Rigid Body - Kinematics

MechanicsPhysics 151Lecture 9Rigid 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)dΩ= ndΦdr = r×dΩ• Behaves as an axial vector (like angular momentum)

Goals For Today• Time derivatives in rotating coordinate system• To calculate velocities and accelerations• Coriolis effect• Express angular velocity using Euler angles• Try to write down Lagrangian for rigid body• Separate rotation from movement of CoM• Define inertia tensor• Will almost get to the equation of motion

Body Coordinates• Consider a rotating rigid body• Define body coordinates (x’, y’, z’)• Between t and t + dt, the bodycoordinates rotate bydΩ= ndΦ• The direction is CCW because weare talking about the coordinate axes• Observed in the space coordinates,any point r on the body moves bydr = dΦ n× r = dΩ×rzz′x x′nrdΦdry′ySign opposite from last lecturebecause the rotation is CCW

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 coordinatesbecause of the rotation of the latter( d ) = ( d ) + ( d )G G G Difference is due to rotationspace body rot• If G is fixed to the body( d G ) body= 0 and ( d ) space( d ) rotG = dΩ×GG = dΩ×G Generally true

Angular Velocity• For any vector Gspace( ) ( )d G = d G + d Ω×Gbodyspacebody⎛dG⎞ ⎛dG⎞⎜ ⎟ = ⎜ ⎟ + ω×G ωdt = dΩ= dΦn⎝ dt ⎠ ⎝ dt ⎠• 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⎛ d ⎞ ⎛ d ⎞⎜ ⎟ = ⎜ ⎟ + ω×⎝dt⎠ ⎝dt⎠srspace coordinatesrotating coordinates

Coriolis Effect⎛ d ⎞ ⎛ d ⎞⎜ ⎟ = ⎜ ⎟ + ω×⎝dt⎠ ⎝dt⎠sr• Imagine a particle observed in a rotating system• e.g. watching an object’s motion on Earth• Velocity: vs= vr+ ω×r• Acceleration:⎛dvs⎞ ⎛dvs⎞as= ⎜ ⎟ = ⎜ ⎟ + ω×v⎝ dt ⎠s⎝ dt ⎠r= a + 2 ω× v + ω× ( ω×r)r• Newton’s equation works in the space (inertial) system,i.e. F= masma = Feff = F− 2 m( ω× v ) − mω× ( ω×r)rrrObject appears to move according to this forces

Coriolis Effectma = Feff = F− 2 m( ω× v ) − mω× ( ω×r)ωr• Last term is centrifugal force• Middle term is Coriolis effectr• Occurs when object is moving in therotating frame− ω×vr• Moving objects on Earth appears to deflect toward right inthe northern hemisphere• Hurricane wind pattern• Foucault pendulumLv r

Coriolis Effect• Free-falling particle• Drop a ball and ignore air resistance• Coriolis effect is − 2 m( ω×v)• Pointing east (out of screen) x axis• Ignoring the centrifugal forcezvθω⎧mx =−2mωv zsinθ⎨⎩mz =−mgxωgtsin θ gt= , z =−3 23 2ωsin θ ( −2 z)x =3 g3• For sinθ = 1 and z = 100 m x = 2.2 cm

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 vectorsω = n φ + n θ + n ′ ψ xzn zζφξηyζ ′zzθxξξ′η′zyz′nn z′ξxzy′ψx′y

Euler Angles• Let’s express ω in x’-y’-z’• Doing this in x-y-z equally easy (or difficult)• Must express n z , n ξ , n z’ in x’-y’-z’• n z An z , n ξ Bn ξ (n z’ is OK)D C Bz ζz η′nζ ′zθηz′yξ yxφξx ξ′nn z′xzy′ψx′y

Euler Angles• From the last lectureA⎡ cosφsinφ0⎤D =⎢sinφcosφ0⎥⎢−⎥C⎢⎣0 0 1⎥⎦⎡1 0 0 ⎤=⎢0 cosθsinθ⎥⎢ ⎥B⎢⎣0 −sinθcosθ⎥⎦⎡ cosψsinψ0⎤=⎢sinψcosψ0⎥⎢−⎥⎢⎣0 0 1⎥⎦⎡ cosψ cosφ − cosθsinφsinψ cosψ sinφ + cosθ cosφsinψ sinψ sinθ⎤=⎢sinψ cosφ cosθsinφcosψ sinψ sinφ cosθ cosφcosψ cosψ sinθ⎥⎢− − − +⎥⎢⎣sinθsinφ −sinθ cosφ cosθ⎥⎦nz⎡0⎤ ⎡sinψsinθ⎤= A⎢0⎥ ⎢cosψsinθ⎥⎢ ⎥=⎢ ⎥⎢⎣1⎥⎦ ⎢⎣ cosθ⎥⎦nξ⎡1⎤ ⎡ cosψ⎤= B⎢0⎥ ⎢sinψ⎥⎢ ⎥=⎢−⎥⎢⎣ 0⎥⎦ ⎢⎣ 0 ⎥⎦nz′⎡0⎤=⎢0⎥⎢ ⎥⎢⎣ 1⎥⎦

Euler Angles⎡ φ sinψ sinθ + θ cosψ⎤⎢⎥ω = n zφ + n ξθ + nz′ψ= φcosψ sinθ − ⎢θ sinψ⎥⎢ φcosθ + ψ ⎥⎣⎦• Remember: this is in the x’-y’-z’ system• Now we know how to express velocities in terms oftime-derivatives of Euler angles• We can write down the Lagrangian

Kinetic Energy• Kinetic energy of multi-particle system is1 1T = Mv + mv′2 2 i i2 2Remember Einstein conventionMotion of CoMMotion around CoM• If we define the body axis from the center of mass1 2 2 2T = M( x + y + z) + T′( φ, θψ , )2• T’ depends only on the angular velocity• Must be a 2 nd order homogeneous function

Potential Energy• Potential energy can often be separated as wellV = V ( x, y, z) + V ( φ, θψ , )1 2• Uniform gravity gV = −gr⋅• Uniform magnetic field B and magnetic dipole moment MV =−M⋅B2• Lagrangian can then be written asL= L ( x, y, z, x, y, z) + L ( φ, θψφθψ , , , , )tr1TranslationalRotationalIt is often possible to separate the translational and rotational motions bytaking 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• v i is given by the rotation ω asL= m r × ( ω×r )i iiv = ω×riL= m ir i× v i2 2⎡mi( ri −xi ) −mx i iyi −mxz⎤i i i2 ⎢2 2⎥= m⎡i ⎣ωri −ri( ri⋅ ω) ⎤⎦ = ⎢ −myx i i imi( ri − yi ) −myzi i i ⎥ ω⎢2 2−mzx i i i−mzy i i imi( ri −zi) ⎥⎣⎦Inertia tensor Ii

Inertia Tensor• Diagonal components are familiarmoment of inertiaI = m ( r − x ) = mr sin Θ2 2 2 2xx i i i i i• What are the off-diagonal components?• I yx produces L y when the object is turned around x axis• Imagine turning something like:rΘxBalancedUnbalancedUnbalanced one has non-zerooff-diagonal components,which represents “wobbliness”of rotation

Inertia Tensor• Using( x, y , z ) → ( x , x , x )i i i i1 i2 i3I2 2⎡mi( ri −xi ) −mx i iyi −mxz⎤i i i⎢2 2⎥= ⎢ −myx i i imi( ri − yi ) −myzi i i ⎥⎢2 2−mzx i i i−mzy i i imi( ri −zi) ⎥⎣⎦(2δ )I = m r −xxjk i i jk ij ik• We can also deal with continuous mass distribution ρ(r)(2δ )Ijk= ∫ ρ()r rjk−xj xkdr

Kinetic Energy(2δ )I = m r −x xjk i i jk ij ik• Kinetic energy due to rotation is• Usingv = ω×rinIn ⋅ 2 1 2T = ω = Iω2 2• NB: n moves with timeiwhere1= mv⋅ v2 i i i1ωωT = miv i⋅ ( ω× ri) = ⋅ mi( ri× vi)= ⋅L2 2 2⋅• Using L=Iω T = ω Iω2• Defining n as the unit vector in the direction of ω ω = ωn• I = I(t) must change accordingly with timeTI = nIn ⋅ = m ⎡⎣r−( r⋅n)2 2i i i⎤⎦Moment of inertia about axis n

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 translatedfrom originr ′i= R+ri I = m ( r × n) = m[( R+ r′) × n]from CoM2 2i i i i2 2i i i i= M( R× n) + m ( r′ × n) + 2 m ( R× n) ⋅ ( r′× n)I of CoMI from 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 momentumand kinetic energy• Next: Equation of motion (finally!)⎛ d ⎞ ⎛ d ⎞⎜ ⎟ = ⎜ ⎟ + ω×⎝dt⎠ ⎝dt⎠s(2δ )I = m r −x xi i jk ij iknIn 2 1 2L=Iω T = ω = Iω2 2r