lecture 23 rigid body dynamics - cs.cmu.edu

Lecture 23Rigid bodydynamics

Newtonianmechanics of aparticle

Newtonianmechanics of asystem of particles

Newtonianmechanics of arigid body

The angular inertiatensor

Euler’s equations

Poinsot’sconstruction

Lecture 23Rigid body dynamics

Matthew T. Mason

Mechanics of Manipulation






Euler’s equations


Today’s outline

Newtonian mechanics of a particle

Newtonian mechanics of a system of particles

Newtonian mechanics of a rigid body

The angular inertia tensor

Euler’s equations

Poinsot’s construction






Euler’s equations


Preview

I We will apply Newton’s second law to derive fourclosely related forms:

Force Moment2D F = mv̇ N = Iω̇3D F = mv̇ N = Iω̇ + ω × Iω

(ω in lower right should be bold. Font problem!)I Bottom right corner is different!

1. Inertia term is a 3× 3 matrix, not a scalar.2. Unexpected (?) term: ω × Iω.3. Zero torque does not imply zero angular acceleration!

I But first, the basics.






Euler’s equations


Newton’s laws

1. Every body continues at rest, or in uniform motion ina straight line, unless forces act upon it.

2. The rate of change of momentum is proportional tothe applied force.

3. The forces acting between two bodies are equal andopposite.

DefinitionDefine momentum to be mass times velocity.






Euler’s equations


Consider a particle

I Consider a particle of mass m,I with position represented by a vector x,I total applied force F,I momentum

p = mv = mdxdt

I so Newton’s second law can be written

F = md2xdt2






Euler’s equations


Impulse, kinetic energy

DefinitionIntegrating Newton’s second law:

p2 − p1 =

∫ t2

t1F dt

stating that the change in momentum is equal to theimpulse.

DefinitionWe can also define kinetic energy T

T =m2|v|2






Euler’s equations


Power

DefinitionDifferentiating kinetic energy yields

dTdt

=m2

ddt

(v · v)

=m2

(dvdt· v + v · dv

dt

)= m

dvdt· v

= F · v

stating that the time rate of change of kinetic energy ispower.






Euler’s equations


Work

DefinitionIntegrating the power over a time interval,

T2 − T1 =

∫ t2

t1F · v dt

or

T2 − T1 =

∫ x2

x1

F · dx

stating that the change in kinetic energy is work.






Euler’s equations


Moment of force

DefinitionRecall definition of moment of force about a point x:

n = x× f

and about a line l through origin with direction l̂

nl = l̂ · n






Euler’s equations


Moment of momentum

DefinitionSimilarly, suppose a particle at x has momentum p.

I Define moment of momentum about the origin

L = x× p

I and about the line l

Ll = l̂ · L






Euler’s equations


Rate of change of moment of momentum

I Differentiating the moment of momentum:

dLdt

=ddt

(x× p)

=ddt

(x×mv)

= m(

dxdt× v + x× dv

dt

)= x×m

dvdt

= x× F= N

which is essentially a restatement of Newton’ssecond law, but using moments of force andmomentum.






Euler’s equations


So, for a particle . . .

I Using either F = dp/dt or N = dL/dt , we have threesecond order differential equations.

I If F or N is uniquely determined by the state (x,v),then there is a unique solution giving x(t) and v(t) forany given initial conditions x(0) = x0, v(0) = v0.






Euler’s equations


For a bunch of particles

For the k th particleI Let mk be the mass,I let xk be the position vector,I and let pk be the momentum.I Let the force be composed of internal force (from

interactions with other particles in the system) andexternal forces Fk = Fi

k + Fek






Euler’s equations


Momentum and force

We define the momentum of the system to be

P =∑

pk

and the total force on the system to be

F =∑

Fek

(The sum of all internal forces is zero, by Newton’s thirdlaw.)






Euler’s equations


Newton’s 2nd law for system of particles

Newton’s 2nd law for k th particle:

dpk

dt= Fe

k + Fik

Summing: ∑ dpk

dt=∑(

Fek + Fi

k

)Hence

dPdt

= F

Newton’s second law extends to the system of particles.






Euler’s equations


Center of mass

Define total mass:M =

∑mk

and the center of mass,

X =1M

∑mkxk

ThenP = M

dXdt

and

F = Md2Xdt2

which means that the center of mass behaves just like asingle particle.






Euler’s equations


Moments for systems of particles

I Define Lk to be the angular momentum of the k thpoint,

I Define the total angular momentum to be the sum,

L =∑

Lk

I Define the total torque,

N =∑

xk × Fek






Euler’s equations


Rate of change of moment of momentum

I Now for the k th particle

dLk

dt= xk × Fe

k + xk × Fik

Summing over all the particles,

dLdt

= N +∑

xk × Fik

By Newton’s third law the sum of the internalmoments is zero, so that the second term vanishes:

dLdt

= N

which is grand, but six equations is not enough todetermine the motion of several particles.






Euler’s equations


Rigid body dynamics

I A rigid body is a bunch of particles, but with alldistances fixed. Six degrees of freedom. Wouldn’t itbe keen if the six equations

F = dP/dtN = dL/dt

were enough?






Euler’s equations


Angular inertia, part one

I For a rigid body, velocity of k th particle is

v = v0 + ω × x

I Substituting into moment of momentum

Lk = mkxk × (v0 + ω × xk )

I Summing to obtain the total angular momentum,

L =∑

mkxk × v0 +∑

mkxk × (ω × xk )

= MX× v0 +∑

mkxk × (ω × xk )

I Place origin at center of mass to eliminate first termon right

L =∑

mkxk × (ω × xk )






Euler’s equations


Angular inertia, part two

I How can we get that pesky ω out of the sum?

L =∑

mkxk × (ω × xk )

I Applying the identity a× (b× c) = (a · c)b− (a · b)c,

L =∑

mk [(xk · xk )ω − xk (xk · ω)]

I Represent each vector as a column matrix, andsubstitute xt

kω for xk · ω:

L =(∑

mk

(|xk |2I3 − xkxt

k

))ω

where I3 is the three-by-three identity matrix.






Euler’s equations


Angular inertia part three

Definition

I Define the angular inertia matrix I:

I =∑

mk

(|xk |2I3 − xkxt

k

)

I Substituting above,

L = Iω

where Newton’s second law gives

N =dLdt






Euler’s equations


Differentiating L = Iω

I I is constant in the body frame, not in an inertialframe. In an inertial frame:

N =d(Iω)

dt(1)

= Idωdt

+dIdtω (2)

= Idωdt

+ ω × (Iω) (3)

I Zero torque implies constant angular momentum.I Zero torque does not imply constant angular velocity.I What can you say about how angular velocity

changes? First we need to look closer at the angularinertia tensor.






Euler’s equations


The inertia tensor

I Let body be continuous with density ρ.

I =∫ρ(|x|2I3 − xxt

)dV (4)

I In components:

I =∫ρ

x22 + x2

3 −x1x2 −x1x3−x1x2 x2

1 + x23 −x2x3

−x1x3 −x2x3 x21 + x2

2

dV (5)






Euler’s equations


Moments of inertia; products of inertia

I Diagonal elements: moments of inertia w.r.t. thecoordinate axes:

I11 =

∫ρ(x2

2 + x23 )dV (6)

etc. (7)

I Off-Diagonal elements: the products of inertia:

I12 = I21 = −∫ρx1x2 dV (8)

etc. (9)

I We could try to understand them, or we could get ridof them . . .






Euler’s equations


Principal axes; principal moments of inertiaI Inertia matrix is symmetric—diagonal in the right

frame. Define A:

AI =

AI11 0 00 AI22 00 0 AI33

(10)

I I in A-coordinates can be obtained by:

AI = AIAT (11)

where matrix A transforms to A-coordinates.I principal axes: coordinate axes of A—eigenvectors

of I.I principal moments: diagonal elements of

AI—eigenvalues of I.I Distinct eigenvalues implies uniquely determined

principal axes.






Euler’s equations


Scalar angular inertia. Radius of gyration.

I Consider moment of momentum L = Iω. When are Land ω parallel?

I Consider rotation about some fixed axis in directionn̂. Scalar angular inertia In is

In = n̂t In̂ (12)

I radius of gyration kn with respect to the axis n̂:

In = Mk2n (13)

I The radius of gyration represents the distance of apoint mass that would give the same angular inertia.






Euler’s equations


Inertia ellipsoid

I Consider the surface described by the equation

rt Ir = a (14)

I In principal coordinates, since moments are positive,we get an ellipsoid:

Ixx r2x + Iyy r2

y + Izzr2z = a (15)

I Let r = r n̂. Then

In = n̂t In̂ =1r2 rt Ir =

ar2 (16)

So distance to ellipsoid surface is inverse of radius ofgyration.

Mk2n =

ar2 (17)






Euler’s equations


Cylinder and its inertia ellipsoid

x y

z

x y

z






Euler’s equations


Principal axes by inspection

I There are theorems:I Any plane of symmetry is perpendicular to a principal

axis.I Any line of symmetry is a principal axis.

I If you start in the principal frame, you know theproducts of inertia are zero, so you can get theinertia tensor by just doing three integrals.






Euler’s equations


Rigid body tumblingI Applying Newton’s second law in principal

coordinates yields:

N =

I1ω̇1I2ω̇2I3ω̇3

+

ω1ω2ω3

× I1ω1

I2ω2I3ω3

(18)

=

I1ω̇1 + (I3ω2ω3 − I2ω2ω3)I2ω̇2 + (I1ω3ω1 − I3ω3ω1)I3ω̇3 + (I2ω1ω2 − I1ω1ω2)

(19)

I If N = 0 we get Euler’s equations:

ω̇1 =I2 − I3

I1ω2ω3 (20)

ω̇2 =I3 − I1

I2ω3ω1 (21)

ω̇3 =I1 − I2

I3ω1ω2 (22)






Euler’s equations


Coordinate frame issue in differentiation

I We differentiated in a fixed frame, instantaneouslycoinciding with body principal frame.

I Euler’s equations will be true only fleetingly in aglobal fixed frame! Cannot integrate them.

I So, transform to moving body frame from coincidentfixed frame. N, I, ω unchanged. New angularacceleration is

dωdt

+ ω × ω

I.e., Euler’s equations work in the body frame.






Euler’s equations


Exploring Euler’s equations

I Consider special cases for Euler’s equations:

ω̇1 =I2 − I3

I1ω2ω3 (23)

ω̇2 =I3 − I1

I2ω3ω1 (24)

ω̇3 =I1 − I2

I3ω1ω2 (25)

I What if ω is along a principal axis?I What if the body has a symmetric mass distribution?

I I1 = I2 = I3?I I1 = I2 6= I3?






Euler’s equations


Poinsot’s construction

polhode

herpolhode

L

I Rigid bodytumbling: inertiaellipsoid rollswithout slippingon a plane.






Euler’s equations


Proof of Poinsot’s constructionI If N is zero, then kinetic energy T is constant:

T =12ωt Iω is constant (26)

That is, ω is on the surface of the inertia ellipsoid.I What is the tangent plane normal at ω?

∇12ωt Iω = ∇1

2(ω2

1I1 + ω22I2 + ω2

3I3) (27)

= (I1ω1, I2ω2, I3ω3) = L (28)

The attitude of the tangent plane is constant!I How far from center of mass to tangent plane?

ω · L|L|

=2T|L|

(29)

which is also constant!I So the tangent plane is fixed: the invariable plane.

The ellipsoid rolls without slipping on the invariableplane.






Euler’s equations


polhodes

I Take an ellipsoid,hold the center afixed distancefrom an inkpad,and roll it around.

I Near the pointyend you get littleloops.

I Near the centerof mass you getlittle loops.

I Near the thirdprincipal axis,you get sentaway.






Euler’s equations


Video of tumbling body

Original author unknown. Copied from Michel andSchnizer, “Simulations in Analytical Mechanics”

lecture 23 rigid body dynamics - cs.cmu.edu

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