Deriving Acceleration from Potential Energy?

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Discussion Overview

The discussion revolves around deriving acceleration components from a given potential energy expression in a three-dimensional space. Participants explore the relationship between potential energy, force, and acceleration, using concepts from classical mechanics, particularly the conservation of energy and gradients of potential energy.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant presents a potential energy expression, V(x,y,z) = α·x + β·y² + γ·z³, and seeks to find the velocity at a specific point using conservation of energy.
  • Another participant questions the squaring of the potential energy term in the velocity equation.
  • It is suggested that the force can be derived from the negative gradient of the potential energy, leading to the acceleration components being determined by dividing the force by mass.
  • A participant discusses the dependence of acceleration on the coordinates x, y, and z, and introduces the use of the chain rule to express acceleration in terms of velocity and position.
  • One participant calculates the z-component of acceleration as az = -(3γ/m)·z² and questions whether this expression is sufficient or needs further reduction.
  • Another participant responds that if the gradient was computed correctly, the expression for az should be adequate for further analysis.

Areas of Agreement / Disagreement

Participants generally agree on the method of deriving acceleration from the gradient of potential energy, but there are differing views on the sufficiency and expression of the acceleration components, particularly regarding the z-component.

Contextual Notes

The discussion includes assumptions about the correctness of the gradient calculation and the implications of variable dependencies on acceleration. There are unresolved questions about the adequacy of the derived expressions for acceleration.

cj
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I have a known potential energy, V, expression:

V(x,y,z) = α·x + β·y2 + γ·z3

I'm given: @(0,0,0), v = v0 and then asked to find v at (1,1,1).

I can determine v from Conservation of Energy:

v2 = v02 - (2/m)·(α + β + γ)2

In general, what is the expression for the accelerations ax, ay, az?

Do I find F from -∇V?

If so, what's next (as far as finding the acceleration's x, y and z-components)?

Thanks!
 
Physics news on Phys.org
1.Why have you squared the potential energy term??

2. Yes, and divide F by m to find the accelerations.
 
cj said:
I have a known potential energy, V, expression:

V(x,y,z) = α·x + β·y2 + γ·z3

I'm given: @(0,0,0), v = v0 and then asked to find v at (1,1,1).

I can determine v from Conservation of Energy:

v2 = v02 - (2/m)·(α + β + γ)2

In general, what is the expression for the accelerations ax, ay, az?

Do I find F from -∇V?

If so, what's next (as far as finding the acceleration's x, y and z-components)?

Thanks!

arildno sort of told you how to start it off. You should have F (and a) from the gradient of V. However, you will notice that "a" has a dependence on x, y, and z. If a is a function of t, then it is trivial to find v. But you don't have that here.

So what you need to do to find v is to use some calculus gymnastics by invoking the chain rule, i.e.

a = dv/dt = (dv_x/dx * dx/dt)i^ + (dv_y/dy * dy/dt)j^ + (dv_z/dz * dz/dt)k^

It is easier to solve this component by component, so for the x-component, you have

a_x = dv_x/dx * v_x (since dx/dt = v_x)

Thus, a_x dx = v_x dv_x

I think you should be able to handle the baby integral here using the initial conditions given. Do the same thing for the other 2 components.

Zz.
 
Thanks a lot Zz.

When solving, for example, az, I arrive at:

az = -(3γ/m)·z2

In general, is this a sufficient expression for az,
or should it be reduced or otherwise expressed differently?

ZapperZ said:
arildno sort of told you how to start it off. You should have F (and a) from the gradient of V. However, you will notice that "a" has a dependence on x, y, and z. If a is a function of t, then it is trivial to find v. But you don't have that here.

So what you need to do to find v is to use some calculus gymnastics by invoking the chain rule, i.e.

a = dv/dt = (dv_x/dx * dx/dt)i^ + (dv_y/dy * dy/dt)j^ + (dv_z/dz * dz/dt)k^

It is easier to solve this component by component, so for the x-component, you have

a_x = dv_x/dx * v_x (since dx/dt = v_x)

Thus, a_x dx = v_x dv_x

I think you should be able to handle the baby integral here using the initial conditions given. Do the same thing for the other 2 components.

Zz.
 
cj said:
Thanks a lot Zz.

When solving, for example, az, I arrive at:

az = -(3γ/m)·z2

In general, is this a sufficient expression for az,
or should it be reduced or otherwise expressed differently?

ASSUMING you did the gradient correctly, that should be a sufficient expression for the a_z to play with.

Zz.
 

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