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(c)2017 van Putten
Center of Mass and Angular Momentum
1
(c)2017 van Putten
Outline
Seesaw balance Center of mass Force, CM and torque
Angular momentum Ice skater - ballerina effect
Spinning top Inertial torque Precession Nutation
Anglicisation of French “ci-ça” (back-and-forth); or from "scie" - French for "saw" with Anglo Saxon "saw." "scie-saw" became “see saw." (Wikipedia)
2
Center of mass
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http://www.pbs.org/opb/circus/classroom/circus-physics/center-mass/
3
Rotation of center of mass about pivots
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b: horizontal distance CM to pivot
b ' b '
unbalancedbalanced
b
double trouble
4
Schematic overview
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buckling
pivot
CM
5
CM defined
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M1r1 +M 2r2 = 0
CM
M1
M 2
CM is relatively close to the heavy mass
6
CM of two bodies is along a line connecting them:
CM defined
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CM
M1
M 2
r1r2
r = r1 − r2
Distances to CM:
r1 =M 2
M1 +M 2
r = q1+ q
r
r2 = − M1
M1 +M 2
r = − 11+ q
r
q = M 2
M1
7
Balance of seesaw (I)
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ll
r1 =q1+ q
r, r2 = − 11+ q
r
CM
M1
M 2
pivot
pivot
Balanced: CM at pivot
q = 1: M1 = M 2
no net torque, no tendency to rotate
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Balance of seesaw (II)
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dEi = Migdhi i = 1,2( )pivot
h1 h2 dh1 = −dh2Symmetric seesaw:
0 = dE = dE1 + dE2 = g M1dh1 +M 2dh2( ) = gdh1 M1 −M 2( )
M1 = M 2
No tendency to rotate (in seeking a lowest E) if
Zero force from virtual displacements if
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Virtual displacement
Imbalanced seesaw
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CM away from pivot:q <1: M1 > M 2
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Asymmetric seesaw (I)
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l1 > l2 :
CM
M1
M 2
pivot
li = ri i = 1,2( ) :
r1 =q1+ q
r, r2 = − 11+ q
r
Pivot coincides with CM if
l2l1
h1
h2
q = M 2
M1
=r1r2
= l1l2
11
Asymmetric seesaw (II)
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dEi = Migdhi i = 1,2( )
dh1l1
= − dh2l2
Asymmetric seesaw:
0 = dE = dE1 + dE2 = g M1dh1 +M 2dh2( ) = gdh1 M1 −l2l1M 2
⎛⎝⎜
⎞⎠⎟
l1M1 = l2M 2
No tendency to rotate (in seeking a lowest E) if
Force-free virtual displacements if
l2l1
h1
h2
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Virtual displacement
Leverage
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creates leverage to lift mass:
M 2 =l1l2M1 > M1
l2l1
h1
h2
dh2 = − l2l1dh1 < dh1
but over reduced heights:
l1 > l2
Potential energies: dUi = gMidhi : dU1 + dU2 = 0
Conservative exchange of potential energy
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Conversion to kinetic energy...
http://en.wikipedia.org/wiki/Seesaw
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Torque about pivot
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Fi = gMi i = 1,2( )
F = gMM = M1 +M 2
CM
F1 F1
pivot
15
Torque about pivot
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F = gM
b
CM at torque arm length b from pivot
T = b × F
T
16
Energy in rotation
Erot = Ek1 + Ek2
Eki =12Mivi
2, vi = liω
ω = dϕdt
l1 l2
ω
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Energy in angular momentum
Erot =12Iω 2
I = M1l12 +M 2l2
2
Total Ek in rotational motion
Moment of inertia
J = IωAngular momentum
Recall Kepler’s orbital motion:J = r × p : J = Mj, j = 2 dA
dt
l1 l2
ω
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is constant
19
Spin in circus acts
Spin carries angular momentum
http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/
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Angular momentum in the trapezium jump
http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/
21(c)2017 van Putten
Spin up and down, conserving J
Erot =12Iω 2 = J 2
2I∝ I −1 ∝ l−2
J = Iω : ω = JI∝ I −1 ∝ l−2
I = M × l2
l
l
http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/
22(c)2017 van Putten
Ballerina effect
IA = M × lA2
IB = M × lB2
ω B
ω A
= lAlB
⎛⎝⎜
⎞⎠⎟
2
>>1
http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/
23(c)2017 van Putten
In the air...
ωC
ω B
= curled upstretched
⎛⎝⎜
⎞⎠⎟2
<<1
Spin down before landing...
ωC
ω Bω A
ω B
ω A
= stretchedcurled up
⎛⎝⎜
⎞⎠⎟
2
>>1
Spin up for the act:
http://www.pbs.org/opb/circus/classroom/circus-physics/angular-momentum/
24(c)2017 van Putten
“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?
Angular momentum carries energy
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Mach’s principle
J
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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”)
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
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The Boy with a Spinning Top Jean-Baptiste Chardin, France (1699-1779)
pivot
CM
F = Mg
b
Ideal pivot: no air friction, conservation of total energy in angular momentum
θ
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b = l sinθ l = CM − pivot
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Distance CM to pivot:
Spinning top
Precession and torque
J motion of J
ω p processional angular velocity
τ
τ =ω pJ
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torque associated with orientation change of an angular momentum vector
Precessing top
T = b × F, e = JJ, τ = d
dte, τ̂ = τ
τ[T ]= energy
T ≡ dJdt
= ddt
Je( ) = dJdte + J d
dte
ω pJ = bFmotion of J=J(t)
precessional angular velocity
θω p
ττ =ω pJ
30(c)2017 van Putten
In magnitude and direction, this torque is absorbed by precession of J:
Angular momentum is a conserved quantity
Feynman Lectures on Physics, Ch20
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Jz = 0 Jz = 0
Torques and counter-torques
ω p
J
JωJ
applied torque to the handle
T
Tcounter-torque
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Jz = 0
Jz = 0
Jz = 0
Horizontally suspended rotating disk
Disk’s weight rests on pivot.
torque by gravity
ωT
ω p
Disk absorbs T by precession.
33(c)2017 van Putten
Gyroscope inertial reaction torques
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Particle trajectories
35
Fapplied = maFapplied = −Freaction Newton’s third law
v
am
p = mv
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Inertial reaction forces
36
F = ma = dpdt
v
am
Force applied to m:
F = −ma = − dpdt
Inertial reaction force:
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Centripetal acceleration
37
F = ma = dpdt
v =ωρ
a = v2
ρ=ω 2ρ
ω
v
ac
p Δp = pΔθΔθ =ωΔt
ρ = pωm
Radius of curvature
m
p = mv
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Rotational motion:
F =ω p
Curved trajectories
38
ωρ
v
acm
Instantaneous orbital plane spanned by p and a
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ρ is local radius of curvature
Rotation induced torque
39
v =ωl
a = v2
l=ω 2l
ω
bl
vF
T
m
F = pω :T = bF = mω 2bl
Inertial pull
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T
F
ρ = l
Inertial forces in a rotating ring
40
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…
l
ω p =dϕdt
C
Absent precession, the ring rotates in a tangent plane V to a cylinder about the z-axis
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V
Bring ring’s CM to the origin
41
r θ ,ϕ( ) = bsinθ cosϕsinθ sinϕcosθ
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟, dθ
dt=ω , dϕ
dt=ω p
x
y
z
ϕ
θ
Ring spins over θ (fast angle)
Precession over φ (slow angle)
mass element in ring
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Some vector calculus
42
δ J = r ×δ p = δm r × v, δm = M2π
δθ
v = ddtr, a = d
dtv, v × v ≡ 0
δT = ddtδ J = δm d
dtr × v + r × d
dtv⎛
⎝⎜⎞⎠⎟ = δm r × a
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Explicit evaluation
43
v = bcosθ cosϕcosθ sinϕ−sinθ
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟ω + b
−sinθ sinϕsinθ cosϕ
0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟ω p ,
a = b−sinθ cosϕ−sinθ sinϕ−cosθ
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟ω 2 + b
−sinθ cosϕ−sinθ sinϕ
0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟ω p
2 + 2b−cosθ sinϕcosθ cosϕ
0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟ωω p
a = −r ω 2 +ω p2( )+ bω p
2 cosθ iz + 2b−cosθ sinϕcosθ cosϕ
0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟ωω p :
δT = δm r × a = δmb ω p2r × iz + 2ωω pr ×
−cosθ sinϕcosθ cosϕ
0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
(c)2017 van Putten
Net torque in a rotating ring
44
r × iz = 12π
r × iz dθ =0
2π
∫b2π
sinθ sinϕ−sinθ cosϕ
0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟dθ =
0
2π
∫ 0
r ×−cosθ sinϕcosθ cosϕ
0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
= 12π
r ×−cosθ sinϕcosθ cosϕ
0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟0
2π
∫ dθ = b2π
−cos2θ cosϕ−cos2θ sinϕ2sinθ cosθ
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟0
2π
∫ dθ = − b2
cosϕsinϕ0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
Average over all mass positions in the ring (fast angle θ)
T = δT0
2π
∫ = M2π
r × ad0
2π
∫ θ = M b ω p2 r × iz + 2ωω p r ×
−cosθ sinϕcosθ cosϕ
0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
T =ωω pMb2 =ω pJ
J =ω I , I = Mb2 :
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Starting precession: dip by gravity’s torque
Precession started by initial dip
45(c)2017 van Putten
(red > blue)
vdown = v0 + ε , v0 =ω 0bvup = v0 − ε , ε = gt
τ = b vdown2
b−vup2
b⎛
⎝⎜⎞
⎠⎟= 4ω 0εb
about z-axis
t = 0+
fnLeft > fn
Right
fnBottom = fn
Top
vdown
vup
Precession in equilibrium
ω pJ = lF, F = MgPrecession subsequently balances
gravity’s torque
46(c)2017 van Putten
fnLeft = fn
Right
fnBottom > fn
Top
t > 0α > 0l
vprorate
vretrograde
(red > blue)
l→σ = l cosα( )
Nutation
J⊥ < 0
balances precessional angular momentum
Jz = Izω p = Mσ 2ω p
σ = l cosα
J = Mb2ω 0
tanα =ω p
Ω⎛⎝⎜
⎞⎠⎟
2
, Ω = gσ
J = J! + J⊥
J⊥ = J sinα
J!J⊥
αJz
Damped motion of CM to steady state
47(c)2017 van Putten
Mgσω p−1 tanα = J cosα( ) tanα = J sinα = Jz = Izω p = Mσ 2ω p
48
Summary
CM and pivots balance Virtual displacements
Angular momentum moment of inertia angular velocity
Spinning top Torque Precession
top related