1. PHY 183 - Program Physics for Scientists and Engineers Chapter 1 KINEMATICS 7/5 Chapter 2 DYNAMICS 7/5 Chapter 3 WORK AND ENERGY 6/4 Chapter 4 ROTATIONAL MOTION 6/4 The first test 40% (2) Chapter 5 PERIODIC MOTION 5/3 Chapter 6 WAVE MOTION 5/3 Chapter 7 FLUIDS AND THERMAL PHYSICS 5/3 Chapter 8 GAS LAWS AND KINETIC THEORY 5/3 Chapter 9 LIQUID PHASE 6/4 The final examination 60%

2. 1. Net force and Newton's first law 2. Newton's second law 3. Newton's third law 4. Frictional forces 5. Gravitation 6. Circular motion 7. Centripetal force 8. Static equilibrium and reference frames

3. Part 4 Frictional Forces

4. Forces: Normal Force Book at rest on table: What are forces on book? Weight is downward System is “in equilibrium” (acceleration = 0  net force = 0) Therefore, weight balanced by another force W FNFN F N = “normal force” = force exerted by surface on object F N is always perpendicular to surface and outward For this example F N = W

5. Learning Check Book at rest on table titled an angle of 30 0 to horizontal How much is normal force ?? W FNFN 30 0 Y

6. Forces: Kinetic Friction Kinetic Friction (a sliding Friction): A force, f k, between two surfaces that opposes relative motion. Magnitude: f k =  k F N *  k = coefficient of kinetic friction a property of the two surfaces FNFN W Ffkfk direction of motion

7. Learning Check Book (50g) move with a= 2m/s 2 on table titled an angle 30 0 to horizontal How much is coefficient of kinetic friction ?? W FNFN 30 0 Y FkFk X

8. Forces: Static Friction W FNFN F fsfs Static Friction: A force, f s, between two surfaces that prevents relative motion. f s ≤ f smax =  s F N force just before breakaway  s = coefficient of static friction a property of the two surfaces

9. Exercise 10N FNFN F fsfs F increases from 30N to 50N ( force just before breakaway) coefficient of static friction is ?? F s changes from ……… to………. (N) What is unit of coefficient of static friction ??

10. Forces: Tension Tension: force exerted by a rope (or string) Magnitude: same everywhere in rope Not changed by pulleys Direction: same as direction of rope. T

11. example: box hangs from a rope attached to ceiling T 5N y  F y = ma y T - W = ma y T = W + ma y In this case a y = 0 So T = W=5N How much is rope tension ?

12. Part 5 Gravitation

14. Gravity is a very tiny force Force between two objects each 1 Kg at a distance of 1 meter is F = G M 1 M 2 /R 2 G = 6.67 x 10 - 11 (Nm 2 /kg 2 ) 1 N is about the weight of one apple (100g) on the earth The reason the effects of gravity are so large is that the masses of the earth, sun, stars, …. are so large -- and gravity extends so far in space

15. Gravity and Weight Force on mass: MeMe ReRe mass on surface of Earth m  g F g  W = mg Calculate the earth’s radius from the gravity of Earth (note:M=5,98.10 24 kg; g=9,81 m/s 2 ) How much weight of a toy of 30g mass ??

16. See movie

17. Part 6 Circular motion

18. What is CM? (Circular Motion) Motion in a circle with: Constant Radius R –Trajectory is circular form R v x y (x,y)

19. How can we describe CM? In general, one coordinate system is as good as any other: – Cartesian: (x,y) [position] (v x,v y ) [velocity] – Polar: (R,  )[position] (v R,  ) [velocity] In CM: – R is constant (hence v R = 0). –  angular (t) is function of time measured by RAD. – Polar coordinates are a natural way to describe CM! R v x y (x,y) 

20. Polar and Cartesian Coordinates x = R cos  y = R sin   2  3  /222 1 0 sin cos R x y (x,y)(x,y)   X 2 + y 2 = R 2 tg  y / x

21. Polar Coordinates... In Cartesian coordinates, we say velocity: dx/dt = v and x = vt + x 0 In polar coordinates, angular velocity: d  /dt = .  =  t +   Has units of radians/second. Displacement s = vt. but s = R  = R  t, so: R v x y s  t v =  R

22. A fighter pilot flying in a circular turn (with 500m in radius) follow the equation:  = 5t+10t 2 (rad). It’s position at time t=2s is: a)1432,3 rad b) 1423,3 degree c) 50 degree The angular velocity at time t=3s is: (a) 650 m (b) 65 rad/s (c)105 rad/s What is velocity V ??? MCQ test

23. Period and Frequency Recall that 1 revolution = 2  radians –frequency (f) = revolutions / second (a) –angular velocity (  ) = radians / second (b) –By combining (a) and (b)  = 2  f Realize that: –period (T) = seconds / revolution –So T = 1 / f = 2  /   = 2  / T = 2  f R v s 

24. Example The Wings of motor spins with frequency f = 3000 rpm. The length of each wings is L = 80cm ? what is angular velocity ? Total path of point A travel after 10 s ?? What is period ?? f L A  = 2  f = 6000  (rad/m) S=  Lt = 2  fL t =1000 .0,8=800  (m) T=1/f=1/50(rps)=0.02 s

25. Polar Coordinates... In Cartesian coordinates, we say acceleration: dv/dt = a and v = at + v 0 In polar coordinates, angular acceleration : d  /dt =  and  =  t +   Has units of radians/second. For vector of angular velocity Has direction of axis looking point A turning to watch loop R v x y s  t

26. A fighter pilot flying in a circular turn (with 500m in radius) follow the equation:  =5t+10t 2 (rad). The angular acceleration at time t=3s is: (a) 650 rad/s 2 (b) 20 rad/s 2 (c) 105 rad/s 2 MCQ test

27. What is UCM? (Uniform Circular Motion) Motion in a circle with: –Constant Radius R –Trajectory is circular form v Constant Speed v = |v| R v x y (x,y)

28. Acceleration in UCM: speedvelocity notEven though the speed is constant, velocity is not constant since the direction is changing: must be some acceleration! v – Consider average acceleration in time  t a av =  v /  t vv2vv2  t vv1vv1 vv1vv1 vv2vv2 vvvv R

29. Acceleration in UCM: This is calledThis is called Centripetal Acceleration. Now let’s calculate the magnitude: vv2vv2 vv1vv1 vv1vv1 vv2vv2 vvvvR RRRR Similar triangles: But  R = v  t for small  t So:

30. Useful Equivalent We know that and v =  R Substituting for v we find that: a =  2 R

31. Centripetal Acceleration The Space Shuttle is in Low Earth Orbit (LEO) about 300 km above the surface. The period of the orbit is about 91 min. What is the acceleration of an astronaut in the Shuttle in the reference frame of the Earth? (The radius of the Earth is 6.4 x 10 6 m.) (a) 0 m/s 2 (b) 8.9 m/s 2 (c) 9.8 m/s 2

32. Example: Newton & the Moon What is the acceleration of the Moon due to its motion around the Earth? What we know (Newton knew this also): –T = 27.3 days = 2.36 x 10 6 s (period ~ 1 month) –R = 3.84 x 10 8 m(distance to moon) –R E = 6.35 x 10 6 m(radius of earth) RRERE

33. Moon... Calculate angular velocity: So  = 2.66 x 10 -6 s -1. Now calculate the acceleration. – a =  2 R = 0.00272 m/s 2 = 0.000278 g a r –direction of a points at the center of the Earth (-r ). ^

34. First calculate the angular frequency  : Exercise: Centripetal Acceleration of ES ?? Given: R O = R E + 300 km = 6.4 x 10 6 m + 0.3 x 10 6 m = 6.7 x 10 6 m Period = 91 min RORO 300 km RERE ES

35. Home work Thesis 1. Hair tension and it’s applications 2. Frictions and their applications 3. Frictional reduction 4. The moon movements 5. Water moving by moon gravitation 6. Solar system movements

36. Calculating accelerations x’x’ O x O h m M   T FNFN W Y w T

37. m 1 > m 2 m 1 m 2 O x T W1W1 T W2W2