Lecture 23 Rigid body dynamics

Newtonian mechanics of a particle

Newtonian Lecture 23 mechanics of a system of particles

Rigid body dynamics Newtonian mechanics of a rigid body

The angular inertia Matthew T. Mason tensor Euler’s equations

Poinsot’s construction Mechanics of Manipulation Lecture 23 Today’s outline Rigid body dynamics

Newtonian mechanics of a Newtonian mechanics of a particle particle

Newtonian mechanics of a Newtonian mechanics of a system of particles system of particles Newtonian mechanics of a rigid body Newtonian mechanics of a rigid body The angular inertia tensor

Euler’s equations The angular inertia tensor Poinsot’s construction Euler’s equations

Poinsot’s construction Lecture 23 Preview Rigid body dynamics

Newtonian mechanics of a I We will apply Newton’s second law to derive four particle Newtonian closely related forms: mechanics of a Force Moment system of particles Newtonian 2D F = mv˙ N = Iω˙ mechanics of a rigid body 3D F = mv˙ N = Iω˙ + ω × Iω The angular inertia (ω in lower right should be bold. Font problem!) tensor Euler’s equations I Bottom right corner is different! Poinsot’s 1. Inertia term is a 3 × 3 matrix, not a scalar. construction 2. Unexpected (?) term: ω × Iω. 3. Zero torque does not imply zero angular acceleration!

I But first, the basics. Lecture 23 Newton’s laws Rigid body dynamics

Newtonian mechanics of a particle 1. Every body continues at rest, or in uniform motion in Newtonian mechanics of a a straight line, unless forces act upon it. system of particles

Newtonian 2. The rate of change of momentum is proportional to mechanics of a the applied force. rigid body The angular inertia 3. The forces acting between two bodies are equal and tensor opposite. Euler’s equations Poinsot’s construction Definition Define momentum to be mass times velocity. Lecture 23 Consider a particle Rigid body dynamics

Newtonian mechanics of a particle

I Consider a particle of mass m, Newtonian mechanics of a I with position represented by a vector x, system of particles Newtonian I total applied force F, mechanics of a rigid body I momentum dx The angular inertia p = mv = m tensor dt Euler’s equations I so Newton’s second law can be written Poinsot’s construction d 2x F = m dt2 Lecture 23 Impulse, kinetic energy Rigid body dynamics

Definition Newtonian mechanics of a Integrating Newton’s second law: particle Newtonian mechanics of a Z t2 system of particles

p2 − p1 = F dt Newtonian t1 mechanics of a rigid body stating that the change in momentum is equal to the The angular inertia tensor

impulse. Euler’s equations

Poinsot’s Definition construction We can also define kinetic energy T m T = |v|2 2 Lecture 23 Power Rigid body dynamics

Newtonian Definition mechanics of a particle Differentiating kinetic energy yields Newtonian mechanics of a system of particles

dT m d Newtonian = (v · v) mechanics of a dt 2 dt rigid body m dv dv = · v + v · The angular inertia 2 dt dt tensor Euler’s equations dv = m · v Poinsot’s dt construction = F · v

stating that the time rate of change of kinetic energy is power. Lecture 23 Work Rigid body dynamics

Newtonian mechanics of a particle

Definition Newtonian mechanics of a Integrating the power over a time interval, system of particles Newtonian t mechanics of a Z 2 rigid body T − T = F · v dt 2 1 The angular inertia t1 tensor or Euler’s equations Z x2 Poinsot’s construction T2 − T1 = F · dx x1 stating that the change in kinetic energy is work. Lecture 23 Moment of force Rigid body dynamics

Newtonian mechanics of a particle

Newtonian Definition mechanics of a system of particles

Recall definition of moment of force about a point x: Newtonian mechanics of a rigid body

n = x × f The angular inertia tensor and about a line l through origin with direction ˆl Euler’s equations Poinsot’s ˆ construction nl = l · n Lecture 23 Moment of momentum Rigid body dynamics

Newtonian mechanics of a particle Definition Newtonian mechanics of a Similarly, suppose a particle at x has momentum p. system of particles Newtonian I Define moment of momentum about the origin mechanics of a rigid body

The angular inertia L = x × p tensor Euler’s equations I and about the line l Poinsot’s construction ˆ Ll = l · L Lecture 23 Rate of change of moment of momentum Rigid body dynamics

I Differentiating the moment of momentum: Newtonian mechanics of a dL d particle = (x × p) Newtonian dt dt mechanics of a d system of particles = (x × mv) Newtonian dt mechanics of a dx dv rigid body = m × v + x × The angular inertia dt dt tensor dv Euler’s equations = x × m Poinsot’s dt construction = x × F = N

which is essentially a restatement of Newton’s second law, but using moments of force and momentum. Lecture 23 So, for a particle ... Rigid body dynamics

Newtonian mechanics of a particle

Newtonian mechanics of a system of particles

I Using either F = dp/dt or N = dL/dt, we have three Newtonian mechanics of a second order differential equations. rigid body

I If F or N is uniquely determined by the state (x, v), The angular inertia tensor

then there is a unique solution giving x(t) and v(t) for Euler’s equations

any given initial conditions x(0) = x0, v(0) = v0. Poinsot’s construction Lecture 23 For a bunch of particles Rigid body dynamics

Newtonian mechanics of a particle

Newtonian For the kth particle mechanics of a system of particles I Let mk be the mass, Newtonian mechanics of a I let xk be the position vector, rigid body The angular inertia I and let pk be the momentum. tensor I Let the force be composed of internal force (from Euler’s equations Poinsot’s interactions with other particles in the system) and construction i e external forces Fk = Fk + Fk Lecture 23 Momentum and force Rigid body dynamics

Newtonian mechanics of a particle We define the momentum of the system to be Newtonian mechanics of a X system of particles P = pk Newtonian mechanics of a rigid body

and the total force on the system to be The angular inertia tensor X e Euler’s equations F = Fk Poinsot’s construction (The sum of all internal forces is zero, by Newton’s third law.) Lecture 23 Newton’s 2nd law for system of particles Rigid body dynamics

Newtonian mechanics of a Newton’s 2nd law for kth particle: particle Newtonian mechanics of a dp system of particles k = Fe + Fi dt k k Newtonian mechanics of a rigid body

Summing: The angular inertia X dp X   tensor k = Fe + Fi dt k k Euler’s equations Poinsot’s Hence construction dP = F dt Newton’s second law extends to the system of particles. Lecture 23 Center of mass Rigid body dynamics

Define total mass: Newtonian X mechanics of a M = mk particle Newtonian and the center of mass, mechanics of a system of particles 1 X Newtonian X = m x mechanics of a M k k rigid body The angular inertia Then tensor dX Euler’s equations P = M Poinsot’s dt construction and d 2X F = M dt2 which means that the center of mass behaves just like a single particle. Lecture 23 Moments for systems of particles Rigid body dynamics

Newtonian mechanics of a particle I Define L to be the angular momentum of the kth Newtonian k mechanics of a point, system of particles Newtonian I Define the total angular momentum to be the sum, mechanics of a rigid body

X The angular inertia L = Lk tensor

Euler’s equations

I Define the total torque, Poinsot’s construction X e N = xk × Fk Lecture 23 Rate of change of moment of momentum Rigid body dynamics

I Now for the kth particle Newtonian mechanics of a particle dLk e i = xk × Fk + xk × Fk Newtonian dt mechanics of a system of particles Summing over all the particles, Newtonian mechanics of a rigid body dL X = N + x × Fi The angular inertia dt k k tensor Euler’s equations

By Newton’s third law the sum of the internal Poinsot’s moments is zero, so that the second term vanishes: construction dL = N dt which is grand, but six equations is not enough to determine the motion of several particles. Lecture 23 Rigid body dynamics Rigid body dynamics

Newtonian mechanics of a particle

Newtonian mechanics of a I A rigid body is a bunch of particles, but with all system of particles

distances fixed. Six degrees of freedom. Wouldn’t it Newtonian mechanics of a be keen if the six equations rigid body

The angular inertia F = dP/dt tensor Euler’s equations N = dL/dt Poinsot’s construction were enough? Lecture 23 Angular inertia, part one Rigid body dynamics

I For a rigid body, velocity of kth particle is Newtonian mechanics of a particle v = v0 + ω × x Newtonian mechanics of a I Substituting into moment of momentum system of particles Newtonian mechanics of a Lk = mk xk × (v0 + ω × xk ) rigid body The angular inertia tensor I Summing to obtain the total angular momentum, Euler’s equations X X Poinsot’s L = mk xk × v0 + mk xk × (ω × xk ) construction X = MX × v0 + mk xk × (ω × xk )

I Place origin at center of mass to eliminate first term on right X L = mk xk × (ω × xk ) Lecture 23 Angular inertia, part two Rigid body dynamics

Newtonian I How can we get that pesky ω out of the sum? mechanics of a particle

X Newtonian L = mk xk × (ω × xk ) mechanics of a system of particles

Newtonian I Applying the identity a × (b × c) = (a · c)b − (a · b)c, mechanics of a rigid body

X The angular inertia L = mk [(xk · xk )ω − xk (xk · ω)] tensor

Euler’s equations

I Represent each vector as a column matrix, and Poinsot’s t construction substitute xk ω for xk · ω:

X  2 t  L = mk |xk | I3 − xk xk ω

where I3 is the three-by-three identity matrix. Lecture 23 Angular inertia part three Rigid body dynamics

Definition Newtonian mechanics of a particle

I Define the angular inertia matrix I: Newtonian mechanics of a system of particles X  2 t  I = mk |xk | I3 − xk xk Newtonian mechanics of a rigid body

The angular inertia tensor

I Substituting above, Euler’s equations

Poinsot’s L = Iω construction

where Newton’s second law gives

dL N = dt Lecture 23 Differentiating L = Iω Rigid body dynamics

I I is constant in the body frame, not in an inertial Newtonian mechanics of a frame. In an inertial frame: particle Newtonian d(Iω) mechanics of a N = (1) system of particles dt Newtonian dω dI mechanics of a = I + ω (2) rigid body dt dt The angular inertia dω tensor = I + ω × (Iω) (3) dt Euler’s equations Poinsot’s construction 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 angular inertia tensor. Lecture 23 The inertia tensor Rigid body dynamics

Newtonian mechanics of a particle I Let body be continuous with density ρ. Newtonian mechanics of a Z system of particles  2 t  Newtonian I = ρ |x| I3 − xx dV (4) mechanics of a rigid body

The angular inertia I In components: tensor Euler’s equations  2 2  Z x2 + x3 −x1x2 −x1x3 Poinsot’s 2 2 construction I = ρ  −x1x2 x1 + x3 −x2x3  dV (5) 2 2 −x1x3 −x2x3 x1 + x2 Lecture 23 Moments of inertia; products of inertia Rigid body dynamics

Newtonian I Diagonal elements: moments of inertia w.r.t. the mechanics of a coordinate axes: particle Newtonian Z mechanics of a 2 2 system of particles I11 = ρ(x2 + x3 ) dV (6) Newtonian mechanics of a etc. (7) rigid body The angular inertia tensor I Off-Diagonal elements: the products of inertia: Euler’s equations Z Poinsot’s construction I12 = I21 = − ρx1x2 dV (8) etc. (9)

I We could try to understand them, or we could get rid of them ... Lecture 23 Principal axes; principal moments of inertia Rigid body dynamics I Inertia matrix is symmetric—diagonal in the right frame. Define A: Newtonian mechanics of a  A  particle I11 0 0 A A Newtonian I =  0 I22 0  (10) mechanics of a system of particles 0 0 AI 33 Newtonian mechanics of a rigid body I I in A-coordinates can be obtained by: The angular inertia tensor A = T I AIA (11) Euler’s equations

Poinsot’s where matrix A transforms to A-coordinates. construction

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. Lecture 23 Scalar angular inertia. Radius of gyration. Rigid body dynamics

Newtonian mechanics of a I Consider moment of momentum L = Iω. When are L particle and ω parallel? Newtonian mechanics of a I Consider rotation about some fixed axis in direction system of particles ˆ Newtonian n. Scalar angular inertia In is mechanics of a rigid body t In = nˆ Inˆ (12) The angular inertia tensor

Euler’s equations I radius of gyration kn with respect to the axis nˆ: Poinsot’s construction 2 In = Mkn (13)

I The radius of gyration represents the distance of a point mass that would give the same angular inertia. Lecture 23 Inertia ellipsoid Rigid body dynamics

I Consider the surface described by the equation Newtonian mechanics of a rt Ir = a (14) particle Newtonian mechanics of a I In principal coordinates, since moments are positive, system of particles Newtonian we get an ellipsoid: mechanics of a rigid body 2 2 2 Ixx r + Iyy r + Izz r = a (15) The angular inertia x y z tensor

Euler’s equations I Let r = rnˆ. Then Poinsot’s construction 1 a I = nˆt Inˆ = rt Ir = (16) n r 2 r 2 So distance to ellipsoid surface is inverse of radius of gyration. a Mk 2 = (17) n r 2 Lecture 23 Cylinder and its inertia ellipsoid Rigid body dynamics

Newtonian mechanics of a particle

Newtonian mechanics of a z z system of particles Newtonian mechanics of a rigid body

The angular inertia tensor

Euler’s equations

Poinsot’s y construction x y x Lecture 23 Principal axes by inspection Rigid body dynamics

Newtonian mechanics of a particle

Newtonian mechanics of a I There are theorems: system of particles Newtonian I Any plane of symmetry is perpendicular to a principal mechanics of a axis. rigid body I Any line of symmetry is a principal axis. The angular inertia tensor I If you start in the principal frame, you know the Euler’s equations products of inertia are zero, so you can get the Poinsot’s construction inertia tensor by just doing three integrals. Lecture 23 Rigid body tumbling Rigid body dynamics I Applying Newton’s second law in principal

coordinates yields: Newtonian mechanics of a       particle I1ω˙ 1 ω1 I1ω1 Newtonian N =  I2ω˙ 2  +  ω2  ×  I2ω2  (18) mechanics of a system of particles I3ω˙ 3 ω3 I3ω3 Newtonian   mechanics of a I1ω˙ 1 + (I3ω2ω3 − I2ω2ω3) rigid body =  I2ω˙ 2 + (I1ω3ω1 − I3ω3ω1)  (19) The angular inertia tensor I3ω˙ 3 + (I2ω1ω2 − I1ω1ω2) Euler’s equations I If N = 0 we get Euler’s equations: Poinsot’s construction

I2 − I3 ω˙ 1 = ω2ω3 (20) I1 I3 − I1 ω˙ 2 = ω3ω1 (21) I2 I1 − I2 ω˙ 3 = ω1ω2 (22) I3 Lecture 23 Coordinate frame issue in differentiation Rigid body dynamics

Newtonian mechanics of a particle

I We differentiated in a fixed frame, instantaneously Newtonian mechanics of a coinciding with body principal frame. system of particles I Euler’s equations will be true only fleetingly in a Newtonian mechanics of a global fixed frame! Cannot integrate them. rigid body The angular inertia I So, transform to moving body frame from coincident tensor

fixed frame. N, I, ω unchanged. New angular Euler’s equations acceleration is Poinsot’s dω construction + ω × ω dt I.e., Euler’s equations work in the body frame. Lecture 23 Exploring Euler’s equations Rigid body dynamics

Newtonian mechanics of a I Consider special cases for Euler’s equations: particle

Newtonian I2 − I3 mechanics of a ω˙ 1 = ω2ω3 (23) system of particles I1 Newtonian I − I mechanics of a 3 1 rigid body ω˙ 2 = ω3ω1 (24) I2 The angular inertia tensor I1 − I2 ω˙ = ω ω (25) Euler’s equations 3 I 1 2 3 Poinsot’s construction 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? Lecture 23 Poinsot’s construction Rigid body dynamics

Newtonian mechanics of a particle

Newtonian mechanics of a system of particles

Newtonian mechanics of a polhode rigid body I Rigid body The angular inertia tumbling: inertia tensor ellipsoid rolls Euler’s equations herpolhode Poinsot’s without slipping construction L on a plane. Lecture 23 Proof of Poinsot’s construction Rigid body dynamics I If N is zero, then kinetic energy T is constant:

1 t Newtonian T = ω Iω is constant (26) mechanics of a 2 particle That is, ω is on the surface of the inertia ellipsoid. Newtonian mechanics of a I What is the tangent plane normal at ω? system of particles Newtonian 1 t 1 2 2 2 mechanics of a ∇ ω Iω = ∇ (ω I + ω I + ω I ) (27) rigid body 2 2 1 1 2 2 3 3 The angular inertia = (I1ω1, I2ω2, I3ω3) = L (28) tensor Euler’s equations

The attitude of the tangent plane is constant! Poinsot’s I How far from center of mass to tangent plane? construction ω · L 2T = (29) |L| |L| which is also constant! I So the tangent plane is fixed: the invariable plane. The ellipsoid rolls without slipping on the invariable plane. Lecture 23 polhodes Rigid body dynamics

I Take an ellipsoid, Newtonian mechanics of a hold the center a particle

fixed distance Newtonian mechanics of a from an inkpad, system of particles and roll it around. Newtonian mechanics of a I Near the pointy rigid body The angular inertia end you get little tensor

loops. Euler’s equations I Near the center Poinsot’s construction of mass you get little loops.

I Near the third principal axis, you get sent away. Lecture 23 Video of tumbling body Rigid body dynamics

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

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