PHY 711 Classical Mechanics and Mathematical Methods 10-10:50 AM MWF Olin 103 Plan for Lecture 5: 1. Chapter 2 – Physics described in an accelerated coordinate frame 9/7/2012 PHY 711 Fall 2012 -- Lecture 5 1 #5 9/7/2012 PHY 711 Fall 2012 -- Lecture 5 2 Special seminar today: 11-11:50 AM Olin 103 SILICENE 9/7/2012 PHY 711 Fall 2012 -- Lecture 5 3 Application of Newton’s laws in a coordinate system which has an angular velocity ω and linear acceleration d 2a 2 dt inertial Newton’s laws; Let r denote the position of particle of mass m: d 2r m 2 Fext dt inertial d 2r d 2a dω dr m 2 Fext m 2 2mω m r mω ω r dt dt body dt body dt inertial Coriolis force 9/7/2012 PHY 711 Fall 2012 -- Lecture 5 Centrifugal force 4 Motion on the surface of the Earth: 2 Fext 7.3 10 5 rad/s GM e m rˆ F ' 2 r Main contributions : d 2r GM e m dω dr ˆ m 2 r F ' 2 m ω m r mω ω r 2 r dt dt earth dt earth 9/7/2012 PHY 711 Fall 2012 -- Lecture 5 5 Non-inertial effects on effective gravitational “constant” d 2r GM e m dω dr ˆ m 2 r F ' 2 m ω m r mω ω r 2 r dt dt earth dt earth d 2r dr For 0 and 2 0, dt earth dt earth GM e m 0 rˆ F ' mω ω r 2 r F ' mg GM e g 2 rˆ ω ω r r r Re GM e 2 2 2 ˆ ˆ R sin r sin cos R θ e e 2 Re 0.03 m/s2 9/7/2012 9.80 m/s2 PHY 711 Fall 2012 -- Lecture 5 6 Foucault pendulum http://www.si.edu/Encyclopedia_SI/nmah/pendulum.htm The Foucault pendulum was displayed for many years in the Smithsonian's National Museum of American History. It is named for the French physicist Jean Foucault who first used it in 1851 to demonstrate the rotation of the earth. 9/7/2012 PHY 711 Fall 2012 -- Lecture 5 7 Equation of motion on Earth’s surface d 2r GM e m dω dr ˆ m 2 r F ' 2mω m r mω ω r 2 r dt dt earth dt earth y (east) z (up) x (south) 9/7/2012 PHY 711 Fall 2012 -- Lecture 5 8 Foucault pendulum continued – keeping leading terms: d 2r GM e m dr ˆ m 2 r F ' 2mω 2 Re dt earth dt earth GM e m rˆ mgzˆ 2 r F ' T sin cos xˆ T sin sin yˆ T coszˆ ω sin xˆ cos zˆ dr ω y cos xˆ x cos z sin yˆ y cos zˆ dt earth 9/7/2012 PHY 711 Fall 2012 -- Lecture 5 9 Foucault pendulum continued – keeping leading terms: d 2r GM e m dr ˆ m 2 r F ' 2mω 2 Re dt earth dt earth mx T sin cos 2my cos my T sin sin 2m x cos z sin mz T cos mg 2my cos Further approximation : 1; z 0; T mg mx mg sin cos 2my cos my mg sin sin 2mx cos Also note that : x sin cos y sin sin 9/7/2012 PHY 711 Fall 2012 -- Lecture 5 10 Foucault pendulum continued – coupled equations: g x 2 cos y g y y 2 cos x x Try to find a solution of the form : x(t ) Xe iqt y (t ) Yeiqt Denote cos q 2 g i 2 q X i 2 q q 2 g Y 0 Non - trivial solutions : q 2 9/7/2012 g PHY 711 Fall 2012 -- Lecture 5 11 Foucault pendulum continued – coupled equations: Solution continued : x(t ) Xe iqt y (t ) Yeiqt q 2 g i 2 q X i 2 q q 2 g Y 0 Non - trivial solutions : q 2 g Amplitude relationship : X iY General solution with complex amplitudes C and D : y (t ) ReCe x(t ) Re iCe i t iDe i t 9/7/2012 i t De i t PHY 711 Fall 2012 -- Lecture 5 12 General solution w ith complex amplitudes C and D : y (t ) ReCe x(t ) Re iCe i t iDe i t i t De i q 2 g 7 10 5 cos rad / s 9/7/2012 t g PHY 711 Fall 2012 -- Lecture 5 13
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