Center of Mass and Angular Momentum

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

http://www.pbs.org/opb/circus/classroom/circus-physics/center-mass/

balanced unbalanced

double trouble

b' b' b b: horizontal distance CM to pivot

buckling

pivot

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

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

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

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

CM away from pivot:

q < 1: M1 > M 2

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

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

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

http://en.wikipedia.org/wiki/Seesaw

CM pivot F = gM

M = M1 + M 2

Fi = gM i (i = 1,2) F F1 1

CM at torque arm T = b × F length b from pivot

F = gM T

Erot = Ek1 + Ek2

l1 l2 1 2 E = M v , v = l ω ki 2 i i i i dϕ ω = ω dt

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

19 Spin carries angular momentum

http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/

l I = M × l 2

http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/

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/

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/

“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?

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”)

“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

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

ω p processional angular velocity

J motion of J τ

τ = ω p J

torque associated with orientation change of an angular momentum vector

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:

Jz = 0 Jz = 0

Feynman Lectures on Physics, Ch20

T applied torque to the handle ω J J ω p counter-torque T

Jz = 0 J Jz = 0

Jz = 0

ω p ω T torque by gravity

Disk’s weight rests on pivot.

Disk absorbs T by precession.

a p = mv m v

Fapplied = ma

Fapplied = −Freaction Newton’s third law

a m v

dp Force applied to m: F = ma = dt dp Inertial reaction force: F = −ma = − dt

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

ρ ω ac

m v

Instantaneous orbital plane spanned by p and a

ρ is local radius of curvature

F F = pω : T = bF = mω 2bl T

dϕ ω = p dt

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…

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 ⎠⎟

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 ⎠⎟

⎛ 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 ⎠⎟ ⎠⎟

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

+ 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

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

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

CM and pivots balance Virtual displacements

Angular momentum moment of inertia angular velocity

Spinning top Torque Precession