Spring with circular motion

In summary, a 3.00 kg object is attached to a spring with a spring constant of 132 N/m and is rotating with an angular speed of 2.00 radians/s. The question is to find the radius at which the mass rotates without moving toward or away from the origin. By applying Newton's 2nd law to the mass, it is determined that the object is accelerating due to circular motion. After solving the equation F_{spring}(R)=k(R-L), the correct solution is obtained.
  • #1
Abarak
28
0

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([tex]\omega[/tex]) at which the mass rotates without moving toward or away from the origin.

Homework Equations



[tex]k(R-L)[/tex]
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...

[tex]F_{spring}(R)=k(R-L)[/tex]

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 [tex]R-L=k(R-L)[/tex] but this does not work.

Any ideas?

-- Abarak
 
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  • #2
Apply Newton's 2nd law to the mass. Is it accelerating?
 
  • #3
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:

[tex]? = k(R-L)[/tex]

-- Abarak
 
  • #4
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.)
 
  • #5
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
 
  • #6
how did you do this problem, because i have the same problem and its been bugging me like crazy.
 

1. What is spring with circular motion?

Spring with circular motion is a type of motion where a spring or elastic object is attached to a rotating object, causing the spring to stretch and compress as the object rotates.

2. How does a spring with circular motion work?

A spring with circular motion works by using the rotational force of the object to cause the spring to stretch and compress. This creates a circular motion of the spring, which can be used to power mechanical devices or create visual effects.

3. What factors affect the motion of a spring with circular motion?

The motion of a spring with circular motion can be affected by factors such as the strength of the spring, the speed of rotation, and the weight of the rotating object. Other factors like air resistance and friction can also impact the motion.

4. What are some real-life applications of spring with circular motion?

Spring with circular motion can be found in various everyday objects, such as wind-up toys, mechanical clocks, and amusement park rides. It is also used in engineering and manufacturing to create precise and controlled movements.

5. How is spring with circular motion related to other types of motion?

Spring with circular motion is related to other types of motion, such as simple harmonic motion and rotational motion. It combines the properties of both types of motion to create a unique and useful effect.

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