1 / 37

Physics 111: Lecture 2 Today’s Agenda

This physics lecture covers the recap of 1-D motion with constant acceleration, 1-D free fall, vectors, and 3-D kinematics. Topics include shoot the monkey, baseball, and the independence of x and y components.

haleyj
Download Presentation

Physics 111: Lecture 2 Today’s Agenda

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Physics 111: Lecture 2Today’s Agenda • Recap of 1-D motion with constant acceleration • 1-D free fall • example • Review of Vectors • 3-D Kinematics • Shoot the monkey • Baseball • Independence of x and y components

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

  3. 12 22 32 42 Recall what you saw:

  4. 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

  5. Ball w/ cup 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!

  6. 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)

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

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

  9. 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

  10. 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

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

  12. Chicago r Urbana 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 Chicago? • Choose origin at Urbana • Choose coordinates of distance (miles), and direction (N,S,E,W) • In this case r is a vector that points 120 milesnorth.

  13. Vectors... • There are two common ways of indicating that something is a vector quantity: • Boldface notation: A • “Arrow” notation: A =

  14. 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| (x,y) y arctan( y / x ) r  x

  15. Vectors... • The magnitude (length) of r is found using the Pythagorean theorem: r y x • The length of a vector clearly does not depend on its direction.

  16. 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

  17. Vector addition: • Consider the vectors A and B. Find A+B. B B A A A C=A+B B • We can arrange the vectors as we want, as long as we maintain their length and direction!!

  18. Vector addition using components: • Consider C=A + B. (a) C = (Ax i + Ayj) + (Bx i + By j) = (Ax + Bx)i + (Ay + By)j (b)C = (Cx i+ Cy j) • Comparing components of (a) and (b): • Cx = Ax + Bx • Cy = Ay + By By C B Bx A Ay Ax

  19. Lecture 2, Act 2Vectors • Vector A = {0,2,1} • Vector B = {3,0,2} • Vector C = {1,-4,2} What is the resultant vector, D, from adding A+B+C? (a){3,5,-1}(b){4,-2,5}(c){5,-2,4}

  20. Lecture 2, Act 2Solution D = (AXi+ AYj+ AZk) + (BXi+ BYj+ BZk) + (CXi+ CYj + CZk) = (AX + BX + CX)i+ (AY + BY+ CY)j+ (AZ + BZ + CZ)k = (0 + 3 + 1)i + (2 + 0 - 4)j+ (1 + 2 + 2)k = {4,-2,5}

  21. 3-D Kinematics • The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as: r = x i + y j + z k v = vx i + vy j + vz k(i,j,kunit vectors ) a = ax i + ay j + az k • We have already seen the 1-D kinematics equations:

  22. 3-D Kinematics • For 3-D, we simply apply the 1-D equations to each of the component equations. • Which can be combined into the vector equations: r = r(t) v = dr / dt a = d2r / dt2

  23. 3-D Kinematics • So for constant acceleration we can integrate to get: • a = const • v = v0 + a t • r = r0 + v0 t + 1/2 a t2 (where a, v, v0, r, r0, are all vectors)

  24. 2-D Kinematics lost marbles • 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

  25. “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.

  26. 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

  27. 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 • Equations of motion are: vx = v0xvy = v0y - gt x = vxt y = y0 + v0y t - 1/ 2 gt2

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

  29. 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!!

  30. 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

  31. 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)

  32. 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!!

  33. 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 ! )

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

  35. 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

  36. 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

  37. Recap of Lecture 2 • Recap of 1-D motion with constant acceleration. (Text: 2-3) • 1-D Free-Fall (Text: 2-3) • example • Review of Vectors (Text: 3-1 & 3-2) • 3-D Kinematics (Text: 3-3 & 3-4) • Shoot the monkey (Ex. 3-11) • Baseball problem • Independence of x and y components • Look at textbook problemsChapter 3: # 35, 39, 69, 97

More Related