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. 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 Centrifugal force 358 Centrifugal force modifies gravity (tangential component towards equator and reduces overall magnitude) ⌦~ ⌦zˆ “ mg On earth, r points from r center (origin) to where we stand Centrifugal force 359
r rzˆ At poles, no “ Ñ ⌦~ r ⌦~ r⌦2 zˆ zˆ zˆ 0 effect at all p q “ p q “ ° ° ° °
⌦~ ⌦zˆ “ mg
r Centrifugal force 360
r rxˆ “ Ñ ⌦~ r ⌦~ r⌦2 zˆ xˆ zˆ r⌦2yˆ zˆ r⌦2xˆ ⌦2r p q “ p q “ “ “ ° ° ° ° ° At equator, centrifugal force is in exact ⌦~ ⌦zˆ “ opposition to mg gravity, with magnitude RΩ2 = r 0.3% of gravity Centrifugal force 361 r rxˆ “ Ñ ⌦~ r ⌦~ r⌦2 zˆ xˆ zˆ r⌦2yˆ zˆ r⌦2xˆ ⌦2r p q “ p q “ “ “ ° ° ° ° ° At equator, centrifugal force is in exact opposition to gravity, with magnitude RΩ2 = 0.3% of gravity θ But off the equator, the direction changes, too ... What is the centrifugal force? 362 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)
θ More with centrifugal force 363
More familiar formulas: v ⌦~ r On earth, r points from “ center (origin) to v ⌦°r “ 2 where we stand Fcf mv r “ { ⌦~ ⌦zˆ r rxˆ “ “ Ñ ⌦~ r ⌦~ r⌦2 zˆ xˆ zˆ r⌦2yˆ zˆ r⌦2xˆ ⌦2r p q “ p q “ “ “ r ryˆ ° ° ° “° Ñ ° ⌦~ r ⌦~ r⌦2 zˆ yˆ zˆ r⌦2xˆ yˆ r⌦2yˆ ⌦2r p q “ p q “´ “ “ ° ° ° ° ° r rzˆ “ Ñ ⌦~ r ⌦~ r⌦2 zˆ zˆ zˆ 0 p q “ p q “ ° ° ° ° Angle between “apparent” and “true” gravity 364 α=Angle between combined force and pure gravity θ F g sin ✓, 0, g cos ✓ grav “p´ ´ q F !2R sin ✓, 0, 0 cen “p q F g sin ✓ !2R sin ✓, 0, g cos ✓ “p´ ` ´ q F F F F cos ↵ ¨ grav “| grav|| | F F g2 sin2 ✓ gR!2 sin2 ✓ g2 cos2 ✓ ¨ grav “ ´ ` F F g2 gR!2 sin2 ✓ ¨ grav “ ´ F g | grav|“ F g2 sin2 ✓ !4R2 sin2 ✓ 2gR!2 sin2 ✓ g2 cos2 ✓ | |“ ` ´ ` b F g2 !4R2 sin2 ✓ 2gR!2 sin2 ✓ | |“ ` ´ b Angle between “apparent” and “true” gravity 365
F g sin ✓, 0, g cos ✓ grav “p´ ´ q F !2R sin ✓, 0, 0 cen “p q F g sin ✓ !2R sin ✓, 0, g cos ✓ θ “p´ ` ´ q F F F F cos ↵ ¨ grav “| grav|| | F F g2 sin2 ✓ gR!2 sin2 ✓ g2 cos2 ✓ ¨ grav “ ´ ` F F g2 gR!2 sin2 ✓ ¨ grav “ ´ F g α=Angle | grav|“ F g2 sin2 ✓ !4R2 sin2 ✓ 2gR!2 sin2 ✓ g2 cos2 ✓ | |“ ` ´ ` b between F g2 !4R2 sin2 ✓ 2gR!2 sin2 ✓ | |“ ` ´ b g2 gR!2 sin2 ✓ combined force cos ↵ ´ “ g g2 !4R2 sin2 ✓ 2gR!2 sin2 ✓ ` ´ and pure gravity 1 2 g2 gR!2 sina 2 ✓ !4R2 sin2 ✓ 2R!2 sin2 ✓ ´ { cos ↵ ´ 1 “ g2 ` g2 ´ g 2 ˆ ˙ 1 2 R! 0.003g g2 gR!2 sin2 ✓ 2R!2 sin2 ✓ ´ { cos ↵ ´ 1 „ „ g2 ´ g ˆ ˙ R!2 sin2 ✓ R!2 sin2 ✓ cos ↵ 1 1 „ ´ g ` g ˆ ˙ˆ ˙ R2!4 sin4 ✓ cos ↵ 1 „ ´ g2 ˆ ˙ Angle between “apparent” and “true” gravity 366
F g sin ✓, 0, g cos ✓ grav “p´ ´ q F !2R sin ✓, 0, 0 cen “p q F g sin ✓ !2R sin ✓, 0, g cos ✓ θ “p´ ` ´ q F F F F cos ↵ ¨ grav “| grav|| | F F g2 sin2 ✓ gR!2 sin2 ✓ g2 cos2 ✓ ¨ grav “ ´ ` F F g2 gR!2 sin2 ✓ ¨ grav “ ´ F g α=Angle | grav|“ F g2 sin2 ✓ !4R2 sin2 ✓ 2gR!2 sin2 ✓ g2 cos2 ✓ | |“ ` ´ ` b between F g2 !4R2 sin2 ✓ 2gR!2 sin2 ✓ | |“ ` ´ b g2 gR!2 sin2 ✓ combined force cos ↵ ´ “ g g2 !4R2 sin2 ✓ 2gR!2 sin2 ✓ ` ´ and pure gravity 1 2 g2 gR!2 sina 2 ✓ !4R2 sin2 ✓ 2R!2 sin2 ✓ ´ { cos ↵ ´ 1 “ g2 ` g2 ´ g 2 ˆ ˙ 1 2 R! 0.003g g2 gR!2 sin2 ✓ 2R!2 sin2 ✓ ´ { cos ↵ ´ 1 „ „ g2 ´ g ˆ ˙ 2 4 2 2 2 2 o R ! R! sin ✓ R! sin ✓ ✓ 45 cos ↵ 1 cos ↵ 1 1 “ Ñ “p ´ 4g2 q „ ´ g ` g ˆ ˙ˆ ˙ x2 R!2 R2!4 sin4 ✓ o cos ↵ 1 cos x 1 ↵ 0.1 „ ´ g2 „ ´ 2 Ñ “ 2g „ ˆ ˙ Coriolis Force 367
This one is a bit more Gaspard-Gustave difficult to picture, but is Coriolis 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
Note: if v=0, no Coriolis force! And if v is parallel to Ω, also no Coriolis force Fcor 2mr9 ⌦ 2mv ⌦ “ “ ° ° Coriolis Force 368
Fcor 2mr9 ⌦ 2mv ⌦ “ “ 5 1 ⌦ 7°.3x10´ s´ ° “ max 5 F F 1.5x10´ v cor { grav “p qp q v 67 km/s for Coriolis force to equal grav force! “ A simple but useful problem 369
Let’s look at example 9.2 together Coriolis Force in a picture 370 As it travels north, orange projectile travels further eastward than the earth beneath it
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 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) On coordinate systems 372 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)
θcol θlat Particle in free-fall 373 Include centrifugal force in g Coriolis term :r g 2~! r9 “ ´ t t ° :rdt g 2~! r9 dt 0 “ 0 ´ ª ª ” ° ı r9 r9 0 gt 2~! r 2~! r 0 ´ p q“ ´ ` p q r9 gt 2~! °r 2~! °r 0 “ ´ ` p q Let’s start particle from° rest (ie dropped° from rest), so v(0) = 0 Let’s plug the last line into the first line ... Particle in free-fall 374
:r g 2~! r9 “ ´ r9 gt 2~! r °2~! r 0 “ ´ ` p q :r g 2~! gt °2~! r °2~! r 0 “ ´ ´ ` p q ° ´ ° ° ¯ :r g 2~! gt 2~! r 0 r “ ´ ` p p q´ q ° ´ ° ¯ small compared to gt :r g 2t~! g “ ´ ° This we can solve 375
:r g 2t~! g “ ´ t t ° :rdt g 2t~! g dt 0 “ 0 ´ ª ª ” ° ı v(0) = 0 r9dt r9 0 gt t2~! g ` p q“ ´ t t ° r9dt gt t2~! g dt “ ´ ª0 ª0 ”1 1° ı r t r 0 gt2 t3~! g p q´ p q“2 ´ 3 1 1 ° r t gt2 t3~! g r 0 p q“2 ´ 3 ` p q ° How to interpret 376 1 1 r t gt2 t3~! g r 0 p q“2 ´ 3 ` p q ° 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), = (0, ,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 1 Deflection in easterly t3~! g !gt3xˆ ´3 “ 3 direction! ° Deflection in the well 377 1 1 r t gt2 t3~! g r 0 p q“2 ´ 3 ` p q ° Deflection still quite small, so t for descent down the well is given by standard formula of
t 2d g 2h g “ { „ { a a 1 1 t3~! g !gt3xˆ ´3 “ 3 ° 1 3 2 3 2 Deflection = !g 2h g { xˆ 8 9g !h { xˆ 3 p { q “ {p q a h = 100m (well) → 2.2 cm h = 2 km (skydiving) → 2 m Another quick exercise 378
Let’s do problem 9.9 together Foucault pendulum 379
At Chicago’s Museum of Science and Industry Foucault pendulum 380 Pendulum with length L z(up)
β m:r mg 2~! r9 T “ ´ ` y(north) x2 y2 z2°L2 ` ` “ T T x L T x „ p { q T T y L x(east) y „ p { q m T T z L z „ p { q T mg mg z „ Foucault pendulum 381 Pendulum with length L z(up)
~! 0, ! sin ✓, ! cos ✓ “p q β x: gx L 2y9! cos ✓ 2z9! sin ✓ “´ { ` ´ y(north) y: gy L 2!x9 cos ✓ “´ { ´ For small oscillations, T expect vz small x(east) x: gx L 2y9! cos ✓ m “´ { ` y: gy L 2x9! cos ✓ mg “´ { ´ !2 g L, ! cos ✓ ! 0 “ { “ z 2 x: 2!zy9 ! x 0 ´ ` 0 “ 2 y: 2!zx9 ! y 0 ` ` 0 “ Solution using earlier trick with complex numbers 382 2 x: 2!zy9 ! x 0 ´ ` 0 “ 2 z(up) y: 2!zx9 ! y 0 ` ` 0 “ ⌘ x iy Sum “ ` β 2 iy: 2i!zx9 i! y 0 ` ` 0 “ y(north) 2 x: iy: 2!z ix9 y9 ! x iy 0 ` ` p ´ q` 0p ` q“ 2 ⌘: 2i!z⌘9 ! ⌘ 0 ` ` 0 “ T i↵t ⌘ t e´ p q“ x(east) ↵2 2i! i↵ !2 0 m ´ ` zp´ q` 0 “ ↵2 2↵! !2 0 ´ ` z ` 0 “ mg ↵2 2↵! !2 0 ´ z ´ 0 “ ↵ 2! 4!2 4!2 2 !0 !z “ z ˘ z ` 0 { °° Ñ ˆ b ˙ ↵ ! ! 2 2 z 0 ↵ !z ! ! „ ˘ „ ˘ z ` 0 b Solution for pendulum 383 ! ! cos ✓ z “ z(up) i↵t ⌘ t e´ p q“ ↵ ! ! β „ z ˘ 0 i! t i! t i! t ⌘ e´ z C e 0 C e´ 0 y(north) “ 1 ` 2 ` ˘ What does this term T x(east) do? m 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 Solution for pendulum 384 z(up) ! ! cos ✓ !0 !z z “ °° i! t ⌘ t x t iy t Ae´ z cos ! t β p q“ p q` p q“ p 0 q y(north) At first, natural frequency much larger than the extra T term in front, but x(east) eventually, that rotates m motion in x direction to y mg direction and back 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! One last word on these subjects 386
Choice of frame Recall from earlier in the course: determines whether F = m(¨r r ˙2) r accelerations ˙ ¨ or forces are F = m(2r ˙ + r ) complicated (alternatively, which one is simple) Homework (due as usual in 1 week) 387
9.2, 9.14, 9.25,9.26, 9.28, 9.29