How Does Angular Velocity Affect the Angle of a Spring-Mass System?

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SUMMARY

The discussion centers on the relationship between angular velocity (ω) and the angle (α) of a spring-mass system. The equation presented is cos(α) = (g / (ω² l))(1 - (m ω² / k)), where g is gravitational acceleration, l is the length of the spring, m is the mass, and k is the spring constant. The equation is validated through the application of Newton's second law and the principles of centripetal acceleration and elastic force. The participants confirm the correctness of the equation and its derivation from fundamental physics principles.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with centripetal acceleration concepts
  • Knowledge of elastic force and Hooke's law
  • Basic trigonometry for angle calculations
NEXT STEPS
  • Study the derivation of centripetal acceleration in rotating systems
  • Explore Hooke's law and its applications in spring mechanics
  • Learn about the implications of angular velocity on dynamic systems
  • Investigate the role of gravitational forces in oscillatory motion
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of spring-mass systems and rotational motion.

Gavroy
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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 [itex]\alpha[/itex] to the vertical axis that is given by
[itex]cos(\alpha)=\frac{g}{\omega^2 l}(1-\frac{m \omega^2}{k})[/itex]
this looks similar to something like [itex]cos(\alpha)=F_{gravitation}/F_{centripetal}-F_{gravitation}/F_{spring}[/itex] 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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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.
 
thank you for this hint. now, i found it by myself.
 

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