Homework 9 # 8.19, 8.24, 8.25

Homework 9: # 8.19, 8.24, 8.25 8.19 The point of suspension of a simple pendulum of length l and mass m is constrained to move on a parabola z = ax2 in the vertical plane. Derive a Hamiltonian governing the motion of the pendulum and its point of suspension. Obtain the Hamilton’s equations of motion. Answer: Let x′ = x + l sin θ ′ z = ax2 − l cos θ Then T = 1 m( ẋ′2 + ż ′2 ) 2 U = mgz ′ Solving in terms of generalized coordinates, x and θ, our Lagrangian is L = T −U = 1 2 ˙ 2 x2 ẋ2 +4axẋl˙ θ sin θ+l2 θ̇2 )−mg (ax2 −l cos θ) m( ẋ2 +2 ẋlcos θθ+4a Using 1 ˜ q̇Tq̇ 2 where q̇ and T are matrices. We can see with T = m(1 + 4a2 x2 ) ml(cos θ + 2axsin θ) ml(cos θ + 2axsin θ) ml 2 1  ẋ˙ θ q̇ =  L = L0 + L0 = −mg (ax2 − l cos θ) The Hamilitonian is H= 1 −1 pT ˜ p − L0 2 Inverting T by a b c d with the algebra, −1 = 1 ad− bc d −b −c a  1 m2 l2 (1 + 4ax2 ) − m2 l2 (cos θ + 2axsin θ)2 1 = ad− bc this is = 1 m2 l2 (sin2 θ + 4ax2 − 4axcos θ sin θ − 4a2 x2 sin2 θ) = 1 m2 l2 (sin2 θ − 4axsin θ cos θ + 4a2 x2 cos2 θ) which I’ll introduce, for simplicity’s sake, Y. = 1 ≡ m2 l2 (sin θ − 2axcos θ)2 1 m2 l 2 Y So now we have T −1 = T −1 1 m2 l 2 Y 1 = ml2 −ml(cos θ + 2axsin θ) −ml(cos θ + 2axsin θ) m(1 + 4a2 x2 )  1 −(cos θ + 2axsin θ)/l 2 2 2 −(cos θ + 2axsin θ) /l (1 + 4a x ) /l mY I want to introduce a new friend, lets call him J J ≡ (cos θ + 2axsin θ) Y ≡ (sin θ − 2axcos θ)2 So, T −1 = 1 1 −J/l −J/l (1 + 4a2 x2 )/l2 mY Proceed to derive the Hamiltonian, H= 1 −1 pT ˜ p − L0 2 2   we can go step by step, 1 T −1 p = and 1 −J/l −J/l (1 + 4a2 x2 /l2 mY   px pθ = 1 mY px − (J/l)pθ (−J/l)px + (1 + 4a2 x2 /l 2 )pθ  1 1 + 4a2 x2 2 J J (p2 − pθ px − pθ) pθ px + mY x l2 l l the full Hamiltonian is ˜ −1 p = pT H= 1 + 4a2 x2 2 1 J (p2 − 2 pθ px + pθ ) + mg (ax2 − l cos θ) 2mY x l2 l plugging in my Y and J H= 1 1 + 4a2 x2 2 cos θ + 2axsin θ (p2 −2 p θ )+mg (ax2 −l cos θ) pθ px + 2m(sin θ − 2axcos θ)2 x l2 l Now to find the equations of motion. They are ẋ = ∂H ∂px θ̇ = ∂H ∂pθ ṗx = − ∂H ∂H ṗθ = − ∂θ ∂x The first two are easy, especially with my substitutions. ẋ = θ̇ = J 1 cos θ + 2axsin θ 1 [px − pθ ] = [px − pθ ] l l mY m(sin θ − 2axcos θ)2 1 + 4a2 x2 1 + 4a2 x2 1 1 [−Jpx + [−(cos θ+2axsin θ)px + pθ ] = mY l ml(sin θ − 2axcos θ)2 l l But the next two are far more involved. I handled the partial with respect to x by taking the product rule between the two main pieces, the fraction out front, and mess inside the parenthesis that has p terms. I then broke each p term and began grouping them. Go slowly, and patiently. After taking the derivative before grouping, my ṗx looked like this: ṗx = − ∂H = ∂x − ∂H ∂x 8a2 x 2 1 −4a sin θ p] pθ px + [ l2 θ l 2m(sin θ − 2axcos θ)2 1 + 4a2 x2 2 cos θ + 2axsin θ −2(−2a cos θ) [p2 − 2 p θ ] + 2mgax pθ px + 2m(sin θ − 2axcos θ)3 x l2 l 3 pθ ] Now start simplifying. Lets group the p terms. 4a(cos θ + 2axsin θ) 2 p 2ml2 [sin θ − 2axcos θ]3 θ 4a cos θ p2 2m[sin θ − 2axcos θ]3 x and the longest one.. 2a [sin2 θ − 2 − 2axcos θ sin θ]pθ px lm[sin θ − 2axcos θ]3 Adding them all up yields, for ṗx : − ∂H =− ∂x cos θ + 2axsin θ 2 2 − sin2 θ + 2axsin θ cos θ 2a [cos θpx2+ p θ− px pθ ]−2mgax m[sin θ − 2axcos θ]3 l2 l Now for the next one, ṗθ : ∂H ∂θ Taking the derivative, you get a monster, of course ṗθ = − 1 2 sin θ 4axcos θ ∂H = [ − ]pθ px l ∂θ 2m[sin θ − 2axcos θ]2 l +[p2x − 2 1 + 4a2 x2 2 −2(cos θ + 2axsin θ) cos θ + 2axsin θ p θ ][ pθ px + + mglsin θ l2 l 2m[sin θ − 2axcos θ]3 separating terms.. −(cos θ + 2axsin θ) 2 p m[sin θ − 2axcos θ]3 x cos θ + 2axsin θ 1 + 4a2 x2 2 pθ l2 m[sin θ − 2axcos θ]3 and the longest one... [ (sin θ − 2axcos θ)2 2(cos θ + 2axsin θ)2 + ]pθ px lm[sin θ − 2axcos θ]3 lm[sin θ − 2axcos θ]3 add them all up for the fourth equation of motion, ṗθ − 1 + 4a2 x2 2 1 ∂H [(cos θ + 2axsin θ)(p2x + pθ ) = l2 m[sin θ − 2axcos θ]3 ∂θ − [(sin θ − 2axcos θ)2 + 2(cos θ + 2axsin θ)2 ] pθ px l 4 Together in all their glory: ṗθ = − 1 + 4a2 x2 2 1 ∂H [(cos θ + 2axsin θ)(p2x + pθ) = l2 m[sin θ − 2axcos θ]3 ∂θ − ṗx = − ∂H =− ∂x [(sin θ − 2axcos θ)2 + 2(cos θ + 2axsin θ)2 ] pθ px l 2 cos θ + 2axsin θ 2 2 − sin θ + 2axsin θ cos θ 2a [cos θpx2+ p θ− px pθ ]−2mgax m[sin θ − 2axcos θ]3 l2 l ẋ = θ̇ = cos θ + 2axsin θ 1 [px − pθ ] l m(sin θ − 2axcos θ)2 1 1 + 4a2 x2 [−(cos θ + 2axsin θ)px + pθ ] ml(sin θ − 2axcos θ)2 l 8.24 A uniform cylinder of radius a and density ρ is mounted so as to rotate freely around a vertical axis. On the outside of the cylinder is a rigidly fixed uniform spiral or helical track along which a mass point m can slide without friction. Suppose a particle starts at rest at the top of the cylinder and slides down under the influence of gravity. Using any set of coordinates, arrive at a Hamiltonian for the combined system of particle and cylinder, and solve for the motion of the system. Answer: My generalized coordinates will be θ, the rotational angle of the particle with respect to the cylinder, and φ the rotational angle of the cylinder. The moment of inertia of a cylinder is 1 1 M a2 = ρπha4 2 2 There are three forms of kinetic energy in the Lagrangian. The rotational energy of the cylinder, the rotational energy of the particle, and the translational kinetic energy of the particle. The only potential energy of the system will be the potential energy due to the height of the particle. The hardest part of this Lagrangian to understand is likely the translational energy due to the particle. The relationship between height and angle of rotational for a helix is I= h = cθ Where c is the distance between the coils of the helix. MathWorld gives a treatment of this under helix. Understand that if the cylinder was not rotating 5 then the rotational kinetic energy of the particle would merely be 2 ma2 θ̇2 , but the rotation of the cylinder is adding an additional rotation to the particle’s position. Lets write down the Lagrangian, L= 1 m 2 [a ( θ̇ + φ̇)2 + c2 ˙θ2 ] + mgcθ I φ̇2 + 2 2 This is L = L0 + 1 ˜ q̇Tq̇ 2 Solve for T. θ  ˙ ˙  ma2 + mc2 ma2 q̇ = 2 2 T = ma I + ma φ previous problem, we Using the same 2 by 2 inverse matrix form from the may solve for T −1 . 1 T −1 = (ma2 + mc2 )(I + ma2 ) − m2 a4 Now we can find the Hamiltonian. H= I + ma2 −ma2 −ma2 m(a2 + c2 )  1 p˜ T−1 p − L0 2 This is H= p2θ (I + ma2 ) − 2ma2 pθ pφ + p2 φ m( a 2[m(a2 + c2 )(I + ma2 ) − m2 a4 ] 2 + c2 ) − mgcθ From the equations of motion, we can solve for the motion of the system. (duh!) Here are the EOM: − ∂H = ṗθ = mgc ∂θ ∂H = ṗφ = 0 − ∂φ (I + ma2 )pθ − ma2 pφ ∂H = θ̇ = ∂pθ m(a2 + c2 )(I + ma2 ) − m2 a4 −ma2 pθ + pφ m(a2 + c2 ) ∂H = φ̇ = m(a2 + c2 )(I + ma2 ) − m2 a4 ∂pφ ˙ = φ̇(0) = 0 To solve for the motion, lets use the boundary conditions.θ(0) leads to pφ (0) = pθ (0) = 0 leads to pθ = mgct pφ = 0 6 Pluggin and chuggin into θ̇ and φ˙ and integrating, yields the motion φ= −m2 a2 gct2 2[m(a2 + c2 )(I + ma2 ) − m2 a4 ] θ= (I + ma2 )mgct2 2[m(a2 + c2 )(I + ma2 ) − m2 a4 ] 1 2 a where M is the mass of the cylinder, we obtain If we plug in I =2 M φ= θ= 2[mc2 −mgct2 + 12 M (a2 + c2 )] (m + 1 2 M )gct2 2[mc2 + 12 M (a2 + c2 )] 8.25 Suppose that in the previous exercise the cylinder is constrained to rotate uniformly with angular frequency ω. Set up the Hamiltonian for the particle in an inertial system of coordinates and also in a system fixed in the rotating cylinder. Identify the physical nature of the Hamiltonian in each case and indicate whether or not the Hamiltonians are conserved. Answer: In the laboratory system, the particle moves through an angle ψ = θ + φ. The cylinder moves uniformly, φ = ωt, so the kinetic energy T = 1 2 1 ma ( θ̇ + φ̇)2 + mc2 θ̇2 2 2 may be expressed 1 2 2 1 ma ψ̇ + mc2 ( ψ̇ − ω)2 2 2 The potential energy may be written T = U = −mgc(ψ − ωt) So we have L= 1 2 m(a2 ψ̇ 2 + c2 ( ˙ψ − ω) ) + mgc(ψ − ωt) 2 ∂L = p = ma2 ˙ψ + mc2 ( ˙ψ − ω) ∂ q̇ and with H= 1 (˜ p − a)T −1 (p̃ − a) − L0 2 7 we find T −1 from 1 ˜ −1 q̇ T q̇ + q̇a+ L0 2 We can see things better if we spread out L L= L= 1 1 1 2 2 2 2 ma2 ψ̇ 2 + mc2 ψ̇ − mc ω ˙ψ + mc ω + mgc(ψ − ωt) 2 2 2 so L0 = 1 mc2 ω 2 + mgc(ψ − ωt) 2 and T = [ma2 + mc2 ] T −1 = 1 m ( a2 + c2 ) Therefore, for our Hamiltonian, we have Hlab = (p − mc2 ω)2 mc2 ω 2 − 2m(a2 + c2 ) 2 − mgc(ψ − ωt) This is dependent on time, therefore it is not the total energy. For the Hamiltonian in the rotating cylinder’s frame, we express the movement in terms of the angle θ this is with respect to the cylinder. ψ = θ + φ = θ + ωt ψ̇ = θ̇ + φ̇ = θ̇ + ω T = L= 1 2 1 2 ma ( θ̇ + ω)2 + mc2 θ̇ 2 2 U = −mgcθ 1 2 ˙ 1 ma ( θ + ω)2 + mc2 θ̇2 + mgcθ 2 2 1 L = ˜q̇Tq̇ + q̇a+ L0 2 Spread out L 1 1 [ma2 + mc2 ] ˙ θ2 + ma2 ˙θω+ ma2 ω 2 + mgcθ 2 2 It becomes clear that L= 8 T = [ma2 + mc2 ] T −1 = L0 = 1 ma2 + mc2 1 ma2 ω + mgcθ 2 Using again, H= 1 (p − a)T −1 (p − a) − L0 2 we may write H= (p − ma2 ω)2 1 ma2 ω − mgcθ − 2 2 m ( a2 + c2 ) This is not explicitly dependent on time, it is time-independent, thus conserved. 9