Work Done in Circle - Why Not Zero? (Calc 3)

In summary, the conversation discusses finding the work done by a force field as an object moves counterclockwise about a circle, both through direct evaluation and using Green's Theorem. The issue at hand is whether the work done is zero, and it is determined that this is not always the case for non-conservative force fields. The conversation also clarifies that if the work is zero over any closed loop, then the field is conservative.
  • #1
Weather Freak
40
0

Homework Statement


Find the work done by the force field F(x,y) = (x+2y^2)j as an object moves once counterclockwise about the circle (x-2^2+y^2=1 by first evaluating the integral directly then use Green's Theorem and evaluate the double integral.

Homework Equations


W = F*D*cos(theta)

The Attempt at a Solution


My problem is not in getting the answer. I haven't gotten the final answer yet, but that's not the issue at hand.

My Calc teacher told me that it is NOT zero, and I am wondering why. If you move around in a circle, there is no displacement, so therefore, work should be equal to zero. So what is it that makes this case different from everything I did in Physics class?
 
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  • #2
If you move around in a circle through a force field that is a gradient of a scalar function, the work done by the force field is zero. Gravity is a conservative force for this reason--gravity is the gradient of gravitational potential. But if the force field is not the gradient of a function, the work done can be nonzero.
 
  • #3
The force fields in physics (gravitational & coulombian) are conservatives. This means that they have the special caracteristic that if the net displacement is 0, then so is the total work. But not all vector fields have this property.
 
  • #4
Weather Freak said:

Homework Statement


Find the work done by the force field F(x,y) = (x+2y^2)j as an object moves once counterclockwise about the circle (x-2^2+y^2=1 by first evaluating the integral directly then use Green's Theorem and evaluate the double integral.

Homework Equations


W = F*D*cos(theta)

The Attempt at a Solution


My problem is not in getting the answer. I haven't gotten the final answer yet, but that's not the issue at hand.

My Calc teacher told me that it is NOT zero, and I am wondering why. If you move around in a circle, there is no displacement, so therefore, work should be equal to zero. So what is it that makes this case different from everything I did in Physics class?
Surely you were not taught that the work done in moving away from and then back to an initial point is always 0? That is true only for a conservative force field. This field is not conservative.
 
  • #5
Okay, I think I get it now, thanks! One more thing though... if the work ends up being zero, does that necessarily imply that the vector field is conservative, or is it only true the other way around?
 
  • #6
If the work is 0 over ANY closed loop, then the field is conservative. But if it is 0 over one particular loop, it does not mean that it is conservative. It could be non-zero around some other loop.
 

1. What is work done in a circle?

Work done in a circle refers to the amount of energy or force required to move an object in a circular path. This concept is commonly studied in physics and engineering, and is also known as circular motion or centripetal force.

2. How is work done in a circle calculated?

Work done in a circle can be calculated using the formula W = Fdcosθ, where W is the work done, F is the force applied, d is the distance traveled, and θ is the angle between the force and the direction of motion. This formula is derived from the dot product of the force and displacement vectors.

3. Why is the work done in a circle not zero?

The work done in a circle is not zero because even though the displacement may be zero (since the object returns to its starting point), there is still a force acting on the object that is causing it to move in a circular path. This force is constantly changing direction and thus, work is being done on the object.

4. Can the work done in a circle be negative?

Yes, the work done in a circle can be negative. This occurs when the applied force is in the opposite direction of the displacement, resulting in a negative value for the work done. This indicates that the force is acting against the motion of the object.

5. What are some real-life examples of work done in a circle?

Some examples of work done in a circle include a car driving on a curved road, a satellite orbiting around the Earth, and a roller coaster moving along a circular track. In all of these scenarios, there is a force acting on the object to keep it moving in a circular path, resulting in work being done.

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