SPH4U1 “Kinematics”

1. SPH4U1 “Kinematics”

2. Fundamental Units • How we measure things! • All things in classical mechanics can be expressed in terms of the fundamental units: • Length: L • Mass : m • Time : t • For example: • Speed has units of L / t (e.g. kilometres per hour). • Force has units of mL/ t2 (kgm/s2 )etc...

3. Units... • SI (Système International) Units: • mks: L = meters (m), M = kilograms (kg), T = seconds (s) • cgs: L = centimeters (cm), M = grams (gm), T = seconds (s) • We will use mostly SI mks units, but you may run across some problems using cgs units. You should know to convert back & forth.

4. Dimensional Analysis • This is a very important tool to check your work • It’s also very easy! • Example: Doing a problem you get the answer distance d = vt 2(velocity x time2) Units on left side = L Units on right side = L / t x t2 = L x t • Left units and right units don’t match, so answer must be wrong!!

5. Vectors • Avectoris a quantity that involves both magnitudeanddirection. • 55 km/h [N35E] • A downward force of 3 Newtons • Ascalaris a quantity that does not involve direction. • 55km/h • 18cm long

6. Vector Notation • Vectors are often identified with arrows in graphics and labeled as follows: We label a vector with a variable. This variable is identified as a vector either by an arrow above itself : Or By the variable being BOLD: A

7. 1m 6.7m 3m Start Finish 5m Displacement • Displacementis an object’s change in position. Distanceis the total length of space traversed by an object. = 500 m = Displacement 0 m = Distance 5 00 m Displacement: Distance:

8. B B B R R R A A A D D C C C E E Vector Addition E D A + B + C + D + E = Distance R = Resultant = Displacement

9. A R B Rectangular Components y Quadrant I Quadrant II  -x x Quadrant III Quadrant IV -y

10. Vectors... • The components (in a particular coordinate system) of r, the position vector, are its (x,y,z) coordinates in that coordinate system • r= (rx ,ry ,rz ) = (x,y,z) • Consider this in 2-D (since it’s easier to draw): • rx = x = r cos • ry = y = r sin where r = |r| (x,y) y arctan( y / x ) r  x

11. Vectors... • The magnitude (length) of r is found using the Pythagorean theorem: r y x • The length of a vector clearly does notdepend on its direction.

12. force: action of one body on another; characterized by its point of application, magnitude, line of action, and sense. • Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force. Resultant of Two Forces • The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs. • Force is a vector quantity.

13. Resultant of Several Concurrent Forces • Concurrent forces: set of forces which all pass through the same point. A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces. • Vector force components: two or more force vectors which, together, have the same effect as a single force vector.

14. Sample Problem • Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the diagonal. Two forces act on a bolt at A. Determine their resultant. • Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.

15. Components • Resolve both forces into their horizontal and vertical components • Px =40 N cos 20 • Py = 40 N sin 20 • Qx = 60 N cos 25 • Qy = 60 N sin 25

16. Add the x components • Add the y components • Use pythagorean theorem to find magnitude • Use the arc tan to solve for direction

17. x trajectory x2 x Vav = slope of line connecting x1 and x2. x1 t1 t2 t t 1-D kinematics • Velocity v is the “rate of change of position” • Average velocity vav in the time t = t2 - t1is:

18. 1-D kinematics... • Consider limit t1 t2 • Instantaneous velocity v is definedas: so v(t2) = slope of line tangent to path at t2. x x2 x x1 t1 t2 t t

19. 1-D kinematics... • Acceleration a is the “rate of change of velocity” • Average acceleration aav in the time t = t2 - t1is: a = v2 –v1 t2-t1

20. Recap • So for constant acceleration we find: x t v t a t

21. Recap: • For constant acceleration:

22. Problem Solving Tips: • Read Carefully! • Before you start work on a problem, read the problem statement thoroughly. Make sure you understand what information is given, what is asked for, and the meaning of all the terms used in stating the problem. • Using what you are given, set up the algebra for the problem and solve for your answer algebraically • USE symbols for quantities you know as needed • Don’t plug in numbers until the end • Watch your units ! • Always check the units of your answer, and carry the units along with your formula during the calculation.

23. 1-D Free-Fall • This is a nice example of constant acceleration (gravity): • In this case, acceleration is caused by the force of gravity: • Usually pick y-axis “upward” • Acceleration of gravity is “down”: y t v t a y t ay =  g

24. Gravity facts: • g does not depend on the nature of the material! • Galileo (1564-1642) figured this out without fancy clocks & rulers! • On the surface of the earth, gravity acts to give a constant acceleration • demo - feather & penny in vacuum • Nominally, g = 9.81 m/s2 • At the equator g = 9.78 m/s2 • At the North pole g = 9.83 m/s2

25. Gravity facts: • It’s a force between two objects, like me and the earth. or earth and moon, or sun and Neptune, etc • Gravitational Force is proportional to product of masses: • Proportional to 1/r2 At the surface of earth gravitational force attracts “m” toward the center of the earth, is approximately constant and equal to mg. The number g=9.81 m/s2 contains the effect of Mearth and rearth.

26. 1000 m Question: • The pilot of a hovering helicopter drops a lead brick from a height of 1000 m. How long does it take to reach the ground and how fast is it moving when it gets there? (neglect air resistance)

27. y Solution: • First choose coordinate system. • Origin and y-direction. • Next write down position equation: • Realize that v0y = 0. 1000 m y = 0

28. y Solution: • Solve for time t when y = 0 given that y0 = 1000 m. • Recall: • Solve for vy: y0= 1000 m y = 0

29. 1D Free Fall • Alice and Bob are standing at the top of a cliff of height H. Both throw a ball with initial speed v0, Alice straight down and Bob straight up. The speed of the balls when they hit the ground are vA and vB respectively. Which of the following is true: (a) vA < vB (b) vA = vB (c) vA > vB Alice v0 Bob v0 H vA vB

30. 1D Free fall • Since the motion up and back down is symmetric, intuition should tell you that v = v0 • We can prove that your intuition is correct: Equation: This looks just like Bill threw the ball down with speed v0, sothe speed at the bottom shouldbe the same as Alice’s ball. H • Bob v0 v= v0 y = 0

31. 2-D Kinematics • Most 2-D problems when acceleration is constant: • Choose y axis to be along direction of acceleration • Choose x axis to be along the “other” direction of motion • Example: Throwing a baseball (neglecting air resistance) • Acceleration is constant (gravity) • Choose y axis up: ay = -g • Choose x axis along the ground in the direction of the throw

32. Reference frame on the moving train. Reference frame on the ground. “x” and “y” components of motion are independent. • A man on a train tosses a ball straight up in the air. • View this from two reference frames:

33. Problem: • David clobbers a fastball toward center-field. The ball is hit 1 m (yo) above the plate, and its initial velocity is 36.5 m/s(v ) at an angle of 30o () above horizontal. The center-field wall is 113 m (D) from the plate and is 3 m (h) high. • What time does the ball reach the fence? • Does David get a home run? v h  y0 D

34. Problem... • Choose y axis up. • Choose x axis along the ground in the direction of the hit. • Choose the origin (0,0) to be at the plate. • Say that the ball is hit at t = 0, x(0) = x0 = 0. y(0) = y0 = 1m • Equations of motion are: vx = v0x vy = v0y - gt x = vxt y = y0 + v0y t - 1/ 2 gt2

35. Problem... • Use geometry to figure out v0x and v0y : g Find v0x = |v| cos . and v0y = |v| sin . y v v0y  y0 v0x remember, we were told that q= 30 deg x

36. Problem... • The time to reach the wall is: t = D / vx (easy!) • We have an equation that tell us y(t) = y0 + v0y t + a t2/ 2 • So, we’re done....now we just plug in the numbers: a = -g • Find: • vx = 36.5 cos(30) m/s = 31.6 m/s • vy = 36.5 sin(30) m/s = 18.25 m/s • t = (113 m) / (31.6 m/s) = 3.58 s • y(t) = (1.0 m) + (18.25 m/s)(3.58 s) - (0.5)(9.8 m/s2)(3.58 s)2 = (1.0 + 65.3 - 62.8) m = 3.5 m • Since the wall is 3 m high, David gets the homer!!

37. Motion in 2D • Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it? • Assume the receiver and QB are the same height (a) D2 = 2D1 (b) D2 = 4D1 (c) D2 = 8D1

38. To figure out “time in air”, consider the equation for the height of the ball: • When the ball is caught, y = y0 (time of catch) two solutions (time of throw) Solution • The distance a ball will go is simply x= (horizontal speed) x (time in air) = v0x t

39. v0,2 v0y ,2 v0,1 ball 2 ball 1 v0y ,1 v0x ,1 v0x ,2 Solution • So the time spent in the air is proportional to v0y: • Since the angles are the same, both v0y and v0x for ball 2are twice those of ball 1. • Ball 2 is in the air twice as long as ball 1, but it also has twicethe horizontal speed, so it will go 4 times as far!!

40. Projectile Motion Range Travel Time: Distance: Height Time to top: Maximum:

41. Example • A golfer hits a golf ball so that it leaves the club with an initial speed v0=37.0 m/s at an initial angle of 0=53.1o. • Determine the position of the ball when t=2.00s. • Determine when the ball reaches the highest point of its flight and find its height, h, at this point. • Determine its horizontal range, R. a) The x-distance: The y-distance:

42. Example • A golfer hits a golf ball so that it leaves the club with an initial speed v0=37.0 m/s at an initial angle of 0=53.1o. • Determine the position of the ball when t=2.00s. • Determine when the ball reaches the highest point of its flight and find its height, h, at this point. • Determine its horizontal range, R.

43. Shooting the Monkey(tranquilizer gun) • Where does the zookeeper aim if he wants to hit the monkey? • ( He knows the monkey willlet go as soon as he shoots ! )

44. Shooting the Monkey... r = r0 • If there were no gravity, simply aim at the monkey r =v0t

45. Dart hits the monkey! Shooting the Monkey... r = r0 - 1/2 g t2 • With gravity, still aim at the monkey! r = v0 t - 1/2 g t2

46. Recap:Shooting the monkey... x= v0 t y = -1/2 g t2 • This may be easier to think about. • It’s exactly the same idea!! • They both have the same Vy(t) in this case x = x0 y = -1/2 g t2

47. Shooting the Monkey