Ch3_malleys

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=Vectors - Motions and Forces in Two Dimensions=

Vectors - Fundamentals and Operations
toc A vector quantity is a quantity that is fully described by both magnitude and direction. On the other hand, a scalar quantity is a quantity that is fully described by its magnitude. Examples of vector quantities that have been [|previously discussed] include [|displacement], [|velocity], [|acceleration], and [|force]. Vector quantities are not fully described unless both magnitude and direction are listed. Vector quantities are often represented by scaled [|vector diagrams]. Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Such diagrams are commonly called as [|free-body diagrams]. An example of a scaled vector diagram is shown in the diagram at the right. The vector diagram depicts a displacement vector. Observe that there are several characteristics of this diagram that make it an appropriately drawn vector diagram. **Conventions for Describing Directions of Vectors** Vectors can be directed due East, due West, due South, and due North. But some vectors are directed //northeast// (at a 45 degree angle); and some vectors are even directed //northeast//, yet more north than east. There are a variety of conventions for describing the direction of any vector. The two conventions that will be discussed and used in this unit are described below: >
 * a scale is clearly listed
 * a vector arrow (with arrowhead) is drawn in a specified direction. The vector arrow has a //head// and a //tail//.
 * the magnitude and direction of the vector is clearly labeled. In this case, the diagram shows the magnitude is 20 m and the direction is (30 degrees West of North).
 * 1) The direction of a vector is often expressed as an angle of rotation of the vector about its "[|tail]" from east, west, north, or south.
 * 2) The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "[|tail]" from due East. Using this convention, a vector with a direction of 30 degrees is a vector that has been rotated 30 degrees in a counterclockwise direction relative to due east.

**Representing the Magnitude of a Vector** The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale.



Vectors can be represented by use of a scaled vector diagram. On such a diagram, a vector arrow is drawn to represent the vector. The arrow has an obvious tail and arrowhead. The magnitude of a vector is represented by the length of the arrow. A scale is indicated and the arrow is drawn the proper length according to the chosen scale. The arrow points in the precise direction. Directions are described by the use of some convention. The most common convention is that the direction of a vector is the counterclockwise angle of rotation which that vector makes with respect to due East.

Vector Addition
Two vectors can be added together to determine the result (or resultant). The //net force// experienced by an object was determined by computing the vector sum of all the individual forces acting upon that object. That is the [|net force] was the result (or [|resultant]) of adding up all the force vectors. There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors. The two methods that will be discussed in this lesson and used throughout the entire unit are:
 * the Pythagorean theorem and trigonometric methods
 * [|the head-to-tail method using a scaled vector diagram]

**The Pythagorean Theorem**   **Using Trigonometry to Determine a Vector's Direction** The direction of a //resultant// vector can often be determined by use of trigonometric functions. The **sine function** relates the measure of an acute angle to the ratio of the length of the side opposite the angle to the length of the hypotenuse. The **cosine function** relates the measure of an acute angle to the ratio of the length of the side adjacent the angle to the length of the hypotenuse. The **tangent function** relates the measure of an angle to the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.

The procedure is restricted to the addition of __two vectors that make right angles to each other__. When the two vectors that are to be added do not make right angles to one another, or when there are more than two vectors to add together, we will employ a method known as the head-to-tail vector addition method. **Use of Scaled Vector Diagrams to Determine a Resultant** The magnitude and direction of the sum of two or more vectors can also be determined by use of an accurately drawn scaled vector diagram. Using a scaled diagram, the **head-to-tail method** is employed to determine the vector sum or resultant. A common Physics lab involves a //vector walk//. Either using centimeter-sized displacements upon a map or meter-sized displacements in a large open area, a student makes several consecutive displacements beginning from a designated starting position. The head-to-tail method involves [|drawing a vector to scale] on a sheet of paper beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins (thus, //head-to-tail// method). The process is repeated for all vectors that are being added. Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to //real// units using the given scale. The [|direction] of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East. A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.
 * 1) Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
 * 2) Pick a starting location and draw the first vector //to scale// in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
 * 3) Starting from where the head of the first vector ends, draw the second vector //to scale// in the indicated direction. Label the magnitude and direction of this vector on the diagram.
 * 4) Repeat steps 2 and 3 for all vectors that are to be added
 * 5) Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as **Resultant** or simply **R**.
 * 6) Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
 * 7) Measure the direction of the resultant using the counterclockwise convention discussed [|earlier in this lesson].

Vector Resolution
Any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components). The process of determining the magnitude of a vector is known as **vector resolution**. **Parallelogram Method of Vector Resolution** The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. A step-by-step procedure for using the parallelogram method of vector resolution is: **Trigonometric Method of Vector Resolution** The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. The method of employing trigonometric functions to determine the components of a vector are as follows:
 * 1) Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) Sketch a parallelogram around the vector: beginning at the [|tail] of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the [|head] of the vector; the sketched lines will meet to form a rectangle (a special case of a parallelogram).
 * 3) Draw the components of the vector. The components are the //sides// of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward velocity component might be labeled vx; etc.
 * 5) Measure the length of the sides of the parallelogram and [|use the scale to determine the magnitude] of the components in //real// units. Label the magnitude on the diagram.
 * 1) Construct a //rough// sketch (no scale needed) of the vector in the indicated direction. Label its magnitude and the angle that it makes with the horizontal.
 * 2) Draw a rectangle about the vector such that the vector is the diagonal of the rectangle. Beginning at the [|tail] of the vector, sketch vertical and horizontal lines. Then sketch horizontal and vertical lines at the [|head] of the vector. The sketched lines will meet to form a rectangle.
 * 3) Draw the components of the vector. The components are the //sides// of the rectangle. The tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward force velocity component might be labeled vx; etc.
 * 5) To determine the length of the side opposite the indicated angle, use the sine function. Substitute the magnitude of the vector for the length of the hypotenuse. Use some algebra to solve the equation for the length of the side opposite the indicated angle.
 * 6) Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

Relative Velocity and Riverboat Problems
On occasion objects move within a medium that is moving with respect to an observer. In such instances as this, the magnitude of the velocity of the moving object with respect to the observer on land will not be the same as the speedometer reading of the vehicle. Motion is relative to the observer. The observer on land, often named (or misnamed) the "stationary observer" would measure the speed to be different than that of the person on the boat. The observed speed of the boat must always be described relative to who the observer is.  **Analysis of a Riverboat's Motion** If a motorboat were to head straight across a river (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s. The resultant velocity of the motorboat can be determined in the same manner as was done for the plane. The resultant velocity of the boat is the vector sum of the boat velocity and the river velocity. Since the boat heads straight across the river and since the current is always directed straight downstream, the two vectors are at right angles to each other. Thus, the [|Pythagorean theorem] can be used to determine the resultant velocity.

Motorboat problems such as these are typically accompanied by three separate questions: The second and third of these questions can be answered using the [|average speed equation] (and a lot of logic). **ave. speed = distance/time**
 * 1) What is the resultant velocity (both magnitude and direction) of the boat?
 * 2) If the width of the river is //X// meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore?

Independence of Perpendicular Components of Motion
Any vector - whether it is a force vector, displacement vector, velocity vector, etc. - directed at an angle can be thought of as being composed of two perpendicular components. These two components can be represented as legs of a right triangle formed by projecting the vector onto the x- and y-axis. The two perpendicular parts or components of a vector are independent of each other. While the change in one of the components will alter the magnitude of the resulting force, it does not alter the magnitude of the other component. Perpendicular components of motion do not affect each other. All vectors can be thought of as having perpendicular components. In fact, any motion that is at an angle to the horizontal or the vertical can be thought of as having two perpendicular motions occurring simultaneously. These perpendicular components of motion occur independently of each other. Any component of motion occurring strictly in the horizontal direction will have no affect upon the motion in the vertical direction. Any alteration in one set of these components will have no affect on the other set. 

=Projectile Motion=

What is a Projectile?
What is a projectile? A projectile is an object upon which the only acting force is gravity. There are many examples of projectiles - an object in free-fall, an object thrown upward, an object thrown vertically upward. It's any object that, once projected, continues in motion by its own inertia.

Is a force required to keep an object in motion? NO! A force is only required to maintain an acceleration. When a projectile is moving upward, there is a downward force and a downward acceleration - it's moving upward and slowing down.

What is a free-body diagram? It is a diagram of a projectile that would show a single force acting downwards, labeled force of gravity. Regardless of whether it is moving upwards, downwards, rightwards, etc, the free-body diagram still looks the same.

What affect does gravity have on projectiles? Gravity is the downward force upon a projectile that influences its vertical motion and causes the parabolic trajectory that is characteristic of projectiles. It causes a vertical acceleration.

What happens without horizontal forces? A projectile remains in motion with a constant horizontal velocity. They are NOT require to keep a projectile moving horizontally!

Characteristics of a Projectile's Trajectory
What is the acceleration of a projectile? The horizontal motion has no acceleration. The vertical motion has a downward acceleration of 9.8 m/s/s (the acceleration of gravity).

Why do projectiles travel with a parabolic trajectory? When an object is thrown up in the air, it must eventually come down. The downward force of gravity would act on the projectile to cause the same vertical motion as before (a downward acceleration). The parabolic trajectory, therefore, is a direct result of gravity's influence.

What are the two components of a projectile's motion? Horizontal and vertical motion are the two components. Because they are perpendicular, they move independently of each other, meaning they must each be discussed separately.

What is the velocity of a projectile? As far as horizontal motion is concerned, the velocity is constant the whole time. As far as vertical motion is concerned, the velocity is changing at a rate of 9.8 m/s every second.

What if there was no gravity? If this was the case, the projectile would travel in a straight line. An object in motion would continue in motion at a constant speed in the same direction, as per Newton's law. This is the case in outer space, where there is no gravity.

Describing Projectiles with Numbers: Horizontal and Vertical Velocity/Horizontal/Vertical Displacement
How will the vertical and horizontal velocity values change over time if the projectile is launched at an upward angle? The horizontal velocity will still remain constant during the course of the trajectory; likewise, the vertical velocity will still change by 9.8 m/s.

What is the vertical displacement of an object affected by? What is the formula to find it? The vertical displacement of a projectile is dependent only on the acceleration of gravity, not the horizontal velocity. The equation to find the vertical displacement would be: y = 0.5g*t^2. (This is the equation for vertical displacement of a horizontally launched projectile.)

What affects the horizontal displacement of a projectile? What is the equation to find it? It is only affected by the speed at which the projectile moves horizontally and the time that it has been moving - it is not affected by gravity. The formula, therefore, is x=Vix*t, where x = displacement, Vix = the speed at which the object moves horizontally, and t = time, of course.

How do you find the vertical displacement of an angled-launched projectile? You can do so by using the equation y = Viy*y+0.5*g*t^2. Viy would equal the initial velocity and g would equal -9.8 m/s/s (the acceleration of gravity).

What causes vertical acceleration? A vertical force causes acceleration (in the case of projectiles, the acceleration would be 9.8 m/s/s). A horizontal force does NOT cause vertical acceleration; indeed, there is no horizontal force acting upon a projectile at all.

=Vector Activity= Graphical

Analytical

Percent Error
 * Conclusion:** Our percent error shows that we weren't very far off in our calculations, especially when it came to the length. The angle had a much larger percent error, relatively speaking. The percent error can probably be attributed to the fact that we likely weren't careful enough in our measurements.

=Ball in Cup Activity= with Danielle Bonnett, Caroline Braunstein, and Andrew Chung
 * Objectives**:
 * Measure the initial velocity of a ball.
 * Apply concepts from two-dimensional kinematics to predict the impact point of a ball in projectile motion.
 * Take into account trial-to-trial variations in the velocity measurement when calculating the impact point.
 * Pre-Lab Questions:**
 * If you were to drop a ball, releasing it from rest, what information would be needed to predict how much time it would take for the ball to hit the floor?
 * The only information that you would need to be given is the distance between the launcher and the floor. Everything else you should already know, or could find out with math.
 * If the ball in question 1 is traveling at a known horizontal velocity when it starts to fall, explain how you would calculate how far it will travel before it hits the ground.
 * We can calculate the horizontal velocity using the equation d=vt, as the horizontal component will move at constant speed.
 * A single Photogate can be used to accurately measure the time interval for an object to break the beam of a Photogate. If you wanted to know the velocity of the object, what additional information would you need?
 * You would need to know the horizontal distance (the range) between the Photogate and the object in question.
 * What data will you need to collect for this lab before you begin?
 * We should find the horizontal distance that the ball travels and the vertical distance between the ground and the launcher. We can use this to find the time and the velocity.
 * How will you analyze your results in terms of precision and/or accuracy?
 * We will use several trials in order to test for precision - by doing so, we can make sure that all of the results are close to each other. To test for accuracy, we will have to go the actual trial to see exactly how accurate it is. We can find the percent error afterwards.

Calculations to find initial velocity/where the ball will hit the ground: Calculations to find where we should put the cup: Percent Error: =Gourd-o-rama Project=
 * Calculations:**
 * Conclusion**: We weren't that far off at the end of the experiment. At the beginning, when we forgot to take into account the height of the cup, our results were much, much different - shown here are the "correct" (relatively speaking, of course) results that we obtained. We should have been more careful in our measurements. Perhaps the launcher was slightly off - we know that it is not always precise. But the biggest factor was probably error in our measurements; again, as with every experiment, we would use a different method of measuring in an ideal world.
 * Pictures:**
 * Calculations:**

=Shoot Your Grade Lab= with Caroline Braunstein, Danielle Bonnett, and Andrew Chung Solving for initial velocity: Solving to find out where to put the cup: Solving to figure out where to place the rings: Calculations for percent error: The cup should have been placed 4.61 meters from the launcher, according to our calculations; however, we never got a chance to test this experimentally. media type="file" key="Movie on 2011-11-09 at 09.13.mov" This video shows us shooting the ball through four hoops.
 * Results**: Our cart moved a total of 1.5 meters in 1.38 seconds. The initial velocity was 2.17 m/s, and its acceleration was -1.57 m/s/s. It weighed a total of 1.41 kg.
 * Conclusion**: If I had to redo our cart, I would definitely make sure the weight was better balanced throughout - there shouldn't have been so much weight on the front. When we sent our cart down the ramp, it was apt to flip over (at least our pumpkin got some sweet air time out of it). It would have been much more useful to make the cart bigger in the back and smaller in the front, and to place the pumpkin towards the back of the cart. I also would have made sure our axles were on properly so that our cart would move straight, as opposed to slightly to one side.
 * Objective**: To launch a ball from the launcher at a given angle and speed setting so that the ball passes consecutively through five rings and lands in a cup on the floor.
 * Our Given Angle**: 20%
 * Materials**: String, tape, rolls of tape, launcher, ball, tape measure, cup, scissors
 * Calculations:**
 * Results**:
 * x-distance (m from the launcher) || theoretical y-distance (m above the ground) || actual/experimental y-distance (m) || percent error ||
 * 1.15 || 1.45 || 1.44 || 0.69% ||
 * 1.75 || 1.46 || 1.45 || 0.69% ||
 * 2.25 || 1.36 || 1.38 || 1.47% ||
 * 2.75 || 1.32 || 1.32 || 0% ||
 * 3.25 || .989 || -- || -- ||


 * Conclusion:** This lab proves that it is, in fact, possible trace the path of a projectile; it is, in other words, a "real-life" application of the things that we have been working on in class. There were many, many, //many// things that could have gone wrong in this lab, especially when it came to measurements. There were so many measurements that we had to keep track of - and they had to be very precise. Going on this, we had to also make sure our math was right, which required checking and double-checking. Our results were surprisingly accurate - we even had 0% error for one of our distances, and our largest percent error was a mere 1.47%. Unfortunately, this level of precision left little time - this is why we couldn't quite finish our experiment, even though we figured out where the fifth hoop and the cup //should// be. However, when it comes to measuring, it is very hard to be 100% accurate all of the time. If I had to redo this lab, I would definitely make sure we held the tape measure more steadily. I also would have tried to find something better than string to hold the rings from the ceiling - even though it was effective, I feel as if something else could have been more effective. The string often slipped from the ceiling, even after being taped down, and the friction from the ceiling, too, caused it to slip. Even though this was not possible, I think it would have been better for our results had we been able to complete the lab in one run, rather then having only fifty minute periods and having other classes use our same materials.