Understanding Rotational Motion in Classical Mechanics

phy 711 classical mechanics and mathematical n.w
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Explore the concepts of rotational motion in classical mechanics, including rigid body motion in body-fixed frames, conversion between reference frames, and symmetric top motion. Dive into moment of inertia tensors, kinetic energy calculations, and angular momentum descriptions in rotating bodies. Understand the complexities of torque equations and solve Euler equations for a deeper grasp of rotational dynamics.

  • Rotational Motion
  • Classical Mechanics
  • Rigid Body Motion
  • Moment of Inertia
  • Angular Momentum
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  1. PHY 711 Classical Mechanics and Mathematical Methods 10-10:50 AM MWF Olin 103 Plan for Lecture 24: Rotational motion (Chapter 5) 1. Rigid body motion in body fixed frame 2. Conversion between body and inertial reference frames 3. Symmetric top motion 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 1

  2. 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 2

  3. Summary of previous results describing rigid bodies rotating about a fixed origin r d dt r = r inertial 2 ( ) 1 2 1 2 = = 2 p Kinetic energy: T m v m r p p p p p 1 2 1 2 I ( ) ( ) = m r r p p p p )( I ) ( m ) 2 ( = m r r r 1 p p p p ( ) p 1 2 2 p p p r r r p = p 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 3

  4. Moment of inertia tensor Matrix notation: I I I I I I I I I xx xy xz ( ) 2 I I m ij p r pi pj r r yx yy yz ij p p zx zy zz 1 2 = For general coordinate system: T I ij i j ij For (body fixed) coordinate system that diagonalizes = = e I e moment of inertia tensor: 1,2,3 I i i i i 1 2 = + + e = 2 i 1 1 e e T I 2 2 3 3 i i 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 4

  5. Continued -- summary of previous results describing rigid bodies rotating about a fixed origin r r d dt = r inertial ( ) = = L p p r v p p r Angular momentum: m m r p p p p ( ) ( ) = L m r r r r p p p p p p ( ) 2 p I 1 p p r r m r = L I p p 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 5

  6. Descriptions of rotation about a given origin -- continued For (body fixed) coordinate system that diagonalizes moment of inertia tensor: I I I d d dt L 1 1 1 2 2 2 3 3 I I I + + e e e = = + + 1 1 e e e I e L e I e 2 2 3 3 i i i = + + L 1 1 e e 1 2 2 2 3 3 3 L = + Time derivative: L dt body d dt = + 3 ( ) ( ) ( ) + + e e e I I I I I I 2 3 3 2 1 3 1 1 3 2 1 2 2 1 3 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 6

  7. Descriptions of rotation about a given origin -- continued Note that the torque equation d d dt dt L L = + = L body is very difficult to solve directly in the body fixed frame. For 0 we can solve the Euler equations: d I I I dt ( 2 3 3 2 1 I I + e ( ( ( 3 1 2 2 I + = = L = = + + + 1 1 e e e 0 1 2 2 2 ) ) ) ) 3 3 3 ( ) ( ) + e e I I I I I 3 1 1 3 2 1 2 2 1 3 + = 0 I I 1 1 2 3 3 2 + = 0 I I I 2 2 3 1 1 3 PHY 711 Fall 2019 -- Lecture 24 1 0 I 10/25/2019 3 I 7

  8. Euler equations rotation for = I body in fixed frame : ( ( ( ) ) ) ~ ~ ~ + 0 I I 1 1 2 3 3 2 ~ ~ ~ + = 0 I I I 2 2 3 1 1 3 ~ ~ ~ + = 0 I I I 3 3 1 2 2 1 = Solution + symmetric for top - - : I I 2 1 ( ( ) ) ~ ~ ~ = 0 I I I 1 1 2 3 3 1 ~ ~ ~ + = 0 I I I 1 2 3 0 1 1 3 ~ ~ = = (constant) I I 3 3 3 ~ ~ = ~ I ~ 1 2 3 1 Define : ~ 3 = 1 I 2 1 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 8

  9. Solution of Euler equations for a symmetric top -- continued ~ 2 2 1 I ~ ~ ~ = = 1 I I ~ 3 1 where 3 1 t ~ = + Solution : ( ) cos( ) A t 1 ~ = + = ( 1 ) sin( 1 + ) t A t ~ 2 1 ~ i = 2 i 2 2 3 T I I A I 1 3 2 2 2 i ~ A ~ t ~ + = + + e e e L I I I 1 1 1 2 2 + 2 3 3 sin 3 ( ( ) ( ) ) ~ = + + e e e cos I t I 1 1 2 3 3 3 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 9

  10. Euler equations rotation for = I body in fixed frame : ( ( ( ) ) ) ~ ~ ~ + 0 I I 1 1 2 3 3 2 ~ ~ ~ + = 0 I I I 2 2 3 1 1 3 ~ ~ ~ + = 0 I I I 3 3 1 2 2 1 Solution asymmetric for top - - : I I I I 3 2 1 I ~ ~ 3 2 Suppose : 0 Define : 3 1 3 I 1 I I ~ 3 1 Define : 2 3 I 2 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 10

  11. Euler equations for asymmetric top -- continued ( ( ( ) ) ) + = 0 I I I 1 1 2 3 3 2 + = 0 I I I 2 2 3 1 1 3 + = 0 I I I 3 3 1 2 2 1 I I I I If 0, Define: 3 2 3 1 3 1 3 2 3 I I 1 2 ~ ~ ~ ~ = = 1 1 2 2 2 1 If and A are both t positive or both negative : 1 2 ( ) ~ + ( ) cos t 1 1 2 ( ) ~ + ( ) sin 2 t A t 1 2 2 1 If and have opposite signs, solution unstable. is 1 2 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 11

  12. Transformation between body-fixed and inertial coordinate systems Euler angles inertial body x y y x http://en.wikipedia.org/wiki/Euler_angles 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 12

  13. 0 3 e ~ = + + 0 3 ' 2 e e e 3 3 e x Need to express all components in body-fixed frame: y y = + + 1 1 e 2 2 e 3 3 e x ' 2 e 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 13

  14. ~ = + + 0 3 ' 2 e e e 3 0 3 e = + ' 2 e e e sin cos 1 2 3 e Matrix representa tion : x cos sin 0 0 sin y = = ' 2 e sin cos 0 1 cos y 0 0 1 0 0 x ' 2 e 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 14

  15. ~ = + + 0 3 ' 2 3 e e 0 e e 3 = + + 0 3 e e e = Matrix representation: cos sin 0 = sin ' cos sin cos ' e 1 3 sin sin + e e cos 1 2 3 3 e sin cos 0 0 1 cos 0 1 0 sin 0 cos 0 0 1 x = 0 3 e 0 y 0 sin y sin sin cos sin x ' 2 e cos 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 15

  16. ~ = + + 0 3 ' 2 e e e 3 sin cos sin 0 ~ = + + sin sin cos 0 cos 0 1 ~ ~ ~ ~ = + + e e e 1 1 2 2 3 3 sin cos sin 0 ~ = cos + + sin sin cos 0 sin 0 1 ( ( cos ) ~ ~ = + + sin cos 1 ) ~ = sin sin cos 2 = + 3 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 16

  17. ~ = + + 0 3 ' 2 e e e 3 0 3 e ( ) ) + = + e sin cos sin 1 ( + e sin sin cos 2 + + e cos 3 e 3 x y y x ' 2 e 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 17

  18. Rotational kinetic energy ( ) 1 1 1 ~ ~ ~ = + + 2 2 2 2 3 , , , , , T I I I 1 1 2 3 2 2 2 1 ( ) 2 = + sin cos sin I 1 2 1 ( ) 2 + + sin sin cos I 2 2 1 2 + + cos I 3 2 = If : I I 1 2 ( ) ( ) 1 1 2 = + + + 2 2 2 , , , , , sin cos T I I 1 3 2 2 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 18

  19. Transformation between body-fixed and inertial coordinate systems Euler angles inertial body x y y x http://en.wikipedia.org/wiki/Euler_angles 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 19

  20. General transformation between rotated coordinates Euler angles inertial = = R R R R V V V R ' = body cos sin 0 cos 0 sin cos sin 0 sin cos 0 0 1 0 sin cos 0 0 0 1 sin 0 cos 0 0 1 x y y x http://en.wikipedia.org/wiki/Euler_angles 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 20

  21. Motion of a symmetric top under the influence of the torque of gravity: Mg ( ) ( ) 1 , = + + 2 2 2 , , , , sin L I 1 2 1 2 + cos cos I Mgl 3 2 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 21

  22. ( ) ( ) 1 , = + + 2 2 2 , , , , sin L I 1 2 1 2 + cos cos I Mgl 3 2 Constants of motion the : L = = 1 + + 2 sin cos cos p I I 3 L = = + cos p I 3 2 p 2 1 ( ) 1 I = + + 2 E I V ( eff 2 3 ) 2 2 cos ( ) p p p 1 1 I = + + 2 , cos L I Mgl 2 2 2 sin 2 I 1 3 ( ) 2 cos p p ( ) = + cos V Mgl eff 2 2 sin I 1 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 22

  23. ( ) 2 2 cos p p p 1 1 I = + + + 2 cos E I Mgl 2 2 2 2 sin I 3 1 p ( ) 2 2 cos p p 1 1 I = = + + 2 ' cos E E I Mgl 2 2 2 2 sin I 3 1 Stable/unstable solutions near =0 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 23

  24. = Suppose and 0 p p ( ) 2 2 cos p p p 1 = = + + 2 ' cos E E I Mgl 1 2 2 2 2 sin I I 3 1 ( 1 ) 2 2 + p 2 ( 1 ) 1 1 1 + + 2 2 ' 2 E I Mgl 1 1 2 2 2 2 I 1 2 p 1 Mgl + + 2 2 I Mgl 1 2 8 2 I 1 solution Stable if 4 p MglI 1 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 24

  25. More general case: ( ) 2 2 cos p p p 1 1 I = = + + 2 ' cos E E I Mgl 2 2 2 2 sin I 3 1 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 25

  26. Constants of motion the : L = = + cos p I 3 L Mg = = 1 + + 2 sin cos cos p I I 3 = 1 + 2 sin cos I p ( ) 2 2 cos p p p 1 1 I = = + + 2 ' cos E E I Mgl 2 2 2 2 sin I 3 1 10/25/2019 PHY 711 Fall 2019 -- Lecture 24 26

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