How Is the Equilibrium Radius Determined in Rotational Spring Motion?

AI Thread Summary
The discussion centers on determining the equilibrium radius for a mass attached to a rotating spring. The mass of 3.00 kg is connected to a spring with a spring constant of 132 N/m and rotates at an angular speed of 2.00 radians/s. Participants emphasize the importance of applying Newton's second law, noting that the mass is indeed accelerating due to circular motion, which affects the forces involved. The key equation involves balancing the spring force with the centripetal force required for circular motion. Ultimately, understanding these dynamics allows for the correct calculation of the radius at which the mass remains in equilibrium.
Abarak
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Homework Statement


An object of mass M = 3.00 kg is attached to a spring with spring constant k = 132 N/m whose unstretched length is L = 0.170 m, and whose far end is fixed to a shaft that is rotating with an angular speed of omega = 2.00 radians/s. Neglect gravity and assume that the mass also rotates with an angular speed of 2.00 radians/s as shown.

http://img337.imageshack.us/img337/4482/6172alq9.jpg

Question:
Given the angular speed of omega = 2.00 radians/s, find the radius R(\omega) at which the mass rotates without moving toward or away from the origin.

Homework Equations



k(R-L)
The amount of force a spring exerts is proportional to the distance it is stretched or compressed with respect to its equilibrium length ( L = 0.170 m in this case).

so...

F_{spring}(R)=k(R-L)

The Attempt at a Solution



"force a spring exerts is proportional to the distance it is stretched or compressed with respect to its equilibrium length" I am having problems with this part. I cannot figure out what the other side of the equation is.

I tried R-L=k(R-L) but this does not work.

Any ideas?

-- Abarak
 
Last edited by a moderator:
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Apply Newton's 2nd law to the mass. Is it accelerating?
 
From what I gather the object is not accelerating so Newton's 2nd law would not apply to this.

"force a spring exerts is proportional to the distance it is stretched or compressed with respect to its equilibrium length" I don't see how this would apply to Newton's second law or the other side of the equation:

? = k(R-L)

-- Abarak
 
Abarak said:
From what I gather the object is not accelerating so Newton's 2nd law would not apply to this.
Sure it's accelerating--it's going in a circle! (Reread the chapter in your text about circular motion.)
 
Oh snap! Talk about a lack of judgment. After applying Newton's 2nd Law everything worked like a charm.

Thanks again for the help Doc.

-- Abarak
 
how did you do this problem, because i have the same problem and its been bugging me like crazy.
 
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