Da DaVinci Coke. Or, Adult Fizzix 1 MidTerm Exam Review 2015

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1 Da DaVinci Coke Or, Adult Fizzix 1 MidTerm Exam Review 2015

2 General List of Stuff: Chapters Covered (#1-10 OpenStax Text) Ch1: Obligatory Not-Important Stuff Ch2: 1-D Kinematics Ch3: 2-D Kinematics Ch4 & 5: Newt s Laws Ch 6: Circular Motion & Gravitation Ch7: Work / Energy / Power Ch8: MV Ch9: Rotational Kinematics (Torque & Statics) Ch10: Rotational Dynamics (L)

3 Ch1 & 2 Basic Crap (Ch 1) Metrix Sci Method 1-D Motion (Ch 2) The FAB FIVE Graffing Motion d a v 2ad v d t v f v t i( f ) t v 2 f i 1 2 at v 2 i 2

4 Ch1 Specifics Measurements SI (MKS) System Scientific Notation Graphing Relationships Linear y mx b

5 Ch1 Specifics Measurements SI (MKS) System Scientific Notation Graphing Relationships Parabolic y ax 2

6 Ch1 Specifics Measurements SI (MKS) System Scientific Notation Graphing Relationships Hyperbolic y a x 2

7 Ch2 Specifics Displacement v. Distance Displacement=CHANGE in position Velocity v. Speed v x t

8 Ch2 Specifics Acceleration a v v f v i t t

9 Ch2 Specifics Free Fall a g m s 2 Fizzoids use10 m s 2

10 All graphs shown are for same object at same time.

11 Shows relationships between possible combos of velocity and acceleration vectors. Splane?

12 Figure 2.14 a) This car is speeding up as it moves toward the right. It therefore has positive acceleration in our coordinate system. b) This car is slowing down as it moves toward the right. Therefore, it has negative acceleration in our coordinate system, because its acceleration is toward the left. The car is also decelerating: the direction of its acceleration is opposite to its direction of motion. c) This car is moving toward the left, but slowing down over time. Therefore, its acceleration is positive in our coordinate system because it is toward the right. However, the car is decelerating because its acceleration is opposite to its motion. d) This car is speeding up as it moves toward the left. It has negative acceleration because it is accelerating toward the left. However, because its acceleration is in the same direction as its motion, it is speeding up (not decelerating).

13 Figure 2.42 (a) A person throws a rock straight up, as explored in Example The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. (b) A person throws a rock straight down from a cliff with the same initial speed as before, as in Example Note that at the same distance below the point of release, the rock has the same velocity in both cases.

14 Figure 2.43 Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with time while displacement increases with time squared. Acceleration is a constant and is equal to gravitational acceleration.

15 Figure 2.47 Graph of displacement versus time for a jet-powered car on the Bonneville Salt Flats.

16 Ch3 Specifics Scalar v. Vector Scalar has ONLY size (magnitude if you like big fancy woids ) Distance = 5 m Mass = 100 kg Speed = 60 MPH

17 Ch3 Specifics Scalar v. Vector Vector has BOTH size and DIRECTION Displacement = 5 m North Weight = 1000 N Downward Velocity = 60 MPH South

18 Ch3 Specifics Vector Addition Add graphically / algebraically Head-to-Tail Boy Scout hike 3km North, then 4 km East Find d and d 4 km d=7 km & d = 5 km 53 o E of N 3 km 5 km

19 Ch3 Specifics Vector Addition with non-right triangles Use LOC 4 km >5 km c 2 a 2 b 3 km 2 2abcos

20 General List of Stuff: 2-D Motion (Ch3) Projectile Stuff TWO individual Probs H & V Vectors vs Scalars

21 Figure 3.6 This shows the motions of two identical balls one falls from rest, the other has an initial horizontal velocity. Each subsequent position is an equal time interval. Arrows represent horizontal and vertical velocities at each position. The ball on the right has an initial horizontal velocity, while the ball on the left has no horizontal velocity. Despite the difference in horizontal velocities, the vertical velocities and positions are identical for both balls. This shows that the vertical and horizontal motions are independent.

22 Ch3 Specifics Projectiles & Relative Motion Range Equation v 2 sin 2 R i g

23 Figure 3.38 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because a x = 0 and v x is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory.

24 Figure 3.40 The trajectory of a rock ejected from the Kilauea volcano.

25 Figure 3.41 Trajectories of projectiles on level ground. (a) The greater the initial speed v 0, the greater the range for a given initial angle. (b) The effect of initial angle θ 0 on the range of a projectile with a given initial speed. Note that the range is the same for 15º and 75º, although the maximum heights of those paths are different.

26 Figure 3.42 Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved.

27 Figure 3.44 A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.

28 Ch3 Specifics Projectiles & Relative Motion Relative Motion

29

30 Ch4 : Newton s Laws Newton s Laws Ch4 1. I 2. Biggee 3. F = -F Friction & motion caused by F F ma

31 Ch4 Specifics Force Push or pull Measured in Newtons 1N 1kg 1m 1s 2

32 Free Body Diagrams Ch4 Specifics Sketch with all forces acting on body Include every force that actually has something to do with acceleration.

33 Figure 4.3 Part (a) shows an overhead view of two ice skaters pushing on a third. Forces are vectors and add like other vectors, so the total force on the third skater is in the direction shown. In part (b), we see a free-body diagram representing the forces acting on the third skater.

34 Figure 4.13 Since motion and friction are parallel to the slope, it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N is perpendicular to the slope and f is parallel to the slope, but w has components along both axes, namely w and w. N is equal in magnitude to w, so that there is no motion perpendicular to the slope, but f is less than w, so that there is a downslope acceleration (along the parallel axis).

35 Figure 4.15 When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force T, that force must be parallel to the length of the rope, as shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). This is an example of Newton s third law. The rope is the medium that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations along the rope.

36 Figure 4.24 (a) A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The free-body diagram for the traffic light is also shown. (d) The forces projected onto vertical (y) and horizontal (x) axes. The horizontal components of the tensions must cancel, and the sum of the vertical components of the tensions must equal the weight of the traffic light. (e) The free-body diagram shows the vertical and horizontal forces acting on the traffic light.

37 Weight of an object Since F=ma, F g =mg F g is WEIGHT Normal Force Ch4 Specifics Force perpendicular to surface Force surface fights back with due to NL3

38 Ch4 Specifics Friction Force that resists motion when two surfaces are in contact moving relative to each other.

39 Ch4 Specifics Friction Kinetic Friction is friction while IN MOTION Static friction is friction while NOT IN MOTION F N f

40 Figure 5.4 The motion of the skier and friction are parallel to the slope and so it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N (the normal force) is perpendicular to the slope, and f (the friction) is parallel to the slope, but w (the skier s weight) has components along both axes, namely w and W //. N is equal in magnitude to w, so there is no motion perpendicular to the slope. However, f is less than W // in magnitude, so there is acceleration down the slope (along the x-axis).

41

42 MAX s

43 Ch6: Circles & Gravitation Rotation & Gravitation Ch5 Kepler s Laws T K R 3 GM s

44 Ch6 Rotation & Gravitation F c ma c F g G m m 1 r 2 2 F c mv r 2

45 Ch6 Specifics Circular Motion & Gravitation Radian : Angle subtending an arc of length r. 2 radians = 360 o l r

46 Figure 6.6 As an object moves in a circle, here a fly on the edge of an old-fashioned vinyl record, its instantaneous velocity is always tangent to the circle. The direction of the angular velocity is clockwise in this case.

47 Ch6 Specifics Circular Motion & Gravitation Angular velocity: t

48 Ch6 Specifics Circular Motion & Gravitation Angular acceleration: t

49 Ch6 Specifics Circular Motion & Gravitation Linear to angular equations v f v i at f i t

50 Ch6 Specifics Circular Motion & Gravitation Linear to angular equations 2ad 2 v v f i f 2 i

51 Ch6 Specifics Circular Motion & Gravitation Centripetal Acceleration v 2 a c r

52

53 Ch6 Specifics Circular Motion & Gravitation Centripetal Force v 2 F m c r

54 Ch6 Specifics Kepler v. Newton Kepler s 3 rd Law K T R 2 3 Orbital Period Orbital Radius

55 Kepler v. Newton F Newton s 4 th Law g F c G M M R Ch6 Specifics K T R M 1 4 GM s v r 2 2

56

57 Figure 6.13 The car on this banked curve is moving away and turning to the left.

58 Figure 6.36 A bicyclist negotiating a turn on level ground must lean at the correct angle the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle θ, the speed v, and the radius of curvature r of the turn similar to that for the ideal banking of roadways.

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60

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64

65 Ch7: Work / Energy Theoremy W Fd cos W Fd ( mg) h PE W Fd mad 1 mv 2 KE 2

66 Work & Energy Springs & Pendulae Power 1 kx 2 Fv PEs t 2 P W t mgh E t

67 Work Ch7 Specifics F and d MUST be in same direction DOT Product

68 Ch7 Specifics Work & Energy Essentially the same thing Work CAUSES a CHANGE in E Both are SCALARS W E

69 Ch7 Specifics Work Energy Theorem Work & Energy must be CONSERVED within a closed system W E W E i i f f

70 Figure 7.8 The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the roller-coaster-earth system s gravitational potential energy is converted to kinetic energy. If work done by friction is negligible, all ΔPE g is converted to KE.

71 Figure 7.9 A marble rolls down a ruler, and its speed on the level surface is measured.

72 Figure 7.10 (a) An undeformed spring has no PEs stored in it. (b) The force needed to stretch (or compress) the spring a distance x has a magnitude F = kx, and the work done to stretch (or compress) it is 1 2 kx2 (c) Because the force is conservative, this work is stored as potential energy (PEs) in the spring, and it can be fully recovered. (d) A graph of F vs. x has a slope of k, and the area under the graph is 1 2 kx2. Thus the work done or potential energy stored is 1 2 kx2.

73

74

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76

77 Ch8: Linear Momentum Momentum Impulse = P Conservation of P is Consequence of NL3 F ma v F m Impulse t Change in P Ft mv

78 Ch8 Specifics Momentum Product of mass and velocity Is a vector MV

79 Ch8 Specifics Momentum is Conserved Immediately before and immediately after collisions. P i P f

80 Ch8 Specifics Two types of collisions Elastic: BOTH P and KE are conserved General hint: Start and end with same number of unhampered objects Inelastic: ONLY P is conserved General Hint: Start with 2 objects end with 1.

81 Figure 8.6 An elastic one-dimensional twoobject collision. Momentum and internal kinetic energy are conserved.

82 Figure 8.8 An inelastic one-dimensional two-object collision. Momentum is conserved, but internal kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward one another at the same speed. (b) The objects stick together (a perfectly inelastic collision), and so their final velocity is zero. The internal kinetic energy of the system changes in any inelastic collision and is reduced to zero in this example.

83 General List of Stuff:

84 Figure 8.11 A two-dimensional collision with the coordinate system chosen so that m 2 is initially at rest and v 1 is parallel to the x -axis. This coordinate system is sometimes called the laboratory coordinate system, because many scattering experiments have a target that is stationary in the laboratory, while particles are scattered from it to determine the particles that make-up the target and how they are bound together. The particles may not be observed directly, but their initial and final velocities are.

85 Ch9&10 Specifics Torque Rotational analog of Force FxR I

86 Torque Ch9&10 Specifics F and R MUST be perpendicular

87 This pencil is in the condition of equilibrium. The net force on the pencil is zero and the total torque about any pivot is zero. Figure 9.11

88 Figure 9.12 If the pencil is displaced slightly to the side (counterclockwise), it is no longer in equilibrium. Its weight produces a clockwise torque that returns the pencil to its equilibrium position.

89 If the pencil is displaced too far, the torque caused by its weight changes direction to counterclockwise and causes the displacement to increase. Figure 9.13

90 Ch10: L Angular Momentum (L) P=MV, L=I MUST be conserved for any isolated system Planets, satellites, spinning tops

91 Centripetal acceleration a c occurs as the direction of velocity changes; it is perpendicular to the circular motion. Centripetal and tangential acceleration are thus perpendicular to each other. Figure 10.5

92 Figure Some rotational inertias.

93 Figure The net force on this disk is kept perpendicular to its radius as the force causes the disk to rotate. The net work done is thus (net F)Δs. The net work goes into rotational kinetic energy.

94 Figure A large grindstone is given a spin by a person grasping its outer edge.

95 Figure (a) (b) An ice skater is spinning on the tip of her skate with her arms extended. Her angular momentum is conserved because the net torque on her is negligibly small. In the next image, her rate of spin increases greatly when she pulls in her arms, decreasing her moment of inertia. The work she does to pull in her arms results in an increase in rotational kinetic energy.

96 Figure Figure (a) shows a disk is rotating counterclockwise when viewed from above. Figure (b) shows the right-hand rule. The direction of angular velocity ω size andangular momentum L are defined to be the direction in which the thumb of your right hand points when you curl your fingers in the direction of the disk s rotation as shown.

97 In figure (a), the torque is perpendicular to the plane formed by r and F and is the direction your right thumb would point to if you curled your fingers in the direction of F. Figure (b) shows that the direction of the torque is the same as that of the angular momentum it produces. Figure 10.29

98

99 Ch16 : Waves Hooke s Law F R is restoring force in spring k is spring constant x is displacement stretched or compressed F kx R

100 Ch 16 : Waves Pendulums (Pendulae?) KE MAx = PE MAX Diagram p445 T 2 L g

101 Ch16 : Waves Amplitude Period Frequency Mass on spring T f 2 1 T m k

102 Ch 16 : Waves Types of Waves Transverse Longitudinal

103 Wave speed Ch 16 : Waves v d t v f T

104 Ch16 : Waves Interference & Transmission

105 Ch16 : Waves Interference & Transmission

106 Reflection Ch16 : Waves

107 Production Ch17 : Sound Vibrating object and a medium

108 Characteristics Ch17 : Sound Longitudinal & NEEDS a physical medium Human: 20 20KHz Travel in all 3-Directions Doppler Effect

109 Ch17 : Sound Doppler Stationary Source Moving Source

110 Ch17 : Sound Sound Intensity P I A P I 4 r 2

111 Ch17 : Sound Relative Intensity (Decibels!) 10log I I o 90dB 10 x 80dB 100 x 70dB 10 9 = 10 x 10 8 = 100 x 10 7

112 Harmonics Ch17 : Sound Fundamental Frequency (1 st Harmonic) f 2 =2f 1... Harmonic Series on string v f n n 2L

113 L=½ L=¼ Ch17 : Sound Open & Closed tubes (Instruments) Open Closed

= o + t = ot + ½ t 2 = o + 2

= o + t = ot + ½ t 2 = o + 2 Chapters 8-9 Rotational Kinematics and Dynamics Rotational motion Rotational motion refers to the motion of an object or system that spins about an axis. The axis of rotation is the line about which the

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w = mg F g = τ = Frsinθ a c v f + at = v i F = ma F f p = mv W = Fd v f 2 = v i 2 + 2ad F t = Δ(mv) Impulse = F t KE = 1 2 mv 2 PE = mgh PE = 1 2 kx 2 Wissahickon Physics Midterm x = x i + vt + 1 2 at 2 v f 2 = v i 2 + 2ad Name v f = v i + at Gm 1 m 2 w = mg F g = r 2 g = 9.8m/s 2 G = 6.67 10 11 (N m 2 /k g 2 ) F = ma F f = µn τ = Frsinθ a c = v 2 p

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