Midterm Feb. 17, 2009 Physics 110B Secret No.=

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1 Midterm Feb. 17, 29 Physics 11B Secret No.= PROBLEM (1) (4 points) The radient operator = x i ê i transforms like a vector. Use ɛ ijk to prove that if B( r) = A( r), then B( r) =. B i = x i x i = x j = x i = x i A k ɛ jki x j A k ɛ jki x i A k ɛ ikj x j x j A k ɛ jki B i = B i = x i x i interchane order of deriv. rename i and j antisymmetry of ɛ PROBLEM (2) (6 points) Assume that there is a full moon on Christmas Eve in San Dieo. Explain why the two hih tides on that day have very different heihts. Draw clear diarams showin the sun, the moon, and the earth illustratin why the heihts are different. Will the hiher hih tide be at niht or durin the day? Full moon means sun and moon are on opposite sides of earth. Christmas Eve mean winter implyin sun is in the southern hemisphere durin the day. Earth s axis is tilted at 23 derees from its orbit. It turns out that the moon is also close to the earth s orbital plane so that the sun and moon are really nearly opposite each other. Durin the day, the sun and moon are 23+32=55 derees off from bein over (under) San Dieo. Durin the niht (as shown in your diaram) they are only off by about 32-23=9 derees, makin the niht-time hih tide hiher. Both the sun and the moon have this effect since they are nearly opposite, however, I don t expect the students to know about the plane of the moon s orbit. I do expect them to know about the 23 deree tilt of the earth s axis relative its orbital plane (the ecliptic) and which way it points near the winter solstice.

2 PROBLEM (3) (8 points) A projectile is shot vertically (vertical determined by plumb bob) upward to a (small) heiht h, at a latitude λ = 3 derees south. Calculate the displacement from the shootin point when it returns to the round, to first order in ω. Give a numerical answer and state clearly what the direction of the displacement is. The projectile oes up and down. It is deflected by the Coriolis force. The centrifual force just chanes by a small amount. We take = 9.8 meters pr second squared. If t is the time taken to et to the heiht h, then we have h = 1 2 t2 2h or t =. We can find the initial velocity since v t = or v = t = 2h. The Coriolis force acts from to 2t = 8h when the projectile hits the round. a Coriolis = 2 ω v ˆ east = 2mωv z cos(3) ˆ east = 3ω(v t) ˆ east = 3ω( 2h t) ˆ east We may interate this to et the velocity as a function of time. The eastward velocity is initially zero so. t v x (t) = a x dt = t 3ω ( 2h t)dt = ( ) 2ht t 2 3ω 2 We may interate the velocity to et the position in the x direction, aain startin from zero. x = 3ω q 8h ( ) 2ht t 2 dt = [ ] q 2h t 2 8h 3ω 2 2 t 3 6 x = [ 2h 4h 3ω 2 3 = 3ω [ 2h 4h 3 ] 4h 2h = 1 ω 2h 4h 3 = 4 2h 2 hω 3 = 4 3 ωh 3 2 This at least has the riht units. We are supposed to et a number (except for the h part). 2 x = 4 3 2π (24)(36) h = 3 π (3)(36) h 3 2 = h 3 2 So for a heiht of 1 meters, the displacement would be 7.59 cm to the west. 4h ] 8h

3 PROBLEM (4) (1 points) A riid body consists of three masses connected by nearly massless rods. m 1 = 5m at (b,, b) m 2 = 3m at ( b,, b) m 3 = 2m at (b, b, ) a) Find the inertia tensor for rotations about the iven oriin. b) The system is rotated about the z axis with anular velocity ω. What is the anular momentum vector in the body frame? a) I ij = [δ ij rα 2 r αi r αj ] α = 5m δ ij 2b 2 b m δ ij 2b 2 b m δ ij 2b 2 b = 8mb 2 δ ij mb 2 δ ij = 2mb 2 8δ ij mb 2 2δ ij = 2mb 2 1δ ij = 2mb = 2mb b) L = I ω = 2mb 2 ω 1 9 = 2mb

4 PROBLEM (5) (1 points) A yroscope of mass M is made of a thin uniform disk of radius R attached to a spindle (perpendicular to the disc) of neliible mass. The center of the disk, which is the center of mass of the yroscope, is a distance h from the pivot point of the spindle. a) Compute the inertia tensor in the principle axis system. b) Assume the yroscope spins in space with no forces actin on it. The anular velocity has a manitude ω and is at an anle φ from the symmetry axis. Find the precession frequency of ω around the symmetry axis in the body system. (Put the results of part (a) in last so we can check part (b) independently.) c) Now assume that the bottom point of the spindle is fixed and that the yroscope rotates in a constant ravitational acceleration. Find the three constants of the motion in terms of the Euler anles and their derivatives. a) The z principle axis is the symmetry axis of the disk. The other two are perpendicular and in the plane of the disk. The density of the disk (per unit area) is ρ = M/πR 2. I (3) = ρ R I (1) = I (2) = ρ r 2 2πrdr = M πr 2 2π R 2π R r 2 cos 2 θdθrdr = M πr 2 π 1 I = MR b) Use the 3 Euler equations to et the precession frequency. r 3 dr = M R4 2π πr2 4 = 1 2 MR2 R I (3) ω 3 I (1) ω 1 ω 2 + I (2) ω 2 ω 1 = r 3 dr = M R 2 R 4 4 = 1 4 MR2 Since I (1) = I (2). ω 3 = I (2) ω 2 + I (1) ω 1 ω 3 I (3) ω 3 ω 1 = I (1) ω 1 I (2) ω 2 ω 3 I (3) ω 3 ω 2 = ω 2 = I(3) I (1) I (2) ω 3 ω 1 Ωω 1 ω 1 = I(3) I (2) I (1) ω 3 ω 2 Ωω 2 ω 2 = Ω ω 1 = Ω 2 ω 2 This precesses with the anular frequency Ω = I(3) I (1) I (2) ω 3 = ω 3. c) There will be two conserved momenta derived from the Laranian plus enery will be conserved. L = 1 2 I(12) ( φ2 sin 2 θ + θ 2 ) Mh cos θ

5 L = 1 8 MR2 ( φ2 sin 2 θ + θ 2 ) MR2 ( Mh cos θ The Laranian does not depend on φ or ψ so p φ = L φ and p ψ = L are conserved. ψ p φ = 1 ( ( ) ) φ 4 MR2 sin 2 θ + 2 φ cos θ + ψ cos θ = 1 ( φ(1 4 MR2 + cos 2 θ) + ψ ) cos θ The other conserved quantity is just the Enery. p ψ = 1 ( ) φ 2 MR2 cos θ + ψ E = 1 2 I(12) ( φ2 sin 2 θ + θ 2 ) + Mh cos θ E = 1 8 MR2 ( φ2 sin 2 θ + θ 2 ) MR2 ( + Mh cos θ

6 t 2 δ L(q i, q i ; t)dt = t 1 f k (q j ; t) = Laranian Dynamics L q j L q j d L dt d L dt q j q j = p j = L + k q j λ k (t) f k q j = ṗ j = L q j H(q k, p k, t) = j p j q j L(q k, q k, t) q j = H p j ṗ j = H q j Vectors and Rotations ê i ê j = δ ij v x = v i x i ( v x) k = v i x j ɛ ijk ɛ 123 = 1 ɛ ijk = ɛ ikj = ɛ jik = ɛ kji ɛ ijk = ɛ kij = ɛ jki ɛ mnj ɛ ikj = δ mi δ nk δ mk δ ni x i = R ij x j R ij = ê i ê j R(θ) = cos(θ) cos( π θ) cos( π) cos(θ) sin(θ) 2 2 cos( π + θ) cos(θ) cos( π) = sin(θ) cos(θ) R 1 = R T 2 2 cos( π) cos( π) cos() Non-Inertial Coordinate Systems V rotatin = V inertial ω V ω = θ atidal GMmoon r ed [2 cos φê d 2 x sin φê y ] F A = F I ma 2m ω v A m d ω x dt A m ω ω x A FA = F I 2m ω v A m ω ω x A Riid Body Rotation I ij α m α [δ ij r 2 α r αi r αj ] T rot = 1 2 I ijω i ω j L i = I ij ω j I ij = R im R jn I mn = R im I RN Rnj T I = RIR T L = I ω Iij I (k) δ ij = I ij ω (k) j = I (k) ω (k) i ˆω (j) ˆω (k) = δ jk 1) Rotate by anle φ 2) Rotate by anle θ 3) Rotate by anle ψ about z I axis. about x axis. about the z B axis. r = r + a I ij = I ij + M [δ ij a 2 a i a j ] cos ψ cos φ cos θ sin φ sin ψ cos ψ sin φ + cos θ cos φ sin ψ sin ψ sin θ R(φ, θ, ψ) = sin ψ cos φ cos θ sin φ cos ψ sin ψ sin φ + cos θ cos φ cos ψ cos ψ sin θ sin θ sin φ sin θ cos φ cos θ ω 1 = φ sin θ sin ψ + θ cos ψ ω 2 = φ sin θ cos ψ θ sin ψ ω 3 = φ cos θ + ( ψ ) I (k) ω k + I (j) ω j ω i ɛ ijk = Γ k I (3) ω 3 + (I (2) I (1) I )ω 2 ω 1 = Ω (3) I (1) ω I (2) 3 L = 1 2 I(12) ( φ2 sin 2 θ + θ 2 ) mh cos θ E = 1 2 I(12) ( φ2 sin 2 θ + θ 2 ) + mh cos θ