Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Coriolis force wikipedia , lookup
Centrifugal force wikipedia , lookup
N-body problem wikipedia , lookup
Mechanics of planar particle motion wikipedia , lookup
Artificial gravity wikipedia , lookup
Relativistic angular momentum wikipedia , lookup
Fictitious force wikipedia , lookup
Weightlessness wikipedia , lookup
SPH3UW: Lecture 1 “Kinematics” 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. miles per hour). Force has units of ML / T2 etc... (as you will learn). 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) British Units: Inches, feet, miles, pounds, slugs... We will use mostly SI units, but you may run across some problems using British units. You should know where to look to convert back & forth. Ever heard of Google Converting between different systems of units Useful Conversion factors: 1 inch = 2.54 cm 1m = 3.28 ft 1 mile = 5280 ft 1 mile = 1.61 km Example: convert miles per hour to meters per second: mi mi ft 1 m 1 hr m 1 1 5280 0.447 hr hr mi 3.28 ft 3600 s s 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!! Dimensional Analysis The period P of a swinging pendulum depends only on the length of the pendulum d and the acceleration of gravity g. Which of the following formulas for P could be correct ? (a) P = 2 (dg)2 (b) d P 2 g (c) P 2 d g Given: d has units of length (L) and g has units of (L / T 2). Solution Realize that the left hand side P has units of time (T ) Try the first equation 2 (a) (a) L L4 L 2 4 T T T P 2 dg 2 (b) Not Right! P 2 d g (c) P 2 d g Solution Realize that the left hand side P has units of time (T ) Try the second equation (b) (a) L T2 T L T2 P 2 dg 2 Not Right! (b) P 2 d g (c) P 2 d g Solution Realize that the left hand side P has units of time (T ) Try the first equation (c) (a) L T2 T L T2 P 2 dg 2 (b) Dude, this is it P 2 d g (c) P 2 d g Time to Drop an Apple Want to ask myself a question: If I drop an Apple from a certain height, h, what then happens to the time, t, it takes for the apple to fall? The time t, must be proportional to the height to the power of some value. The time t may be proportional to the mass of the Apple to the power of some value. The time t, may be proportional to the acceleration due to gravity, g to some power t Ch m g 1 L T C L M 2 T There is no M on Left side: 0 There is no L on Left side: 0 There is T to power 1 on Left side: 1 2 1 1 Therefore 2 2 Dimensional Analysis to the Rescue 1 2 t Ch m 0 g C h h g 1 2 Time for Apple to Drop t h Dimensional Analysis tells us, that if we increase the height by a factor of 100, then to time will increase by the square root of 100, or by a factor of 10 Let’s verify this by dropping an apple from 3 m then from 1.5 m and compare the times x y e E x y 2 2 2 2 2 1.417 781 551 0.006 2 Since we double the height, the time should be 2 1.4142 longer. The experiment shows: 1.417±0.006 781ms 1.417 551ms Vectors A vector is a quantity that involves both magnitude and direction. 55 km/h [N35E] A downward force of 3 Newtons A scalar is a quantity that does not involve direction. 55 km/h 18 cm long 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 : A Or By the variable being BOLD: A Displacement Displacement is an object’s change in position. Distance is the total length of space traversed by an object. 1m 6.7m 3m Start = 500 m Finish 5m Displacement: 6m 3m 2 2 6.7m Distance: 5m 3m 1m 9m Displacement = 0 m Distance = 500 m Vector Addition A R E B R E C B A D D E D R C A C B A + B + C + D + E = Distance R = Resultant = Displacement Rectangular Components Quadrant II R A B 2 2 R sin y A R -x A opp sin R hyp B adj cos R hyp A opp tan B adj Quadrant I B R cos Quadrant III Quadrant IV -y x 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): where r = |r | rx = x = r cos ry = y = r sin (x,y) y arctan( y / x ) r x Vectors... The magnitude (length) of r is found using the Pythagorean theorem: r y r r x2 y 2 x The length of a vector clearly does not depend on its direction. Vector Example 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) Resultant of Two Forces • 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. • The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs. • Force is a vector quantity. Vectors • Vector: parameters possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations. • Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature P Q • Vector classifications: - Fixed or bound vectors have well defined points of application that cannot be changed without affecting an analysis. - Free vectors may be freely moved in space without changing their effect on an analysis. - Sliding vectors may be applied anywhere along their line of action without affecting an analysis. P -P • Equal vectors have the same magnitude and direction. • Negative vector of a given vector has the same magnitude and the opposite direction. Addition of Vectors • Trapezoid rule for vector addition • Triangle rule for vector addition P • Law of cosines, Q R 2 P 2 Q 2 2PQ cos B Q R PQ P P Q • Law of sines, sin A sin B sin C Q R A • Vector addition is commutative, PQ Q P -Q Q P-Q P • Vector subtraction P Q P Q Addition of Vectors Q S • Addition of three or more vectors through repeated application of the triangle rule P Q S P • The polygon rule for the addition of three or more vectors. • Vector addition is associative, PQ S PQ S P Q S P 2P -1.5P • Multiplication of a vector by a scalar increases its length by that factor (if scalar is negative, the direction will also change.) 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. 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. Sample Problem Solution R Q P R 98 N 35 • 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. Sample Problem Solution • Trigonometric solution From the Law of Cosines, R 2 P 2 Q 2 2 PQ cos B 40N 60N 2 40N 60N cos155 2 2 R 97.73N From the Law of Sines, sin A sin B Q R sin A sin B Q R sin155 A 15.04 20 A 35.04 60N 97.73N Sample Problem • Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge axis and length proportional to 5000 N. A barge is pulled by two • Find a trigonometric solution by tugboats. If the resultant of the applying the Triangle Rule for vector forces exerted by the tugboats addition. With the magnitude and is 5000 N directed along the direction of the resultant known and the axis of the barge, determine directions of the other two sides parallel to the ropes given, apply the Law of a) the tension in each of the Sines to find the rope tensions. o ropes for α = 45 , Sample Problem • Graphical solution Parallelogram Rule with known resultant direction and magnitude, known directions for sides. T1 3700 N T2 2600 N T1 30 5000N 45 • Trigonometric solution Triangle Rule with Law of Sines 30 45 T2 T1 T2 5000 N sin 45 sin 30 sin105 5000N 45 T2 30 105 T1 T1 3700 N T2 2600 N Rectangular Components of a Force: Unit Vectors • May resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle. Fx and Fy are referred to as rectangular vector components • Define perpendicular unit vectors iˆ and ˆj which are parallel to the x and y axes. • Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components. F Fxiˆ Fy ˆj Fx and Fy are referred to as the scalar components of Fx and Fy F Addition of Forces by Summing Components • Wish to find the resultant of 3 or more concurrent forces, P Py j S Sy j Qx i Px i S xi Qy j Q R PQS • Resolve each force into rectangular components Rxi Ry j Pxi Py j Qxi Qy j S xi S y j Px Qx S x i Py Qy S y j Ry j R Rx i • The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces. Ry Py Qy S y Rx Px Qx S x Fx Fy • To find the resultant magnitude and direction, R R R 2 x 2 y tan 1 Ry Rx Sample Problem Plan: Four forces act on bolt A as shown. Determine the resultant of the force on the bolt. • Resolve each force into rectangular components. • Determine the components of the resultant by adding the corresponding force components. • Calculate the magnitude and direction of the resultant. Sample Problem Solution • Resolve each force into rectangular components. force mag x comp y comp F1 150 129.9 75.0 F2 80 27.4 75.2 F3 110 0 110.0 F4 100 96.6 25.9 Rx 199.1 Ry 14.3 • Determine the components of the resultant by adding the corresponding force components. • Calculate the magnitude and direction. R 199.1 14.3 2 2 tan 14.3 N 199.1N R 199.6N 4.1 Equilibrium of a Particle • When the resultant of all forces acting on a particle is zero, the particle is in equilibrium. • Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line. • Particle acted upon by two forces: - equal magnitude - same line of action - opposite sense • Particle acted upon by three or more forces: - graphical solution yields a closed polygon R F 0 - algebraic solution Fx 0 Fy 0 Free-Body Diagrams TAB 50 Space Diagram: A sketch showing the physical conditions of the problem. A TAC 30 736N Free-Body Diagram: A sketch showing only the forces on the selected particle. Sample Problem Plan of Attack: • Construct a free-body diagram for the particle at the junction of the rope and cable. In a ship-unloading operation, a 3500-lb automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope? • Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle. • Apply trigonometric relations to determine the unknown force magnitudes. Sample Problem SOLUTION: • Construct a free-body diagram for the particle at A. • Apply the conditions for equilibrium in the horizontal and vertical directions. TAB horizontal 2 A cos 88 TAB cos 30 TAC 0 TABy Tcar TAC y 0 30 TAC 3500lb TABx TACx 0 Vertical cos 2 T 3500lb sin 30 T 0 AB AC Sample Problem cos 88TAB cos 30TAC 0 cos 2TAB 3500lb sin 30TAC 0 TAB TAC cos 88 TAB cos 30 TAB sin 30 TAC 3500lb cos 2 2 TAC A 30 cos 88 sin 30 TAC 3500lb cos 30 cos 2 0.02016015TAC 141.12105lb 0.979839TAC 141.12105lb TAC 144lb 3500lb Sample Problem • Solve for the unknown force magnitudes using Sine Law. TAB TAC 3500lb sin120 sin 2 sin 58 TAB 3570lb TAC 144lb TAC 58 120 TAB 3500lb 2 Sometimes the Sine Law / Cosine Law is faster than component vectors. Intuition should tell you which is best. Sample Problem PLAN OF ATTACK: • Choosing the hull as the free body, draw a free-body diagram. It is desired to determine the drag force at a given speed on a prototype sailboat hull. A model is placed in a test channel and three cables are used to align its bow on the channel centerline. For a given speed, the tension is 40 lb in cable AB and 60 lb in cable AE. Determine the drag force exerted on the hull and the tension in cable AC. • Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero. • Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions. Sample Problem SOLUTION: • Choosing the hull as the free body, draw a free-body diagram. 7 ft tan 1.75 4 ft 60.25 TAC TAB=40 60.26 69.44 A FD 1.5 ft 0.375 4 ft 20.56 tan • Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero. TAE=60 R TAB TAC TAE FD 0 Sample Problem TAC cos 69.44 TAB 40sin 60.26 TAC TAC sin 69.44 40cos 60.26 A FD • Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions. TAE TAC sin 69.44 40cos 60.26 60 0.936305TAC 19.842597 TAE=60 TAC 42.889 FD TAC cos 69.44 40sin 60.26 0.351188TAC 34.73142 0.351188 42.889 34.73142 19.66 lb Sample Problem This equation is satisfied only if all vectors when combined, complete a closed loop. Rectangular Components in Space • The vector F is contained in the plane OBAC. • Resolve Fh into • Resolve F into horizontal and vertical rectangular components components. Fh F sin y Fy F cos y Fx Fh cos F sin y cos Fy Fh sin F sin y sin Rectangular Components in Space • With the angles between F and the axes, Fx F cos x Fy F cos y Fz F cos z F Fxi Fy j Fz k F cos x i cos y j cos z k F cos xi cos y j cos z k of F • is a unit vector along the line action of F and cos x , cos y , and cos z are the direction cosines for F Rectangular Components in Space Direction of the force is defined by the location of two points, M x1, y1, z1 and N x2 , y2 , z2 d vector joining M and N d xi d y j d z k d x x2 x1 d y y2 y1 d z z2 z1 F F d is the length of the vector F 1 d xi d y j d z k d Fd y Fd x Fd Fx Fy Fz z d d d Sample Problem PLAN of ATTACK: The tension in the guy wire is 2500 N. Determine: a) components Fx, Fy, Fz of the force acting on the bolt at A, b) the angles x, y, z defining the direction of the force • Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B. • Apply the unit vector to determine the components of the force acting on A. • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. Sample Problem SOLUTION: • Determine the unit vector pointing from A towards B. AB 40m i 80m j 30m k AB 40m 80m 30m 2 2 94.3 m 40 80 30 i j k 94.3 94.3 94.3 0.424 i 0.848 j 0.318k • Determine the components of the force. F F 2500 N 0.424 i 0.848 j 0.318k 1060 N i 2120 N j 795 N k 2 Sample Problem • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. cos x i cos y j cos z k 0.424 i 0.848 j 0.318k x 115.1 y 32.0 z 71.5 Motion in 1 dimension In 1-D, we usually write position as x(t). Since it’s in 1-D, all we need to indicate direction is + or . Displacement in a time t = t2 - t1 is x = x(t2) - x(t1) = x2 - x1 x x some particle’s trajectory in 1-D x 2 x 1 t1 t t2 t 1-D kinematics Velocity v is the “rate of change of position” Average velocity vav in the time t = t2 - t1 is: vav x(t2 ) x(t1 ) x t2 t1 t x x trajectory x 2 Vav = slope of line connecting x1 and x2. x 1 t1 t2 t t 1-D kinematics... Consider limit t1 t2 Instantaneous velocity v is defined as: v (t ) dx (t ) dt so v(t2) = slope of line tangent to path at t2. x x x 2 x 1 t1 t2 t t 1-D kinematics... Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is: v(t2 ) v(t1 ) v aav t2 t1 t And instantaneous acceleration a is defined as: dv(t ) d 2 x(t ) a(t ) dt dt 2 Calculus way of saying t gets very very small dx (t ) using v (t ) dt Recap If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time! x x x( t ) dx dt dv d 2x a dt dt 2 v v a Calculus (don’t worry you will understand this in next year.) t t t 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: t2 x(t2 ) x(t1 ) v(t )dt t1 Graphically, this is adding up lots of small rectangles: v(t) + +...+ = displacement t Recap So for constant acceleration we find: x 1 x x0 v 0 t at 2 2 v v 0 at v t a const a t t Motion in One Dimension Question 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) Both v = 0 and a = 0. (b) v 0, but a = 0. (c) v = 0, but a 0. y Solution Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. x 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. v t t a t Recap: This is just for constant acceleration! For constant acceleration: 1 x x0 v 0 t at 2 2 v v 0 at a const From which we know: v 2 v 02 2a(x x0 ) v av 1 (v 0 v) 2 Recap: For constant acceleration: 1 2 x x0 v0t at 2 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 Invent 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. 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). 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” y Acceleration of gravity is “down”: ay g v y = v 0y - gt t v t 1 2 y y0 v0 y t g t 2 a y ay = g t Gravity facts: Penny & feather 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 More on gravity in a few lectures! Gravity facts: Actually, gravity is a “fundamental force”. Other fundamental forces: electric force, strong and weak forces 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: F(1 acting on 2) proportional to M1 times M2 F(2 acting on 1) proportional to M1 times M2 too! Proportional to 1/r2 r is the separation of the 2 masses For gravity on surface of earth, r = radius of earth Example of Gauss’s Law (more on this later) 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. 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) 1000 m Solution: First choose coordinate system. Origin and y-direction. Next write down position equation: y y 0 v 0y t 1 2 gt 2 1000 m Realize that v0y = 0. 1 2 y y0 gt 2 y y=0 Solution: Solve for time t when y = 0 given that y0 = 1000 m. 1 y y0 - gt 2 2 t 2 y0 2 1000m 14.3s 2 g 9.81 m s y0 = 1000 m Recall: vy2 - v02y 2a( y - y0 ) Solve for vy: y v y 2 gy0 140 m / s y=0 1D Free Fall (a) 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: vA < vB (b) vA = vB Alice v0 (c) vA > vB Bob v0 H vA vB 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: 2 2 v v 0 2( g ) H H 0 Equation: Bob v0 H v = v0 This looks just like Bill threw the ball down with speed v0, so the speed at the bottom should be the same as Alice’s ball. y=0 Does motion in one direction affect motion in an orthogonal direction? For example, does motion in the y-direction affect motion in the x-direction? It depends…. For simple forces, like gravitational and electric forces, NO For more complicated forces/situations, like magnetism, YES In any case, vectors are the mathematical objects that we need to use to describe the motion Vectors have Magnitude Units (like meters, Newtons, Volts/meter, meter/sec2…) Direction Recall Vectors: 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 vector r in 2 dimensions. Example: Where is Waterloo? Choose origin at Toronto Choose coordinates of distance (km), and direction (N,S,E,W) In this case r is a vector that points 120 km north. Waterloo r Toronto A vector is a quantity with a magnitude and a direction 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 “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: Reference frame on the moving train. Reference frame on the ground. Problem: David Eckstein 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 y0 h D 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 x = vxt vy = v0y - gt y = y0 + v0y t - 1/ 2 gt2 Problem... Use geometry to figure out v0x and v0y : y g v y0 Find and v0x = |v| cos . v0y = |v| sin . v0y v0x remember, we were told that = 30 deg x 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, Eckstein gets the homer!! Thinking deeper: Can you figure out what angle gives the longest fly ball? To keep things simple, assume y0 = 0, and go from there… 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 Solution 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 1 y y v t g 0 0y equation for the height of the ball: 2 When the ball is caught, y = y0 1 t v0 y g 2 1 v0 y t g t 2 0 2 t2 t 0 two solutions t2 v0 y g t 0 (time of catch) (time of throw) Solution v0 y So the time spent in the air is proportional to v0yt: 2 Since the angles are the same, both v0y and v0x for ball 2 are twice those of ball 1. v0,2 ball 1 v0x ,1 v0,1 g v0y ,2 ball 2 v0y ,1 v0x ,2 Ball 2 is in the air twice as long as ball 1, but it also has twice the horizontal speed, so it will go 4 times as far!! Projectile Motion As you can see, it can become difficult to solve problems that involve motion in both the x and y axis. Lucky for you, people from all over the world have had the same difficulties. Therefore a complete set of equations have been created that will help solve these problems. These are known as ballistic formulas. They assume launch height and landing height are the same. Range Distance: vi2 sin 2 d x g Travel Time: t 2vi sin g Height Maximum: v sin h i 2g 2 Time to top: vi sin t g Motion in 2D Again 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 2vo sin 2 d x g (c) D2 = 8D1 2 vi2 sin 2 4 g vi2 sin 2 d x g 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. a) Determine the position of the ball when t=2.00s. b) Determine when the ball reaches the highest point of its flight and find its height, h, at this point. c) Determine its horizontal range, R. a) v0 x v0 cos 0 37.0m / s cos 53.1 22.2m / s v0 y v0 sin 0 37.0m / s sin 53.1 29.6m / s The x-distance: x v0 xt 22.2m / s 2.00s 44.4m 1 2 1 2 2 y v t gt 29.6 m / s 2.00 s 9.80 m / s 2 .0 0 s 39.6m 0y The y-distance: 2 2 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. a) Determine the position of the ball when t=2.00s. b) Determine when the ball reaches the highest point of its flight and find its height, h, at this point. c) Determine its horizontal range, R. h vi sin 2 2g 37.0m / s sin 53.1 2 9.80m / s 2 2 44.7m vi sin g 37.0m / s sin 53.1 3.02s 9.80m / s 2 t vi2 sin 2 d x g 37.0m / s 134m 2 sin 2 53.1 9.80m / s 2 Shooting the Monkey (tranquilizer gun) Where does the zookeeper aim if he wants to hit the monkey? ( He knows the monkey will let go as soon as he shoots ! ) Shooting the Monkey... If there were no gravity, simply aim at the monkey r =v0t r = r0 Shooting the Monkey... With gravity, still aim at the monkey! r = v0 t - 1/ 2g t2 r = r0 - 1 / 2 g t 2 Dart hits the monkey! 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 Feeding the Monkey Kinematics Flash Review