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

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SUMMARY

The maximum kinetic energy (Tmax) for a particle in a bounded region is determined by the potential energy function V(r) = Eo[(a/r)^3 - (a/r)]. After analyzing the equilibrium points, it is established that r = sqrt(3)*a is a stable point, while r = -sqrt(3)*a is unstable. The correct formula for Tmax is derived as Tmax = Vmax - Vmin, resulting in Tmax = 4Eo/(3sqrt(3)). This conclusion clarifies the relationship between kinetic and potential energy in this context.

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  • Understanding of potential energy functions in classical mechanics
  • Knowledge of equilibrium points and stability analysis
  • Familiarity with kinetic energy calculations
  • Basic calculus for taking derivatives
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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.
 
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I got Tmax = Vmax - Vmin = Eo[ 2/(3sqrt(3)) - ( -2(3sqrt(3)))
Tmax = 4Eo/(3sqrt(3))

Is this correct?
 

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