PHY5200 F06

Chapter 4: Energy

Reading

Taylor 4.1-4.2 (today) and 4.2-4.3 (Monday).

Recall

Relations for linear and angular momentum from chapter 3.

Kinetic Energy and Work

We define the kinetic energy of a single particle as T=½mv², where m is the particle's mass and v is the magnitude of its velocity. The time derivative of the kinetic energy is dT/dt = ½m d(v⋅v)/dt = mv[dot]⋅v = ma⋅v = F⋅v. Multiply both sides by dt and note that vdt = dr to get a relation known as the Work-Energy Theorem

dT = F⋅dr
In this form, both sides of the equation can be integrated along the path followed by the particle. To integrate along a path from r1 to r2, first imagine dividing the path up into lots of small steps of length Δri, then adding up along the path to get
∑ΔTi = ΔT = T2 - T1 = ∑Fi⋅Δri
Take the limit as the Δr's go to zero, and the sum goes over to a line integral
∑Fi⋅Δri --> ∫12 F⋅dr.
A line integral is a one-dimensional integral performed over a path rather than a single coordinate like x. The normal way to calculate such an integral is to parameterize the path in terms of another variable, such as t, then integrate over that variable. Sometimes it is easy to break a single path into several segments, where each segment can be easily integrated and the results summed together.

Example: Calculating Line Integrals

Evaluate the work done

W = ∫PQ F⋅dr = ∫PQ (Fx dx + Fy dy)

by the two-dimensional force F=(-yn, xn) along the three paths joining the points P = (1,0) and Q = (0,1): (a) the path that goes from P to the origin, O = (0,0), and then to Q; (b) the path that goes from P to Q along a straight line (express y as a function of x and rewrite the integral as an integral over x); and (c) the quarter circle from P to Q centered on the origin, O.

For each case, we want to express the path in terms of a single variable to perform the integral, or part of the integral. Let's tackle the these one at a time.

(a) This is the easiest case. We can break the entire path into two segments, one running along x with y constant, and the other along y with x constant. When moving along x with y constant, dr = dx xhat, and when moving along y with x constant, dr = dy yhat. Then F⋅dr will reduce to Fxdx = -yndx when moving along x with y constant, and Fydy = xndy when moving along y with x constant. These are integrated along the appropriate segment and summed to get the work,

W = ∫PO Fxdx + ∫OQ Fydy = ∫x=10 -yndx(y=0) + &inty=01 xndy(x=0) = 0 + 0 = 0

(b) This integral is done by parameterizing the path in terms of one variable. The obvious way is to express y as a function of x, y=1-x. Then dy = -dx, so we can replace y and dy with x and dx and perform the integral as follows

W = ∫PQ (-yndx + xndy) = -∫10 ((1-x)n +xn)dx = 1/(n+1)[1n+1 + 1n+1] = 2/(n+1)

(c) This is an example of the most general case where you have to parameterize the path in terms of a new variable. Normally, in mechanics, we know r(t), so the natural parameter is t. In this particular example, the natural parameter is an angle φ going from 0 at P to π/2 at Q, where x=cosφ and y=sinφ. We also need to replace dx and dy with dx=-sinφ dφ and dy=cosφ dφ. The integral for the work becomes

W = ∫0π/2 (sinn+1φ + cosn+1φ)dφ

This is a difficult integral to do in the general case, so let's choose a couple of easy examples. If n=0, then the integral is W = -cos(π/2) + cos(0) + sin(π/2) - sin(0) = 2. If n=1, then the integrand becomes sin²&phi + cos²φ = 1, and the integral is W = π/2.

The force in the integral for the work is the net force on a particle, Ftot = F1 + F2 +...+ Fn = ∑i=1n Fi. The work integral over the sum of forces can be changed to a sum over integrals of the forces:

W = ∫12 ∑i Fi⋅dr = ∑i ∫12 Fi⋅dr
So the total work done can be calculated by finding the work done by each force separately and summing.

If the total work done on an object is zero, then the kinetic energy is the same at the beginning and end of the path (but maybe not in between), and the magnitude of the velocity is the same. If the net force on an object is zero, then the work done along any path is zero, the kinetic energy is constant, and the speed of the particle (magnitude of the velocity) is constant.

© Robert Harr 2006