Center of Mass and Angular MomentumAngular momentum carries energy (c)2017 van Putten 25 Mach’s principle J (c)2017 van Putten 26 Absent external torques, angular momentum is constant

(c)2017 van Putten

Center of Mass and Angular Momentum

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(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)

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Center of mass

(c)2017 van Putten

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

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

Rotation of center of mass about pivots

(c)2017 van Putten

b: horizontal distance CM to pivot

b ' b '

unbalancedbalanced

b

double trouble

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Schematic overview

(c)2017 van Putten

buckling

pivot

CM

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CM defined

(c)2017 van Putten

M1r1 +M 2r2 = 0

CM

M1

M 2

CM is relatively close to the heavy mass

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CM of two bodies is along a line connecting them:

CM defined

(c)2017 van Putten

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

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Balance of seesaw (I)

(c)2017 van Putten

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)

(c)2017 van Putten

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

(c)2016 van Putten

CM away from pivot:q <1: M1 > M 2

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Asymmetric seesaw (I)

(c)2016 van Putten

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

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Asymmetric seesaw (II)

(c)2017 van Putten

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

(c)2017 van Putten

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

14(c)2017 van Putten

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

Torque about pivot

(c)2016 van Putten

Fi = gMi i = 1,2( )

F = gMM = M1 +M 2

CM

F1 F1

pivot

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Torque about pivot

(c)2017 van Putten

F = gM

b

CM at torque arm length b from pivot

T = b × F

T

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Energy in rotation

Erot = Ek1 + Ek2

Eki =12Mivi

2, vi = liω

ω = dϕdt

l1 l2

ω


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


is constant

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Spin in circus acts

Spin carries angular momentum

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



Angular momentum in the trapezium jump




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




Ballerina effect

IA = M × lA2

IB = M × lB2

ω B

ω A

= lAlB

⎛⎝⎜

⎞⎠⎟

2

>>1




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:




“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


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”)

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


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

(c)2017 van Putten

Distance CM to pivot:

Spinning top

Precession and torque

J motion of J

ω p processional angular velocity

τ

τ =ω pJ


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


In magnitude and direction, this torque is absorbed by precession of J:

Angular momentum is a conserved quantity

Feynman Lectures on Physics, Ch20


Jz = 0 Jz = 0

Torques and counter-torques

ω p

J

JωJ

applied torque to the handle

T

Tcounter-torque


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.


Gyroscope inertial reaction torques


Particle trajectories

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Fapplied = maFapplied = −Freaction Newton’s third law

v

am

p = mv

(c)2017 van Putten

Inertial reaction forces

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F = ma = dpdt

v

am

Force applied to m:

F = −ma = − dpdt

Inertial reaction force:

(c)2017 van Putten

Centripetal acceleration

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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

(c)2017 van Putten

Rotational motion:

F =ω p

Curved trajectories

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ωρ

v

acm

Instantaneous orbital plane spanned by p and a

(c)2017 van Putten

ρ is local radius of curvature

Rotation induced torque

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v =ωl

a = v2

l=ω 2l

ω

bl

vF

T

m

F = pω :T = bF = mω 2bl

Inertial pull

(c)2017 van Putten

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

(c)2017 van Putten

V

Bring ring’s CM to the origin

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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

(c)2017 van Putten

Some vector calculus

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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

(c)2017 van Putten

Explicit evaluation

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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

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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 :

(c)2017 van Putten

Starting precession: dip by gravity’s torque

Precession started by initial dip


(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


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


Mgσω p−1 tanα = J cosα( ) tanα = J sinα = Jz = Izω p = Mσ 2ω p

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Summary

CM and pivots balance Virtual displacements

Angular momentum moment of inertia angular velocity

Spinning top Torque Precession

Documents

Center of Mass and Angular MomentumAngular momentum carries energy (c)2017 van Putten 25 Mach’s principle J (c)2017 van Putten 26 Absent external torques, angular momentum is constant