Summary for Lecture 3

1. Summary for Lecture 3 4.5 Projectile motion 4.6 Range of projectile Dynamics 4.8 Relative motion in 1D 4.9 Relative motion in 3-D 4.7 Circular motion (study at home) 5.2 Newton 1 5.3-5.5 Force, mass, and Newton 2 55 ProblemsChap. 4 24, 45, 53,

2. First-year Learning Centre Tutors are available to help you from 12.30 pm to 2.30 pm on Tuesday – Friday In the First-year Learning Centre

3. Notices Physics Phrequent Phlyers Clarify the physics Extend your knowledge Ask the lecturer Room pp211 Physics Podium Thursdays 12 – 2 pm Representative for SSLC (Staff-Student Liaison Cttee)

4. Projectiles

5. Gallileo Parabola Max range 45o 2 angles for any range Running out of “impetus” Projectiles

6. What do we know from experience? The trajectory depends on: Initial velocity Projection angle Anything else? Air resistance Ignore for now

7. y r(t) y(t) x(t) x Projectile Motion To specify the trajectory we need to specify every point (x,y) on the curve. That is, we need to specify the displacement vector r at any time (t). r(t) = ix(t) + jy(t)

8. -g y vo v0sinq  v0cosq x What do we know? Usually we know the initial vector velocity vo. v0 = ivox + jvoy = iv0cosq + jv0sinq We know acceleration is constant = -g. accel is a vectora = iax + jay = i0 - jg

9. Finding an expression for Projectile motion The vertical component of the projectile motion is the same as for a falling object The horizontal component is motion at constant velocity

10. -g y vo  v0cosq r x To define the trajectory, we need r(t) That is:- we need x(t) and y(t) • Consider Horizontal motion • velocity 0 since ax=0 vx(t) = v0x + axt = v0 cosq  const • displacement Use x - xo = v0t + ½ at2 x(t) = v0xt + ½axt2 0 since ax=0 x(t) = v0 cosq t

11. y ay= -g vo v0sinq  Recall r x vertical component of displacement Use y - yo = ut + ½ at2 y(t) = v0yt + ½ ayt2 = v0 sinq t – ½ gt2 Re-arranging gives Height (y) as a function of distance (x) for a projectile.

12. y y = kx2 x y = -k1 x2 + k2 x

13. y x vo range y = -k1x2 + k2x

14. Range where i.e. where x = 0 or x=0 Maximum when 2 = 900. i.ewhen = 450 sine angle 0 For what x values does y = 0? R

15. http://www.colorado.edu/physics/phet/web-pages/simulations-base.htmlhttp://www.colorado.edu/physics/phet/web-pages/simulations-base.html

16. Range up a slope y Gradient of slope R’  x R

17. True forx = 0 or

18. x’ y’  R y y’ = mx’ R’ x If m = 0 (on level ground) R’= x’ = R

19. -g Time to travel distance d is h vo y  d For collision, x and y for dart and monkey must be the same at instant t y for darty - y0 = v0t + ½ at2 yo = 0

20. y for dart at impact (ie at time ) g h vo y  d How far has monkey fallen? dist = v0monkt + ½ at2 Therefore the height of the monkey is y = h -

21. WHY things move Dynamics Kinematics HOW things move Dynamics

22. Q P 40% . R

23. A body only moves if it is driven. In the absence of a FORCE a body is at rest 350 BC In the absence of a FORCE A body at rest WILL REMAIN AT REST If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO 1660 AD

24. Dynamics Aristotle For an object to MOVE we need a force. Newton For an object to CHANGE its motion we need a force Newtons mechanics applies for motion in an inertial frame of reference! ???????

25. Inertial reference frame Newton clarified the mechanics of motion in the “real world”. He believed that there existed an absolute (not accelerating) reference frame, and an absolute time. Inertial reference frame Inertial reference frame His laws applied only when measurements were made in this reference frame….. …or in any other reference frame that was at rest or moving at a constant velocity relative to this absolute frame. The laws of physics are always the same in any inertial reference frame.

26. Relativity Newton believed that there existed an absolute (not accelerating) reference frame, and an absolute time. Einstein recognised that all measurements of position and velocity (and time) are relative. There is no absolute reference frame. The laws of physics are always the same in any inertial reference frame.

27. y p z y’ z’ o x o x’ Frames of Reference In mechanics we need to specify position, velocity etc. of an object or event. This requires a frame of reference The reference frames for the same object may be different The connection between inertial reference frames is the “Gallilean transformation”. The reference frames may have a constant relative velocity We need to be able to relate one set of measurements to the other

28. Ref. Frame G (ground) VPG(const) xPG xTP T (Lunch Trolley) xTG Gallilean transformations Ref. Frame P (my seat in Plane) xTG=xTP+xPG Vel. = d/dt(xTG)vTG=vTP+ vPG 0 Accel = d/dt (vTG)aTG = aTP+ aPG In any inertial frame the laws of physics are the same

29. GROUND AIR N N’ P rPA vAG rPG E’ rAG E vPG=(rPG)=vPA+ vAG aPG=(vPG)=aPA+ aAG Looking from above rPG = rPA+ rAG In any inertial frame the laws of physics are the same 0

30. Ground Speed of Plane AG PA AG PA VPG = VPA + VAG VPA = 215 km/h to East VAG = 65 km/h to North Tan  = 65/215  = 16.8o

31. Ground Speed of Plane AG PA AG PA VPA = 215 km/h to East VAG = 65 km/h to North In order to travel due East, in what direction relative to the air must the plane travel? (i.e. in what direction is the plane pointing?), and what is the plane’s speed rel. to the ground? Do this at home and then do sample problem 4-11 (p. 74)

32. N A motorboat with its engine running at a constant rate travels across a river from Dock A, constantly pointing East. Flow X I will quiz you on this next lecture A Y Z Compare the times taken to reach points X, Y, or Z when the river is flowing and when it is not. Your answer should include an explanation of what might affect the time taken to cross the river, and a convincing justification of your conclusion.

33. Here endeth the lesson lecture No. III

34. Isaac Newton 1642-1727

35. Newton’s 1st Law If a body is at rest, and no force acts on it, IT WILL REMAIN AT REST. If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO If a body is moving at constant velocity, we can always find a reference frame where it is AT REST. At rest  moving at constant velocity

36. 1 a = F m F = a m Inertial mass Force The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed. If things do not need pushing to move at constant velocity, what is the role of FORCE??? An applied forcechangesthe velocity of the body a F a = F/m The more massive a body is, the less it is accelerated by a given force.

37. =ma vectorSUM of ALLEXTERNAL forces acting on the body vectorSUM of ALLEXTERNAL forces acting on the body a _______ = m Newton’s 2nd Law a = SF/m SF = ma

38. Newton’s 2nd Law SF=ma F = iFx + jFy + kFz a = iax + jay + kaz SFx=max SFy=may SFz=maz Applies to each component of the vectors

39. cos= cos47  =280 Fy = 0 220 170 FA+ FB+ FC= 0 since a= SF/mand a = 0 F = 0 Fx = 0Fy = 0 220N FCcos - FAcos47= 0 170N 170 cos - 220 cos47= 0 (280) FA sin 47 + FCsin – FB = 0 FB = FA sin 47 +FCsin = 220 sin 47 +170 sin28 = 160 +80  240 N ? N

41. N M2 M1 T m2g T m1g For mass m1 Vertical (only) m1g - T = m1a Apply F = ma to each body i.e Fx = max and Fy = may m1g - m2a = m1a For m2 Vertically Fy = -N + m2g = m2ay = 0 ÞN = m2g m1g = m2a + m1a = a(m2 + m1) Horizontally Fx = T = m2a Analyse the equation!

42. 2.5 x 103 N 2.5 x 103 N 3.9 x 104 N 41% 26% 33%

43. 2.5 x 103 N 2.5 x 103 N 3.9 x 104 N 3.9 x 104 N T a 5.0 x 103 N F = ma m is mass of road train F = ma a = F/m a = (39 – 5) x 103/4 x 104 a = 0.85 m s-2

44. Don’t care! a = 0 2.5 x 103 N 2.5 x 103 N T T 2.5 x 103 N F = ma m is mass of trailer! Fis net force on TRAILER So F= 0 a = 0 T - 2.5 x 103 = 0 T = 2.5 x 103 N

45. 2.5 x 103 N 2.5 x 103 N 3.9 x 104 N v = 0u = 20 m s-1, F = ma m is mass of roadtrain No driving force so F = -5.0 x 103 N a = - 5.0 x 103/4.0 x 104 = - 0.125 m s -2 Use v2 = vo2 + 2a(x – x0)  x – x0 = 400 m