Obtaining potential from force

Similarly, if I tell you that \frac{\partial f(x,y,z)}{\partial y}=g(x,y,z) then f(x,y,z)=\int g(x,y,z) dy + k(x,z). So when you saym A \omega^2 cos(\omega t) x = U_xYou are missing the contribution from h(y,z), since it should actually bem A \omega^2 cos(\omega t) x = U_x + h(y,z)So what is h(y,z)?Well, if x=Acos(\omega t) and we know x=x(t), then shouldn't y and z be constants?We were given a trajectory in fixed plane, so I am assuming that means
  • #1
csnsc14320
57
1

Homework Statement


A particle of mass m moves in a fixed plane along the trajectory [tex] \vec{r} = \hat{i} A cos(3 \omega t) + \hat{j} A cos(\omega t) [/tex].
(a) Sketch the trajectory of the particle.
(b) Find the force acting on the particle
(c) find its potential energy
(d) find its total energy as functions of time
(e) is the motion periodic? if so, find the period

Homework Equations


The Attempt at a Solution


Part (a) I won't worry about for now
Part (b): I think this is correct:
[tex] \vec{r} = \hat{i} A cos(\omega t) + \hat{j} A cos(3 \omega t) [/tex].
[tex] \vec{v} = -\hat{i} A \omega sin(\omega t) - \hat{j} 3 A \omega sin(3 \omega t) [/tex]
[tex] \vec{a} = -\hat{i} A \omega^2 cos(\omega t) - \hat{j} 9 A \omega^2 cos(3 \omega t}) [/tex]

[tex] \vec{F} = m \vec{a} [/tex]
[tex] \vec{F} = m (-\hat{i} A \omega^2 cos(\omega t) - \hat{j} 9 A \omega^2 cos(3 \omega t})) [/tex]

So, for part (b), this is where I'm wondering if I am correct:
[tex] F = -\frac{d U}{d t} [/tex]
[tex] U = m( A \omega sin(\omega t) + 3 A \omega sin(3 \omega t)) [/tex]

And, if this is correct, would this be the kinetic energy as a function of time?
[tex] K(t) = \frac{1}{2} m (-\hat{i} A \omega sin(\omega t) - \hat{j} 3 A \omega sin(3 \omega t))^2 [/tex]
Which would make part (c):
[tex] E = \frac{1}{2} m (-\hat{i} A \omega sin(\omega t) - \hat{j} 3 A \omega sin(3 \omega t))^2 + m( A \omega sin(\omega t) + 3 A \omega sin(3 \omega t)) [/tex]

Thanks
 
Last edited:
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  • #2
csnsc14320 said:
[tex] \vec{r} = \hat{i} A cos(\omega t) + \hat{j} A cos(\omega t) [/tex].
[tex] \vec{v} = -\hat{i} A \omega sin(\omega t) - \hat{j} 3 A \omega sin(\omega t) [/tex]
[tex] \vec{a} = -\hat{i} A \omega^2 cos(\omega t) - \hat{j} 9 A \omega^2 cos(\omega t}) [/tex]

Where are the 3 and 9 coming from?

So, for part (b), this is where I'm wondering if I am correct:
[tex] F = -\frac{d U}{d t} [/tex]

The time derivative of potential energy doesn't even have units of force...Isn't Force the negative gradient of U?:wink:
 
  • #3
Where are the 3 and 9 coming from?

Oops, I wrote the equation wrong. The original equation should be:

[tex] \vec{r} = \hat{i} A cos(\omega t) + \hat{j} A cos(3 \omega t) [/tex]

The time derivative of potential energy doesn't even have units of force...Isn't Force the negative gradient of U?:wink:

So how would you get the potential from the force? I don't see how you can reverse the gradient
 
  • #4
Well, in Cartesian coords, the gradient of some scalar function [itex]f[/itex] is of course

[tex]\vec{\nabla}f=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{j}+\frac{\partial f}{\partial z} \hat{k}[/tex]

So if I tell you that [tex]\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}=-\vec{\nabla}f[/tex], what can you tell me about [itex]A_x[/itex], [itex]A_y[/itex] and [itex]A_z[/itex]?
 
  • #5
gabbagabbahey said:
Well, in Cartesian coords, the gradient of some scalar function [itex]f[/itex] is of course

[tex]\vec{\nabla}f=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{j}+\frac{\partial f}{\partial z} \hat{k}[/tex]

So if I tell you that [tex]\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}=-\vec{\nabla}f[/tex], what can you tell me about [itex]A_x[/itex], [itex]A_y[/itex] and [itex]A_z[/itex]?

so...

[tex] A_x = -\frac{\partial f}{\partial x} [/tex]
[tex] A_y = -\frac{\partial f}{\partial y}[/tex]
[tex] A_z = -\frac{\partial f}{\partial z}[/tex]

If [tex] \vec{F} = -\vec{\nabla}U [/tex]
and [tex] \vec{\nabla}U=\frac{\partial U}{\partial x} \hat{i}+\frac{\partial U}{\partial y} \hat{j}+\frac{\partial U}{\partial z} \hat{k} [/tex]

I have

[tex]\vec{F} =- \vec{\nabla}U = m (-\hat{i} A \omega^2 cos(\omega t) - \hat{j} 9 A \omega^2 cos(3 \omega t})) = -\frac{\partial U}{\partial x} \hat{i}-\frac{\partial U}{\partial y} \hat{j}-\frac{\partial U}{\partial z} \hat{k} [/tex]

[tex]m A \omega^2 cos(\omega t) = \frac{\partial U}{\partial x} [/tex]
[tex]m 9 A \omega^2 cos(3 \omega t}) = \frac{\partial U}{\partial y} [/tex]
[tex]0 = \frac{\partial U}{\partial z} \hat{k} [/tex]

Would I integrate the left side with respect to x and y to get

[tex]m A \omega^2 cos(\omega t) x = U_x [/tex]
[tex]m 9 A \omega^2 cos(3 \omega t) y = U_y [/tex]

or just

[tex] U = m A \omega^2 cos(\omega t) x + m 9 A \omega^2 cos(3 \omega t) y [/tex]
 
  • #6
csnsc14320 said:
[tex]m A \omega^2 cos(\omega t) = \frac{\partial U}{\partial x} [/tex]
[tex]m 9 A \omega^2 cos(3 \omega t}) = \frac{\partial U}{\partial y} [/tex]
[tex]0 = \frac{\partial U}{\partial z} \hat{k} [/tex]

Correct :approve:

[tex] U = m A \omega^2 cos(\omega t) x + m 9 A \omega^2 cos(3 \omega t) y [/tex]

There are two things you need to be careful of here:

(1) [itex]m A \omega^2 cos(\omega t)[/itex] is not necessarily independent of [itex]x(t)[/itex] and [itex]m 9 A \omega^2 cos(3 \omega t})[/itex] is not necessarily independent of [itex]y(t)[/itex]. So in order to integrate them properly, you need to determine their functional dependence on [itex]x[/itex] and [itex]y[/itex] first.

Hint: As always, [tex]\vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}[/tex] so compare that to the trajectory you were given and find out what [itex]x(t)[/itex] and [itex]y(t)[/itex] are.

(2)If I tell you that [tex]\frac{\partial f(x,y,z)}{\partial x}=g(x,y,z)[/tex] Then that means [tex]f(x,y,z)=\int g(x,y,z) dx + h(y,z)[/tex] where [itex]h(y,z)[/itex] is some unknown function of [itex]y[/itex] and [itex]z[/itex]. It does not just mean that [tex]f(x,y,z)=\int g(x,y,z) dx[/tex].
 

Related to Obtaining potential from force

1. What is potential energy?

Potential energy is the energy that an object possesses due to its position, shape, or state. It is a measure of the stored energy that can be converted into other forms, such as kinetic energy.

2. How is potential energy related to force?

Potential energy is related to force through the concept of work. Work is the force applied to an object multiplied by the distance the object moves in the direction of the force. The work done on an object is equal to the change in its potential energy.

3. What is the formula for calculating potential energy from force?

The formula for calculating potential energy from force is PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

4. Can potential energy be negative?

Yes, potential energy can be negative. This occurs when an object has a lower potential energy than its reference point. For example, an object at the bottom of a hill has a lower potential energy than an object at the top of the hill, thus having a negative potential energy.

5. How is potential energy transformed into other forms of energy?

Potential energy can be transformed into other forms of energy, such as kinetic energy, through the application of a force. When a force is applied to an object, it can do work on the object and convert its potential energy into another form of energy.

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