Local Minimum of Potential Function of Spherical Pendulum

[ovidiupetre]

Homework Statement

http://img13.imageshack.us/img13/5793/84188411.jpg

Homework Equations

Find a condition on b such that x = 0 is a local minimum of the potential function.

The Attempt at a Solution

To find local minimum, potential function (V) of the system should be written. V must be positive definite and derivative of V must be negative semi definite. I tried to write hundreds of potential functions that provide local minimum constraints but i can't get rid of sin and cos terms from derivative of V so i couldn't find a condition for b to show x = 0 is a local minimum of the potential function.

 

[fzero]

Which derivative of V must be negative? Also remember that if you write the equation of motion in terms of the force, you can get a nice expression for the potential.

 

[ovidiupetre]

Which derivative of V must be negative? Also remember that if you write the equation of motion in terms of the force, you can get a nice expression for the potential.

First derivative of course. I couldn't get a nice expression for the potential, my equations are not simplifiable.

 

[praharmitra]

The equations of motion are of the form
$$\ddot{x} = -\frac{\partial V}{\partial x}$$
From this you can read off V. Now that you have the function V(x), what are the conditions for such a function to have a minimum at x=0 ?

 

[ovidiupetre]

The equations of motion are of the form
$$\ddot{x} = -\frac{\partial V}{\partial x}$$
From this you can read off V. Now that you have the function V(x), what are the conditions for such a function to have a minimum at x=0 ?

V is a candidate Lyapunov function and it must be positive definite. Moreover, first derivative of V must be negative semi definite in order to x = 0 be local minimum.

 

[fzero]

The first derivative vanishes at an extremum. The second derivative is used to distinguish between local maxima and minima. You don't need to solve for V explicitly to do this problem, but it's not that hard to do so.

 

[ovidiupetre]

The first derivative vanishes at an extremum. The second derivative is used to distinguish between local maxima and minima. You don't need to solve for V explicitly to do this problem, but it's not that hard to do so.

Problem is not about solving V, the problem is "construction of V". If i construct V, i can find a condition to make x = 0 local minimum by looking negative definiteness of the derivative of V.

 

[fzero]

Problem is not about solving V, the problem is "construction of V". If i construct V, i can find a condition to make x = 0 local minimum by looking negative definiteness of the derivative of V.

You're given the first derivative of V. This is enough information to compute the second derivative and find the condition.

 

[ovidiupetre]

You're given the first derivative of V. This is enough information to compute the second derivative and find the condition.

First derivative of potential function is not given. Construction of potential function (V) is the aim or the step that is needed to pass. When you construct potential function, the rest is easy.

 

[fzero]

First derivative of potential function is not given. Construction of potential function (V) is the aim or the step that is needed to pass. When you construct potential function, the rest is easy.

As I explained in post #2, you can relate the acceleration to the potential. praharmitra gave you the formula up to a factor of the mass of the particle in post #4.