# SPH4UI: Lecture 1 “Kinematics” Course Info & Advice Course has several components: Course has several components: Lecture: (me talking, demos and you.

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SPH4UI: Lecture 1 “Kinematics”

Course Info & Advice Course has several components: Course has several components: Lecture: (me talking, demos and you asking questions) Lecture: (me talking, demos and you asking questions) Discussion sections (tutorials, problem solving, quizzes) Discussion sections (tutorials, problem solving, quizzes) Homework Web based Homework Web based Labs: (group exploration of physical phenomena) Labs: (group exploration of physical phenomena) What happens if you miss a lab or class test What happens if you miss a lab or class test Read notes from online Read notes from online What if you are excused?? (What you need to do.) What if you are excused?? (What you need to do.) That topic of your exam will be evaluated as your test That topic of your exam will be evaluated as your test The first few weeks of the course should be review, hence the pace is fast. It is important for you to keep up! The first few weeks of the course should be review, hence the pace is fast. It is important for you to keep up! Then, watch out…. Then, watch out….

How we measure things! How we measure things! All things in classical mechanics can be expressed in terms of the fundamental units: All things in classical mechanics can be expressed in terms of the fundamental units: Length: L Length: L Mass : M Mass : M Time :T Time :T For example: For example: Speed has units of L / T (e.g. miles per hour). Speed has units of L / T (e.g. miles per hour). Force has units of ML / T 2 etc... (as you will learn). Force has units of ML / T 2 etc... (as you will learn). Fundamental Units

Units... SI (Système International) Units: SI (Système International) Units: mks: L = meters (m), M = kilograms (kg), T = seconds (s) mks: L = meters (m), M = kilograms (kg), T = seconds (s) cgs: L = centimeters (cm), M = grams (gm), T = seconds (s) cgs: L = centimeters (cm), M = grams (gm), T = seconds (s) British Units: British Units: Inches, feet, miles, pounds, slugs... 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. 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: Useful Conversion factors: 1 inch= 2.54 cm 1 inch= 2.54 cm 1 m = 3.28 ft 1 m = 3.28 ft 1 mile= 5280 ft 1 mile= 5280 ft 1 mile = 1.61 km 1 mile = 1.61 km Example: convert miles per hour to meters per second: Example: convert miles per hour to meters per second:

Dimensional Analysis This is a very important tool to check your work This is a very important tool to check your work It’s also very easy! It’s also very easy! Example: Example: Doing a problem you get the answer distance d = vt 2 (velocity x time 2 ) Units on left side = L Units on right side = L / T x T 2 = L x T Left units and right units don’t match, so answer must be wrong!! 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. 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 ? Which of the following formulas for P could be correct ? (b) (c) Given: d has units of length (L) and g has units of (L / T 2 ). (a) P = 2  (dg) 2

Solution Solution Realize that the left hand side P has units of time (T ) Realize that the left hand side P has units of time (T ) Try the first equation Try the first equation (a)(b)(c) (a) Not Right!

Solution Solution Realize that the left hand side P has units of time (T ) Realize that the left hand side P has units of time (T ) Try the second equation Try the second equation (a)(b)(c) (b) Not Right!

Solution Solution Realize that the left hand side P has units of time (T ) Realize that the left hand side P has units of time (T ) Try the first equation Try the first equation (a)(b)(c) (c) Dude, this is it

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 : 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. 5m 3m 1m Distance: Displacement: 6.7m Start Finish = 500 m m m 005 Distance 0 Displacement = =

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

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

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

Vector Example Vector A = (0,2,1) Vector A = (0,2,1) Vector B = (3,0,2) Vector B = (3,0,2) Vector C = (1,-4,2) 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) (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. 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. Equal vectors have the same magnitude and direction. Negative vector of a given vector has the same magnitude and the opposite direction. Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature P Q P+Q P -P

Addition of Vectors Trapezoid rule for vector addition Law of cosines, Law of sines, Vector addition is commutative, Vector subtraction P Q P+Q P Q Triangle rule for vector addition P P+Q Q P Q -Q P-Q

Addition of Vectors Addition of three or more vectors through repeated application of the triangle rule The polygon rule for the addition of three or more vectors. Vector addition is associative, Multiplication of a vector by a scalar increases its length by that factor (if scalar is negative, the direction will also change.) P Q S Q+S P+Q+S P Q S P 2P -1.5P

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 Two forces act on a bolt at A. Determine their resultant. 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. 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 Q P R 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, From the Law of Sines,

Sample Problem a)the tension in each of the ropes for α = 45 o, A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5000 N directed along the axis of the barge, determine 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. Find a trigonometric solution by applying the Triangle Rule for vector addition. With the magnitude and direction of the resultant known and the directions of the other two sides parallel to the ropes given, apply the Law of Sines to find the rope tensions.

Sample Problem Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides. Trigonometric solution - Triangle Rule with Law of Sines T1T1 T2T2 5000N T2T2 T1T1

Rectangular Components of a Force: Unit Vectors Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components. F x and F y are referred to as the scalar components of May resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle. are referred to as rectangular vector components Define perpendicular unit vectors which are parallel to the x and y axes.

Addition of Forces by Summing Components Wish to find the resultant of 3 or more concurrent forces, Resolve each force into rectangular components The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces. To find the resultant magnitude and direction, P Q S R

Sample Problem Four forces act on bolt A as shown. Determine the resultant of the force on the bolt. Plan: Resolve each force into rectangular components. Calculate the magnitude and direction of the resultant. Determine the components of the resultant by adding the corresponding force components.

Sample Problem Solution Resolve each force into rectangular components. Calculate the magnitude and direction. Determine the components of the resultant by adding the corresponding force components.

Equilibrium of a Particle When the resultant of all forces acting on a particle is zero, the particle is in equilibrium. 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 -algebraic solution 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.

Free-Body Diagrams Space Diagram: A sketch showing the physical conditions of the problem. Free-Body Diagram: A sketch showing only the forces on the selected particle. T AB T AC 736N A

Sample Problem 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? Plan of Attack: Construct a free-body diagram for the particle at the junction of the rope and cable. 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. A T AB 3500lb T AC horizontal Vertical

A T AB 3500lb Sample Problem

Solve for the unknown force magnitudes using Sine Law. 3500lb T AB T AC Sometimes the Sine Law / Cosine Law is faster than component vectors. Intuition should tell you which is best. Sample Problem

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. PLAN OF ATTACK: Choosing the hull as the free body, draw a free-body diagram. 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. Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero. T AB =40 T AC FDFD T AE =60 A

Sample Problem Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions. T AB T AC FDFD T AE =60 A

Sample Problem This equation is satisfied only if all vectors when combined, complete a closed loop.

Rectangular Components in Space The vector is contained in the plane OBAC. Resolve into horizontal and vertical components. Resolve into rectangular components

Rectangular Components in Space With the angles between and the axes, is a unit vector along the line of action of and are the direction cosines for

Rectangular Components in Space Direction of the force is defined by the location of two points, d is the length of the vector F

Sample Problem The tension in the guy wire is 2500 N. Determine: a) components F x, F y, F z of the force acting on the bolt at A, b) the angles  x,  y,  z defining the direction of the force PLAN of ATTACK: 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. Determine the components of the force.

Sample Problem Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

Motion in 1 dimension In 1-D, we usually write position as x(t). In 1-D, we usually write position as x(t). Since it’s in 1-D, all we need to indicate direction is + or . Since it’s in 1-D, all we need to indicate direction is + or .  Displacement in a time  t = t 2 - t 1 is  x = x(t 2 ) - x(t 1 ) = x 2 - x 1 t x t1t1 t2t2  x  t x1x1 x2x2 some particle’s trajectory in 1-D

1-D kinematics Velocity v is the “rate of change of position” Velocity v is the “rate of change of position” Average velocity v av in the time  t = t 2 - t 1 is: Average velocity v av in the time  t = t 2 - t 1 is: t x t1t1 t2t2  x x1x1 x2x2 trajectory  t V av = slope of line connecting x 1 and x 2.

1-D kinematics... Consider limit t 1 t 2 Consider limit t 1 t 2 Instantaneous velocity v is defined as: Instantaneous velocity v is defined as: so v(t 2 ) = slope of line tangent to path at t 2. t x t1t1 t2t2  x x1x1 x2x2  t

1-D kinematics... Acceleration a is the “rate of change of velocity” Acceleration a is the “rate of change of velocity” Average acceleration a av in the time  t = t 2 - t 1 is: Average acceleration a av in the time  t = t 2 - t 1 is: l And instantaneous acceleration a is defined as: using Calculus way of saying gets very very small

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! 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 a v t t t Calculus (don’t worry you will understand this in next year.)

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

Recap So for constant acceleration we find: So for constant acceleration we find: x a v t 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? 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 x a v t t t l Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. l Since the velocity is continually changing there must continually changing there must be some acceleration. be some acceleration.  In fact the acceleration is caused by gravity (g = 9.81 m/s 2 ).  (more on gravity in a few lectures) l The answer is (c) v = 0, but a  0.

Recap: l For constant acceleration: l From which we know: This is just for constant acceleration!

Recap: l For constant acceleration:

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

Gravity facts: g does not depend on the nature of the material! g does not depend on the nature of the material! Galileo (1564-1642) figured this out without fancy clocks & rulers! Galileo (1564-1642) figured this out without fancy clocks & rulers! On the surface of the earth, gravity acts to give a constant acceleration On the surface of the earth, gravity acts to give a constant acceleration demo - feather & penny in vacuum demo - feather & penny in vacuum Nominally, g = 9.81 m/s 2 Nominally, g = 9.81 m/s 2 At the equatorg = 9.78 m/s 2 At the equatorg = 9.78 m/s 2 At the North poleg = 9.83 m/s 2 At the North poleg = 9.83 m/s 2 More on gravity in a few lectures! More on gravity in a few lectures! Penny & feather Penny & feather

Gravity facts: Actually, gravity is a “fundamental force”. Actually, gravity is a “fundamental force”. Other fundamental forces: electric force, strong and weak forces 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 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: Gravitational Force is proportional to product of masses: F(1 acting on 2) proportional to M 1 times M 2 F(1 acting on 2) proportional to M 1 times M 2 F(2 acting on 1) proportional to M 1 times M 2 too! F(2 acting on 1) proportional to M 1 times M 2 too! Proportional to 1/r 2 Proportional to 1/r 2 r is the separation of the 2 masses r is the separation of the 2 masses For gravity on surface of earth, r = radius of earth For gravity on surface of earth, r = radius of earth Example of Gauss’s Law (more on this later) 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/s 2 contains the effect of M earth and r earth.

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) 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. First choose coordinate system. Origin and y-direction. Origin and y-direction. Next write down position equation: Next write down position equation: Realize that v 0y = 0. Realize that v 0y = 0. 1000 m y = 0 y

Solution: Solve for time t when y = 0 given that y 0 = 1000 m. Solve for time t when y = 0 given that y 0 = 1000 m. Recall: Recall: Solve for v y : Solve for v y : y 0 = 1000 m y y = 0

1D Free Fall Alice and Bob are standing at the top of a cliff of height H. Both throw a ball with initial speed v 0, Alice straight down and Bob straight up. The speed of the balls when they hit the ground are v A and v B respectively. Which of the following is true: Alice and Bob are standing at the top of a cliff of height H. Both throw a ball with initial speed v 0, Alice straight down and Bob straight up. The speed of the balls when they hit the ground are v A and v B respectively. Which of the following is true: (a) v A v B v0v0v0v0 v0v0v0v0 BobAlice H vAvAvAvA vBvBvBvB

1D Free fall Since the motion up and back down is symmetric, intuition should tell you that v = v 0 Since the motion up and back down is symmetric, intuition should tell you that v = v 0 We can prove that your intuition is correct: We can prove that your intuition is correct: v0v0v0v0 Bob H v = v 0 Equation: This looks just like Bill threw the ball down with speed v 0, 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? For example, does motion in the y-direction affect motion in the x-direction? It depends…. It depends…. For simple forces, like gravitational and electric forces, NO For simple forces, like gravitational and electric forces, NO For more complicated forces/situations, like magnetism, YES 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 In any case, vectors are the mathematical objects that we need to use to describe the motion Vectors have Vectors have Magnitude Magnitude Units (like meters, Newtons, Volts/meter, meter/sec 2 …) Units (like meters, Newtons, Volts/meter, meter/sec 2 …) Direction Direction 2 ball drop

Vectors: In 1 dimension, we could specify direction with a + or - sign. For example, in the previous problem a y = -g etc. In 1 dimension, we could specify direction with a + or - sign. For example, in the previous problem a y = -g etc. In 2 or 3 dimensions, we need more than a sign to specify the direction of something: 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. To illustrate this, consider the position vector r in 2 dimensions. Example Example: Where is Waterloo? Toronto  Choose origin at Toronto  Choose coordinates of distance (km), and direction (N,S,E,W) r  In this case r is a vector that points 120 km north. Waterloo Toronto r 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: 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 y axis to be along direction of acceleration Choose x axis to be along the “other” direction of motion Choose x axis to be along the “other” direction of motion Example: Throwing a baseball (neglecting air resistance) Example: Throwing a baseball (neglecting air resistance) Acceleration is constant (gravity) Acceleration is constant (gravity) Choose y axis up: a y = -g Choose y axis up: a y = -g Choose x axis along the ground in the direction of the throw Choose x axis along the ground in the direction of the throw

“x” and “y” components of motion are independent. l 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 (y o ) above the plate, and its initial velocity is 36.5 m/s (v ) at an angle of 30 o (  ) above horizontal. The center-field wall is 113 m (D) from the plate and is 3 m (h) high. David Eckstein clobbers a fastball toward center-field. The ball is hit 1 m (y o ) above the plate, and its initial velocity is 36.5 m/s (v ) at an angle of 30 o (  ) 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? What time does the ball reach the fence? Does David get a home run? Does David get a home run?  v h D y0y0

Problem... Choose y axis up. Choose y axis up. Choose x axis along the ground in the direction of the hit. Choose x axis along the ground in the direction of the hit. Choose the origin (0,0) to be at the plate. Choose the origin (0,0) to be at the plate. Say that the ball is hit at t = 0, x(0) = x 0 = 0. y(0) = y 0 = 1m Say that the ball is hit at t = 0, x(0) = x 0 = 0. y(0) = y 0 = 1m Equations of motion are: Equations of motion are: v x = v 0x v y = v 0y - gt x = v x t y = y 0 + v 0y t - 1 / 2 gt 2

Problem... Use geometry to figure out v 0x and v 0y : Use geometry to figure out v 0x and v 0y : y x g  v v 0x v 0y Find v 0x = |v| cos . and v 0y = |v| sin . y0y0 remember, we were told that  = 30 deg

Problem... The time to reach the wall is: t = D / v x (easy!) The time to reach the wall is: t = D / v x (easy!) We have an equation that tell us y(t) = y 0 + v 0y t + a t 2 / 2 We have an equation that tell us y(t) = y 0 + v 0y t + a t 2 / 2 So, we’re done....now we just plug in the numbers: a = -g So, we’re done....now we just plug in the numbers: a = -g Find: Find: v x = 36.5 cos(30) m/s = 31.6 m/s v x = 36.5 cos(30) m/s = 31.6 m/s v y = 36.5 sin(30) m/s = 18.25 m/s v y = 36.5 sin(30) m/s = 18.25 m/s t = (113 m) / (31.6 m/s) = 3.58 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/s 2 )(3.58 s) 2 y(t) = (1.0 m) + (18.25 m/s)(3.58 s) - (0.5)(9.8 m/s 2 )(3.58 s) 2 = (1.0 + 65.3 - 62.8) m = 3.5 m = (1.0 + 65.3 - 62.8) m = 3.5 m Since the wall is 3 m high, Eckstein gets the homer!! 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 y 0 = 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 30 o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D 1 from the thrower, how far away from the thrower D 2 will the receiver of ball 2 be when he catches it? Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30 o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D 1 from the thrower, how far away from the thrower D 2 will the receiver of ball 2 be when he catches it? Assume the receiver and QB are the same height Assume the receiver and QB are the same height (a) D 2 = 2D 1 (b) D 2 = 4D 1 (c) D 2 = 8D 1

Solution l The distance a ball will go is simply x = (horizontal speed) x (time in air) = v 0x t l To figure out “time in air”, consider the equation for the height of the ball: l When the ball is caught, y = y 0 (time of throw) (time of catch) two solutions

Solution l So the time spent in the air is proportional to v 0y : l Since the angles are the same, both v 0y and v 0x for ball 2 are twice those of ball 1. ball 1 ball 2 v 0y,1 v 0x,1 v 0y,2 v 0x,2 v 0,1 v 0,2 4 l 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: Travel Time: Height Maximum: Time to top:

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

Example A golfer hits a golf ball so that it leaves the club with an initial speed v 0 =37.0 m/s at an initial angle of  0 =53.1 o. 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) The x-distance: The y-distance:

Example A golfer hits a golf ball so that it leaves the club with an initial speed v 0 =37.0 m/s at an initial angle of  0 =53.1 o. 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.

Shooting the Monkey (tranquilizer gun) l 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... l If there were no gravity, simply aim at the monkey r = r 0 r =v 0 t

Shooting the Monkey... rvg r = v 0 t - 1 / 2 g t 2 l With gravity, still aim at the monkey! rg r = r 0 - 1 / 2 g t 2 Dart hits the monkey!

Recap: Shooting the monkey... x = x 0 x = x 0 yg y = - 1 / 2 g t 2 l This may be easier to think about. It’s exactly the same idea!! They both have the same V y (t) in this case x = v 0 t x = v 0 t yg y = - 1 / 2 g t 2

Feeding the Monkey

Kinematics Flash Review

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