Introduction to Mechanics Motion in 2 Dimensions Lana heridan De Anza College Oct 17, 2017
Last time vectors and trig
Overview wrap up vectors introduction to motion in 2 dimensions constant velocity in 2 dimensions relative motion
Vectors Properties and A 1 B Operations 5 B 1 A Properties of Addition addition construction in Figure 3.8, is known as the commutative law of addition: Draw B, then add A. A (3.5) B ector B is e resultant R is rom the tail of A + B = B + A (commutative) D B C A (A + B) + C = A + (B + C) (associative) Figure 3.7 Geometric construction for summing four vectors. The shows that A 1 B 5 B 1 A or, in Figure 3.8 This construction 3.3 ome Properties of Vectors resultant vector R is by definition other words, that vector addition is the one that completes the polygon. Add B and C commutative. ; Add A and B Figure 3.9 Geo ; then add the then add C tions for verifyin to law of addition. result to A. the result. A ( A B C ) B C D C B B C A ( ) B C B A B R B A B A A C B B Draw A, then add B. A A
Thinking about Vectors What can you say about two vectors that add together to equal zero?
Thinking about Vectors What can you say about two vectors that add together to equal zero? When can a nonzero vector have a zero horizontal component?
The negative of the vector A is defined as the vector that when added to A gives zero for the vector sum. That is, A 1 12 A 2 5 0. The vectors A and 2 A have the Vectors same Properties magnitude but point in and opposite Operations directions. Negation ubtracting Vectors The operation of vector subtraction makes use of the definition of the negative of a vector. We define the operation A 2 B as vector 2 B added to vector A : If u = v then u has the same magnitude as v but points in the opposite direction. Figure 3.10a. ubtraction A A B = A + ( B) A 2 B 5 A 1 12 B 2 (3.7) The geometric construction for subtracting two vectors in this way is illustrated in Another way of looking at vector subtraction is to notice that the difference 2 B between two vectors A and B is what you have to add to the second vector would draw We B here if we were adding it to A. A B Vector C A B is the vector we must add to B to obtain A. A B a B Adding B to A is equivalent to subtracting B from A. b B A C A B Figure 3.10 vector B fro tor 2 B is eq vector B an site directio looking at ve
Vectors Properties and Operations There are several different multiplicative operations on vectors. For right now, we will only talk about how to multiply a vector by a scalar.
Vectors Properties and Operations There are several different multiplicative operations on vectors. For right now, we will only talk about how to multiply a vector by a scalar. Multiplication by a scalar uppose we want to multiply a scalar, like the number 5, by the vector: v = 2 i + 1 j The result is: 5v = (5 2) i + (5 1) j = 10 i + 5 j Each component is multiplied by the scalar. The direction of the vector doesn t change, but its magnitude increases by a factor of 5.
Quick review of Vector Expressions Let a, b, and c be vectors. Let l, m, and n be scalars. Could this possibly be a valid equation? a = b (A) yes (B) no
Quick review of Vector Expressions Let a, b, and c be vectors. Let l, m, and n be scalars. Could this possibly be a valid equation? a = n (A) yes (B) no
Quick review of Vector Expressions Let a, b, and c be vectors. Let l, m, and n be scalars. Could this possibly be a valid equation? a = n (A) yes (B) no
Quick review of Vector Expressions Let a, b, and c be vectors. Let l, m, and n be scalars. Could this possibly be a valid equation? a = b + n (A) yes (B) no
Quick review of Vector Expressions Let a, b, and c be vectors. Let l, m, and n be scalars. Could this possibly be a valid equation? a = n b (A) yes (B) no
Motion in 2 Dimensions o far we have looked at motion in 1 dimension only, motion along a straight line. However, motion on a plane (2 dimensions), or through space (3 dimensions) obeys the same equations. We will now focus on 2 dimensional motion.
from Equations 4.3 and 4.6, which give Motion in 2 directions Imagine v 5 d an r 5air dx hockey puck moving with horizontally constant dt dt i^ 1 dy dt j^ 5 v x i^ 1 v y j^ (4.7) velocity: vectors, y elocity, in ure, s that nsions two ns in ctions. a b y x x Figure across a table at directio in the y puck, th ponent ponent in the p If it experiences a momentary upward (in the diagram) acceleration, it will have a component of velocity upwards. The horizontal motion remains unchanged! 1 Figure from erway & Jewett, 9th ed.
Direction and Motion When we say something is moving, we mean that it is moving relative to something else. In order to describe measurements of where something is how fast it is moving we must have reference frames. In 2 dimensions we need to choose a pair of perpendicular directions to be our x and y axes.
nts Motion of Vectors in 2 directions: Components of velocity can be resolved into horizontal and Motion in perpendicular directions can be analyzed separately. ents. A vertical force (gravity) does not affect horizontal motion. The horizontal component of velocity is constant. 1 Drawing by Paul Hewitt, via Pearson.
Constant Velocity in 2 Dimensions and Consider a turtle that moves with a constant velocity. as the x and y equation y d = v 0 t y = d sin θ O θ = 25 x = d cos θ x We can find the distance it travels (a) by using the equation d = v 0 t. FIGURE 4 1 Constant velocity A turtle walks from the origin with a speed of v 0 = 0.26 m/s. (a) In a 1 Figure thusfrom the xwalker, and y displacements Physics. are x = d cos u, y = d sin u. (b) Equiv
Constant Velocity in 2 Dimensions and However, if know how far the turtle has traveled in the x direction and y direction separately, this gives us the components of the the turtle s displacement vector, r. as the x and y equations of m y d = v 0 t y = d sin θ O θ = 25 x = d cos θ x v 0y = v We can find the distance it travels (a) in the x-direction using trigonometry: x = d cos θ. FIGURE 4 1 Constant velocity A turtle walks from the origin with a speed of v 0 = 0.26 m/s. (a) In a time t th thus the x and y displacements are x = d cos u, y = d sin u. (b) Equivalently, and v 0y = v 0 sin u; hence x = v 0x t and y = v 0y t. And in the y-direction: y = d sin θ. 1 Figure from Walker, Physics.
Constant Velocity in 2 Dimensions x and y equations of motion. y = y 0 + v 0y t y 4 2 d sin θ y = v 0y t v 0y = v 0 sin θ O v 0 θ = 25 v 0x = v 0 cos θ x = v 0x t x Or, we can find the distance it travels in the x-direction by considering what is its rate of change of x-position with time! 0.26 m/s. (a) In a time t the turtle moves through a straight-line distance of d = v 0 t; = d sin u. (b) Equivalently, the turtle s x and y components of velocity are v 0x = v 0 cos u (b) v 0x = x t = v 0 cos θ x = (v 0 cos θ)t And in the y-direction: v 0y = y t = v 0 sin θ y = (v 0 sin θ)t 1 Figure from Walker, Physics.
Relative Motion We can use the notion of motion in 2 dimensions to consider how one object moves relative to something else. All motion is relative. Our reference frame tells us what is a fixed position. An example of a reference for time and space might be picking an object, declaring that it is at rest, and describing the motion of all objects relative to that.
Intuitive Example for Relative Velocities tors ocity relative to ds on the relative to the d s velocity. 1 Figure by Paul Hewitt.
Intuitive Example Now, imagine an airplane that is flying North at 80 km/h but is blown off course by a cross wind going East at 60 km/h. How fast is the airplane moving? In which direction? ketch: 1 Figure by Paul Hewitt.
Intuitive Example Now, imagine an airplane that is flying North at 80 km/h but is blown off course by a cross wind going East at 60 km/h. How fast is the airplane moving? In which direction? ketch: 1 Figure by Paul Hewitt.
Intuitive Example Now, imagine an airplane that is flying North at 80 km/h but is blown off course by a cross wind going East at 60 km/h. How fast is the airplane moving? In which direction? ketch: Hypothesis: It will travel to the North-East, at a speed greater than 80 km/h, but less than 80+60 = 140 km/h. 1 Figure by Paul Hewitt.
Intuitive Example
Relative Motion Example v BA dt A boat crossing a wide river frame moves B, which with moves ato speed the right of 10.0 km/h with a constant velocity relative to the water. The water in the river v BA. The has a uniform speed of vector r PA is the particle s position 5.00 km/h due east relative vector relative the Earth. to A, and If r Pthe B is its boat heads due position vector relative to north, determine the velocity of the boat relative B. to an observer standing on either bank. 1 ketch: ore, we conclude that a P A 5 a P B hat is, the acceleration of the partirence is the same as that measured city relative to the first frame. r Figure 4.20 A particle located at P is described by two observers, one in the fixed frame of reference A and the other in the peed of ver has a e Earth. locity of bank. s a river will not end up W N vbr E you relau vre vbe W N E vbr v re a b 2 Page Figure 97, 4.21 erway (Example & Jewett 4.8) (a) A boat aims directly across a u vbe
Another Boat Question You are riding in a boat whose speed relative to the water is 6.1 m/s. The boat points at an angle of 25 upstream on a river flowing at 1.4 m/s. What is your velocity relative to the ground? 1 Figure from Walker, Physics.
Another Boat Question You are riding in a boat whose speed relative to the water is 6.1 m/s. The boat points at an angle of 25 upstream on a river flowing at 1.4 m/s. What is your velocity relative to the ground? ative to the water is 6.1 m/s. The boat points at an angle of 25 upstream our velocity relative to the ground? ketch:, and the y axis to boat relative to the he water relative to ts in the negative y v bw = 6.1 m/s v wg ction in which it is the boat will move of the boat we use g to the water (w), O 25 v bg v wg = 1.4 m/s x : s in the v! bg = v! bw + v! wg 1 Figurev! wg from= Walker, 1-1.4 m/s2yn Physics.
Another Boat Question You are riding in a boat whose speed relative to the water is 6.1 m/s. The boat points at an angle of 25 upstream on a river flowing at 1.4 m/s. What is your velocity relative to the ground? ative to the water is 6.1 m/s. The boat points at an angle of 25 upstream our velocity relative to the ground? ketch:, and the y axis to boat relative to the he water relative to ts in the negative y v bw = 6.1 m/s v wg ction in which it is the boat will move of the boat we use g to the water (w), O 25 v bg v wg = 1.4 m/s x : Hypothesis: Less than 6.1 m/s, but more than 6.1-1.4 = 4.7 m/s. The angle the velocity makes to the x-axis will be less than 25. v! bg = v! bw + v! wg s in the 1 Figurev! wg from= Walker, 1-1.4 m/s2yn Physics.
m/s. Another The boat points Boatat Question an angle of 25 upstream the ground? y v bw = 6.1 m/s v wg v bg O 25 v wg = 1.4 m/s x v! bw + v! wg 1 Figure from Walker, Physics.
3 6 RELATIVE MOTION 71 Another Boat Question m/s. The boat points at an angle of 25 upstream the ground? y v bw = 6.1 m/s v wg v bg O 25 v wg = 1.4 m/s x v! bw + v! wg 1-1.4 m/s2yn 1 Figure from Walker, Physics.
ummary vectors motion in 2-dimensions motion with constant velocity relative motion Homework Walker Physics: Ch 4, onward from page 100. Conceptual Questions: 5; Problems: 3, 5, 7, 9