What Determines the Maximum Kinetic Energy for a Particle in a Bounded Region?

[Nusc]

say I'm givin the potential energy fxn

V(r) = Eo [ (a/r)^3 - (a/r) ]

Eo a constant

After taking the derivatives to find the equilibrium pts and checking for stability we get

r = sqrt(3)*a and this pt is stable

r = -sqrt(3)*a and this pt is unstable

Now let's pose the question

b) What is the max kinetic energy that the particle can have and still remain bounded (ie. remain within a region of finite spatial extent)

Tmax = Vmax - Vmin right?

I am still all little confused about the concept from "stable eq I" Only because that question didn't have 2 equilibrium pts.

 

[Nusc]

I got Tmax = Vmax - Vmin = Eo[ 2/(3sqrt(3)) - ( -2(3sqrt(3)))
Tmax = 4Eo/(3sqrt(3))

Is this correct?

 

[quicknote]

First, let's review the concept of stable equilibrium. In a system with stable equilibrium, the potential energy function has a minimum at the equilibrium point. This means that any small disturbance from the equilibrium point will result in a restoring force that brings the system back to the equilibrium point. In contrast, an unstable equilibrium point has a maximum potential energy, meaning any small disturbance will cause the system to move away from the equilibrium point.

In this case, the potential energy function given is V(r) = Eo [ (a/r)^3 - (a/r) ]. By taking the derivatives and setting them equal to zero, we find two equilibrium points: r = sqrt(3)*a and r = -sqrt(3)*a. The first point is a stable equilibrium point because it is a minimum of the potential energy function. The second point, however, is an unstable equilibrium point because it is a maximum of the potential energy function.

Now, let's consider the question of maximum kinetic energy. The maximum kinetic energy that the particle can have and still remain bounded is equal to the difference between the maximum and minimum potential energies. In this case, the maximum potential energy is V(sqrt(3)*a) = -2Eo and the minimum potential energy is V(-sqrt(3)*a) = 2Eo. Therefore, the maximum kinetic energy is Tmax = Vmax - Vmin = -2Eo - 2Eo = -4Eo.

In comparison to the concept of stable equilibrium in the previous question, this scenario has two equilibrium points, but the concept remains the same. The stable equilibrium point is the minimum of the potential energy function, and the unstable equilibrium point is the maximum of the potential energy function. The only difference is that in this case, we are also considering the maximum kinetic energy that the particle can have and still remain bounded.