# Obtaining potential from force

1. Feb 17, 2009

### csnsc14320

1. The problem statement, all variables and given/known data
A particle of mass m moves in a fixed plane along the trajectory $$\vec{r} = \hat{i} A cos(3 \omega t) + \hat{j} A cos(\omega t)$$.
(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

2. Relevant equations

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

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

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

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

Thanks

Last edited: Feb 17, 2009
2. Feb 17, 2009

### gabbagabbahey

Where are the 3 and 9 coming from?

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

3. Feb 17, 2009

### csnsc14320

Where are the 3 and 9 coming from?

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

$$\vec{r} = \hat{i} A cos(\omega t) + \hat{j} A cos(3 \omega t)$$

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

So how would you get the potential from the force? I don't see how you can reverse the gradient

4. Feb 17, 2009

### gabbagabbahey

Well, in Cartesian coords, the gradient of some scalar function $f$ is of course

$$\vec{\nabla}f=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{j}+\frac{\partial f}{\partial z} \hat{k}$$

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

5. Feb 17, 2009

### csnsc14320

so...

$$A_x = -\frac{\partial f}{\partial x}$$
$$A_y = -\frac{\partial f}{\partial y}$$
$$A_z = -\frac{\partial f}{\partial z}$$

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

I have

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

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

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

$$m A \omega^2 cos(\omega t) x = U_x$$
$$m 9 A \omega^2 cos(3 \omega t) y = U_y$$

or just

$$U = m A \omega^2 cos(\omega t) x + m 9 A \omega^2 cos(3 \omega t) y$$

6. Feb 17, 2009

### gabbagabbahey

Correct

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

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

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

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