The Study of Motion

1. The Study of Motion • Galileo's axiom: ignorato motu, ignoratur natura " not to know motion is not to know nature “ . • So We shall begin with the key topic in mechanics, the laws governing some of the simpler motions of objects : • Mathematics and the Description of Motion • Galileo and the Kinematics of Free Fall • Projectile Motion

2. Constant velocity • According to Galileo and Descartes, the natural state of a physical object is rest or motion along a straight line, at constant speed. • How the physicist describes such motion in a way that can easily be generalized to other kinds of motion. • Suppose we are watching the motion of a car traveling along a perfectly straight and smooth road, and that we wish to find the exact relationship between the distance covered and the time required to cover it, such as table 6. Galileo (1564-1642) Descartes (1596-1650)

3. the motion can be written s ~ t, where the symbol ~ means "is proportional to. “ But this is equivalent to the equation s = kt, where k is a constant that depends on neither s nor t. ……….What look like is the graph ? In analytic geometry, the equation s = kt represents a straight line passing through the origin of coordinates. The constant k is called the slopek = s/t This ratio of distance covered to time elapsed is, by definition, the speed; the fact that it is equal to a constant of the car travels with constant or uniform speed.

4. Motion in 1 dimension • In 1-D, we usually write position as s(t1) or x(t1 ). • Since it’s in 1-D, all we need to indicate direction is + or . • Displacement in a time t = t2 - t1isx = x(t2) - x(t1) = x2 - x1 x some particle’s trajectoryin 1-D x2 x x1 t1 t2 t t

5. 1-D kinematics • Velocity v is the “rate of change of position” • Average velocity vav in the time t = t2 - t1is: x trajectory x2 x Vav = slope of line connecting x1 and x2. x1 t1 t2 t t

6. 1-D kinematics... • Considerlimit t1 t2 • Instantaneous velocity v is definedas: x sov(t2) = slope of line tangent to path at t2. x2 x x1 t1 t2 t t

7. 1-D kinematics... • Acceleration a is the “rate of change of velocity” • Average acceleration aavin the time t = t2 - t1is: • Andinstantaneous acceleration a is definedas: using

8. Recap • If the position x is known as a function of time, then we can find both velocity vand acceleration a as a function of time! x t v t a t

9. More 1-D kinematics • We saw that v = dx / dt • In “calculus” language we would write dx = v dt, which we can integrate to obtain: • Graphically, this is adding up lots of small rectangles: v(t) + +...+ = displacement t

10. 1-D Motion with constant acceleration • High-school calculus: • Also recall that • Since a is constant, we can integrate this using the above rule to find: • Similarly, since we can integrate again to get:

11. Solving for t: Useful Formula • Plugging in for t:

12. Recap: • For constant acceleration: x t v • From which we know: t a

13. Lecture 1, Act 2Motion in One Dimension • When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? (a)Bothv = 0anda = 0. (b)v  0, but a = 0. (c) v = 0, but a  0. y

14. Lecture 1, Act 2Solution • Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. • Since the velocity is continually changing there must be some acceleration. • In fact the acceleration is caused by gravity (g = 9.81 m/s2). • (more on gravity in a few lectures) • The answer is (c) v = 0, but a  0. x t v t a t

15. vo ab x = 0, t = 0 Problem 1 • A car is traveling with an initial velocity v0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab

16. Problem 1... • A car is traveling with an initial velocity v0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab. At what time tf does the car stop, and how much farther xf does it travel? v0 ab x = 0, t = 0 v = 0 x = xf , t = tf

17. Problem 1... • Above, we derived: v = v0 + at • Realize thata = -ab • Also realizing that v = 0at t = tf : find 0 = v0 - ab tfor tf = v0 /ab

18. Problem 1... • To find stopping distance we use: • In this case v = vf = 0, x0 = 0 and x = xf

19. Problem 1... • So we found that • Suppose that vo = 65 mi/hr = 29 m/s =104.6 km/hr • Suppose also thatab = g = 9.81 m/s2 • Find that tf = 3 s and xf = 43 m

20. Tips: • Read ! • 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. • Watch your units ! • Always check the units of your answer, and carry the units along with your numbers during the calculation. • Understand the limits ! • Many equations we use are special cases of more general laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration).

21. Review: • For constant acceleration we found: x t v • From which we derived: t a t

22. 12 22 32 42 Recall what you saw:

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: Penny & feather • g does not depend on the nature of the material! • Galileo (1564-1642) figured this out without fancy clocks & rulers! • 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 • More on gravity in a few lectures!

25. 1000 m Problem: • 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)

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

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

28. Lecture 2, Act 11D free fall • Alice and Bill are standing at the top of a cliff of heightH. Both throw a ball with initial speedv0, Alice straightdownand Bill straightup. The speed of the balls when they hit the ground arevAandvBrespectively.Which of the following is true: (a)vA < vB(b) vA = vB (c) vA > vB Alice v0 Bill v0 H vA vB

29. Lecture 2, Act 11D 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. • Bill v0 v= v0 H y = 0

30. Lecture 2, Act 11D Free fall • We can also just use the equation directly: Alice: same !! Bill: Alice v0 Bill v0 y = 0

31. 2-D Kinematics • Most 3-D problems can be reduced to 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. G. Merapi r Jogjakarta Vectors (review): • In 1 dimension, we could specify direction with a + or - sign. For example, in the previous problem ay = -g etc. • In 2 or 3 dimensions, we need more than a sign to specify the direction of something: • To illustrate this, consider the position vectorr in 2 dimensions. • Example: Where is G. Merapi? • Choose origin at Jogjakarta • Choose coordinates ofdistance (kilometers), anddirection (N,S,E,W) • In this case r is a vector thatpoints 35 kmnorth.

33. û Unit Vectors: • A Unit Vector is a vector having length 1 and no units • It is used to specify a direction • Unit vector u points in the direction of U • Often denoted with a “hat”: u = û • Useful examples are the Cartesian unit vectors [i, j, k] • point in the direction of the x, yand z axes U y j x i k z

34. Vectors... • The components of r are its (x,y,z) coordinates • 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| ; arctan( y / x ) (x,y) y r  x • The length of a vector clearly does not depend on its direction.

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

37. From eq. (8.1) and (8.2)

38. 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 ! )

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

40. Dart hits the monkey! Shooting the Monkey... r= r0 - 1/2 g t2 • With gravity, still aim at the monkey! r= v0t - 1/2 g t2

41. 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!! x = x0 y= -1/2 g t2

43. To summarize

44. Problem: • Mark McGwire 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 Mark get a home run? v h  y0 D

45. 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 = x0 = 0 • Distance s x • Equations of motion are: vx = v0xvy = v0y - gt x = vxt y = y0 + v0y t - 1/ 2 gt2

46. Problem... • Use geometry to figure out v0x and v0y : g Find v0x = |v| cos . and v0y = |v| sin . y v v0y  y0 v0x x

47. 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: • 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 = 3.5m • Since the wall is 3 m high, Mark gets the homer!!

48. Lecture 2, Act 3Motion 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?(a) D2 = 2D1 (b) D2 = 4D1 (c) D2 = 8D1

49. Lecture 2, Act 3Solution • The distance a ball will go is simply x= (horizontal speed) x (time in air) = v0x t • 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)

50. v0,2 v0y ,2 v0,1 ball 2 ball 1 v0y ,1 v0x ,1 v0x ,2 Lecture 2, Act 3Solution x = v0x t • 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!!