On to a new topic. Non-inertial frames

Transcription

1 On to a new topic 322 Non-inertial frames

2 On to a new topic 323 For fans of xkcd and 007

3 Starting from non-rotating non-inertial frames 324 Standing on the ground (in inertial frame): L φ Amtrak V/A m:r 0 F The velocity v0 of the pendulum as seen from the ground: v 0 v ` V 9v 0 9v ` 9V 9v 9v 0 A Note minus sign v is velocity as seen in the train m 9v m 9v 0 ma m 9v F ma Ñ F inertial ma

4 Starting from non-rotating non-inertial frames 325

5 Starting from non-rotating non-inertial frames 326

6 Starting from non-rotating non-inertial frames 327

7 A fun side problem 328 Helium balloon A Air gets accelerated too... pressure gradient pushes balloon to the right

8 Let s work out Example First let s follow the book... and then what happens in the Lagrangian formulation?

9 Let s work out Example 9.1 in an alternate way L φ Amtrak V/A x l sin ` 1 2 at2 y lp1 cos q 9x l cos 9 ` at 9x 2 l 2 cos ` a 2 t 2 2atl cos 9 9y l sin 9 9y 2 l 2 sin L m 2 p 9x2 ` 9y 2 q mgy L m 2 pl2 9 2 ` a 2 t 2 2atl cos 9 q mglp1 cos q d dt BL B 9 ml2 9 matl cos BL B matl sin 9 mgl sin ml 2 9 matl cos matl sin 9 mgl sin

10 Let s work out Example 9.1 in an alternate way 331 L φ Amtrak V/A d ml 2 9 matl cos matl sin dt 9 mgl sin ml 2 : ` matl sin 9 mal cos matl sin 9 mgl sin ml 2 : mal cos mgl sin l 2 : al cos gl sin : pa{lqcos pg{lqsin

11 What s the equilibrium? 332 L φ Amtrak V/A : 0 pa{lq cos pg{lq sin Ñ g sin a cos tan a{g

12 Let s move on to something more tricky - the tides 333

13 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

14 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

15 In other words Gravitational attraction towards moon 336 Moon Moon Difference between force at CoM

16 Let s calculate this using a non-inertial frame m 9v F ma m:r F ma standard attraction to earth Origin O mgˆr GM m m d 2 ˆd 337 A GM m ˆd0 d 2 0 standard grav. attraction to moon Centripetal acceleration of frame m:r mgˆr GM m m ˆd d 2 ` GM mm ˆd 0 d 2 0

17 Let s calculate this using a non-inertial frame 338 m:r mgˆr GM m m ˆd d 2 ` GM mm ˆd 0 d 2 0 F tid GM m m ˆd0 d 2 0 ˆd d 2

18 What is the potential energy associated with the tidal force? F tid GM m m ˆd0 d 2 0 ˆd d x Is this clear? Ocean surface is equipotential F tid ru tid ˆ x U tid GM m m d 2 0 ` 1 d h is height difference between high and low tide U tid pp q`u grav pp q U tid pqq`u grav pqq U tid pp q U tid pqq U grav pqq U grav P mgh

19 What is the potential energy associated with the tidal force? x(q) = 0 That clear? UpP q UpQq mgh ˆ x U tid GM m m ` 1 d d 2 b 0 b At point Q, d d 2 0 ` r2 d 2 0 ` R2 e x 1 U tid pqq GM m m d 2 ` a 0 d 2 0 ` Re 2 U tid pqq GM m m U tid pqq GM mm d 0 U tid pqq GM mm d 0 U tid pqq GM mm d 0 1 a d 2 0 ` R 2 e 1 a 1 `pre {d 0 q 2 `1 `pre {d 0 q 2 1{2 ˆ 1 R2 e 2d

20 What is the potential energy associated with the tidal force? UpP q UpQq mgh ˆ x U tid GM m m ` 1 d d At point P, d d 0 R e,x R e ˆ Re 1 U tid pp q GM m m d 2 ` 0 d 0 R e U tid pp q GM mm ˆ Re 1 ` d 0 d 0 1 R e {d 0 ˆ Re ` 1 ` R e {d 0 `pr e {d 0 q 2 q U tid pp q GM mm d 0 d 0 U tid pp q GM mm d 0 `1 `pre {d 0 q 2 q

21 Putting it together UpP q UpQq mgh UpP q UpQq mgh UpP q GM mm d 0 UpQq GM mm d 0 mgh GM mm d 0 ˆR2 e p1 ` R2 e d 2 0 q p1 R2 e 2d 2 q 0 d 2 0 ` R2 e 2d gh GM mr 2 e d 3 0 p1 ` 1{2q h 3GM mr 2 e 2gd 3 0 g GM e {R 2 e Ñ h 3 2 M m M e R 4 e d 3 0

22 Plugging in the numbers 343 h 3 2 M m M e R 4 e d 3 0 h = 54 cm (moon) h = 25 cm (sun) Of course, this is a simplification, including land, seasonal and depth effects (not to mention wind)

23 On to frames with rotation 344

24 On to frames with rotation 345 Don t forget the righthand rule ~! ~!!~u ~! ~u! Can imagine situations where any of these are constant, or functions of time

25 On to frames with rotation 346 r pacos,asin,hq v 9r p a 9 sin,a 9 cos, 0q ~! ẑ 9 v ~! r v!r if ~! r 0

26 We can use this to say more 347 v ~! r dr dt Ñ de dt ~! e For any vector e, including unit vectors

27 Addition of angular velocities 348 vij = velocity of frame i relative to frame j v 31 v 32 ` v 21 ~! 31 r ~! 32 r ` ~! 21 r ~! 31 r p~! 32 ` ~! 21 q r Ñ ~! 31 ~! 32 ` ~! 21 ωij = angular velocity of frame i relative to frame j Vectors add just as translational vectors

28 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 Ω 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 = s 1

29 Time derivatives in a rotating frame 350 ˆdQ rate of change of Q relative to inertial frame S 0 dt S 0 ˆdQ rate of change of Q relative to rotating frame S dt S Q Q 1 e 1 ` Q 2 e 2 ` Q 3 e 3 valid in both frames, though ei are constant in S but not in S0 i 3 ÿ i 1 Q i e i ˆdQ dt S ÿ i dq i dt e i Since ei are constant in S

30 Time derivatives in a rotating frame ˆdQ 351 dq dt S 0 ÿ i dq ÿ dt S 0 i ˆdQ ÿ dt S 0 i ˆdQ ÿ dt S 0 i rate of change of Q relative to inertial frame S 0 dt S 0 ˆdQ rate of change of Q relative to rotating frame S dt S dq i dt e i ` ÿ dq i dt e i ` ÿ dq i dt e i ` ~ ÿ i i i Q i p ~ e i q dq i dt e i ` ~ Q Q i ˆdei dt Q i e i ˆdQ dt S 0 S 0 Recall: ˆdQ dt de dt ~ e S ` ~ Q

31 A quick exercise 352 Let s do Problem 9.7 together

32 Now we can move back to Newton s Laws ˆdQ ˆdQ ` dt dt ~ Q S 0 S 353 ˆd2 r dt 2 S 0 ˆd2 r m dt 2 F S 0 ˆdr ˆdr ` dt S 0 dt ~ r ˆd2 S ˆ r d ˆdr dt 2 S 0 dt S 0 dt S ˆd2 ˆ 0 r d ˆdr dt 2 ` S 0 dt S 0 dt ~ r S ˆ d ˆdr ` dt dt ~ ˆdr r ` ~ dt S S S ` ~ r

33 Need some simplification 354 ˆd2 r dt 2 S 0 ˆ d dt S ˆdr dt S ` ~ r Evaluate in frame where Ω const, so d/dt(ω) = 0 Dots are with respect to rotating frame S ˆd2 r :r ` S0 ~ 9r ` ~ dt 2 ˆd2 r dt 2 S 0 ` ˆdr ~ dt S ` ~ r 9r ` ~ ı r :r ` 2 ~ 9r ` ~ p ~ rq

34 Putting it together 355 m ˆd2 r dt 2 S 0 F m:r ` 2m ~ 9r ` m ~ p ~ rq Changing cross product order cancels minus signs... m:r F ` 2m9r ~ ` mp ~ rq ~ Coriolis Force Centrifugal Force

35 Coriolis force and centrifugal force 356 m:r F ` 2m9r ~ ` mp ~ rq ~ Coriolis Force = 0 if v = 0 Centrifugal Force m:r F ` 2m9r ~ ` mp ~ rq ~ V ~ rω ~ speed of rotation on earth ~1000 mi/h F cor mv F cf mr 2 F cor F cf v r v V v ~ v as seen on rotating earth

36 A quick exercise 357 Let s do problem 9.8 together. Tricky! Let s define a coordinate system. x = easterly, y = northerly, z = up (ie radially out). What is Ω at equator? y direction What is Ω near (but not at) the north pole? Mostly z direction, but a little y direction What is r? Always z direction

37 Centrifugal force 358 Centrifugal force modifies gravity (tangential component towards equator and reduces overall magnitude) ~ ẑ r mg On earth, r points from center (origin) to where we stand

38 Centrifugal force 359 r rẑ Ñ p ~ rq ~ r 2 pẑ ẑq ẑ 0 At poles, no effect at all ~ ẑ mg r

39 Centrifugal force 360 r rˆx Ñ p ~ rq ~ r 2 pẑ ˆxq ẑ r 2 ŷ ẑ r 2ˆx 2 r ~ ẑ r mg At equator, centrifugal force is in exact opposition to gravity, with magnitude RΩ 2 = 0.3% of gravity

40 Centrifugal force r rˆx Ñ p ~ rq ~ r 2 pẑ ˆxq ẑ r 2 ŷ ẑ r 2ˆx 2 r 361 θ At equator, centrifugal force is in exact opposition to gravity, with magnitude RΩ 2 = 0.3% of gravity But off the equator, the direction changes, too...

41 What is the centrifugal force? Some might refer to it as fictitious but that is a bit unfair to it! It s a result of the inertia of the system as it is continually accelerated in a rotating system. It draws the system away from the center of rotation (ie away from Earth, for example) 362 θ

42 More with centrifugal force 363 More familiar formulas: v ~ r v r F cf mv 2 {r r rˆx Ñ On earth, r points from center (origin) to where we stand ~ ẑ p ~ rq ~ r 2 pẑ ˆxq ẑ r 2 ŷ ẑ r 2ˆx 2 r r rŷ Ñ p ~ rq ~ r 2 pẑ ŷq ẑ r 2ˆx ŷ r 2 ŷ 2 r r rẑ Ñ p ~ rq ~ r 2 pẑ ẑq ẑ 0

43 Angle between apparent and true gravity 364 θ F α=angle between combined force and pure gravity F grav p g sin, 0, g cos q F cen p! 2 R sin, 0, 0q F p g sin `! 2 R sin, 0, g cos q F F grav F grav F cos F F grav g 2 sin 2 gr! 2 sin 2 ` g 2 cos 2 F F grav g 2 gr! 2 sin 2 F grav g b g 2 sin 2 `! 4 R 2 sin 2 2gR! 2 sin 2 ` g 2 cos 2 b F g 2 `! 4 R 2 sin 2 2gR! 2 sin 2

44 Angle between apparent and true gravity 365 θ α=angle between combined force and pure gravity R! g F grav p g sin, 0, g cos q F cen p! 2 R sin, 0, 0q F p g sin `! 2 R sin, 0, g cos q F F grav F grav F cos F F grav g 2 sin 2 gr! 2 sin 2 ` g 2 cos 2 F F grav g 2 gr! 2 sin 2 F grav g b F g 2 sin 2 `! 4 R 2 sin 2 2gR! 2 sin 2 ` g 2 cos 2 b F g 2 `! 4 R 2 sin 2 2gR! 2 sin 2 cos cos g2 gr! 2 sin 2 g 2 g 2 gr! 2 sin 2 g a g 2 `! 4 R 2 sin 2 2gR! 2 sin 2 ˆ 1 `!4 R 2 sin 2 cos g2 gr! 2 sin 2 cos g 2 ˆ 1 R!2 sin 2 g cos g 2 2R!2 sin 2 g ˆ 1 2R!2 sin 2 g ˆ 1 ` R!2 sin 2 g ˆ 1 R2! 4 sin 4 g 2 1{2 1{2

45 Angle between apparent and true gravity 366 θ α=angle between combined force and pure gravity R! g 45 o Ñ cos p1 R2! 4 4g 2 q cos x 1 x2 2 Ñ R!2 2g 0.1o F grav p g sin, 0, g cos q F cen p! 2 R sin, 0, 0q F p g sin `! 2 R sin, 0, g cos q F F grav F grav F cos F F grav g 2 sin 2 gr! 2 sin 2 ` g 2 cos 2 F F grav g 2 gr! 2 sin 2 F grav g b F g 2 sin 2 `! 4 R 2 sin 2 2gR! 2 sin 2 ` g 2 cos 2 b F g 2 `! 4 R 2 sin 2 2gR! 2 sin 2 cos cos g2 gr! 2 sin 2 g 2 g 2 gr! 2 sin 2 g a g 2 `! 4 R 2 sin 2 2gR! 2 sin 2 ˆ 1 `!4 R 2 sin 2 cos g2 gr! 2 sin 2 cos g 2 ˆ 1 R!2 sin 2 g cos g 2 2R!2 sin 2 g ˆ 1 2R!2 sin 2 g ˆ 1 ` R!2 sin 2 g ˆ 1 R2! 4 sin 4 g 2 1{2 1{2

46 Coriolis Force 367 This one is a bit more difficult to picture, but is again due to the fact that we are in a rotating frame, and our intuition about inertia only holds in a non-rotating inertial frame Gaspard-Gustave Coriolis Note: if v=0, no Coriolis force! And if v is parallel to Ω, also no Coriolis force Fcor 2m9r 2mv

47 Coriolis Force 368 F cor 2m9r 2mv 7.3x10 5 s 1 F max cor {F grav p1.5x10 5 qpvq v 67 km/s for Coriolis force to equal grav force!

48 A simple but useful problem 369 Let s look at example 9.2 together

49 Coriolis Force in a picture As it travels north, orange projectile travels further eastward than the earth beneath it 370 As it travels north, yellow projective travels less eastward than the ground beneath it Curve to the right in northern hemisphere, to left in southern hemisphere

50 Coriolis Force in another picture 371 What about red projectile? It is traveling faster now than the ground beneath it and thus it will want to fly out/off the earth (inertial Coriolis force is up)

51 On coordinate systems In these discussion, we sometimes need to be careful about our choice of latitude vs colatitude. Subtle but important. When the book uses θ, it s typically using colatitude, ie the angle from the z axis, which runs from 0 (north pole) to 180 (south pole). Geographers typically use the latitutde, which runs from +90 (north pole) to -90 (south pole) 372 θcol θlat

52 Particle in free-fall Include centrifugal force in g ª t 0 :r g 2~! 9r ª t :rdt g 2~! ı 9r dt 0 9r 9rp0q gt 2~! r ` 2~! rp0q 9r gt 2~! r ` 2~! rp0q Coriolis term Let s start particle from rest (ie dropped from rest), so v(0) = 0 Let s plug the last line into the first line

53 Particle in free-fall 374 :r g 2~! 9r 9r gt 2~! r ` 2~! rp0q :r g 2~! gt 2~! r ` 2~! rp0q :r g 2~! gt ` 2~! prp0q rq small compared to gt :r g 2t~! g

54 This we can solve 375 v(0) = 0 ª t 0 ª t 0 :r g 2t~! g ª t :rdt g 2t~! ı g dt 0 9rdt ` 9rp0q gt t 2 ~! g ª t 9rdt gt t 2 ~! ı g dt 0 rptq rp0q 1 2 gt2 1 3 t3 ~! g rptq 1 2 gt2 1 3 t3 ~! g ` rp0q

55 How to interpret rptq 1 2 gt2 1 3 t3 ~! g ` rp0q 376 Define coordinates with x = easterly, y = northerly, z = up (ie radially out) Assume we re on the r = (x,y,z), g = (0,0,-g), ω = (0,ω,0) equator If we drop a particle down a well over depth h r(0) = (0,0,R+h) where R is radius of earth 1 3 t3 ~! g 1 3!gt3ˆx Deflection in easterly direction!

56 Deflection in the well rptq 1 2 gt2 1 3 t3 ~! g ` rp0q 377 Deflection still quite small, so t for descent down the well is given by standard formula of t a 2d{g a 2h{g 1 3 t3 ~! g 1 3!gt3ˆx Deflection = 1 3!gp2h{gq3{2ˆx a 8{p9gq!h 3{2ˆx h = 100m (well) 2.2 cm h = 2 km (skydiving) 2 m

57 Another quick exercise 378 Let s do problem 9.9 together

58 Foucault pendulum 379 At Chicago s Museum of Science and Industry

59 Foucault pendulum 380 Pendulum with length L m:r mg 2~! 9r ` T x 2 ` y 2 ` z 2 L 2 T x T px{lq T y T py{lq T z T pz{lq T z mg β x(east) z(up) y(north) T m mg

60 Foucault pendulum 381 Pendulum with length L z(up) ~! p0,! sin,! cos q :x gx{l ` 2 9y! cos 2 9z! sin :y gy{l 2! 9x cos For small oscillations, expect vz small :x gx{l ` 2 9y! cos :y gy{l 2 9x! cos β x(east) y(north) T m mg! 2 0 g{l,! cos! z :x 2! z 9y `! 2 0x 0 :y ` 2! z 9x `! 2 0y 0

61 Solution using earlier trick with complex numbers Sum :x 2! z 9y `! 2 0x 0 :y ` 2! z 9x `! 2 0y 0 x ` iy i:y ` 2i! z 9x ` i! 2 0y 0 :x ` i:y ` 2! z pi 9x 9yq`! 2 0px ` iyq 0 : ` 2i! z 9 `! ptq e i t 2 ` 2i! z p i q`! ` 2! z `! ! z! ˆ b 2! z 4! z 2 ` 4! 0 2 {2! z b! z 2 `! 0 2 β x(east) z(up)! 0! z Ñ! z! 0 T m y(north) mg 382

62 Solution for pendulum ptq e i t! z! cos! z! 0 e i! zt `C 1 e i! 0t ` C 2 e i! 0t β z(up) y(north) 383 What does this term do? x(east) mg Let s check (as in the book) what happens if at t=0, v=0, x=a, y=0 (so that C1 = C2 = A/2 - let s see why). Take a close look at Fig 9.17 T m

63 Solution for pendulum 384! z! cos! 0! z z(up) ptq xptq`iyptq Ae i! zt cos p! 0 tq At first, natural frequency much larger than the extra term in front, but eventually, that rotates motion in x direction to y direction and back β x(east) y(north) T m mg

64 Rotation of the plane of the pendulum 385 At first, natural frequency z(up) much larger than the extra term in front, but β eventually, that rotates y(north) motion in x direction to y direction and back. T Chicago co-latitude = x(east) 48.2 degrees, so ωz = m ωcos(48.2 deg)=2/3 (360 mg degrees/day) = 240 degrees / 24 hours 90 degree rotation (complete shift to y direction) in just 9 hours!

65 One last word on these subjects 386 Recall from earlier in the course: F r = m( r r 2 ) F = m(2ṙ + r ) Choice of frame determines whether accelerations or forces are complicated (alternatively, which one is simple)

66 Homework (due as usual in 1 week) , 9.14, 9.25,9.26, 9.28, 9.29