Calculating Theta for Fixed Spring Motion

AI Thread Summary
The discussion revolves around deriving an expression for the angle theta that a spring makes with the vertical when it rotates horizontally with a mass attached. The setup includes a spring with spring constant k, unstretched length L, and a mass m moving with angular velocity omega. The initial approach involved using equations related to work and centripetal force, leading to a preliminary expression for cos(theta). However, the user expressed uncertainty about the correctness of this expression. Another participant suggested applying Newton's second law to analyze the vertical and horizontal forces acting on the mass, proposing equations involving tension in the spring and the radius of the circular motion. This method indicates that solving for theta may be complex but feasible. The discussion highlights the challenges of combining physics principles with mathematical approaches in this context.
vladimir69
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hello there
consider a spring that is fixed to a point (i have tried to draw a picture, so to make it clearer as to what i am talking about) that rotates anti clockwise in a horizontal fashion. there is a ball on the end of the spring with mass m, the spring has spring constant k and unstretched length L and the ball moves with angular velocity omega
i am trying to find an expression for theta, the angle that the spring makes with the vertical, as a function of L, w, k, m and g.
me being a maths student and not a physics student just treated this as simaltaneous equations and plugging some stuff together using

W=F*s *cos(theta)
F=m*v^2/L where v is the velocity perpendicular to the centripetal force
F=-k*theta

after a bit of algebra we get that
cos(theta)=-k/(L^2*m*omega^2)
of course one can arrange for theta but i am not sure if this expression is correct, more than likely it is wrong. so if someone could shed some lighton the matter it would be greatly appreciated.

thanks
 

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Here's how I would approach this problem. Apply Newton's 2nd law to the vertical and horizontal forces on the mass:
T sin\theta = mg, where T is the tension in the spring
T cos\theta = m\omega^2 R, where R is the radius of circle
Combine this with what you know about the spring: its length is L + T/k, thus R = (L + T/k)sin\theta.

You should be able to solve this for for \theta, but it looks messy.
 
thanks for the helpful advice
 
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