Calculate for each field the work done in moving a particle

[jlmac2001]

Problem:

Which, if either, of the two force fields F1=-yi+xj+zk, F2=yi+xj+zk is conservative? Calculate for each field the work done in moving a particle counterclockwise around the circle x=cos(t), y=sin(t) in the (x,y) plane.


Answer:

I got the part in telling if the force fields are conservative. F1 is not conservative because d/dx(Fy)-d/dy(Fx) is 2. F2 is conservative because everything is equal to zero.

How do I calculate the work done in each field? I don't get it.

 

[HallsofIvy]

If a force field is conservative, then work done is the difference in potential energy at the two end points. In particular, the work done in moving around a closed path is 0. (The work done in lifting a mass against gravity can be regained by dropping it back down.)

If a force field is not conservative, then you will have to integrate the dot product of the force with ds around the path.

In this case, x= cos(t), y= sin(t), z= 0, so ds= dxi+ dyj+ dzk=
-sin(t)i+ cos(t)j+ 0k. F . ds= (-yi+xj+zk).(-sin(t)i+ cos(t)j+ 0k)
= (-sin(t)i+ cos(t)j+0k).(-sin(t)i+ cos(t)j+ 0k)= sin²2(t)+ cos²(t)= 1. The work done is just the integral of the constant 1 around the circle and so is just the circumference of the circle.

If you did integrate F2 in the same way, F2.ds= (yi+xj+zk).(-sin(t)i+ cos(t)j+ 0k)= (sin(t)i+cos(t)j).(-sin(t)i+ cos(t)j+ 0k)= cos²(t)- sin²(t)= cos(2t) and the integral of that around the circle (t going from 0 to 2π) is 0.

 

[M98Ranger]

To calculate the work done in each field, we can use the formula W = ∫ F · dr, where F is the force field and dr is the displacement vector. Essentially, we are finding the dot product of the force and displacement vectors and integrating it over the path of the particle.

For F1, we can write the force as F1 = -y i + x j + z k, and the displacement vector dr as dr = dx i + dy j + dz k, where dx and dy represent the infinitesimal changes in x and y coordinates as the particle moves along the circle, and dz is zero since the particle is moving in the (x,y) plane.

Substituting these values into the formula, we get W = ∫ (-y dx + x dy) = ∫ x dy, since the y and z components of F1 are zero. We can then substitute the parametric equations for x and y, x = cos(t) and y = sin(t), into the integral to get W = ∫ cos(t) dy = ∫ cos(t) cos(t) dt = ∫ cos^2(t) dt.

Using the identity cos^2(t) = (1 + cos(2t))/2, we can rewrite the integral as W = ∫ (1 + cos(2t))/2 dt. Integrating this, we get W = (t/2) + (sin(2t)/4) + C.

Since we are moving counterclockwise around the circle, the limits of integration for t would be from 0 to 2π. Plugging these values in, we get W = π/2.

For F2, we can use the same process, but since F2 is conservative, the work done will not depend on the path taken. So, we can just calculate the work done at a single point on the circle, say (1,0), and that will be the same for all points on the circle.

Substituting x = 1 and y = 0 into the formula, we get W = ∫ (0 dx + 0 dy) = 0.

Therefore, the work done in moving a particle counterclockwise around the circle in F1 is π/2, while the work done in F2 is 0. This further confirms that F2 is conservative, while F1 is not.