
  • Basic grasp of angular velocity concepts
NEXT STEPS
  • Calculate the spring constant (k) using Hooke's Law with the given mass and extension
  • Apply the kinetic energy equations to find the maximum angular velocity (ω)
  • Explore the relationship between linear and angular motion in spring systems
  • Investigate the effects of varying mass and moment of inertia on angular velocity
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Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators looking for practical examples of spring motion and energy conservation principles.

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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
 
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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.
 

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