On to a new topic 322 Non-inertial frames On to a new topic 323 For fans of xkcd and 007 m:r0 F “ Starting from non-rotating non-inertial frames 324 Standing on the ground L ! V/A (in inertial frame): Amtrak m:r0 F “ The velocity v0 of the v is velocity as pendulum seen in the train as seen from the ground: mv9 mv9 0 mA v0 v V “ ´ “ ` mv9 F mA v9 0 v9 V9 “ ´ “ ` Note minus v9 v9 0 A sign Ñ “ ´ F mA inertial “´ Starting from non-rotating non-inertial frames 325 Starting from non-rotating non-inertial frames 326 Starting from non-rotating non-inertial frames 327 A fun side problem 328 Helium A balloon Air gets accelerated too... pressure gradient pushes balloon to the right Let’s work out Example 9.1 329 First let’s follow the book... and then what happens in the Lagrangian formulation? Let’s work out Example 9.1 in an alternate way 330 1 x l sin ✓ at2 “´ ` 2 y l 1 cos ✓ “ p ´ q x9 l cos ✓✓9 at “´ ` x9 2 l2 cos2 ✓✓92 a2t2 2atl cos ✓✓9 “ ` ´ y9 l sin ✓✓9 L ! V/A “ 2 Amtrak y92 l2 sin ✓✓92 m“ x9 2 y92 mgy L “ 2 p ` q´ m l2✓92 a2t2 2atl cos ✓✓9 mgl 1 cos ✓ L “ 2 p ` ´ q´ p ´ q BL ml2✓9 matl cos ✓ ✓9 “ ´ B BL matl sin ✓✓9 mgl sin ✓ ✓ “ ´ d B ml2✓9 matl cos ✓ matl sin ✓✓9 mgl sin ✓ dt ´ “ ´ ´ ¯ Let’s work out Example 9.1 in an alternate way 331 L V/A ! Amtrak d ml2✓9 matl cos ✓ matl sin ✓✓9 mgl sin ✓ dt ´ “ ´ ml2✓: ´matl sin ✓✓9 mal¯cos ✓ matl sin ✓✓9 mgl sin ✓ ` ´ “ ´ ml2✓: mal cos ✓ mgl sin ✓ ´ “´ l2✓: al cos ✓ gl sin ✓ “ ´ ✓: a l cos ✓ g l sin ✓ “p { q ´p { q What’s the equilibrium? 332 L V/A ! Amtrak ✓: 0 a l cos ✓ g l sin ✓ “ “p { q ´p { q Ñ g sin ✓ a cos ✓ “ tan ✓ a g “ { Let’s move on to something more tricky - the tides 333 Tides (incorrect) 334 This picture is wrong! There are ~2 high tides per day. Tides are not due to the moon “pulling” at the water Tides (correct) 335 Tides are due to the differential force between the moon/sun and the earth’s center of mass vs the moon/sun and the water, which is not at the center of mass. But CoM and the water are both accelerating! In other words, tides are due to the difference in inertial force vs position In other words 336 Gravitational attraction towards moon Moon Moon Difference between force at CoM Let’s calculate this using a non-inertial frame 337 standard attraction to earth mv9 F mA “ ´ m:r F mA mgˆr “ ´ Origin O GM m ´ m dˆ d2 standard grav. attraction to moon Centripetal dˆ0 A GMm 2 acceleration of frame “´ d0 dˆ dˆ0 m:r mgˆr GMmm 2 GMmm 2 “ ´ d ` d0 Let’s calculate this using a non-inertial frame 338 dˆ dˆ0 m:r mgˆr GMmm 2 GMmm 2 “ ´ d ` d0 dˆ0 dˆ Ftid GMmm 2 2 “ ˜ d0 ´ d ¸ What is the potential energy associated with the tidal force? 339 dˆ0 dˆ Ftid GMmm 2 2 “ ˜ d0 ´ d ¸ F U tid “´r tid x x 1 U GM m tid “´ m d2 ` d Is this clear? ˆ 0 ˙ Ocean surface is equipotential h is height difference between high and low tide U P U P U Q U Q tidp q` gravp q“ tidp q` gravp q U P U Q U Q U P mgh tidp q´ tidp q“ gravp q´ grav “ What is the potential energy associated with the tidal force? 340 U P U Q mgh p q´ p q“´ x 1 U GM m tid “´ m d2 ` d ˆ 0 ˙ At point Q, d d2 r2 d2 R2 “ 0 ` „ 0 ` e b b x(Q) = 0 x 1 Utid Q GMmm 2 2 2 p q“´ ˜d0 ` d Re ¸ That clear? 0 ` 1 a Utid Q GMmm 2 2 p q“´ ˜ d Re ¸ 0 ` GMmm a 1 Utid Q 2 p q“´ d0 ˜ 1 Re d0 ¸ `p { q GMmm a 2 1 2 Utid Q 1 Re d0 ´ { p q“´ d0 `p { q GM `m R2 ˘ U Q m 1 e tidp q“´ d ´ 2d2 0 ˆ 0 ˙ What is the potential energy associated with the tidal force? 341 U P U Q mgh p q´ p q“´ x 1 U GM m tid “´ m d2 ` d ˆ 0 ˙ At point P, d d R ,x R “ 0 ´ e “´ e R 1 U P GM m ´ e tidp q“´ m d2 ` d R ˆ 0 0 ´ e ˙ GM m R 1 U P m ´ e tidp q“´ d d ` 1 R d 0 ˆ 0 ´ e{ 0 ˙ GM m R U P m ´ e 1 R d R d 2 tidp q“´ d d ` ` e{ 0 `p e{ 0q q 0 ˆ 0 ˙ GMmm 2 Utid P 1 Re d0 p q“´ d0 `p { q q ` ˘ Putting it together 342 U P U Q mgh p q´ p q“´ U P U Q mgh p q´ p q“´ 2 GMmm Re U P 1 2 p q“´ d0 p ` d0 q 2 GMmm Re U Q 1 2 p q“´ d0 p ´ 2d0 q GM m R2 R2 mgh m e e “ d d2 ` 2d2 0 ˆ 0 0 ˙ 2 GMmRe gh 3 1 1 2 “ d0 p ` { q 2 3GMmRe h 3 “ 2gd0 4 2 3 Mm Re g GMe Re h 3 “ { Ñ “ 2 Me d0 Plugging in the numbers 343 4 3 Mm Re h 3 “ 2 Me d0 h = 54 cm (moon) h = 25 cm (sun) Of course, this is a simplification, including land, seasonal and depth effects (not to mention wind) On to frames with rotation 344 On to frames with rotation 345 Don’t forget the right- ~! hand rule ~! !~u “ ~! ~u Can imagine situations where any of ! these are constant, or functions of time On to frames with rotation 346 r a cos ✓,asin ✓,h “p q v r9 a✓9 sin ✓,a✓9 cos ✓, 0 “ “p´ q ~! ✓9zˆ “ v ~! r “ ° v !r if ~! r 0 | |“ ¨ “ We can use this to say more 347 dr v ~! r “ “ dt de° ~! e Ñ dt “ ° For any vector e, including unit vectors Addition of angular velocities 348 vij = velocity of frame i relative to frame j v v v 31 “ 32 ` 21 ~! r ~! r ~! r 31 “ 32 ` 21 ~! ° r ~! ° ~! °r 31 “p 32 ` 21q ° ~! ~! ~! ° Ñ 31 “ 32 ` 21 ωij = angular velocity of frame i relative to frame j Vectors add just as translational vectors Time derivatives in a rotating frame 349 Consider an inertial frame of references defined by S0 and a second frame (of interest) S with shared origin, but rotating with respect to S0 with angular velocity " For example (Taylor) O has origin at enter of earth, S0 is axes fixed to distant stars, S is non-inertial (earth rotates) Earth’s angular speed of rotation: Ω = 7.292 ×10−5 s−1 € Time derivatives in a rotating frame 350 dQ rate of change of Q relative to inertial frame S dt “ 0 ˆ ˙S0 dQ rate of change of Q relative to rotating frame S dt S “ ˆ ˙ i 3 “ Q Q1e1 Q2e2 Q3e3 Qiei “ ` ` “ i 1 ÿ“ dQ dQ i e dt dt i S “ i ˆ ˙ ÿ valid in both frames, though ei are constant Since ei are in S but not in S0 constant in S Time derivatives in a rotating frame 351 dQ rate of change of Q relative to inertial frame S dt “ 0 ˆ ˙S0 dQ rate of change of Q relative to rotating frame S dt “ ˆ ˙S dQ dQi dei de ei Qi Recall: ⌦~ e dt “ dt ` dt S0 i i S0 dt “ ˆ ˙ ÿ ÿ ˆ ˙ dQ dQ ° i e Q ⌦~ e dt “ dt i ` ip iq S0 i i ˆ ˙ ÿ ÿ ° dQ dQ i e ⌦~ Q e dt “ dt i ` i i S0 i i ˆ ˙ ÿ ° ÿ dQ dQ i e ⌦~ Q dt “ dt i ` S0 i ˆ ˙ ÿ ° dQ dQ ⌦~ Q dt “ dt ` S0 S ˆ ˙ ˆ ˙ ° A quick exercise 352 Let’s do Problem 9.7 together Now we can move back to Newton’s Laws 353 dQ dQ ⌦~ Q dt “ dt ` S0 S ˆ ˙ ˆ ˙ ° d2r m F dt2 “ ˆ ˙S0 dr dr ⌦~ r dt “ dt ` S0 S ˆ ˙ ˆ ˙ ° d2r d dr dt2 “ dt dt ˆ ˙S0 ˆ ˙S0 ˆ ˙S0 d2r d dr ⌦~ r dt2 “ dt dt ` S0 S0 S ˆ ˙ ˆ ˙ „ˆ ˙ ° ⇢ d2r d dr dr ⌦~ r ⌦~ ⌦~ r dt2 “ dt dt ` ` dt ` S0 S S S ˆ ˙ ˆ ˙ „ˆ ˙ ° ⇢ ° „ˆ ˙ ° ⇢ Need some simplification 354 d2r d dr dr ⌦~ r ⌦~ ⌦~ r dt2 “ dt dt ` ` dt ` S0 S S S ˆ ˙ ˆ ˙ „ˆ ˙ ° ⇢ ° „ˆ ˙ ° ⇢ Evaluate in frame where Ω const, so d/dt(Ω) = 0 Dots are with respect to rotating frame S d2r :r ⌦~ r9 ⌦~ r9 ⌦~ r dt2 “ ` ` ` S0 ˆ ˙ ° ° ” ° ı d2r :r 2⌦~ r9 ⌦~ ⌦~ r dt2 “ ` ` p q S0 ˆ ˙ ° ° ° Putting it together 355 d2r m F m:r 2m⌦~ r9 m⌦~ ⌦~ r dt2 “ “ ` ` p q S0 ˆ ˙ ° ° ° Changing cross product order cancels minus signs... m:r F 2mr9 ⌦~ m ⌦~ r ⌦~ “ ` ` p q ° ° ° Coriolis Force Centrifugal Force Coriolis force and centrifugal force 356 m:r F 2mr9 ⌦~ m ⌦~ r ⌦~ “ ` ` p q ° ° ° Coriolis Force = 0 Centrifugal Force if v = 0 m:r F 2mr9 ⌦~ m ⌦~ r ⌦~ “ ` ` p q F°cor mv⌦ ° °v ~ v as seen V ~ rΩ ~ „ 2 speed of Fcf mr⌦ on rotating „ earth rotation on Fcor v v earth ~1000 Fcf „ r⌦ „ V mi/h A quick exercise 357 Let’s do problem 9.8 together.