2. Physics 151: Lecture 6, Pg 1 Announcement: l LABS start this week ! l Homework #2 : due Fri. (Sept. 15) by 5.00 PM on webassign Problems from Chapter 3 and 4.

3. Physics 151: Lecture 6, Pg 2 Physics 151: Lecture 6 Today’s Agenda l Today’s topics çFinish circular motion çDiscuss relative motion çtext sections 4.4-4.6

4. Physics 151: Lecture 6, Pg 3 Review: Uniform Circular Motion (UCM) l Motion in a circle with: çConstant Radius R v ç Constant Speed v = |v| çacceleration ? R v x y (x,y) See text: 4-4 = 0a a = const.a

5. Physics 151: Lecture 6, Pg 4 Polar Coordinates... l In Cartesian co-ordinates we say velocity dx/dt = v. ç x = vt In polar coordinates, angular velocity d  /dt = .   =  t   has units of radians/second. l Displacement s = vt. but s = R  = R  t, so: R v x y s  t frequency (f) =  period (T) = 1 / f = 2  /  v = Rv = R

6. Physics 151: Lecture 6, Pg 5 Lecture 6, ACT 1 Uniform Circular Motion l A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ?

7. Physics 151: Lecture 6, Pg 6 Acceleration in UCM: speedvelocitynot l Even though the speed is constant, velocity is not constant since the direction is changing: must be some acceleration !  Consider average acceleration in time  t vv2vv2R vv1vv1 See text: 4-4 vvvv vv1vv1 vv2vv2 v a av =  v /  t

8. Physics 151: Lecture 6, Pg 7 Acceleration in UCM: l This is called l This is called Centripetal Acceleration. l Now let’s calculate the magnitude: vv1vv1 vv2vv2 vvvv vv2vv2 vv1vv1 R RRRR Similar triangles: But  R = v  t for small  t So:

9. Physics 151: Lecture 6, Pg 8 Centripetal Acceleration l UCM results in acceleration:  Magnitude:a = v 2 / R =   R r çDirection:- r (toward center of circle) R a  ^ See text: 4-4

10. Physics 151: Lecture 6, Pg 9 Lecture 6, ACT 1 Uniform Circular Motion l A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ? (a) 20 m (b) 200 m (c) 2,000 m (d) 20,000 m

11. Physics 151: Lecture 6, Pg 10 Example: Newton & the Moon l What is the acceleration of the Moon due to its motion around the earth? l What we know (Newton knew this also): çT = 27.3 days = 2.36 x 10 6 s (period ~ 1 month) çR = 3.84 x 10 8 m(distance to moon) çR E = 6.35 x 10 6 m(radius of earth) RRERE a = 0.00272 m/s 2

12. Physics 151: Lecture 6, Pg 11 Moon... l So we find that a moon / g =.000278 l Newton noticed that R E 2 / R 2 moon =.000273 RRERE a moon g l This inspired him to propose that F Mm  1 / R 2 ç(more on gravity later)

13. Physics 151: Lecture 6, Pg 12 A satellite is in a circular orbit 600 km above the Earth’s surface. The acceleration of gravity is 8.21 m/s 2 at this altitude. The radius of the Earth is 6400 km. Determine the speed of the satellite, and the time to complete one orbit around the Earth. Lecture 6, ACT 2 Uniform Circular Motion Answer: 7,580 m/s 5,800 s

14. Physics 151: Lecture 6, Pg 13 A stunt pilot performs a circular dive of radius 800 m. At the bottom of the dive (point B in the figure) the pilot has a speed of 200 m/s which at that instant is increasing at a rate of 20 m/s 2. What acceleration does the pilot have at point B ? Lecture 6, ACT 3 Uniform Circular Motion

15. Physics 151: Lecture 6, Pg 14 Lecture 6, ACT 3 The Pendulum 1m  = 30° A) v r = 0 a r = 0 v   0 a   0 B) v r = 0 a r = 0 v  = 0 a   0 C) v r = 0 a r  0 v  = 0 a   0 Which statement best describes the motion of the pendulum bob at the instant of time drawn ? the bob is at the top of its swing. which quantities are non-zero ?

16. Physics 151: Lecture 6, Pg 15 Lecture 6, ACT 3 The Pendulum Solution aa NOT uniform circular motion : Is circular motion so must be a r  0 Speed is increasing so a  not zero 1m  = 30° a O What are components of v and a at  = 0 ? C) v r = 0 a r  0 v  = 0 a   0 arar Animation At the top of the swing, the bob temporarily stops, so v = 0.

17. Physics 151: Lecture 6, Pg 16 Inertial Reference Frames: Reference Frame l A Reference Frame is the place you measure from. ç It’s where you nail down your (x,y,z) axes! l An Inertial Reference Frame (IRF) is one that is not accelerating. ç We will consider only IRF’s in this course. l Valid IRF’s can have fixed velocities with respect to each other. ç More about this later when we discuss forces. ç For now, just remember that we can make measurements from different vantage points. Animation

18. Physics 151: Lecture 6, Pg 17 Lecture 6 – ACT 4 Relative Motion an airplane flying on a windy day. l Consider an airplane flying on a windy day. è A pilot wants to fly from New Haven to Bradley airport. She knows that Bradley is 120 miles due north of New Haven and there is a wind blowing due east at 30 mph. She takes off from New Haven Airport at noon. Her plane has a compass and an air-speed indicator to help her navigate. She uses her compass at the start to aim her plane north, and her air speed indicator tells her she is traveling at 120 mph with respect to the air. Text Ch 4, sect 6 After one hour she is:(A) at Bradley (B) east of Bradley (C) southeast of Bradley

19. Physics 151: Lecture 6, Pg 18 l The plane is moving north in the IRF attached to the air: V ç V p, a is the velocity of the plane w.r.t the air. Air V V p,a See example 4-9,10 (Boat Crossing a River) See text: Ex. 4.9 and 4.10 Lecture 6 – ACT 4 Relative Motion

20. Physics 151: Lecture 6, Pg 19 l But the air is moving east in the IRF attached to the ground. V ç V a,g is the velocity of the air w.r.t the ground (i.e. wind). V V a,g Air V V p,a See text: Ex. 4.9 and 4.10 Lecture 6 – ACT 4 Relative Motion

21. Physics 151: Lecture 6, Pg 20 l What is the velocity of the plane in an IRF attached to the ground? V ç V p,g is the velocity of the plane w.r.t the ground. V V p,g See text: Ex. 4.9 and 4.10 Lecture 6 – ACT 4 Relative Motion

22. Physics 151: Lecture 6, Pg 21 Relative Motion... è VVV è V p,g = V p,a + V a,g Is a vector equation relating the airplanes velocity in different reference frames. V V p,g V V a,g V V p,a See text: Ex. 4.9 and 4.10 The north component of vp,g is just her air speed. So she gets far enough north. The wind takes her further east. Answer is (B) – due east.

23. Physics 151: Lecture 6, Pg 22 Lecture 6, ACT 5 Relative Motion l You are swimming across a 50m wide river in which the current moves at 1 m/s with respect to the shore. Your swimming speed is 2 m/s with respect to the water. You swim across in such a way that your path is a straight perpendicular line across the river. çHow many seconds does it take you to get across? 2m/s 1m/s50m a) b) c) d)

24. Physics 151: Lecture 6, Pg 23 Lecture 6, ACT 5 Solution l The time taken to swim straight across is (distance across) / (v y ) Choose x axis along riverbank and y axis across river y x l Since you swim straight across, you must be tilted in the water so that your x component of velocity with respect to the water exactly cancels the velocity of the water in the x direction: 1m/s m/s y x 2m/s

25. Physics 151: Lecture 6, Pg 24 Lecture 6, ACT 5 Solution l So the y component of your velocity with respect to the water ism/s y x 50m m/s Answer (c) l So the time to get across is

26. Physics 151: Lecture 6, Pg 25 Recap for today: l Circular Motion (Text Ch. 4.4) l Relative Motion (Text Ch 4.6) l Reading assignment for Wed. çRead about Forces: Ch 5.1-3 l Homework #2 : due Fri. (Sept. 15) by 5.00 PM on webassign Problems from Chapter 3 and 4