Mechanics Physics 151

Transcription

1 Mechanics Physics 151 Lecture 11 Rigid Body Motion (Chapter 5)

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3 What We Did Last Time 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 Analyze it today I 1 I3 L= ( θ + φ sin θ) + ( φcos θ + ψ ) Mglcosθ ρ 1 I y

4 Heavy Top Top is spinning on a fixed point Lagrangian is I 1 I3 L= ( θ + φ sin θ) + ( φcos θ + ψ ) Mglcosθ φ and ψ are cyclic Symmetry p φ and p ψ are conserved z θ y φ ψ x

5 Conserved Momenta L I 1 I3 = ( θ + φ sin θ) + ( φcos θ + ψ ) Mglcosθ L pψ = = I3( φcosθ + ψ ) = I3ω3 = const. Ia 1 ψ L pφ = = I 1φsin θ + I3cos θ( φcosθ + ψ ) = const. Ib 1 φ Solve them for φ and ψ b acosθ 1 = ψ sin θ I3 φ We need θ(t) to get φ(t) and ψ(t) Ia cos cos b a θ = θ sin θ Got rid of degrees of freedom

6 Energy Conservation φ = b acosθ sin θ E I 1 I3 = ( θ + φ sin θ) + ( φcos θ + ψ ) + Mglcosθ 1 Middle term is I ω 3 3 conserved I3ω3 I 1θ I1 ( b acos θ) E = E = + + Mglcosθ sin θ We ve got a 1-dim equation of motion of θ It looks like a particle of mass I 1 under a potential I1 b acosθ V( θ ) = + Mglcosθ sinθ conserved conserved

7 1-D Equation of Motion Simplify the equation of motion by defining α E I3ω3 Mgl and β I I 1 1 EoM becomes Switch variable from θ to u = cosθ EoM Integrate t = b acosθ α = θ + + β cosθ sinθ u = (1 u )( α βu) ( b au) ut () u(0) du (1 u )( α βu) ( b au) Elliptic integral

8 Qualitative Behavior Try to extract qualitative behavior Same way as we did with central force problem Consider the RHS of the last equation u = f( u) (1 u )( α βu) ( b au) 3 = βu ( α + a ) u + ( ab β) u+ ( α b ) Physical range is f( u) = u 0 and 1 u 1 f(u) is a cubic function of u with f( ± 1) = ( b au) 0 β > Mgl I 1 0 These conditions constrain the shape of f(u) u = cosθ

9 Shape of f(u) f(u) = 0 has 3 roots Solution for u = f( u) is bounded inside u u u θ oscillates between arccos(u 1 ) and arccos(u ) φ and ψ determined by 1 u u 1 u u1 u 1 f ( u) u 3 u b acosθ φ = sin θ Ia 1 cos cos b a θ ψ = θ I sin θ 3

10 Nutation b acosθ b au φ = = sin θ 1 u φ changes sign at u = u = b/ a Consider the sign of u < u or u > u u < u < u 1 1 φ is monotonous φ switches direction locus θ θ φ φ

11 Initial Condition Suppose the figure axis is initially at rest Spin the top, then release it quietly θ t = 0 = 0 f( u t = 0) = 0 ut= 0 = u1 or u φ = = b au t = 0 = 0 ut= 0 = u t 0 0 t = 0 φ θ Initially, the figure axis falls It then picks up precession in φ How does it know which way to go?

12 Origin of Precession Angular momentum conservation p ψ L = = I3ω3 ψ L p φ = = I 1φ sin θ + I3ω3cosθ φ ω 3 is constant As the figure axis falls, ω 3 s contribution to p φ decreases φ must start precessing to make up for it Direction of precession is same as that of spin

13 Uniform Precession Can we make a top precess without bobbing? i.e. θ = 0, φ = const We need to have a double root for f(u) = 0 f( u ) = (1 u )( α βu ) ( b au ) = f ( u ) = u ( α βu ) β(1 u ) + a( b au ) = β = a φ φ u Combine Ia Iω β Mgl I 1 0 Mgl = φ( I ω I φcos θ ) φ = u 0 b acosθ sin θ 1 f( u) u 3 u

14 Uniform Precession Mgl = φ( I ω I φcos θ ) For any given value of ω 3 and cosθ 0, you must give exactly the right push in φ to achieve uniform precession Quadratic equation solutions Same top can do fast or slow precession For the solutions to exist I ω > 4MglI cosθ ω > MglI cosθ I Uniform precession is achieved only by a fast top

15 Magnetic Dipole Moment Consider a rigid body made of charged particles Mass m i, charge q i, position r i, velocity v i If there is uniform magnetic field B Each particle feels force F = q v B i i i If CoM is at rest and q i /m i = const q F= qivi B= mivi B= m 0 No sum over i! No net force How about the torque? q Ν= ri Fi = qiri ( vi B) = miri ( vi B) m

16 Magnetic Dipole Moment ω Using v i = ω r i q q N = miri ( i ) mi( i)( i ) m v B = ω m r r B B Explicit calculation using polar coordinates ω r r B sinφ 0 Take time average Assume rotation is fast q ( sin ) q N = mi ri θ ω B= L B m m θ Θ φ ( i)( i ) = ωrb i sinθ cos φ (sinθ cosφsin Θ+ cosθ cos Θ) r i

17 Magnetic Dipole Moment N = q m L B Magnetic dipole M in B feels the torque Fast spinning charged rigid body has a magnetic moment M = γ L dl = γ L B dt This makes L to precess around B Equation of motion γ = Angular velocity of precession is q ωprecess = γ B= B m q m Larmor frequency N = M B gyromagnetic ratio B L

18 Elementary Particles Particles such as electrons or protons have Spin, or intrinsic angular momentum, s Magnetic moment µ q µ = s m Differs from classical charged object by factor gq 1 classical object Particle physicists say µ = s g = m Dirac particle Dirac equation for a spin-1/ particle predict g = for electron and muon Dirac particles g =.8 for proton, 1.9 for neutron composite particles

19 Anomalous Magnetic Moment µ of electron and muon known very accurately g electron = ± g muon = ± Not pure Dirac particles, but surrounded by thin cloud of virtual particles due to quantum fluctuation Measurement uses spin precession Store particles with known spin orientation in B field Measure spin direction after time t gq ωprecess = B Need to know B very accurately m

20 Muon g Experiment BNL E-81 muon storage ring g muon = ±

21 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/m 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