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Center of Mass and Angular Momentum Center of Mass and Angular Momentum (c)2017 van Putten 1 Outline Seesaw balance Anglicisation of French “ci-ça” (back-and-forth); or from "scie" - French for "saw" with Anglo Saxon Center of mass "saw." "scie-saw" became “see saw." (Wikipedia) Force, CM and torque Angular momentum Ice skater - ballerina effect Spinning top Inertial torque Precession Nutation (c)2017 van Putten 2 Center of mass http://www.pbs.org/opb/circus/classroom/circus-physics/center-mass/ (c)2017 van Putten 3 Rotation of center of mass about pivots balanced unbalanced double trouble b' b' b b: horizontal distance CM to pivot (c)2017 van Putten 4 Schematic overview buckling pivot CM (c)2017 van Putten 5 CM defined M 2 M1 CM CM of two bodies is along a line connecting them: M1r1 + M 2r2 = 0 CM is relatively close to the heavy mass (c)2017 van Putten 6 CM defined M q = 2 M 2 M1 Distances to CM: M1 CM M 2 q r1 = r = r M1 + M 2 1+ q M1 1 r r2 = − r = − r 2 M M 1 q r1 1 + 2 + r = r1 − r2 (c)2017 van Putten 7 Balance of seesaw (I) q 1 pivot r = r, r = − r 1 1+ q 2 1+ q Balanced: CM at pivot l l M 2 q = 1: M1 = M 2 pivot M1 CM no net torque, no tendency to rotate (c)2017 van Putten 8 Balance of seesaw (II) Virtual displacement dEi = M igdhi (i = 1,2) pivot Symmetric seesaw: dh1 = −dh2 h1 h2 Zero force from virtual displacements if 0 = dE = dE1 + dE2 = g(M1dh1 + M 2dh2 ) = gdh1 (M1 − M 2 ) No tendency to rotate (in seeking a lowest E) if M1 = M 2 (c)2017 van Putten 9 Imbalanced seesaw CM away from pivot: q < 1: M1 > M 2 (c)2016 van Putten 10 Asymmetric seesaw (I) Pivot coincides with CM if li = ri (i = 1,2) : q 1 h2 r = r, r = − r 1 1+ q 2 1+ q h1 l l 2 1 l1 > l2 : M M r l 2 q = 2 = 1 = 1 pivot M1 r2 l2 M1 CM (c)2016 van Putten 11 Asymmetric seesaw (II) Virtual displacement dEi = M igdhi (i = 1,2) dh1 dh2 h2 Asymmetric seesaw: = − l1 l2 h1 l2 l1 Force-free virtual displacements if ⎛ l2 ⎞ 0 = dE = dE1 + dE2 = g(M1dh1 + M 2dh2 ) = gdh1 ⎜ M1 − M 2 ⎟ ⎝ l1 ⎠ No tendency to rotate (in seeking a lowest E) if l1M1 = l2 M 2 (c)2017 van Putten 12 Leverage l1 > l2 creates leverage to lift mass: l1 M 2 = M1 > M1 l2 h2 but over reduced heights: h1 l2 l2 l1 dh2 = − dh1 < dh1 l1 Potential energies: dUi = gM idhi : dU1 + dU2 = 0 Conservative exchange of potential energy (c)2017 van Putten 13 Conversion to kinetic energy... http://en.wikipedia.org/wiki/Seesaw (c)2017 van Putten 14 Torque about pivot CM pivot F = gM M = M1 + M 2 Fi = gM i (i = 1,2) F F1 1 (c)2016 van Putten 15 Torque about pivot b CM at torque arm T = b × F length b from pivot F = gM T (c)2017 van Putten 16 Energy in rotation Erot = Ek1 + Ek2 l1 l2 1 2 E = M v , v = l ω ki 2 i i i i dϕ ω = ω dt (c)2017 van Putten 17 Energy in angular momentum l1 l2 ω Total Ek in rotational motion 1 E = Iω 2 rot 2 2 2 Moment of inertia I = M1l1 + M 2l2 Angular momentum J = Iω Recall Kepler’s orbital motion: dA J = r × p : J = Mj, j = 2 is constant dt (c)2017 van Putten 18 Spin in circus acts 19 Spin carries angular momentum http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/ (c)2017 van Putten 20 Angular momentum in the trapezium jump http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/ (c)2017 van Putten 21 Spin up and down, conserving J 1 J 2 E = Iω 2 = ∝ I −1 ∝ l −2 rot 2 2I J J = Iω : ω = ∝ I −1 ∝ l −2 I l l I = M × l 2 http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/ (c)2017 van Putten 22 Ballerina effect 2 I A = M × lA 2 I B = M × lB 2 ω B ⎛ lA ⎞ = ⎜ ⎟ >> 1 ω A ⎝ lB ⎠ http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/ (c)2017 van Putten 23 In the air... ω A ω B Spin up for the act: 2 ω B ⎛ stretched ⎞ ω C = ⎜ ⎟ >> 1 ω A ⎝ curled up⎠ Spin down before landing... 2 ω C ⎛ curled up⎞ = ⎜ ⎟ << 1 ω B ⎝ stretched ⎠ http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/ (c)2017 van Putten 24 Angular momentum carries energy “What I do is very difficult. It is very hard on the body. If I am lucky I have a good ten years ahead of me, if I am lucky.” Alex Cortes, Big Apple Circus Q: Why do you think this is so? (c)2017 van Putten 25 Mach’s principle J Absent external torques, angular momentum is constant in magnitude and orientation relative to distance stars(*) (*) in flat space-time, in the absence of frame dragging (“dragging of spacetime by matter”) (c)2017 van Putten 26 Mach’s principle “He believed that inertia arises from the presence of all matter in the Universe.” “Distant stars” represent most of the mass. If the philosopher is good enough, after some time he may come back and say, “I understand. We really do not have such a thing as absolute rotation; we are really rotating relative to the stars, you see. And so some influence exerted by the stars on the object must cause the centrifugal force.” Feynman Lectures on Physics, Vol I, Ch16 (c)2017 van Putten 27 Spinning top CM pivot The Boy with a Spinning Top b Jean-Baptiste Chardin, France (1699-1779) θ Distance CM to pivot: b = l sinθ l = CM − pivot Ideal pivot: F = Mg no air friction, conservation of total energy in angular momentum (c)2017 van Putten 28 Precession and torque ω p processional angular velocity J motion of J τ τ = ω p J torque associated with orientation change of an angular momentum vector (c)2017 van Putten 29 Precessing top J d τ T = b × F, e = , τ = e, τˆ = J dt τ [T ] = energy dJ d dJ d T ≡ = (Je) = e + J e dt dt dt dt θ precessional angular velocity ω p ω p J = bF motion of J=J(t) τ In magnitude and direction, this τ = ω p J torque is absorbed by precession of J: (c)2017 van Putten 30 Angular momentum is a conserved quantity Jz = 0 Jz = 0 Feynman Lectures on Physics, Ch20 (c)2017 van Putten 31 Torques and counter-torques T applied torque to the handle ω J J ω p counter-torque T Jz = 0 J Jz = 0 Jz = 0 (c)2017 van Putten 32 Horizontally suspended rotating disk ω p ω T torque by gravity Disk’s weight rests on pivot. Disk absorbs T by precession. (c)2017 van Putten 33 Gyroscope inertial reaction torques (c)2017 van Putten 34 Particle trajectories a p = mv m v Fapplied = ma Fapplied = −Freaction Newton’s third law (c)2017 van Putten 35 Inertial reaction forces a m v dp Force applied to m: F = ma = dt dp Inertial reaction force: F = −ma = − dt (c)2017 van Putten 36 Centripetal acceleration ω ac m v = ωρ v v2 p = mv a = = ω 2 ρ ρ dp F = ma = dt p Δp = pΔθ Rotational motion: Δθ = ωΔt p F = ω p Radius of curvature ρ = ωm (c)2017 van Putten 37 Curved trajectories ρ ω ac m v Instantaneous orbital plane spanned by p and a ρ is local radius of curvature (c)2017 van Putten 38 Rotation induced torque ω ρ = l m Inertial pull v v = ωl F l v2 b a = = ω 2l l T F F = pω : T = bF = mω 2bl T (c)2017 van Putten 39 Inertial forces in a rotating ring dϕ ω = p dt V l C Absent precession, the ring rotates in a tangent plane V to a cylinder about the z-axis Rotating V forces particle trajectories with curvature in both V and the horizontal plane. This is a problem of two angles… exploit translation invariance of rotation induced torques… (c)2017 van Putten 40 Bring ring’s CM to the origin z Ring spins over θ (fast angle) Precession over φ (slow angle) θ mass element in ring ϕ y x ⎛ sinθ cosϕ ⎞ ⎜ ⎟ dθ dϕ r(θ,ϕ) = b sinθ sinϕ , = ω, = ω p ⎜ ⎟ dt dt ⎝⎜ cosθ ⎠⎟ (c)2017 van Putten 41 Some vector calculus M δ J = r ×δ p = δ m r × v, δ m = δθ 2π d d v = r, a = v, v × v ≡ 0 dt dt d ⎛ d d ⎞ δT = δ J = δ m r × v + r × v = δ m r × a dt ⎝⎜ dt dt ⎠⎟ (c)2017 van Putten 42 Explicit evaluation ⎛ cosθ cosϕ ⎞ ⎛ −sinθ sinϕ ⎞ v b⎜ ⎟ b⎜ ⎟ , = ⎜ cosθ sinϕ ⎟ ω + ⎜ sinθ cosϕ ⎟ ω p ⎝⎜ −sinθ ⎠⎟ ⎝⎜ 0 ⎠⎟ ⎛ −sinθ cosϕ ⎞ ⎛ −sinθ cosϕ ⎞ ⎛ − cosθ sinϕ ⎞ a b⎜ ⎟ 2 b⎜ ⎟ 2 2b⎜ ⎟ = ⎜ −sinθ sinϕ ⎟ ω + ⎜ −sinθ sinϕ ⎟ ω p + ⎜ cosθ cosϕ ⎟ ωω p ⎝⎜ − cosθ ⎠⎟ ⎝⎜ 0 ⎠⎟ ⎝⎜ 0 ⎠⎟ ⎛ − cosθ sinϕ ⎞ a r 2 2 b 2 cos i 2b⎜ ⎟ : = − (ω +ω p ) + ω p θ z + ⎜ cosθ cosϕ ⎟ ωω p ⎝⎜ 0 ⎠⎟ ⎛ ⎛ − cosθ sinϕ ⎞ ⎞ ⎜ 2 ⎟ T m r a mb r i 2 r ⎜ ⎟ δ = δ × = δ ⎜ω p × z + ωω p × ⎜ cosθ cosϕ ⎟ ⎟ ⎝⎜ ⎝⎜ 0 ⎠⎟ ⎠⎟ (c)2017 van Putten 43 Net torque in a rotating ring Average over all mass positions in the ring (fast angle θ) ⎛ sinθ sinϕ ⎞ 1 2π b 2π r × i = r × i dθ = ⎜ ⎟ dθ =0 z ∫ z ∫ ⎜ −sinθ cosϕ ⎟ 2π 0 2π 0 ⎝⎜ 0 ⎠⎟ ⎛ ⎞ ⎛ ⎞ ⎛ 2 ⎞ ⎛ ⎞ − cosθ sinϕ 2π − cosθ sinϕ 2π − cos θ cosϕ cosϕ ⎜ ⎟ 1 ⎜ ⎟ b ⎜ ⎟ b ⎜ ⎟ r × cosθ cosϕ = r × cosθ cosϕ dθ = − cos2 θ sinϕ dθ = − sinϕ ⎜ ⎟ 2π ∫ ⎜ ⎟ 2π ∫ ⎜ ⎟ 2 ⎜ ⎟ ⎜ ⎟ 0 ⎜ ⎟ 0 ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎝ 2sinθ cosθ ⎠ ⎝ 0 ⎠ ⎛ ⎛ ⎞ ⎞ 2π 2π − cosθ sinϕ M ⎜ 2 ⎜ ⎟ ⎟ T = δ T = r × adθ = M b ω p r × iz + 2ωω p r × cosθ cosϕ ∫ 2π ∫ ⎜ ⎜ ⎟ ⎟ 0 0 ⎜ ⎜ ⎟ ⎟ ⎝ ⎝ 0 ⎠ ⎠ J = ω I, I = Mb2 : 2 T = ωω p Mb = ω p J (c)2017 van Putten 44 Starting precession: dip by gravity’s torque + vup t = 0 (red > blue) Left Right vdown fn > fn Bottom Top fn = fn vdown = v0 + ε, v0 = ω 0b vup = v0 − ε, ε = gt Precession started 2 2 ⎛ vdown vup ⎞ by initial dip τ = b⎜ − ⎟ = 4ω 0εb ⎝ b b ⎠ about z-axis (c)2017 van Putten 45 Precession in equilibrium vretrograde t > 0 0 l α > (red > blue) vprorate Left Right fn = fn Bottom Top fn > fn ω J = lF, F = Mg Precession p subsequently balances (l → σ = l cosα ) gravity’s torque (c)2017 van Putten 46 Nutation 2 J = Mb ω 0 Damped motion of CM to Jz α steady state J ! J 0 J⊥ ⊥ < balances precessional J = J! + J⊥ angular momentum J⊥ = J sinα 2 Jz = Izω p = Mσ ω p σ = l cosα −1 2 Mgσω p tanα = (J cosα )tanα = J sinα = Jz = Izω p = Mσ ω p 2 ⎛ ω ⎞ g tanα = p , Ω = ⎝⎜ Ω ⎠⎟ σ (c)2017 van Putten 47 Summary CM and pivots balance Virtual displacements Angular momentum moment of inertia angular velocity Spinning top Torque Precession 48.
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