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MIT 8 01T - Experiment 09: Angular momentum

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Experiment 09: Angular momentumExperiment 09: Angular momentumGoals Investigate conservation of angular momentum and kinetic energy in rotational collisions. Measure and calculate moments of inertia. Measure and calculate non-conservative work in an inelastic collision.Apparatus Connect output of phototransistor to channel A of 750. Connect output of tachometer generator to channel B of 750. Connect power supply. Red button is pressed: Power is applied to motor. Red button is released: Rotor coasts: Read output voltage using DataStudio.  Use black sticker or tape on white plastic rotor for generator calibration.4Calibrate tachometer-generator Spin motor up to full speed, let it coast. Measure and plot voltages for 0.25 s period. Sample Rate: 5000 Hz, and Sensitivity: Low. Average the output voltage over the same 10 periods. Time 10 periods to measure ω. Then calculate ω for 1 V output.Measure rotor IR Start DataStudio and let the weight drop. Plot only the generator voltage for rest of experiment. Use a 55 gm weight to accelerate the rotor. Settings:¾ Sensitivity: Low ¾ Sample rate 500 Hz.¾ Delayed start: None¾ Auto Stop: 4 secondsUnderstand graph output to measure IR Generator voltage while measuring IR. What is happening:1. Along line A-B ?2. At point B ?3. Along line B-C ? How do you use this graph to find IR?Measure IRresults Measure and record αupand αdown.  For your report, calculate IR:upup down()Rmr g rIααα−=−downfRIτα=Fast collisionFind ω1(before) and ω2(after), estimate δt for collision.CalculateFalls below 0.5V1 sec200 HzLowAuto StopDelayed StartSample RateSensitivity2212()WoiImrrω=+Slow collision Find ω1and ω2, measure δt , fit or measure to find ac. Keep a copy of your results for the homework problem.Kepler Problem and Planetary Motion8.01tNov 15, 2004Kepler’s Laws1. The orbits of planets are ellipses; and the sun is at one focus2. The radius vector sweeps out equal areas in equal time3. The period T is proportional to the radius to the 3/2 powerT~ r3/2Kepler Problem• Find the motion of two bodies under the influence of a gravitational force using Newtonian mechanics()121,22ˆmmrGr=−FrGReduction of Two Body Problem• Reduce two body problem to one body of mass µ moving about a central point under the influence of gravity with position vector corresponding to the vector from mass m2to mass m1()21,22ˆFddtµ=rrrGG1212mmmmµ=+Solution of One Body Problem• Solving the problem means finding the distance from the origin and angle as functions of time• Equivalently, finding the distance from the center as a function of angle• Solution: ()rt()tθ()rθ01cosrrεθ=−Constants of the Motion• Velocity• Angular Momentum• Energyˆˆdr drdt dtθ=+vr θG21212Gm mEvrµ=−2tangentialdLrv rdtθµµ==222dr dvrdt dtθ⎛⎞⎛ ⎞=+⎜⎟⎜ ⎟⎝⎠⎝ ⎠221212Gm mdr dErdt dt rθµ⎡⎤⎛⎞⎛ ⎞=+−⎢⎥⎜⎟⎜ ⎟⎝⎠⎝ ⎠⎢⎥⎣⎦22122122Gm mdr LEdt r rµµ⎛⎞=+−⎜⎟⎝⎠Reduction to One Dimensional Motion• Reduce the one body problem in two dimensions to a one body problem moving only in the r-direction but under the action of a repulsive force and a gravitational forcePRS QuestionSuppose the potential energy of two particles (reduced mass µ) is given by where r is the relative distance between the particles. The force between the particles is1. attractive and has magnitude2. repulsive and has magnitude3. attractive and has magnitude4. repulsive and has magnitude221()2LUrrµ=23LFrµ=23LFrµ=22LFrµ=22LFrµ=One Dimensional Description•Energy • Kinetic Energy• Effective Potential Energy• Repulsive Force• Gravitational force221221122effectiveGm mdr LEKUdt r rµµ⎛⎞=+−=+⎜⎟⎝⎠212drKdtµ⎛⎞=⎜⎟⎝⎠21222effectiveGm mLUrrµ=−22232centrifugaldL LFdr r rµµ⎛⎞=− =⎜⎟⎝⎠122gravitationalgravitationaldUGm mFdr r=− =−Energy Diagram21222effectiveGm mLUrrµ=−Case 4: Circular Orbit E = EoCase 3: Elliptic Orbit Eo < E < 0Case 2: Parabolic Orbit E = 0Case 1: Hyberbolic Orbit E > 0PRS QuestionThe radius and sign of the energy for the lowest energy orbit (circular orbit) is given by1. energy is positive, 2. energy is positive, 3. energy is negative, 4. Energy is negative,2012LrGm mµ=20122LrGm mµ=20122LrGm mµ=2012LrGm mµ=PRS Answer: Circular Orbit• The lowest energy state corresponds to a circular orbit where the radius can be found by finding the minimum of effective potential energy• Energy of circular orbit()()0212022effectiverrGm mEULµ===−212320effectivedUGm mLdr r rµ==−+2012LrGm mµ=PRS QuestionIf the earth slows down due to tidal forces will the moon’s angular momentum1. increase2. decrease3. cannot tell from the information givenPRS QuestionIf the earth slows down due to tidal forces will the radius of the moon’s orbit1. increase2. decrease3. cannot tell from the information givenOrbit Equation• Solution:where the two constants are• radius of circular orbit• eccentricity01cosrrεθ=−2012LrGm mµ=()12221221ELGm mεµ⎛⎞=+⎜⎟⎜⎟⎝⎠Energy and Angular Momentum• Energy:where E0is the energy of the ‘ground state’• Angular momentumwhere r0is the radius of the ‘ground state’()12201EEε=−()()0212022effectiverrGm mEULµ===−()1/2012LrGmmµ=Properties of Ellipse: ()001maximumrrrθε===−()01minimumrrrθπε===+()00 012211221112maximum minimumrr rGm mar rEεε ε⎛⎞=+=+==−⎜⎟−+ −⎝⎠Semi-Major axislocation of the center of the ellipseSemi-Minor axisArea0021maximumrxr a aεεε=−==−()22 1/21/200baxar=−=3/2 1/20Aab a rππ==Kepler’s Laws: Equal Area• Area swept out in time ∆t• Equal Area Law:()122rArrrtt tθθ∆∆∆ ∆⎛⎞=+⎜⎟∆∆ ∆⎝⎠212dA drdt dtθ=2dLdt rθµ=2dA Lconstantdtµ==Kepler’s Laws: Period•Area • Integral of Equal Area Law•Period• Period squared proportional to cube of the major axis but depends on both masses3/2 1/20Aabarππ==02TorbitdA dtLµ=∫∫3/2 1/2022arTALLµπµ==()223232230212 1 2444aaTarLGm m G m mµπµππ===+Two Body Problem Revisited• Elliptic Case: Each mass orbits around center of mass with()21 211 2 211 112 12 2cmmmmmm mm mµ−+′=− =− = =++rrrrrrR r rGGGGGGG G


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MIT 8 01T - Experiment 09: Angular momentum

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