Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) Last time we looked at position s. time and acceleration s. time graphs. Since the instantaneous elocit is lim t 0 t the (instantaneous) elocit can be found from the slope of the position s. time graph. In a similar wa, since a lim t 0 t the acceleration can be found from the slope of the elocit s. time graph. (B acceleration we refer to instantaneous acceleration.) ssuming that the acceleration is constant, we can find the four (er) useful relations. f i a t This is a rearrangement of the definition of a. No displacement in the equation! 1 ( ) t The displacement equals the aerage elocit times the elapsed time. No acceleration in the equation! f i t i 1 ) a ( t The displacement equals the sum of the displacement due to the initial elocit and the displacement due to the acceleration. No final speed! f a i The difference in the squares of the elocities equals twice the product of the acceleration and the displacement. No time! er common situation with constant acceleration is free fall. If we neglect air resistance, all objects fall with the same acceleration, regardless of the object s mass or state of motion. Surprised? Man people were surprised when Galileo discoered this fact. ideo demonstration: http://www.outube.com/watch?=e43-cfukegs Consider the following. What if heaier objects accelerated faster? (From http://en.wikipedia.org/wiki/galileo%7s_leaning_tower_of_pisa_eperiment): Galileo arried at his hpothesis b a famous thought eperiment outlined in his book On Motion. Imagine two objects, one light and one heaier than the other one, are connected to each other b a string. Drop this sstem of objects from the Lesson 3, page 1
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) top of a tower. If we assume heaier objects do indeed fall faster than lighter ones (and conersel, lighter objects fall slower), the string will soon pull taut as the lighter object retards the fall of the heaier object. But the sstem considered as a whole is heaier than the hea object alone, and therefore should fall faster. This contradiction leads one to conclude the assumption is false. The acceleration due to grait is denoted b g and has the alue g 9.8m/s The direction is downwards, towards the center of the earth. In fact, it is how we define down! To create the equations of ertical motion, we use our original equations and replace with. To create equations for free fall, replace the a with g. original -direction free-fall a t a t gt f i ( ) t t f f i 1 t f i 1 1 ( f i ) t ( f i ) i 1 ) a ( t a i t f i 1 ) a ( t a But in general, motion is not just along a line. How do we deal with that? i Chapter 3 Motion in a Plane f i it f 1 g( t) g Vectors and scalars Vectors hae direction as well as magnitude. The are represented b arrows. The arrow points in the direction of the ector and its length is related to the ector s magnitude. Scalars onl hae magnitude. We write = B if the ectors hae the same magnitude and point in the same direction. Scalars can hae magnitude, algebraic sign, and units. dding scalars is er familiar. You add 10 grams to 15 grams and get 5 grams. You hae $0 and gie $5 to friend and ou hae $15 remaining. Vector addition is different since ectors hae direction as well as magnitude. How do we add ectors? We alread know how to add ectors in one dimension (along the -ais for eample). i Lesson 3, page
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) B +B What happened? The ector B is positioned so that the tail of B is positioned at the head of. The ector sum is drawn from the tail of to the head of B. If is 8 m long and B is 10 m long, the magnitude of +B is 18 m. What if B is reersed? B +B What is the magnitude of +B? Here we see a hint of the problem. Vectors do not add like scalars. How do we add ectors that do not point along the same direction? 1. Draw the first ector in the correct direction and with the appropriate magnitude.. Draw the second ector with the correct direction and magnitude so that its tail is placed at the head of the first ector. 3. If there is a third ector, draw it with the correct direction and magnitude to that its tail is placed at the head of the second ector. 4. When finished with all the ectors, find the ector sum b drawing a ector that starts at the tail of the first ector and ends at the head of the last ector. How do we subtract ectors? Use B = +( B). What is a reasonable definition for B? The negatie of a ector has the same magnitude as the original ector but points in the opposite direction. The idea of ectors is built from the idea of displacement. In the diagram aboe, imagine that ou are in a forest. is our walk to a tree and B is our walk from the first tree to a friend ou see across the forest. +B is our net displacement from our starting point. Lesson 3, page 3
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) nother eample: This procedure is called the graphical addition of ectors. You need to understand this procedure. Howeer, it is too slow and imprecise to be used in soling problems. We add ectors b taking their components. The process is summarized in this figure. We are adding two ectors that are not collinear. We replace each ector with two ectors (called its components). We then add like components together, giing the components of the ector sum. What happens net? C C C Lesson 3, page 4
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) We add C and C to find C. Since the - and -aes are perpendicular, we can find the magnitude of C from the Pthagorean theorem, C C C The direction is normall measured counterclockwise from the +-ais. For this ector in the nd quadrant, first find and then add to 180º. (Wh?) C arctan C How do we find the components of a ector? s an eample suppose has magnitude 0 N and it points at 40º. s the following diagram shows, we are dealing with a right triangle. To find the components we need to use trigonometr. Recall, cos adjacent hpotenuse sin opposite hpotenuse tan opposite adjacent =40 o Using the definitions of cosine and sine, cos sin adjacent hpotenuse cos (0 N)cos 40 opposite hpotenuse sin (0 N)sin 40 15.3N 1.9 N Wh is >? When are the equal? Lesson 3, page 5
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) Suppose we had this picture. What would ou do? =50 o (0 N)sin50 (0 N)cos50 15.3N 1.9 N Usuall we hae cosine associated with -components and sine associated with -components, but not alwas. You hae to look at the diagram. ( er common remark for this semester!) Problem-Soling Strateg: Finding the - and -components of a Vector from its Magnitude and Direction (page 6) 1. Draw a right triangle with the ector as the hpotenuse and the other two sides parallel to the - and -aes.. Determine one of the unknown angles in the triangle. 3. Use trigonometric functions to find the magnitudes of the components. Make sure our calculator is in degree mode to ealuate trigonometric functions of angles in degrees and in radian mode for angles in radians. 4. Determine the correct algebraic sign for each component. Problem-Soling Strateg: Finding the Magnitude and Direction of a Vector from its - and -components (page 6-63) 1. Sketch the ector on a set of - and -aes in the correct quadrant, according to the signs of the components.. Draw a right triangle with the ector as the hpotenuse and the other two sides parallel to the - and -aes. 3. In the right triangle, choose which of the unknown angles ou want to determine. 4. Use the inerse tangent function to find the angle. The lengths of the sides of the triangle represent and. If is opposite the side parallel the side perpendicular to the -ais, then tan = opposite/adjacent = /. If is opposite the side parallel the side perpendicular to the -ais, then tan = opposite/adjacent = /. If our calculator is in degree mode, then the result of the inerse tangent will be in degrees. [In general, the inerse tangent has has two possible alues between 0 and 360º because tan = tan ( + 180º). Howeer, when the inerse tangent is used to find one of the angles in a right triangle, the result can neer be greater than 90º, so the alue the calculator returns is the one ou want. 5. Interpret the angle: specif whether it is the angle below the horizontal, or the angle west of south, or the angle clockwise from the negatie -ais, etc. 6. Use the Pthagorean theorem to find the magnitude of the ector. Lesson 3, page 6
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) Problem-Soling Strateg: dding Vectors Using Components (page 63) 1. Find the - and -components of each ector to be added.. dd the -components (with their algebraic signs) of the ectors to find the -component of the sum. (If the signs are not correct, the sum will not be correct. 3. dd the -components (with their algebraic signs) of the ectors to find the -component of the sum. 4. If necessar, use the - and -components of the sum to find the magnitude and direction of the sum. Een when using the component method to add ectors, the graphical method is an important first step. Graphical addition gies ou a mental picture of what is going on. problem can be made easier to sole with a good choice of aes. Common choices are -ais horizontal and -ais ertical, when the ectors all lie in the ertical plane; -ais east and -ais north, when the ectors lie in a horizontal plane; and -ais parallel to an inclined surface and -ais perpendicular to it. We can use unit ectors to write ectors in a compact wa. We define (read hat ) as the unit ector in the direction and similarl for. The components of can be written as and ˆ and ˆ ˆ ˆ Using unit ectors, the sum of and B can be written as B ˆ ˆ B ˆ B ˆ B ˆ B ˆ Notice that b factoring out the and, the -components are added together and the - components are added together. The unit ector notation can be er helpful but it will not be used in this course. I hae included it since it might be used in PHY054. Lesson 3, page 7
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) We do not deal with ectors, we deal with their components. Here is an algorithm for adding ectors. The diagram is Figure 3.9 on page 63 C B Gien, and B, B B B B Find components* = cos, = sin B = B cos B, B = B sin B C dd like components C = + B, C C = + B B B Return to magnitude and direction format C C C C C tan 1 ** C C C C C C *Be careful with the angles gien. The equations hold for angles measured counterclockwise from the +-ais. **Be careful with tan -1 function on our calculator. If the -component is negatie, add 180 o to the alue found b our calculator. Lesson 3, page 8
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) Now let s use the concept of ectors to etend the kinematical ariables to more dimensions. erage elocit is the displacement oer the time, a r t Instantaneous elocit is r lim t 0 t The elocit is tangent to the path of the particle. The aerage acceleration is Instantaneous acceleration is a a t a lim t 0 t For straight-line motion the acceleration is alwas along the same line as the elocit. For motion in two dimensions, the acceleration ector can make an angle with the elocit ector because the elocit ector can change in magnitude, in direction, or both. The direction of the acceleration is the direction of the change in elocit during a er short time. (page 68) The aboe definitions look good, but the are not useful. We call these formal definitions. The are not used in soling problems. Instead we need a set of equations for the - and - components. The basic rule is WE DO NOT DEL WITH VECTORS. WE DEL WITH THEIR COMPONENTS. Lesson 3, page 9
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) For the elocit, we hae, a, t a t with similar definitions for the other parameters. (See pages 66-68.) We can now generalize the equations at the top of these notes to two dimensions. It is generall easiest to choose the aes so that the acceleration has onl one non-zero component. (page 69) We choose the -ais along the direction of acceleration. This means a = 0. The first equation, f i a t becomes two equations: 0 and a t. f i 1 ( ) t f i becomes two equations as well: t and ( f i ) t 1. t i 1 ) a ( t becomes t and t. i 1 ) a ( t becomes f f a i 0 and a. i Summar (see page 69): -ais : a = 0 -ais: constant a Equation 0 a t (3-19) f t 1 ( f i ) t (3-0) 1 ) t a ( t (3-1) i i Lesson 3, page 10 f i a (3-) f i
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) Projectiles are a good eample of this tpe of motion. Here a = g. This motion is simultaneousl constant elocit in the -direction with constant acceleration in the -direction. The motion in the direction is independent of the motion in the direction. Lesson 3, page 11
Lesson 3: Free fall, Vectors, Motion in a plane (sections.6-3.5) Problem: ball is thrown horizontall from a 30 m tall tower. The ball hits the ground 50 m from the base of the tower. What is the speed of the ball when it is released? Solution: Treat each component separatel. Find the time it takes for the ball to hit the ground and use that to find the initial elocit. i The time it takes to hit the ground is found from the equation for motion in the -direction. The ball is 50 m from the base. t 0 i t 1 g.47s 1 a ( t) g ( t) 30m 9.8m / s t t 50 m.47s 0. m/s Lesson 3, page 1