Mechanics Rigid Body Motion 2, Lecture Notes - Physics, Study notes of Mechanics

Download Mechanics Rigid Body Motion 2, Lecture Notes - Physics and more Study notes Mechanics in PDF only on Docsity! Mechanics Physics 151 Lecture 9 Rigid Body Motion (Chapter 4, 5) What We Did Last Time Discussed 3-dimensional rotation Preparation for rigid body motion Movement in 3-d + Rotation in 3-d = 6 coordinates Looked for ways to describe 3-d rotation Euler angles one of the many possibilities Euler’s theorem Defined infinitesimal rotation dΩ Commutative (unlike finite rotation) Behaves as an axial vector (like angular momentum) d d= ×r r Ω d d= ΦΩ n General Vectors Now consider a general vector G How does it move in space/body coordinates? i.e. what’s the time derivative dG/dt ? Movement dG differs in space and body coordinates because of the rotation of the latter If G is fixed to the body ( ) ( ) ( )space body rotd d d= +G G G Difference is due to rotation ( )body 0d =G ( )spaced d= ×G Ω Gand ( )rotd d= ×G Ω G Generally true Angular Velocity For any vector G ω = instantaneous angular velocity Direction = n = instantaneous axis of rotation Magnitude = dΦ/dt = instantaneous rate of rotation Since this works for any vector, we can say ( ) ( )space bodyd d d= + ×G G Ω G space body d d dt dt ⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ G G ω G dt d d= = Φω Ω n s r d d dt dt ⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ω space coordinates rotating coordinates Mechanics Physics 151 Lecture 9 Rigid Body Motion (Chapter 4, 5) What We Did Last Time Discussed 3-dimensional rotation Preparation for rigid body motion Movement in 3-d + Rotation in 3-d = 6 coordinates Looked for ways to describe 3-d rotation Euler angles one of the many possibilities Euler’s theorem Defined infinitesimal rotation dΩ Commutative (unlike finite rotation) Behaves as an axial vector (like angular momentum) d d= ×r r Ω d d= ΦΩ n General Vectors Now consider a general vector G How does it move in space/body coordinates? i.e. what’s the time derivative dG/dt ? Movement dG differs in space and body coordinates because of the rotation of the latter If G is fixed to the body ( ) ( ) ( )space body rotd d d= +G G G Difference is due to rotation ( )body 0d =G ( )spaced d= ×G Ω Gand ( )rotd d= ×G Ω G Generally true Angular Velocity For any vector G ω = instantaneous angular velocity Direction = n = instantaneous axis of rotation Magnitude = dΦ/dt = instantaneous rate of rotation Since this works for any vector, we can say ( ) ( )space bodyd d d= + ×G G Ω G space body d d dt dt ⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ G G ω G dt d d= = Φω Ω n s r d d dt dt ⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ω space coordinates rotating coordinates Coriolis Effect Imagine a particle observed in a rotating system e.g. watching an object’s motion on Earth Velocity: Acceleration: Newton’s equation works in the space (inertial) system, i.e. s r d d dt dt ⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ω s r= + ×v v ω r 2 ( ) s s s s s r r r d d dt dt ⎛ ⎞ ⎛ ⎞= = + ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + × + × × v va ω v a ω v ω ω r sm=F a eff 2 ( ) ( )r rm m m= = − × − × ×a F F ω v ω ω r Object appears to move according to this force Euler Angles Use angular velocity ω to calculate particles’ velocities Use Euler angles to describe the rotation of rigid bodies How are they connected? Infinitesimal rotations can be added like vectors z x y ξ η ζ z x y ξ ′ η′ζ ′ φ θ z x y z′ x′ y′ ψ z zξφ θ ψ′= + +ω n n n zn ξn z′n Euler Angles Let’s express ω in x’-y’-z’ Doing this in x-y-z equally easy (or difficult) Must express nz, nξ, nz’ in x’-y’-z’ nz Anz, nξ Bnξ (nz’ is OK) z x y ξ η ζ z x y ξ ′ η′ζ ′ φ θ z x y z′ x′ y′ ψ zn ξn z′n D C B Euler Angles From the last lecture cos sin 0 sin cos 0 0 0 1 φ φ φ φ ⎡ ⎤ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎣ ⎦ D 1 0 0 0 cos sin 0 sin cos θ θ θ θ ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥−⎣ ⎦ C cos sin 0 sin cos 0 0 0 1 ψ ψ ψ ψ ⎡ ⎤ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎣ ⎦ B cos cos cos sin sin cos sin cos cos sin sin sin sin cos cos sin cos sin sin cos cos cos cos sin sin sin sin cos cos ψ φ θ φ ψ ψ φ θ φ ψ ψ θ ψ φ θ φ ψ ψ φ θ φ ψ ψ θ θ φ θ φ θ − +⎡ ⎤ ⎢ ⎥= − − − +⎢ ⎥ ⎢ ⎥−⎣ ⎦ A 0 sin sin 0 cos sin 1 cos z ψ θ ψ θ θ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ n A 1 cos 0 sin 0 0 ξ ψ ψ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ n B 0 0 1 z′ ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ n Potential Energy Potential energy can often be separated as well Uniform gravity g Uniform magnetic field B and magnetic dipole moment M Lagrangian can then be written as 1 2( , , ) ( , , )V V x y z V φ θ ψ= + 1V = − ⋅g r 2V = − ⋅M B ( , , , , , ) ( , , , , , )t rL L x y z x y z L φ θ ψ φ θ ψ= + Translational Rotational It is often possible to separate the translational and rotational motions by taking the center of mass as the origin of the body coordinate axes Rotational Motion We concentrate on the rotational part Translational part same as a single particle Easy Consider total angular momentum vi is given by the rotation ω as i i im= ×L r v i i= ×v ω r 2 2 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i m m r x m x y m x z m r m y x m r y m y z m z x m z y m r z = × × ⎡ ⎤− − − ⎢ ⎥⎡ ⎤= − ⋅ = − − −⎢ ⎥⎣ ⎦ ⎢ ⎥− − −⎣ ⎦ L r ω r ω r r ω ω Inertia tensor I Inertia Tensor Diagonal components are familiar moment of inertia What are the off-diagonal components? Iyx produces Ly when the object is turned around x axis Imagine turning something like: 2 2 2 2( ) sinxx i i i i iI m r x m r= − = Θ x r Θ Balanced Unbalanced Unbalanced one has non-zero off-diagonal components, which represents “wobbliness” of rotation Shifting Origin Origin of body axes does not have to be at the CoM It’s convenient – Separates translational/rotational motion If it isn’t, I can be easily translated i i′= +r R rfrom origin from CoM 2 2 2 2 ( ) [( ) ] ( ) ( ) 2 ( ) ( ) i i i i i i i i I m m M m m ′= × = + × ′ ′= × + × + × ⋅ × r n R r n R n r n R n r n I from CoMI of CoM Summary 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 Next: Equation of motion (finally!) s r d d dt dt ⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ω ( )2i i jk ij ikm r x xδ= −I =L Iω 2 21 2 2 T Iω ω= =nIn