Mechanics Rigid Body Motion 3, Lecture Notes - Physics, Study notes of Mechanics

Download Mechanics Rigid Body Motion 3, Lecture Notes - Physics and more Study notes Mechanics in PDF only on Docsity! Mechanics Physics 151 Lecture 10 Rigid Body Motion (Chapter 5) 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 s r d d dt dt    = + ×        ω ( )2jk i i jk ij ikI m r x xδ= − =L Iω 21 2 2 T Iω⋅= =ω Iω Finding Principal Axes ! Consider unit vectors n1, n2, n3 along principal axes ! To find the principal axes and principal moments: ! Express I in in any body coordinates ! Solve eigenvalue equation ! Eigenvectors point the principal axes ! Use them to re-define the body coordinates to simplify I ! You can often find the principal axes by just looking at the object i i iI=In n 1 2 3 0 0 0 0 0 0 D I I I    = =      I RIR! ni is an eigenvector of I with eigenvalue Ii ( ) 0λ− =I r 0λ− =I 1 2 3, ,I I Iλ = Rotational Equation of Motion ! Concentrate on the rotational motion ! Newtonian equation of motion gives ! Take the principal axes as the body axes s d dt   =    L N “space” axes b d dt   + =    L ω×L N “body” axes 1 1 1 1 2 2 2 2 3 3 3 3 0 0 0 0 0 0 I I I I I I ω ω ω ω ω ω            = =                 L = Iω s b d d dt dt    = + ×        ω Euler’s Equation of Motion ! Special cases: ! ! 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 I I N d I I N dt I I N ω ω ω ω ω ω ω ω ω                + × =                       1 1 2 3 2 3 1 2 2 3 1 3 1 2 3 3 1 2 1 2 3 ( ) ( ) ( ) I I I N I I I N I I I N ω ω ω ω ω ω ω ω ω − − = − − = − − = " " " Euler’s equation of motion for rigid body with one point fixed 2 3 0ω ω= = 1 1 1I Nω =" 2 3I I= 1 1 1I Nω =" Inertia Ellipsoid I = nρ x′ y′ z′ 11 I 21 I 31 I ρ Inertia ellipsoid represents the moment of inertia of a rigid body in all directions 2 2 2 1 1 2 2 3 31 I I Iρ ρ ρ= + + Usefulness of this definition will become apparent soon … Inertia Ellipsoid ! 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 ! Using I = nρ I = ⋅n In 2 2 2 1 1 2 2 3 3( ) 1F I I Iρ ρ ρ≡ ⋅ = + + =ρ ρ Iρ F = 1 ρ F∇ρ 2F∇ =ρ Iρ 2I T= =ρ n ω 2F T ∇ =ρ L Surface of the inertia ellipsoid is perpendicular to L Invariable Plane ! As ρ moves, inertia ellipsoid must rotate to satisfy this condition continuously ! Consider the projection of ρ on L ! The ellipsoid is touching a fixed plane perpendicular to L 2F T ∇ =ρ L Surface of inertia ellipsoid at ρ is perpendicular to the angular momentum L 2 2 T L LL T ⋅ ⋅= =ρ L ω L constant ρ L 2T L Invariable plane L is conserved! Simple Cases ! Inertia ellipsoid is a sphere (I1 = I2 = I3) ! ρ is constant and parallel to L ! Stable rotation ρ L Simple Cases ! Initial axis is close to one of the principal axes ! Assume I1 > I2 > I3 ! Stable rotation around I1 and I3 ! Not so obvious around I2 ! If ω1 = ω3 = 0, ω2 is constant ! Small deviation leads to instability 11 I 31 I 1 1 2 3 2 3 2 2 3 1 3 1 3 3 1 2 1 2 ( ) 0 ( ) 0 ( ) 0 I I I I I I I I I ω ω ω ω ω ω ω ω ω − − = − − = − − = " " " Simple Cases ! Since I1 > I2 > I3, distance 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 21 I Rotation Under Torque ! We introduce torque ! Things get messy ! Consider a spinning top ! Define Euler angles 1 1 2 3 2 3 1 2 2 3 1 3 1 2 3 3 1 2 1 2 3 ( ) ( ) ( ) I I I N I I I N I I I N ω ω ω ω ω ω ω ω ω − − = − − = − − = " " " z y x φ ψ θ Lagrangian ! Assume I1 = I2 ≠ I3 ! Kinetic energy given by ! Use Euler angles 2 2 2 1 1 2 3 3 1 1( ) 2 2 T I Iω ω ω= + + sin sin cos cos sin sin cos φ ψ θ θ ψ φ ψ θ θ ψ φ θ ψ  +  = −   +  ω " " " " " " 2 2 2 231 ( sin ) ( cos ) 2 2 IIT θ φ θ φ θ ψ= + + +" " " " Lagrangian ! Potential energy is given by the height of the CoM ! Lagrangian is ! 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 … l θ cosV Mgl θ= 2 2 2 231 ( sin ) ( cos ) cos 2 2 IIL Mglθ φ θ φ θ ψ θ= + + + −" " " "