Kinetic Energy and Work

Reviewing line integrals / path integrals from Calculus III:

• Scalar line integrals
• Vector line integrals (and calculating Work)

Kinetic energy, $T$, of a moving particle: $$T=\frac{1}{2}mv^2 = \frac{1}{2}m(\myv{v} \cdot \myv{v})$$

A particle moves, not necessarily on a straight path, from $\myv r_1$ to $\myv{r}_1 + d \myv{r}$. How is its kinetic energy changing? $$\begineq\frac{dT}{dt} =& \frac{1}{2} m \frac{d}{dt}(\myv{v} \cdot \myv{v})\\ =& \frac{1}{2} m (\dot{\myv{v}} \cdot \myv{v} + \myv{v} \cdot \dot{\myv{v}}) = m \dot{\myv{v}} \cdot \myv{v}\\ =& \myv{F} \cdot\myv{v}= \myv{F} \cdot \frac{d\myv{r}}{dt}\endeq$$

Multiplying both sides of this by $dt$, we're left with: $dT = \myv{F} \cdot d\myv{r}$

We can add up the contributions along the path of the particle's motion: $T_2 - T_1 = \int_1^2 \myv{F} \cdot d\myv{r} \equiv W(1 \rightarrow 2)$.

The integral above is a new beast called a line integral or path integral.

$W(1 \rightarrow 2)$ is the work done by the force on the particle between positions 1 and 2.

In general, that integral may very well be different depending on which path is taken from 1 to 2.

Evaluating path integrals

One way to evaluate that path integral is by splitting the dot product into components: $$\begineq \int_1^2 \myv{F} \cdot d\myv{r} =& \int_1^2 (F_x dx + F_y dy + F_z dz) \\ =&\int_{x_1}^{x_2} F_x dx + \int_{y_1}^{y_2}F_y dy + \int_{z_1}^{z_2}F_z dz\endeq$$

where the components $F_x$, etc, depend on the path we take.

Example: Problem 4.2

For a two-dimensional force $\myv{F} = (x^2,2xy)$, what's the work done by the force along the two paths a) and b) shown in the figure?

Path a

$$\begineq W =& \int_0^P \myv{F} \cdot d \myv{r} = \int_0^P (F_x dx + F_y dy)\\ =&\int_0^Q F_x dx + \int_Q^P F_y dy\endeq$$

Along path $OQ$: $F_x = x^2$, and the there is no change in $y$, so we only need to evaluate $\int_O^Q F_x\,dx$.

Along path $QP$: we have $x=1$, and so $F_y=2xy = 2(1)y=2y$. There's no change in $x$ along this path:

$$\begineq W=&\int_{x=0}^1 x^2 dx + \int_{y=0}^1 2y dy \\ =& \left. \frac{x^3}{3} \right|_0^1 +\left. 2\frac{y^2}{2} \right|_0^1 = \frac{1}{3} + 1 = 4/3\endeq$$

Path b

Path b is along the path where $y=x^2$. $$\begineq W=&\int_0^P \myv{F} \cdot d \myv{r} = \int_0^P(F_x dx +F_y dy)\\ =& \int_{x=0}^1 x^2\, dx + \int_{y=0}^1 2xy\,dy\\ \endeq$$

The first integral is $1/3$. For the second integral above, we're not finished because $x$ is not a constant along this path. Actually, $x=x(y)=\sqrt y$. So, we can re-write the integral to be a function of $y$ alone: $$\begineq W=&\frac 13 +\int_{y=0}^1 2\sqrt y y\,dy=\frac 13 +\int_{y=0}^1 2y^{3/2}\,dy\\ =&\frac 13 +\left. 2 \frac 25 y^{5/2}\right|_0^1=\frac 13+\frac 45=\frac{17}{15} \endeq$$

An alternate way to evaluate the integral along the path (this better give the same result!):

• change variables of integration from $y$ to $x$:
• Since $y=x^2 \Rightarrow dy = 2x\, dx$.
• Since we're integrating over $x$ instead of over $y$, we should change the limits of integration to the limits of $x$ along path b: as $y$ goes $0 \rightarrow 1$, $x$ is going $0 \rightarrow 1$.
• Finally we'll substitute $y=y(x)=x^2$:

$$\begineq W =& \frac{1}{3} + \int_{x=0}^{1} 2 x (x^2) 2x dx = \frac{1}{3} + \int_{x=0}^{1} 4x^4 dx \\ =& \frac{1}{3} + \left. 4\frac{x^5}{5} \right|_0^1 = \frac{1}{3}+\frac{4}{5}= \frac{17}{15}\endeq$$

Conservative forces

A special class of conservative forces lets us define a corresponding potential energy.

The conditions for a force to be conservative are:

What about...

1. Our velocity-dependent air resistance forces?
2. $$F_{grav}=m \myv{g} \Rightarrow W(1 \rightarrow 2)=mg(h_1-h_2)$$

In Calculus III we found a bunch of inter-related characteristics of "Conservative vector fields":

• Path independence $$\oint_{\text{all } C} \myv F=0$$
• The vector field is "conservative".
• The curl of the force field vanishes. $$\myv \grad \times \myv F = 0$$

  [Here are my notes on "Differential calculus as plumbing" and the curl. The curl measures "swirliness" at a point. A field $\myv F$ that has zero curl everywhere has "no swirliness" or "no circulation" or is "irrotational".]

• Cross derivatives equal (2-d: Using a common convention that $\myv F=P(x,y)\uv x+Q(x,y)\uv y$, that is $P$ is the $x$ component of the force and $Q$ is the $y$ component of the force: $$ \frac{\partial}{\partial y}P = \frac{\partial}{\partial x}Q $$
• The vector field is the gradient of some potential function $f$, $$\myv F=\myv \grad f.$$

Potential energy

For conservative forces we can define a potential energy

$$U(\myv{r}) \equiv -W(\myv{r}_0 \rightarrow \myv{r}) = -\int_{\myv{r}_0}^{\myv{r}} \myv{F}(\myv{r}') \cdot d\myv{r}'.$$

Because of the way the limits on the integral work, we can say, for a trip through three points 0, 1, and 2, that: $$W(0 \rightarrow 2) = W(0 \rightarrow 1) + W(1 \rightarrow 2)$$ $$\begineq \Rightarrow \Delta T =& W(1 \rightarrow 2) = W(0 \rightarrow 2) - W(0 \rightarrow 1)\\ =& -[U(\myv r_2 ) - U(\myv{r}_1)] = - \Delta U.\endeq$$

We can re-arrange that to read: $$\Delta(T+U) = 0$$

That is to say: between any two points (1 and 2 are nothing special) this quantity does not change (is conserved).

This motivates the definition of the "mechanical energy", $E$:

$$E \equiv KE + PE = T+U(\myv{r})$$

$E$ is a scalar quantity which does not change as long as the force acting on the particle is conservative.

If several different conservative forces (for example, both gravity, and a spring) are acting on the particle at the same time, the forces add as vectors to give a net, conservative force, and the potential energies add such that we can state this principle of conservation of energy more generally:

Non-conservative forces

The work-energy relationship $\Delta T = W$ still holds whether the forces doing work are conservative or not. But it is useful to separate out the work into that done by conservative forces and that done by non-conservative forces. $$\Delta T = W = W_{cons} + W_{nc}$$

Since $W_{cons}= -\Delta U$, we could write the change in mechanical energy as: $$\Delta E = \Delta (T+U) = W_{nc}$$

Now the mechanical energy can change--but exactly by the work done by the non-conservative forces. Often these are frictional forces directed opposite the motion.

Homework