Deriving Acceleration from Potential Energy?

AI Thread Summary
The discussion centers on deriving acceleration components from a given potential energy expression, V(x,y,z) = α·x + β·y² + γ·z³. The user seeks to find the velocity at a specific point using conservation of energy and is guided to calculate forces from the gradient of V. It is clarified that acceleration components ax, ay, and az can be derived by dividing the force by mass and applying the chain rule for differentiation. The user successfully derives az = -(3γ/m)·z² and inquires about its sufficiency, to which it is confirmed that it is adequate for further analysis. The conversation emphasizes the importance of careful calculus application in solving for motion parameters.
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!
 
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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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