Chapter 8 Appendices 8.A: Independence of Path for Work Done by a Conservative Force A worked example introduced in Section 8.3. 8.B: Work Done on a System of Two Particles An algebraic derivation of an important result in Section 8.4. 8.C: Energy Changes Near the Surface of the Earth An outline of the derivation of the “obvious” result used in Section 8.6. 8.D: Gravitational Potential Energy Due to a Spherical Shell Another way to show that the mass of a spherical shell may be considered to be concentrated at its center. Appendix 8A: Independence of Path for Work Done by a Conservative Force Section 8.3 of the text addressed the path-independence of the line integral of a conservative force, specifically that if path 1 c 1 path 2 c 2(1) , (2)BBAAWdW=⋅ = ⋅∫∫Fr Fdr, (8.A.1) path 1 path 2WW=. (8.A.2) The force represented in Figure 8.4 of the text was ˆˆyx=+Fij, so that The two paths were the circular arcs shown in Figure 8.A.1, reproduced here along with a third: dydxxd⋅= +Fry9/30/2008 1Figure 8.A.1 This Appendix will show that the integrals along the three paths shown (as well as a path, not shown in the figure, along the axes) give the same line integral, and further suggest that all line integrals between any two specified points in the x-y plane will give a result independent of the path joining those points. The four paths from the point ()1, 0A = to the point ()0,1B = considered are: 1. The upper arc (blue if seen in color) in Figure 8.A.1, a quarter of the circle of radius 1 centered at the origin. 2. The lower arc (green if seen in color) in Figure 8.A.1, a quarter of the circle of radius 1 centered at the point ()11,. 3. The two straight-line segments from ()10, to ()11, and from ()11, to ( (red if seen in color). )01, 4. The two straight-line segments from ()10, to ()00, and from to (00,)()01, (along the axes, and so not shown in the figure). Path 1: The circular path is best parameterized by polar coordinates, cossinsincosxydx ddy dθθθθθθ===−= (8.A.3) 9/30/2008 2where the polar angle varies from to 0 /2π. The integral is ()()/210path 1/2/200sin sin cos sin1cos2 sin 220.Wd ddπππdθθθ θ θθθθ θ=⋅= − +=− =−=∫∫∫Fr (8.A.4) Path 2: This circular arc is represented in terms of a different angle, 1sin1coscossinxuyudx ududy udu=−=−=−= (8.A.5) where the parameter u varies from to 0 /2π (u may be taken as the angle between the vertical and the vector from the point ()11, to the point on the arc). The integral is then (8.A.6) ()( )()(()/220path 2/22201cos cos 1sin sinsin cos cos sin .W d u udu u uduuu u uduππ=⋅= − − +−=−+−∫∫∫Fr) While the integrals are not hard, we’ll use symmetry arguments to note that (8.A.7) /2 /200/2 /22200sin cossin cosudu uduudu uduππππ==∫∫∫∫ and hence to see that 2path 20Wd=⋅=∫Fr. Path 3: From to (), (10,)11,1x= , 0dx= and ddy⋅=Fr; integration from to 0y = 1y= contributes to the line integral. From 1()11, to ()01,, 1y=, 0dy= and ; dd⋅=Frx9/30/2008 3integration from 1x= to contributes 0x =1− to the integral, with the result that . 3path 30Wd=⋅=∫Fr Path 4: From to , and (10,)()00,0y = 0dy=, so 0d⋅=Fr. From ()00, to , and , so again, and ()01, 0x =0dx =0d⋅=Fr4path 40Wd=⋅=∫Fr. By now a pattern should be emerging. For an arbitrary path in the plane, (),dydydxxdyyxddxdyx dxdx⎛⎞⋅= + =+⎜⎟⎝⎡⎤=⎢⎥⎣⎦Frx⎠ (8.A.8) so that any line integral of this force is found to be ()()()0000,00,ffffxyxxffxxxydWd xydxxyxydx⎡⎤=⋅= ==−∫∫⎢⎥⎣⎦Frxy (8.A.9) This example was chosen for simplicity of calculation, and it may seem that it’s artificial. Not quite. For this case, 22Fxyr==+=F and is a special case of a two-dimensional quadrupole field. A more realistic example (but still two-dimensional) of a quadrupole field is one that would have and have the work integral proportional to the difference between the function 31/F= r4/xyr evaluated at the endpoints of the path. A part of this force field is shown in Figure 8.A.2. Figure 8.A.2 9/30/2008 4The plot is in the third quadrant to show some similarity with Figure 8.A.1. Note the displaced origin (to avoid the singularity at ( ) and expanded scale; because of the strong variation of with , graphing in a larger region would merely show . 0, 0)31/F= r r0F ∼ Appendix 8.B: Work Done on a System of Two Particles In Section 8.4 we asserted that the work done by an internal conservative force in changing a system of two particles of masses and respectively from an initial state 1m2mA to a final state B is equal to (22c12BAWvμ=−)v (8.B.1) where is the square of the relative velocity in state2BvB, 2Av is the square of the relative velocity in state A, and ()12 1 2/mm m mμ=+ is a quantity known as the reduced mass of the system. Proof: We will not need to assume that the force is conservative, and so will drop the subscript “c” for the work done. Newton’s Second Law applied to body 1 is given expressed as 211,2 12dmdt=rF (8.B.2) and to body 2 as 222,1 22dmdt=rF. (8.B.3) Divide each side of Equation (8.B.2) by , 1m 21,2121dmdt=Fr (8.B.4) and divide each side of Equation (8.B.3) by , 2m 9/30/2008 522,1222dmdt=Fr. (8.B.5) Subtract Equation (8.B.5) from Equation (8.B.4) yielding 2221,2 2,1 1,21222 212dddm m dt dt dt−= − =FF rrr (8.B.6) where as in Section 8.4 of the text, 12 1 2,=−rrr. Use Newton’s Third Law, on the left hand side of Equation 1,2 2,1=−FF(8.B.6) to obtain 2221,2121,2221211dddm m dt dt dt⎛⎞+=−=⎜⎟⎝⎠rrrF2. (8.B.7) The quantity is the relative acceleration of body 1 with respect to body 2. 21,2/ddr2t Define 12111mmμ≡+; (8.B.8) as stated above, the quantity μ is known as the reduced mass of the system. Equation (8.B.7) now takes the form 21,21,22ddtμ=rF. (8.B.9) The total work done in the system in displacing the two masses from an initial state to a final state AB is given by 1,2 1 2,1 2BBAAWd d=⋅+ ⋅∫∫FrFr, (8.B.10) as shown in Equation (8.4.3) of the text. Recall that in this usage, the labels and AB denote states of the system, not paths. Except for trivial cases, the two bodies will not
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