Prove the work done by F is zero for a curve on a sphere

In summary, the homework statement is trying to find a way to show that if an object moves along a curve on the sphere, the work done by the force is zero. I tried to show that the curl of the force was zero, but realized that this was impossible as the function f could be anything. I'm looking for a hint in the right direction to solve this problem.
  • #1
SolidSnake
18
0

Homework Statement


Prove that if an object moves along any smooth simple curve C that lies on the sphere [tex]x^2 + y^2 + z^2 = a^2[/tex] in the force field [tex]F(x,y,z) = f(x,y,z)(xi + yj +zk)[/tex] where [tex]f[/tex] is a continuous function, then the work done by [tex]F[/tex] is zero.


Homework Equations



The Attempt at a Solution



I tried to show the curl of F was zero but realized that since f can be anything it'd be impossible to show that the curl was zero, atleast i think so :confused:. Not really sure how else to approach this problem. Looking for a hint in the right direction.
 
Physics news on Phys.org
  • #2
Off the top of my head i'd say to do it in generality you'd be looking at something like Greens theorem or Stokes theorem (one of them deals with conservative fields can't remember which)
 
  • #3
We've yet to learn Stokes Theorem, though we have learned Green's theorem. But I don't see how Green's theorem would apply here as it is just a specific case of Stokes Theorem for two dimension (correct?).
 
  • #4
At a point on the surface of the sphere, think about what direction your force field points.
 
  • #5
Doesn't that depend on the function f(x,y,z) ? since the force field will change depending on what that function is.
 
  • #6
f(x,y,z) is a scalar. The direction is going to be parallel to the vector, so...
 
  • #7
At any point the force would be the vector (f(x,y,z)x , f(x,y,z)y, f(x,y,z)z). Hmm the light bulb still hasn't gone on :(
 
  • #8
If f was constant what direction would your force field be? Remember f can only change the length of the vector or reverse its direction.
 
  • #9
The force would be going straight outwards, wouldn't it? with the length modified by f(x,y,z). which would mean it was perpendicular to the curve right?
 
  • #10
Bingo! Well, almost bingo. It might be pointed inward if f was negative at a point. All the scalar f can do is vary the length but the force must always point inward or outward. Now think about what direction you are moving along your surface curve in comparison to the force.
 
  • #11
Well if the force is perpendicular at one point on the surface curve, and the surface is sphere wouldn't the force always be perpendicular to the curve regardless of which curve is chosen.
 
  • #13
Well the work done by F would be zero then since it doesn't effect the "particle" when moving from point a on the curve to point b.

I now understand it intuitively. Thank you :) How would i go about proving this though. Would i need to parametrize a curve along the sphere and show that it is perpendicular to the force vector?
 
  • #14
SolidSnake said:
Well the work done by F would be zero then since it doesn't effect the "particle" when moving from point a on the curve to point b.

I now understand it intuitively. Thank you :) How would i go about proving this though. Would i need to parametrize a curve along the sphere and show that it is perpendicular to the force vector?

Aye, there's the rub. The pesky equations. What you have is a curve on the sphere:

[tex]\vec R(t) = \langle x(t),y(t),z(t)\rangle[/tex]

with

[tex]|\vec R(t)| = 1[/tex]

and a force that can be written as

[tex]\vec F = f(\vec R)\vec R[/itex]

And you need to show

[tex]W = \int_C \vec F \cdot d\vec R = 0[/tex]

I'm going leave you to think about that now. Express it in terms of t and see if you can figure out how to get that = 0. You obviously need to use

[tex]|\vec R(t)| = 1[/tex]

somehow in your argument, eh? Good luck. Sack time here.
 
  • #15
Thanks for the help. Good night :)
 

1. How can we prove that the work done by F is zero for a curve on a sphere?

To prove that the work done by F is zero for a curve on a sphere, we can use the fundamental theorem of line integrals. This theorem states that if a vector field F is conservative, then the work done by F along any closed curve is equal to zero. Therefore, if we can show that F is a conservative vector field, we can conclude that the work done by F is zero for any curve on a sphere.

2. What is a conservative vector field?

A conservative vector field is a type of vector field where the work done by the field along any closed curve is independent of the path taken. In other words, the work done by a conservative vector field only depends on the starting and ending points of the curve. This property is also known as path independence.

3. How do we determine if a vector field is conservative?

To determine if a vector field is conservative, we can use the curl test. If the curl of the vector field is equal to zero, then the vector field is conservative. This means that the field has no rotational or swirling behavior, and the work done by the field is path independent.

4. Can the work done by a vector field on a curve on a sphere ever be non-zero?

No, the work done by a vector field on a curve on a sphere will always be zero if the vector field is conservative. This is because the surface of a sphere is a closed surface, meaning that any curve on the sphere is a closed curve. Therefore, the fundamental theorem of line integrals applies, and the work done by a conservative vector field is always zero.

5. What are some real-life applications of proving the work done by F is zero for a curve on a sphere?

Understanding and proving the work done by a vector field on a curve on a sphere can have various applications in physics and engineering. One example is in the study of fluid dynamics, where the work done by a conservative vector field can help determine the flow of fluids in a closed system. This concept is also used in the study of electric and magnetic fields, where conservative fields play a crucial role in understanding the behavior of charged particles.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
829
  • Calculus and Beyond Homework Help
Replies
8
Views
348
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
877
  • Calculus and Beyond Homework Help
Replies
21
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
3K
  • Calculus and Beyond Homework Help
Replies
10
Views
734
  • Calculus and Beyond Homework Help
Replies
4
Views
937
  • Calculus and Beyond Homework Help
Replies
1
Views
787
Back
Top