Grf Andrea Karinthy Frigyes Gimnzium, Budapest Description of - PowerPoint PPT Presentation

Gróf Andrea Karinthy Frigyes Gimnázium, Budapest

Description of motions PHYSICS GEOGRAPHY • Choice of reference • Idea of reference frame frame emphasized. not addressed. • But inertial frames used • "Natural" reference frame is exclusively. non-inertial. • Students corrected by • Explanations refer to teacher if they talk about centrifugal centrifugal force. and Coriolis forces.

 1. Survey on the understanding of the physics behind geography 2. A possible introduction to inertial  ! forces: merry-go-round example treated quantitatively  3. Applications in physics and geography

1. Survey on the physics (mechanics) behind geography MCQ questions on timekeeping, the shape of the Earth, motions of air and the seas, tides, etc. 215 students (16 and 17-year olds) Background: 1 year of physical geography and 1 year of physics

Questions involving inertial forces Q Geography: oblate Earth explained in terms of the centrifugal force. Physics problems on rotating objects: such forces not considered. What is the difference? A A.F centrifugal << F centripetal for fast rotations 11% B. F centrifugal only exists for astronomical sizes 10% C. Observer does/does not rotate along 17% D.F centrifugal always present, therefore not felt 49% (no answer 13%)

Q ? N

Q ? S ?

A A.South clockwise, north counterclockwise 56% B. North clockwise, south counterclockwise 21% C. Same sense 8% D.Anything may happen 10% (no answer 5%)

https://www.youtube.com/watch?v=4llVfoDuVlw

2. Describing motions on a playground roundabout The motion of A as seen by B:     2 r 2 1 . 5 m    v 3 . 14 T 3 s A 2 v m     2 a r 6 . 58 2 r s  ma    F 20 5 . 68 132 N B

Tangential motion as described by the inertial observer B m    3 . 14 0 . 5 3 . 64 s 2 2  ( v u ) 3 . 64 m    a 8 . 84 2 r 1 . 5 s      F F ma 0 . 02 8 . 84 0 . 177 N net merry For rotating observer A: 2 2 u 0 . 5 m    0 . 17 , a 2 r 1 . 5 s  ma     F 0 . 02 0 . 17 0 . 003 N 0.177N net What other force is there?

  (inwards) F 0 . 177 N merry cf    (outwards) F 0 . 132 N F 0 . 003 N net (inwards)     needed: 0 . 003 0 . 177 0 . 132  (outwards) 0 . 042 N F merry = 0.177N F coriolis F cf

Algebraically: For B  2 2 2 ( v u ) mv 2 mvu mu       F F ma m net merry r r r r 2 2 mu mv 2 mvu    For A  ma  F F net merry r r r 0.003 N = 0.177 N – 0.132 N – 0.042 N . 2 mvu v        m 2 u m 2 u r r    2 0 . 02 3 . 14 0 . 5  F merry = 0.177N 0 . 042 N 1 . 5 F coriolis F cf

For B a = 0 A B

The same motion as seen by A : Radial speed constant  r  v Tangential speed  t increases  t       t  t        s r t t ω  t 1       2 a ( t ) r t  2  r      a 2 2 v  t

3. Applications: Inertial forces Sports events? on the rotating Earth (a) Gravitational, centrifugal acceleration and free fall acceleration at the Equator? (b) Athlete can jump to 8 metres at the poles. How far can he jump on the Equator? Ω Which way is "down"? (local) vertical Latitude of Budapest: φ = 47.5º F cf (a) Magnitude and direction of F grav m g the centrifugal acceleration in Budapest. (b) Magnitude and direction of the acceleration of free fall in Budapest?

Coriolis force at the Equator Air moving at u = 20 m/s towards the west. Magnitude of acceleration towards the centre of the Earth? (a) according to an inertial observer?   2       2 5 6 ( R u ) ( 7 . 29 10 )( 6 . 38 10 ) 20    2 a 0 . 0369 m/s  6 R 6 . 38 10 (b) according to an Earth-based observer? 2 2 20 u      5 2 a 6 . 27 10 m/s 6  R 6 . 38 10 (c) What is the magnitude and direction of the Coriolis acceleration?           5 3 2 2 2 ( 7 . 29 10 ) 20 2 . 92 10 m/s (verticall y down) a C u

Coriolis force elsewhere Foucault pendulum, Paris: φ = 48.8 ° What is the local angular speed? A 2 π ·sin φ φ A P P P Paris φ     sin      a C 2 sin v http://enggar.net/page/12/?s /

In Paris: φ = 48.8 °              5 5 sin 7 . 29 10 sin 48 . 8 5 . 49 10 / s One period of the Panthéon pendulum is 16.4 sec. (a) How much does it turn in an hour?         5 t 5 . 49 10 3600 0 . 197 rad 11 . 3 (b) Displacement between two successive swings on a circle of radius 3 m?         5 4 t 5 . 49 10 16 . 4 9 . 01 10 rad        4 9 . 01 10 3 2 . 7 mm t r

Is the Coriolis force important? r = 2 cm, v = 10 cm/s (a) Find the acceleration towards the centre. (b) What is the contribution of the Coriolis force to this? 2 2 v 0 . 1    2 a 0 . 5 m/s r 0 . 02    2          5 5 2 a C v sin 2 0 . 1 ( 7 . 3 10 ) sin 47 . 5 1 10 m/s No No

Jupiter: T = 9.8 hours, R = 71 900 km (equatorial) Great Red Spot φ = 22 ° (S), wind v ≈ 100 m/s Find (a) radius of spot 1 ° corresponds to 8    2 R 1 . 4 10    6 1 . 2 10 m 360 360 ≈ 9 ° means r ≈ 1.1·10 7 m (b) acceleration of gas m    4 a 9 10 2 s (c) Coriolis acceleration Ye Ye m    4 www.celestiamotherlode.net/catalog/jupiter.php a 1 . 3 10 C 2 s s

Shinkansen train v = 200 km/h Tokyo to Osaka, both N55 ° 200            5 a C 2 v sin 2 7 . 3 10 sin 35 3 . 6 m  2  0 . 0047 0 . 0005 g s No No http://www.japantimes.co.jp/news/2014/06/18/national/shinkansen-tops-list-100-innovative-postwar-technologies/#.Va84VPkmHpE

Golfer in Scotland (N55 ° ) can hit the ball to 300 m at 45 ° angle. What is the deviation of the ball owing to the Coriolis force?  2    2 v sin 2 v sin cos 2 v sin      0 0 0 t v cos 0 g g g 1 1    2 2 v 2 0 v 2 2    0 300 m 9 . 8 9 . 8   m 2 54 sin 45      v 300 9 . 8 54 , t 8 . 7 s 0 s 9 . 8        a 2 sin v cos C 0 m           5 2 7 . 3 10 sin 55 54 cos 45 0 . 0046 2 No No s 1 1 2 2      d a t 0 . 0046 8 . 7 17 cm C 2 2

Artillery missile, N50 ° , v 0 = 700 m/s towards the East, at 45 ° angle. Deviation owing to the Coriolis force? 1 1    2 2 700    2 2 v sin 2 v sin cos 2 2       0 0 Range v cos 50 km 0 9 . 8 g g 2      4 sin 1 1 2 v sin          2          3 2 0 v sin cos d a t ( 2 sin v cos )   C 0 0 2   2 2 g g  5    4 7 . 3 10 sin 50   3  2     700 sin 45 cos 45 280 m 2 9 . 8 Ye Ye s

An interesting kind of motion Frictionless and horizontal ice rink, 30 m wide, Puck given an initial velocity. Coriolis force only: circular motion Find: (a) speed needed in Budapest (N47.5 ° ) 2 v v     2 sin v Note:      2 sin 2 r local r mm             5 v 2 sin r 2 7 . 3 10 sin 47 . 5 15 0 . 61 s (b) What is the radius if the speed is 1 m/s? At 10 ° latitude? At 80 ° ? v 1    r 9 . 3 km (39 km, 7.0 km)        5 2 sin 2 7 . 3 10 sin 47 . 5

Buoy in the Baltic Sea, SE of Stockholm, N57 °

Both forces to consider Design a space station: • Cylindrical shape • Artificial gravity is provided by centrifugal force owing to spinning about the axis. • Coriolis force on crew walking at 1 m/s is no greater than http://www.astronautix.com/craft/span1984.htm 0.05 mg .  2 r  g g 10    r 160 m    1 2 2 m m 0 . 25          2 1 0 . 05 10 0 . 25 2   s s s

References Anders O. Persson The Coriolis Effect: Four centuries of conflict between common sense and mathematics, History of Meteorology 2 (2005)

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