# Spring with circular motion

1. Jul 15, 2007

### Abarak

1. The problem statement, all variables and given/known data
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.

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.

2. Relevant 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...

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

3. 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

2. Jul 15, 2007

### Staff: Mentor

Apply Newton's 2nd law to the mass. Is it accelerating?

3. Jul 15, 2007

### Abarak

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

4. Jul 15, 2007

### Staff: Mentor

5. Jul 15, 2007

### Abarak

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. Jan 30, 2008

### chenzo89

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