Mechanics Physics 151

Transcription

1 Mechanics Physics 151 Lecture 10 Rigid Body Motion (Chapter 5)

2 What We Did Last Time! Found the velocity due to rotation! Used it to find Coriolis effect! Connected ω with the Euler angles! Lagrangian " translational and rotational parts! Often possible if body axes are defined from the CoM! Defined the inertia tensor! Calculated angular momentum and kinetic energy d d = + ω dt dt s ( 2 δ ) I = m r x x jk i i jk ij ik ω Iω 1 2 L= Iω T = = Iω 2 2 r

3 Diagonalizing Inertia Tensor m r x mx y mxz I = = 2 2 i( i i ) i i i i i i myx i i i mi( ri yi ) myz i i i mi ri jk xijxik 2 2 mzx i i i mzy i i i mi( ri zi )! Inertia tensor I can be made diagonal ( δ ) I1 0 0 ID = RIR! = 0 I2 0 R is a rotation matrix 0 0 I 3 I i > 0 Kinematical properties of a rigid body are fully described by its mass, principal axes, and moments of inertia

4 Principal Axes! Consider a rigid body with body axes x-y-z! Inertia tensor I is (in general) not diagonal! But it can be made diagonal by! Rotate x-y-z by R " New body axes x -y -z! In x -y -z coordinates ω = Rω L = RL = RIω = RIRRω! = I ω D RR! = 1 How do I find them? I = D RIR! One can choose a set of body axes that make the inertia tensor diagonal " Principal Axes

5 Finding Principal Axes! Consider unit vectors n 1, n 2, n 3 along principal axes In = I n i i i! To find the principal axes and principal moments:! Express I in in any body coordinates! Solve eigenvalue equation ( I λ) r = 0 λ = 0! Eigenvectors point the principal axes 1 = = 0 I2 0 I RIR! 0 0! Use them to re-define the body coordinates to simplify I 0 0! You can often find the principal axes by just looking at the object D I n i is an eigenvector of I with eigenvalue I i λ = I, I, I I I 3

6 Rotational Equation of Motion! Concentrate on the rotational motion! Newtonian equation of motion gives dl = N dl dt + ω L = N s dt b d d = + ω dt dt s b space axes body axes! Take the principal axes as the body axes I1 0 0 ω1 I1ω1 L=Iω= 0 I 0 ω = I ω I 3 ω 3 I3ω 3

7 Euler s Equation of Motion I1ω1 ω1 I1ω1 N1 d I ω + ω I ω = N dt I3ω 3 ω 3 I3ω 3 N 3! Special cases:! ω2 = ω3 = 0 I " 1ω 1 = N1! I " ω ω ω ( I I ) = N I " ω ω ω ( I I ) = N I " ω ωω ( I I ) = N Euler s equation of motion for rigid body with one point fixed I 2 = I3 I1ω 1 = N1 "

8 Torque-Free Motion! No linear force " Conservation of linear momentum! No torque " Conservation of angular momentum! Try N = 0 in Euler s equation of motion I " 1ω1 ω2ω3( I2 I3) = 0 I " 2ω2 ω3ω1( I3 I1) = 0 I " ω ωω ( I I ) = ! We will do something more intuitive (hopefully)! Geometrical trick by L. Poinsot Integrating these equation will give us energy and angular momentum conservation

9 Inertia Ellipsoid! For any direction n, i ij j! If we express n using principal axes x -y -z I = I n = I n + I n + I n i I = nin = i ni n Moment of inertia about axis n! Consider a vector ρ = n I = Iiρi = I1ρ1 + I2ρ2 + I3ρ3 Inertia ellipsoid

10 Inertia Ellipsoid ρ = n I 1 = I ρ + I ρ + I ρ I 3 z ρ Inertia ellipsoid represents the moment of inertia of a rigid body in all directions x 1 I 1 1 I 2 y Usefulness of this definition will become apparent soon

11 Inertia Ellipsoid ρ = n I I = nin F( ρ) ρ Iρ = I ρ + I ρ + I ρ = 1! Inertia along axis n is ! F is a function (like potential) defined in the ρ space! F(ρ) = 1 " Inertia ellipsoid! Normal of the ellipsoid given by the gradient F = 2Iρ ρ ρ F ρ! Using F = ρ ρ = n I = ω 2 L T 2T Surface of the inertia ellipsoid is perpendicular to L F = 1

12 Invariable Plane L is conserved! F = ρ 2 L T Surface of inertia ellipsoid at ρ is perpendicular to the angular momentum L! As ρ moves, inertia ellipsoid must rotate to satisfy this condition continuously! Consider the projection of ρ on L ρ L L ω L = = L 2T 2T L constant! The ellipsoid is touching a fixed plane perpendicular to L Invariable plane ρ L 2T L

13 Invariable Plane! Inertia ellipsoid touches the invariable plane! Distance between the center and the plane is! Touching point = instantaneous axis of rotation! Tip of the ρ vector is momentarily at rest in space dρ ω ω = ω ρ = = 0 dt 2T! i.e. it s not sliding, but rolling without slipping on the invariable plane Determined by the initial conditions ρ 2T L

14 Invariable Plane! Inertia ellipsoid rolls on the invariable plane! Rotation of ellipsoid gives the rotation of the body! Direction of ρ gives the direction of ω in space! Touching point draws curves! Curve drawn on the ellipsoid = polhode! Curve drawn on the invariable plane = herpolhode! Let s examine a few simple cases

15 Simple Cases! Inertia ellipsoid is a sphere (I 1 = I 2 = I 3 )! ρ is constant and parallel to L! Stable rotation ρ L

16 Simple Cases! Initial axis is close to one of the principal axes! Assume I 1 > I 2 > I 3! Stable rotation around I 1 and I 3! Not so obvious around I 2! If ω 1 = ω 3 = 0, ω 2 is constant I " 1ω1 ω2ω3( I2 I3) = 0 I " 2ω2 ω3ω1( I3 I1) = 0 I " ω ωω ( I I ) = ! Small deviation leads to instability 1 I 3 1 I 1

17 Simple Cases! Since I 1 > I 2 > I 3, distance 1 I 2 allows a polhode that wraps around the inertia ellipsoid! Rotation around a principal axis is stable except for the one with the intermediate moment of inertia

18 Simple Cases! Inertia ellipsoid is symmetric around one axis! I 1 = I 2 I 3! ρ draws a cone (space cone) on the invariable plane! ρ draws a cone (body cone) in the inertia ellipsoid! Body cone rolls on the space cone ρ L

19 Precession! I 1 = I 2 turns the Euler s equation of motion to I " 1ω1 = ω2ω3( I1 I3) I " 1ω2 = ω3ω1( I3 I1) I " 3ω3 = 0 I3 I1 " ω1 = Ω ω2, " ω2 =Ωω1 Ω ω3 I ω 3 is constant Consider it as a given initial condition 1 ω = Acos Ω t, ω = Asin Ωt 1 2! ω precesses around the I 3 axis! Draws the body cone ω3 A ω

20 Rotation Under Torque! We introduce torque! Things get messy! Consider a spinning top! Define Euler angles z θ I " ω ω ω ( I I ) = N I " ω ω ω ( I I ) = N I " ω ωω ( I I ) = N y φ ψ x

21 Lagrangian! Assume I 1 = I 2 I 3! Kinetic energy given by T = I1( ω1 + ω2) + I3ω3 2 2! Use Euler angles " φsinψ sinθ + " θ cosψ ω = " φcosψ sinθ " θsinψ " φcosθ + ψ " I 1 I T = (" θ + " φ sin θ ) + (" φ cos θ + ψ" )

22 Lagrangian! Potential energy is given by the height of the CoM V = Mglcosθ! Lagrangian is I I3 2 L= ( " θ + " φ sin θ ) + ( " φ cos θ + ψ" ) Mglcosθ 2 2! Finally we are in real business!! How we solve this?! Note φ and ψ are cyclic! Can define conjugate momenta that conserve! To be continued θ l

23 Summary! Discussed rotational motion of rigid bodies! Euler s equation of motion! Analyzed torque-free rotation! Introduced the inertia ellipsoid! It rolls on the invariant plane! Dealt with simple cases! Started discussing heavy top 1 I 3 z x 1 I 1! Found Lagrangian " Will solve this next time I I3 2 L= ( " θ + " φ sin θ ) + ( " φ cos θ + ψ" ) Mglcosθ 2 2 ρ 1 I 2 y