STOKES THEOREM (fundamental theorem of calculus for curls) ∬(∇ x V) • dS = ∮ V • dl where C is the boundary path of the surface S δ = ∂S boundary path/surface is always closed → Sign convention (open surface): Right-hand thumb rule ↑ (+ve) → two surface same boundary (imp special case) Now ∮ V • dl = 0 when ∇ x V = 0 (conservative force concept) ie V = ∇T T is a scalar field → Now what happens when we shrink the boundary then RHS→0 thus also LHS→0. Hence LHS for a closed surface is zero (always) Gauss + Stokes from these we can prove ∇ x (φV) = 0 and ∇ • (∇ x A) = 0 ∬ curl (V) dS = 0 S CURVILINEAR COORDINATE SYSTEMS Plane polar coordinates x = r cos(theta) const r: circles y = r sin(theta) const theta: halflines Now delta x = cos(theta) delta r - r sin(theta) delta theta delta y = sin(theta) delta r + r cos(theta) delta theta thus r_hat = cos(theta) i_hat + sin(theta) j_hat (radial direction) Now keep delta r = 0 then delta x = -r sin(theta) delta theta delta y = r cos(theta) delta theta theta_hat = -sin(theta) i_hat + cos(theta) j_hat Now we can represent them in matrix form (r_hat, theta_hat) = (cos(theta), sin(theta), -sin(theta), cos(theta)) (i_hat, j_hat) Now r_hat and theta_hat are not const vectors unlike i_hat and j_hat wrt the position. (a minor drawback) r_hat dot theta_hat = 0 curvilinear but still orthogonal Elemental distance: dl sr r_hat + r delta theta theta_hat from cartesian (dimensionally correct) dl = delta x i_hat + delta y j_hat h_r = sqrt((partial x / partial r)^2 + (partial y / partial r)^2) = 1 h_theta = sqrt((partial x / partial theta)^2 + (partial y / partial theta)^2)^(1/2) = r Now scale factor is not unity for curvilinear coordinate systems. Scale factor gives a measure of how much a change in the coordinates changes the position of a point from comparison: ∇ • T(r,θ) = (∂T/∂r) * r̂ + (1/r) * (∂T/∂θ) * θ̂ → Velocity Components gradient in polar coordinates v = d/dt (r * r̂) = ṙ * r̂ + r * ṙ̂ r̂ = cosθ * î + sinθ * ĵ note: r̂ is not a constant (important) (r̂, θ̂, θ̂) = θ̇ * (-sinθ cosθ, -cosθ -sinθ) * (x̂, ŷ) dr̂/dt = (-sinθ * î + cosθ * ĵ) * dθ/dt = θ̂ * dθ/dt = θ̂ * θ̇ thus by substitution ↓ (r̂, θ̂, θ̂) = θ̇ * (θ̂, -r̂) v = dr/dt = ṙ * r̂ + (rθ̇) * θ̂ vr vθ Vr: radial velocity component Vθ: circumferential comp Now acceleration components. a = dv/dt = r̈ * r̂ + ṙ * r̂ + ṙ * r̂ + r * r̈̂ a = (r̈ - rθ̇²) * r̂ + (rθ̇ + 2ṙθ̇) * θ̂ Radial acceleration Circumferential "if the force on the particle is central; then which quantity is conserved?" → live lecture assignment Live Lecture (2) week2 (29/3/2022) vector analysis: integral calculus, line integral Surface integral Curvilinear coordinates. Dis path independence obvious? Nope: not all vectors can be represented as ∇T (gradient) path independent if and only if ∇f = ∇S f(x,y) = xy^3 + xy^2 ∫∇f · dl v = 7S S is scalar field A(0,0) → B(1,2) (HW) RECORDED LECTURE-3 s = sqrt(x^2 + y^2) and φ = tan^-1(y/x) ŝ = (xî + yĵ)/s and ṧ = sŝ and Ḟ = d/dφ(ŝ) ŝ(φ) = cos φ î + sin φ ĵ dr = dsŝ + s dφḞ area element = dxdy = dsŝ + s dφḞ dA = [s ds dφ] ẑ dĀ = (dsŝ × s dφḞ) dř = dxî + dyĵ acceleration = ā = (Ḋ - sḞ^2)ŝ + (sḞ + 2ḊḞ)Ḟ = asŝ + aφḞ nos: d^2r/dt^2 along r: -sφ^2 = -rw^2 (centrifugal) -s sφ = αγ = tangential acc 2sφ: coriolis (due to torque) (due to radial force)

Stokes Theorem

Related Documents

Recommended Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock