Chapter 3. Motion in Two or Three dimensions. 3.1 Position and Velocity Vectors

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1 Chapter 3 Motion in Two Three dimensions 3.1 Position and Velocity Vects Extra dimensions. We now generalize the results of previous section to motion in me than one (spacial) dimension. In this chapter we will only concentrate on motions in two and three dimensions (often abbreviated as 2D and 3D) which is what we typically observer by naked eye. However, this does not mean that at very large (cosmological) at very small (high energy) scales the effect of other (usually called extra) dimensions are not present. In fact there are very strong arguments suppting the ideas of extra dimensions base on string they. The reason the extra dimensions (usually six and compact) had to be introduced in such theies is in an attempt to make the physical theies well defined mathematically, but we are still far from accomplishing this task. However, even if there are extra dimensions, but the motion along those extra dimensions is small, then we need not wry about them. This is similar how we did not have to wry about all three dimensions in the previous chapter by considering motion along only a single axis. Position. Position vect r specifies position of an object in three dimensions and in a given codinate systems it it is described by three numbers, r =(x, y, z) =xî + yĵ + zˆk (3.1) One can imagine an arrow connecting the igin with position of object to describe r. If the object is in motion, then the position changes with time the position vect becomes a function (actually three functions) of time r(t) =(x(t),y(t),z(t)) = x(t)î + y(t)ĵ + z(t)ˆk. (3.2) 28

2 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 29 These three functions do depend on the codinate system, but as was already noted the physically measurable quantities should not depend on this choice. Velocity. Similarly to 1D case (see Eqs. (2.5) and(2.8)) in 3D we can define average velocity vect as v avg r t = r(t 2) r(t 1 ) t 2 t 1, (3.3) (where r is now the displacement vect) and instantaneous velocity vect r v(t) lim v avg = lim t 0 t 0 t = d r(t). (3.4) Then we can use the components representation of position vect Eq.(3.2) to obtain v = d r(t) ( dx(t) =, dy(t), dz(t) ) = dx(t) î + dx(t) ĵ + dx(t) ˆk (3.5) and if we denote then v (v x,v y,v z ) (3.6) v x = dx(t) v y = dy(t) v z = dz(t). (3.7) Example 3.1. Aroboticvehicle,rover,isexplingthesurfaceofMars. The stationary Mars lander is the igin of codinates, and the surrounding Martian surface lies in the xy-plane. The rover, which we represent as a point, has x- and y-codinates that vary with time: x(t) = 2.0 m ( 0.25 m/s 2) t 2 y(t) = (1.0 m/s) t + ( m/s 3) t 3 z(t) = 0 m (3.8) (a) Find the rover s codinates and distance from the lander att =2.0 s. (b) Find the rover s displacement and average velocity vects f the interval t =0.0 stot =2.0 s. (c) Find a general expression f the rover s instantaneous velocity vect v. Express v at t =2.0 sincomponentsfromandin

3 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 30 terms of magnitude and direction. Step 1: Codinate system: Was already chosen f you with igin at Mars lander and Martian surface in the xy-plane. Step 2: Initial conditions: Easy to calculate r(0) = (x(0),y(0),z(0)) = (2.0 m, 0m, 0m) ( dx(0) v(0) =, dy(0), dz(0) ) = (0 m/s, 1.0 m/s, 0 m/s) (3.9) Step 3: Final conditions: (a) Find the rover s codinates and distance from the lander at t =2.0 s. Using Eq. (3.8) we get codinates x(2.0s) = 2.0m ( 0.25 m/s 2) (2.0s) 2 =1.0m y(2.0s) = (1.0m/s)(2.0s)+ ( m/s 3) (2.0s) 3 =2.2m z(t) = 0m (3.10) and distance from igin r(2.0s)= r (2.0s) = x (2.0s) 2 + y (2.0s) 2 + z (2.0s) 2 = m. (3.11) (b) Find the rover s displacement and average velocity vects f the interval t =0.0 stot =2.0 s. Displacement is r = r(2.0s) r(0.0s) = (1.0 m, 2.2 m, 0m) (2.0 m, 0m, 0m) = ( 1.0 m, 2.2 m, 0 m) (3.12)

4 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 31 and thus, average velocity is v avg = r =( 1.0 m)î +(2.2 m)ĵ (3.13) ( 1.0 m)î +(2.2 m)ĵ 2.0s =( 0.5 m/s)î +(1.1 m/s)ĵ. (3.14) (c) Find a general expression f the rover s instantaneous velocity vect v. Express v at t =2.0 s in components from and in terms of magnitude and direction. By plugging Eq. (3.8) into Eq. (3.5) we get and at t =2.0s v x (t) = dx(t) v y (t) = dy(t) v z (t) = dy(t) v x (2.0s)= dx(2.0s) v y (2.0s)= dy(2.0s) v z (2.0s)= dz(2.0s) = 0.5 m/s 2 t = 1.0 m/s+ ( m/s 3) t 2 = 0 m/s (3.15) = 0.5 m/s 2 (2.0s)= 1.0m/s = 1.0 m/s+ ( m/s 3) (2.0s) 2 =1.3m/s = 0 m/s. (3.16) v(2.0s)=( 1.0m/s)î +(1.3m/s)ĵ. (3.17) Then the magnitude of velocity ( speed) is v = v = ( 1.0m/s) 2 +(1.3m/s) 2 =1.6 m/s (3.18) and direction is arctan v y v x =arctan 1.3m/s 1.0m/s = 52 (3.19) which is off by 180 because of the way arctan function is defined (range is from 90 to +90 ). Thus the direction (measure from x-axis in contraclockwise direction) is α = =128. (3.20)

5 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS The Acceleration Vect Acceleration. Similarly to 1D case (see Eqs. (2.10) and(2.12)) in 3D we can define average acceleration vect as and instantaneous velocity vect a avg v t = v(t 2) v(t 1 ) t 2 t 1, (3.21) a(t) lim a v avg = lim t 0 t 0 t = d v(t). (3.22) Then we can use the components representation of velocity vect Eq.(3.5) to obtain a = d v(t) ( dvx (t) =, dv y(t), dv ) z(t) = dv x(t) î + dv y(t) ĵ + dv z(t) ˆk (3.23) and if we denote then using Eq. (3.7) a (a x,a y,a z ) (3.24) a x = dv x(t) a y = dv y(t) a z = dv z(t) = d2 x(t) 2 = d2 y(t) 2 = d2 z(t) 2. (3.25) Example 3.2. Let s return to motion of the Mars from the previous section. (a) Find the components of the average acceleration f the interval t =0.0 s to t =2.0 s. (b) Find the instantaneous acceleration at t =2.0 s.

6 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 33 Step 1: Codinate system: As befe the codinate system was chosen f us. Step 2: Initial conditions: In addition to initial conditions f position and velocity (see Eq. (3.9)), we can now calculate the initial conditions f acceleration ( ) d 2 x(0) a(0) =, d2 y(0), d2 z(0) = ( 0.5 m/s 2, 0m/s 2, 0m/s ) (3.26) Step 3: Final conditions: (a) Find the components of the average acceleration f the interval t =0.0 stot =2.0 s. By substituting initial and final velocities from Eq. (3.9) and(3.16) into Eq. (3.21) we get v(2.0s) v(0.0s) a avg = 2.0s 0.0s ( 1.0m/s, 1.3m/s, 0m/s) (0 m/s, 1.0 m/s, 0m/s) = 2.0s = ( 0.5m/s 2, 0.15 m/s 2, 0m/s 2) (3.27) a avg = ( 0.5 m/s 2) î + ( 0.15 m/s 2) ĵ. (3.28) (b) Find the instantaneous acceleration at t = 2.0 s. Using the definition of Eq. (3.23) and instantaneous velocity that was already calculated above (see

7 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 34 Eq. (3.15)) we get a general expression and at a specific time t =2.0 s a(t) = dv x(t) î + dv y(t) ĵ + dv z(t) ˆk = ( 0.5m/s 2) î + ( 0.15 m/s 3) tˆk (3.29) a(2.0s) = ( 0.5m/s 2) î + ( 0.15 m/s 3) (2.0s)ˆk = ( 0.5m/s 2) î + ( 0.3m/s 2) ˆk. (3.30) Parallel and Perpendicular Components. It is sometimes useful to think of components of a vect not in a fixed codinate system but with respect to changing codinate system. A particularly useful example if f a moving codinate system in 2D when one of the axis point in the direction of the velocity vect v. Then we can decompose the acceleration vect into component parallel a and perpendicular a to the direction of motion a = a + a. (3.31) After a little bit of calculus one can show that the magnitude of velocity vect changes as dv(t) = d v(t) v(t) v(t) = v(t) d v(t) = v(t) ( a (t)+ a (t) ) = ˆv (t) a (t) (3.32) and the direction of the velocity vect changes as dˆv(t) ( ) v(t) = v(t) d v(t) v(t) dv(t) v(t) = v(t) 2 ( a (t)+ a (t) ) v(t) (ˆv (t) a (t) ) v(t) = d v(t) 2 (3.33) Evidently, the parallel component is responsible f changes in magnitude of the velocity vect (i.e. speed), but not the direction + a v speed is increasing v a = 0 speed is not changing (3.34) a v speed is decreasing. v = a (t) v(t). and the perpendicular component is responsible f changes in direction of velocity,but not in magnitude. If there is only a perpendicular component then the object is in circular motion.

8 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS Projectile Motion Projectile motion: is a motion which is completely determined by gravitationally acceleration and air resistance starting from some initial condition determined by position and velocity. (The fact that you have to always specify an even number of initial data is a consequence of the fact that equations of motion are (almost always) of the second der in time.) Such motion is always confined to a 2D plane determined by two vects: gravitational acceleration vect and initial velocity vect. It is convenient to choose a codinate system so that one of the axis (usually y-axis pointing upward) is vertical, the other one (usually x-axis) is hizontal and since there is no motion in z-direction it does not really matter. Then, Then from the equations of motion we get a =(0, g) = gĵ (3.35) v(t) = v 0x î +(v 0y gt)ĵ r(t) = (x 0 + v 0x t) î +(y 0 + v 0y t 1 2 gt2 )ĵ (3.36) Example 3.6. Amotcyclestuntriderridesofftheedgeofacliff. Just at the edge his velocity is hizontal with magnitude 9.0 m/s. Find the motcycle s position, distance from the edge of the cliff, and velocity 0.50 s after it leaves the edge of the cliff. Step 1: Codinate System: Let s choose y-axis to point vertically and upward, and x-axis to point hizontally in the direction of the motcycle s motion befe it rides off the cliff. The igin can be at the edge of the cliff. Step 2: Initial Conditions: The initial position and velocity is given by r (0) = (x 0,y 0 )=(0, 0) v(0) = (v 0x,v 0y )=(9.0m/s, 0m/s). (3.37)

9 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 36 Step 3: Final Conditions: Find the motcycle s position, distance from the edge of the cliff, and velocity 0.50 s after it leaves the edge of the cliff. Using Eq. (3.36) we get velocity and position v(0.50 s) = (9.0m/s)î +( 9.8m/s s)ĵ =(9.0m/s)î (4.9m/s)ĵ r(0.50 s) = (9.0m/s 0.50 s) î +( m/s2 (0.50 s) 2 )ĵ =(4.5m)î +( 1.2m)ĵ (3.38) and thus the distance from the edge is r(0.50 s) = (4.5m) 2 +( 1.2m) 2 =4.7m. (3.39) 3.4 Motion in a Circle Unifm motion - is a motion when the direction of velocity vect changes, but magnitude ( speed) does not change, i.e. This already implies that dv dˆv = 0 0, (3.40) v d v = 0 (3.41) but if the magnitude of acceleration also does not change, a(t) =a (t) = const. (3.42) then 0= d a2 = d ( a a) = d ( d v d v ) =2 d2 v d v 2 By combining Eqs. (3.41) and(3.43) in 2D we must conclude that (3.43) v d2 v 2. (3.44) whose solutions are sines and cosines (you will see this differential equation over and over in physics courses). Radial ( centripetal) acceleration. With little me algebra (that we shall skip) one can show the such motion gives rise to circular motion described by equation R = (x(t) X) 2 +(y(t) Y ) 2 (3.45)

10 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 37 where (X, Y ) is at the center of the circle and R is its radius. By choosing the igin of codinates at the center of circle, i.e. making we can simplify Eq. (3.45) to a simple fm (X, Y )=(0, 0) (3.46) r = x(t) 2 + y(t) 2 = R (3.47) By differentiating it with respect to time once we get 0= dx(t) and by differentiating it twice we get x(t)+ dy(t) y(t) (3.48) r v = 0 (3.49) ( ) 2 ( ) ( ) dx(t) d 2 2 ( ) x(t) dy(t) d 2 y(t) 0= + x(t)+ + y(t) (3.50) 2 2 v 2 = a r = a r. (3.51) But since the two vects a and r always point in opposite directions we have v 2 = a R (3.52) a = v2 R. (3.53) Periodic motion. Since the motion is circular the object must come to were it started at some finite time. This time is called period, T,anhe motion is called also periodic. To understand what the period is it is useful to write an exact solution f periodic motion around igin ( r(t) = R sin (2π tt ) + φ,rcos (2π tt )) + φ (3.54) where R, T and φ are some constants. (Check f yourself that this solution indeed satisfies Eq. (3.47).) We already discussed R as its describes the radius of motion, but the other two constants have special meanings as well. Note that as time goes from t to t + T the arguments of the sin and cos

11 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 38 functions change by 2πand thus the values of these functions as well as the position vect does not change r(t) = r(t + T ). (3.55) Do not wry about constant φ f now (it is known as phase). It is now straightfward to figure out the dependance of velocity on the period, which we could have done even without writing the explicit solution. Evidently, an object in circular motion is traveling a distance 2πR with a constant speed v and thus the period must be T = 2πR v (3.56) v = 2πR T. (3.57) By combining Eqs. (3.53) and(3.55) weget a = 4π2 R T 2. (3.58) Example Passengers on a carnival ride move at a constant speed in ahizontalcircleofradius5.0 m, making a complete circle in 4.0 s. What is their acceleration? Step 1: What is the codinate system? Let choose a (moving) codinate system with direction in the direction of motion and direction in the radial direction. Step 2: What is given? Period and radius T =4.0s R =5.0m Step 3: What do we have to find? Then accding to (3.58) wehave a = 4π2 (5.0m) (4.0s) 2 =12m/s 2 =1.3 g and since the motion is unifm a =0.

12 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 39 Non-unifm motion. If an object is in circular motion, but the magnitude of its velocity ( speed) changes, then such motion is called a nonunifm circular motion. Then in addition to perpendicular ( centripetal) acceleration there is a parallel ( tangential) acceleration. 3.5 Relative velocity a (t) = v(t)2 R a (t) = dv(t). (3.59) Reference frame. We have already mentioned how the choice of codinate system is imptant f calculations, but should not matter f physically observable quantities. We have also discussed (in context of 2D motions) how it is sometime useful to wk with respect to a moving codinate system. In particular the question of how one can go from a static codinate system to a moving codinate system. F example, consider a passenger in a train. One can talk about the passenger s position with respect to the ground with respect to the train. Clearly the two things are not the same. Me generally, let object A move with respect to a codinate system ( reference frame) of object B, described by position vect r A/B (t) (3.60) and let object B move with respect to a codinate system ( reference frame) of object C, described by position vect r B/C (t) (3.61) then we say that to describe object A with respect to a codinate system ( reference frame) of object C, we use the following rule r A/C (t) = r A/B (t)+ r B/C (t). (3.62) (Note that this rule wks very well only f small velocities, but fails miserably when velocities approach the speed of light which is the realm of the so-called special they of relativity.) By taking a time derivative of Eq. (3.62) we get a rule f adding velocities d r A/C (t) = d r A/B(t) + d r B/C(t) v A/C (t) = v A/B (t)+ v B/C (t). (3.63)

13 CHAPTER 3. MOTION IN TWO OR THREE DIMENSIONS 40 Example There is a 100 km/h wind from west to east. (a) If an airplane s compass indicates that it is headed due nth, and its airspeed indicat shows that it is moving through the air at 240 km/h, what is the velocity of the airplane relative to earth? (b) What direction should the pilot head with speed 240 km/h to travel due nth? Step 1: Choose codinate system. Let x-axis to point east and y-axis to point nth. Step 2: What is given? v A/E =100km/hî Step 3: What do we have to find? (a) The airspeed indicat tells us that and by using (3.63) weget v P/A =240km/hĵ v P/E = v P/A + v A/E =100km/hî +240km/hĵ (b) In general velocity of airplane relative to air is and (3.63) weget v P/E = v P/A + v A/E = where But since v P/A = x î + y ĵ ( ) x î + y ĵ +100km/hî =(x +100km/h)î + y ĵ. x +100km/h=0 v P/A = x 2 + y 2 =240km/h we have two equations with two unknowns with solution x = 100 km/h y = 218km/h. and thus v P/A = 100 km/h î +218km/hĵ.