Is there any way to prove that a given force is conservative?

[alphabeta1720]

Is there any way to prove that a given force is conservative?

 

[f95toli]

Is there any way to prove that a given force is conservative?

Yes, a force is conservative if the curl is zero.
Or, equivalently, the net work done is zero if you move a particle in a path that starts and ends in the same place.

 

[Idoubt]

Mathematically a vector is said to be conservative if it is the gradient of a potential

This means that for the space considered there must be a potential at every point ( can be zero ) and the work done from moving from one point to another depends only on the difference between the potentials at these two points. So the path has no say

If you want to know if a force is conservative, and if you have the equation for the force ( in terms of catesian or polar co ordinates ) then you have use the mathematical curl operation (del cross f) where del is the vector derivative with respect to space. If the force f is conservative this term will be zero.

The reason for that is that a gradient can be represented as (del V ) where V is a scalar potential.

Then the curl operation becomes ( del cross del V ) since these 2 vectors point in the same direction the cross product has to be zero.


more on del operator
http://hyperphysics.phy-astr.gsu.edu/hbase/vecal.html

 

[sub_zero]

i'm not sure from your question, but you can prove this by analyzing a few simple/isolated systems such as an object hung to a string on a frictionless surface, a simple pendulum, and many others examples, those are simple, and you can further study about the conservation of energy for more understanding.

 

[alphabeta1720]

Thank you for that
but can anyone help me solving these problems (Sorry, but I don't know about 'curl' and 'del' operators)

Q1 Consider the two dimensional force F = f(x,y) i + g(x,y) j . Is it possible to determine whether this is a conservative force without any additional information? What if f(x,y) = f(x)
and g(x,y) = g(y)

Q2 which of the following forces is conservative?

A) F = y i - x j B) F = xy i - xy j
C) F = y i + x j D) F = xy i + xy j
E) F = 3x i + 4y j
(There may be more than 2 correct answers)

Thank You

 

[HallsofIvy]

Thank you for that
but can anyone help me solving these problems (Sorry, but I don't know about 'curl' and 'del' operators)

Q1 Consider the two dimensional force F = f(x,y) i + g(x,y) j . Is it possible to determine whether this is a conservative force without any additional information? What if f(x,y) = f(x)
and g(x,y) = g(y)

Yes, there is. The work done in moving an object using that force along a path in the plane by \int_p \vec{F}\cdot d\vec{s}= \int_p f(x,y)dx+ g(x,y) dy depends only on the beging and ending points and is independent of the specific path. Once can show that that is true if and only if there exist some "potential function" G(x,y) (numerical valued, not vector valued) such that \nabla G= \vec{F} which just means that
f(x,y)= \frac{\partial G}{\partial x}
and
g(x,y)= \frac{\partial G}{\partial y}
Assuming there is such a function G, then
\frac{\partial^2 G}{\partial y\partial x}= \frac{\partial f(x,y)}{\partial y}
and
\frac{\partial^2 G}{\partial x\partial y}= \frac{\partial g(x,y)}{\partial x}

Since, as long as the derivatives are continuous, the "mixed derivatives" are continuous, we can say that a force, \vec{F}= f(x,y)\vec{i}+ g(x,y)\vec{j} is "conservative" if and only if it satifies the "cross condition"
\frac{\partial f}{\partial y}= \frac{\partial g}{\partial x}

Q2 which of the following forces is conservative?

A) F = y i - x j B) F = xy i - xy j
C) F = y i + x j D) F = xy i + xy j
E) F = 3x i + 4y j
(There may be more than 2 correct answers)

Thank You

Apply the condition
\frac{\partial f}{\partial y}= \frac{\partial g}{\partial x}
to each of those.

(I can't help but think that if you are asked a question like this, you are expected to have learned all this before!)

 

[alphabeta1720]

(I can't help but think that if you are asked a question like this, you are expected to have learned all this before!)

OK! next I am going to do is to learn about them