Mechanics Rigid Body Motion 4, Lecture Notes - Physics, Study notes of Mechanics

Download Mechanics Rigid Body Motion 4, Lecture Notes - Physics and more Study notes Mechanics in PDF only on Docsity! Mechanics Physics 151 Lecture 11 Rigid Body Motion (Chapter 5) Administravia Please fill out the midterm evaluation form Critical feedback for me to evaluate how well (or badly) I’m teaching to adjust the level of the course according to your needs to receive your suggestions for improvements Be critical, and be specific I can’t fix them if you don’t tell me what’s wrong It’s anonymous and confidential Thank you! Conserved Momenta Solve them for and We need θ(t) to get φ(t) and ψ(t) 2 2 2 231 ( sin ) ( cos ) cos 2 2 IIL Mglθ φ θ φ θ ψ θ= + + + − 2 1 3 1sin cos ( cos const.) L Ip I bIφ φ θ θ φ θ ψφ ∂ = = + + = ≡ ∂ 3 3 3 1( co cons st.)p I a L I Iψ φ θ ψ ωψ ∂ = = + = = ≡ ∂ φ ψ 2 cos sin b a θφ θ − = 1 2 3 coscos sin I a b a I θψ θ θ − = − Got rid of 2 degrees of freedom conserved Energy Conservation Middle term is We’ve got a 1-dim equation of motion of θ It looks like a particle of “mass” I1 under a potential 2 2 2 231 ( sin ) ( cos ) cos 2 2 IIE Mglθ φ θ φ θ ψ θ= + + + + 21 3 32 I ω conserved 2 2 2 3 3 1 1 2 ( cos ) cos 2 2 2 sin I I I b aE E Mglω θ θ θ θ −′ = − = + + conserved 2 1 cos( ) cos 2 sin I b aV Mglθθ θ θ −⎛ ⎞= +⎜ ⎟ ⎝ ⎠ 2 cos sin b a θφ θ − = 1-D Equation of Motion Simplify the equation of motion by defining Switch variable from θ to u = cosθ Integrate 2 3 3 1 1 2 2andE I Mgl I I ωα β−≡ ≡ 2 2 cos cos sin b a θα θ β θ θ −⎛ ⎞= + +⎜ ⎟ ⎝ ⎠ 2 2 2(1 )( ) ( )u u u b auα β= − − − − ( ) 2 2(0) (1 )( ) ( ) u t u dut u u b auα β = − − − − ∫ Elliptic integral EoM becomes EoM Nutation Consider the sign of changes sign at 2 2 cos sin 1 b a b au u θφ θ − − = = − φ /u u b a′= = 1 2 or u u u u′ ′< > 1 2u u u′< < φ is monotonous φ switches direction φ θlocus φ θ Initial Condition Suppose the figure axis is initially at rest Spin the top, then release it “quietly” 0 0tθ = = 0( ) 0tf u = = 0 1 2or tu u u= = 0 0tφ = = 0 0tb au =− = 0tu u= ′= φ θt = 0 Initially, the figure axis falls It then picks up precession in φ How does it know which way to go? Origin of Precession Angular momentum conservation ω3 is constant As the figure axis falls, ω3’s contribution to pφ decreases φ must start precessing to make up for it Direction of precession is same as that of spin 2 1 3 3sin cos L I Ipφ φ θ ω θφ ∂ = = + ∂3 3 p L Iψ ωψ ∂ = = ∂ Magnetic Dipole Moment Consider a rigid body made of charged particles Mass mi, charge qi, position ri, velocity vi If there is uniform magnetic field B Each particle feels force If CoM is at rest and qi/mi = const How about the torque? i i iq= ×F v B 0i i i i qq m m = × = × =F v B v B No sum over i! No net force ( ) ( )i i i i i i i i qq m m Ν = × = × × = × ×r F r v B r v B Magnetic Dipole Moment Using Explicit calculation using polar coordinates Take time average Assume rotation is fast ( ) ( )( )i i i i i i q qm m m m = × × = × ⋅N r v B ω r r B i i=v ω×r ω B ir Θ φ 2 sin ( )( ) sin cos (sin cos sin cos cos ) 0 i i ir B φ ω θ φ θ φ θ −⎡ ⎤ ⎢ ⎥× ⋅ = Θ + Θ⎢ ⎥ ⎢ ⎥⎣ ⎦ ω r r B ( )2sin 2 2i i q qm r m m θ= =N ω×B L×B θ Magnetic Dipole Moment Magnetic dipole M in B feels the torque Fast spinning charged rigid body has a magnetic moment Equation of motion This makes L to precess around B Angular velocity of precession is d dt γ=L L×B precess 2 q m γ= − = −ω B B 2 q m =N L×B =N M×B γ=M L 2 q m γ = gyromagnetic ratio B L Larmor frequency Muon g–2 Experiment BNL E-821 muon storage ring muon 2.0023318404 0.0000000030 g = ± Summary Analyzed the motion of a heavy top Reduced into 1-dimensional problem of θ Qualitative behavior Precession + nutation Initial condition vs. behavior Magnetic dipole moment of spinning charged object M = γL, where γ = q/2m is the gyromagnetic ratio L precesses in magnetic field by ω = –γB γ of elementary particles contains interesting physics Done with rigid bodies Next: Oscillation