Motion in 2- and 3-dimensions Examples: HPTER 3 MOTION IN TWO & THREE DIMENSIONS General properties of vectors the displacement vector position and velocity vectors acceleration vector equations of motion in 2 and 3-dimensions non-linear motion (circles, planetary orbits, etc.) flight of projectiles (shells, golf balls, etc.) Motion in 2- and 3-dimensions means we have to deal with VETORS... Projectile motion path of projectile time of flight range HOORY What are you so happy about? What the heck is a vector?
asically there are two main types of physical quantities we have to deal with: SLRS VETORS Size Size Direction Scalars have size (magnitude) only, e.g., time, and they can be added, subtracted, multiplied and divided according to the rules of simple math, i.e., (+,,, ). Other examples: ordinary numbers mass distance (NOT displacement) speed (NOT velocity) Examples of vectors: displacement (The result of going E W is not the same as going W E.) force (The result of pushing something UP is not the same as pushing it DOWN.) velocity acceleration You cannot use the simple rules of math with vectors Let s look more closely at some examples... Vectors have the additional complication of direction to take into account.
The displacement vector Party at Spanish River Park Y all invited Follow the map... Military Trail D Spanish River lvd. US1 F 1 Military Trail US1 1 Glades Rd. D Spanish River lvd. F Palmetto Park Rd. E Glades Rd. 35 N of E Palmetto Park Rd. Displacement (Resultant) 7.3km 6.0km E 4.2km Vector equation lternatively: 55 E of N D 35 N of E Vector equation Note that: + = 10.2km but + = Note that: D + D =
Military Trail D Spanish River lvd. US1 F 1 ll parallel vectors of the same length are equivalent. Glades Rd. Note, yet another alternate route will give the same final displacement Palmetto Park Rd. F E E + EF + F = Vector equation Is the distance travelled the same as before? E DDITION SUTRTION D but these two are not equivalent... WHY? + = + ( ) = D i.e., = D
Multiplying a vector by a scalar: vector ( ) multiplied by a scalar (n) is another vector ( ), parallel to but with magnitude (size) n times as large, = 2 + = 4 + + + i.e., = n and = n. means magnitude Working with vectors: Draw them to scale, or Use components and trigonometry y y In component form: = ( x, y ), where = 2 x + 2 y. θ The components are: x = cosθ and y = sin θ, x where tanθ = y x. x
nswer. X Y Z D Question 3.1: Which of the vectors,,,, D represents the vector sum ( X + Y + Z )? X Y Z X + Y + Z = D D
dding vectors in component form: + = y y y θ x x In component form: = ( x, y ) and = ( x, y ) The resultant is = ( x, y ), where = ( x + x ) 2 + ( y + y ) 2. x Question 3.2: bear walks north-east for 12 m then walks east for another 12 m. (a) Draw the displacement graphically and find the resultant displacement. (b) If the bear walks at 0.50 m/s, what is his average velocity? lso x = ( x + x ) and y = ( y + y ) and tanθ = y x. (Similarly in 3-dimensions)
N + = y x θ (a) In component form: = ( x, y ). x = cos45" x = 12 m 0.707 = 8.49 m. y = sin 45" y = 12 m 0.707 = 8.49 m. Similarly: = ( x, y ) = ( 12 m,0). If = ( x, y ), then x = x + x = 20.49 m y = y + y = 8.49 m. θ = tan 1 y = tan 1 8.49 = 22.5 " (N of E) x 20.49 and = 2 x + 2 y = 491.9 = 22.2 m. E Notice the distance walked by the bear, + = 24 m, is greater than the displacement, = 22.2 m. (b) verage velocity displacement total time time for is 12 m = 24.0 s, 0.5 m/s = t. and time for is 12 m = 24.0 s. 0.5 m/s t = 48.0 s. v av = 22.2 m = 0.46 m/s. 48.0 s s we will see below, velocity is a vector and so to be precise, the average velocity is 0.46 m/s in a direction 22.5 " N of E.
Often when we use vectors in component form, we will use unit vectors : ˆ j unit vectors ˆ i So ˆ i = ˆ j = ˆ k = 1. They indicate direction only y y x Question 3.3: If x = xˆ i y = yˆ j x what are (a) D + 2 E and (b) D + 2 E? D = (6ˆ i + 3ˆ j ˆ k ) and E = (4ˆ i 5ˆ j + 8 ˆ k ), Use in vector addition... if = ( x, y ) and = ( x, y ), then = xˆ i + yˆ j and = xˆ i + yˆ j. So: ± = ( x ± x )ˆ i + ( y ± y )ˆ j + ( z ± z ) k ˆ in 3-dimensions
(a) D + 2 E = (6ˆ i + 3ˆ j ˆ k ) + 2(4ˆ i 5ˆ j + 8ˆ k ) = 6ˆ i + 3ˆ j ˆ k + 8ˆ i 10ˆ j +16ˆ k = (6 + 8)ˆ i + (3 10)ˆ j + ( 1+16)ˆ k = 14ˆ i 7ˆ j +15ˆ k. (b) D + 2 E = (14) 2 + ( 7) 2 + (15) 2 Question 3.4: If = 3.8ˆ i +1.5ˆ j and = 1.4ˆ i + 3.5ˆ j, what is the angle between and? = 196 + 49 + 225 = 470 = 21.7.
From earlier, tanθ = y x, where θ is the angle between and the ˆ i direction (x-axis). θ = tan 1 y x = tan 1 1.5 ( 3.8 ) = 21.5 ". Similarly, from the ˆ i direction (x-axis), θ = tan 1 y x = tan 1 3.5 ( 1.4 ) = 68.2 ". ˆ j 4 3 2 1 68.2 " So, the angle between and is θ θ = 68.2 " 21.5 " = 46.7 ". I will show you another way to solve this problem later. 21.5 " 0 1 2 3 4 ˆ i Position and velocity vectors x ˆ i z ˆ k r 2 Δ r particle with position vector at any instant r = xˆ i + yˆ j + zˆ k moves from P 1 (t 1 ) to P 2 (t 2 ). Then, if Δ r = r 2 r 1 and Δt = t 2 t 1 the average velocity vector is: v av = Δ r Δt. The instantaneous velocity at any point r is: Δ r v = Limit = d r Δt 0 Δt dt = dx ˆ i + dy ˆ dz j + k ˆ dt dt dt i.e., v = v xˆ i + v yˆ j + vz k ˆ r 1 P 2 (t 2 ) P 1 (t 1 ) y ˆ j
r = xˆ i + yˆ j + zˆ k x ˆ i z ˆ k r 2 Δ r r 1 P 2 (t 2 ) P 1 (t 1 ) Since the velocity vector is v Δ r = Limit, the velocity Δt 0 Δt at any point r : v = v xˆ i + v yˆ j + vz k ˆ is the tangent at the point r, i.e., parallel to the instantaneous direction of motion at all points. The instantaneous speed is magnitude of the instantaneous velocity, v. y ˆ j cceleration vector v 2 Δ v = v 2 v 1 v 2 P 2 (v 2, t 2 ) v 1 v 1 particle with velocity vector at any instant v = v xˆ i + v yˆ j + vz k ˆ moves from P 1 (v 1,t 1 ) to P 2 (v 2, t 2 ). Then, if Δ v = v 2 v 1 and Δt = t 2 t 1 the average acceleration vector is: a av = Δ v Δt. P 1 (v 1, t 1 ) Thus, the instantaneous acceleration is: Δ v a = Limit = d v Δt 0 Δt dt = dv x ˆ i + dv y ˆ dv j + z k ˆ dt dt dt i.e., a = a xˆ i + a yˆ j + az k ˆ a is parallel to Δ v
v a Special case of circular motion v 2 r 2 P 2 t 2 θ r 1 r 1 = r 2 = r ( ) v 1 P 1 t 1 ( ) v 1 = v 2 = v θ v 1 v 2 Δ v Δ v = v 2 v 1 Δ v = 2v sin( θ 2) Since a is parallel to Δ v, the direction of the acceleration is towards the inside of a curve. Note: on a curve or turn, there is always an acceleration even though the magnitude of the velocity (i.e., speed) may be constant. v a Shown here are the position and velocity vector diagrams of an object in uniform circular motion, i.e., motion at constant speed (v) and constant radius (r). The average acceleration from P 1 to P 2 is: a av = Δ v Δt ( ) = 2vsin θ 2, Δt where Δt is the time to travel from P 1 to P 2. Now, Δt = arc length P 1 P 2 v = rθ v. ( ) a av = 2v2 sin θ 2. rθ
The instantaneous acceleration at the point P 1 occurs when Δt 0, i.e., as θ 0. Then the radial acceleration is 2v 2 sin ( θ a r = Limit 2) θ 0 rθ = 2v2 sin ( θ Limit 2) θ 0 r θ. Put φ = θ 2, then a sin φ r = Limit φ 0 2φ 1 sin φ = Limit φ 0 = 1 2 φ 2. DISUSSION PROLEM [3.1]: ski-jumper flies off the edge of a cliff. What is the direction of his acceleration at,, X, Y? a r = 2v2 r 1 2 = v2 r. Note that the direction of the acceleration is parallel to Δ v, i.e., it is radial. It is called the centripetal (i.e., centerdirected) acceleration. X Y We ll return to this topic in chapter 5.
Question 3.5: n object experiences a constant acceleration of a = 6ˆ i + 4ˆ j ( ) m/s 2. t time t = 0, it is at rest and its position vector is r = 10ˆ i m. (a) Find the velocity and position vectors at any time t. (b) Find the equation describing the path of the particle in the xyplane and sketch the path. When t = 0: r " = (x ",y " ) = 10ˆ i m = (10 m,0), v " = (v "x,v "y ) = (0,0) m/s, a = (a x,a y ) = (6,4) m/s 2 = 6ˆ i + 4ˆ j (a) From h 2: v = v " + at v = v " + a t v = v " + a t = (6ˆ i + 4ˆ j )t = 6tˆ i + 4tˆ j ( ) m/s 2. ( ) m/s. r r " = v " t + 1 a t 2 r = r " + v " t + 1 a t 2 2 2 i.e., r = 10ˆ i + 1 2 (6ˆ i + 4ˆ j )t 2 = (10 + 3t 2 )ˆ i + 2t 2ˆ j m. (b) ut r = (x, y) = xˆ i + yˆ j. x = 10 + 3t 2 and y = 2t 2. Hence t 2 = y 2, so x = 10 + 3 ( y 2 ), i.e., y = ( 2 3)x 6.67 y(m) 10 Slope = 2 3 6.67 m 0 5 10 20 x(m)
This is a relative velocity problem. N θ 520 km Question 3.6: small plane departs from point heading for an airport 520 km due north at point. The airspeed of the plane, i.e., relative to the air, is 240 km/h. There is a steady wind blowing at 50 km/h from west to east. (a) What is the proper heading to make the trip from to? (b) How long does the journey take? (c) If the pilot had failed to take account of the wind, how far due east of the airport would he have been? (a) We draw the velocity vector diagram. The resultant flight path () is the result of the plane setting out along but being redirected by a wind along. i.e., v = v + v. Since v is perpendicular to sin θ = v, we have v 50 km/h = v 240 km/h, i.e., θ = 12.0 " (W of N). (b) lso, v = v cosθ = (240 km/h)(0.978) = 234.8 km/h. This is the effective speed of the plane.
So, the time to travel from to is 520 km 520 km = = 2.21 h (~2h 13min). v 234.8 km/h (c) y not allowing for the wind, the velocity vector diagram is modified: D 520 km Question 3.7: skier starts from rest at and slides down the slope to and then to. If the track is frictionless, (a) what is the skier s speed at? (b) What is the skier s speed at? ssume the skier turns all the corners smoothly with no change in speed. Then D = v w v, where v w is the wind speed and v is the 10 m 45 30 5 m aircraft speed in still air. D = 50 km/h (520 km) =108 km. 240 km/h
10 m (a) For motion down the slope: v 2 = v 2 + 2a 1 ( ), where, from hapter 2, a 1 = gsin 45 = 9.81 m/s 2 ( ) 0.707 = 6.94 m/s 2. Since triangle D is isosceles, D = D. = Since v = 0, v 2 = 2 6.94 m/s 2 ( 10 m) 2 + ( 10 m) 2 = 14.14 m. ( ) 14.14 m i.e., v = D 45 30 E ( ) = 196.3 ( m/s) 2, 196.3 ( m/s) 2 = 14.0 m/s. (b) For motion up the slope: v 2 = v 2 + 2a 2 ( ), where a 2 = gsin 30 = 9.81 m/s 2 5 m ( ) 0.50 = 4.91 m/s 2, and = E sin 30 = 5.0 m 0.5 =10.0 m. 10 m Since v = 14.0 m/s, D 45 30 E ( ) 10.0 m v 2 = ( 14.0 m/s) 2 + 2 4.91 m/s 2 = 97.9 ( m/s) 2, ( ) i.e., v = 97.9 ( m/s) 2 = 9.89 m/s. 5 m This problem can also be solved using the conservation of mechanical energy, which we will show in hapter 7.
Projectile motion thrown objects (baseballs, arrows, shells) bouncing balls nalysis of projectile motion: " v " v v v y x v x v y " v y v y v x " v v y v o v y = 0 θ (x, y ) v x x We can consider the x-y components separately: The initial velocity components are: Important concept Investigated first by Galileo. Historically important for the military. v x = v cosθ and v y = v sin θ The acceleration components are: a x = 0 and a y = g ** We ignore air resistance/drag forces **
ssume no component of acceleration in the x-direction (i.e., no air resistance/drag/added thrust), then: ut in the y-direction: * v x (t) constant = v x. * v y (t) = v y gt. The components of the displacement are: * x(t) = x + v x t * y(t) = y + v y t 1 2 gt2. v y " v v x Δt Δt Δt Δt Δt Δt Equal vertical distances Δy traveled in the same time interval Δt. (These two equations represent a parabola ). These four equations (*) are the equations of motion for a projectile (with no air resistance). Flash photographs taken at equal intervals of time show that these two objects fall at the same rate even though one has a horizontal velocity component. NOTE: the equations of motion for the horizontal (x) and vertical (y) components are separate The only physical quantity that links them is... time (t). The x- and y-components of the motion can be treated quite separately... they are connected only through time.
Here s some really weird stuff NG NG 1 2 gt2 1 If a rock is dropped from the same height as a rifle at the same time it is fired, no matter how powerful the rifle, in the absence of air resistance, the bullet and rock will fall the same distance in the same time... i.e.,... a distance 1 2 gt2 in t seconds, 1 2 gt2 2 gt2 Even though the squirrel jumps off the branch at the instant you pull the trigger, you should still aim directly at the target You do not need to allow for the fall of the target since all objects (bullet and squirrel) fall at the same vertical rate under gravity... NG 1 2 gt2 so if the ground is flat, they will hit the ground at the same time... even if you fire upward or downward
If the take-off and landing are in the same horizontal v yˆ j θ v xˆ i " v plane the horizontal range is: R = v x T where T is the total flight time. R = (v cosθ ) 2 v sin θ g Range = v 2 2sin θ cosθ. g t every point on the path: tanθ = v y v x. t maximum height: v y = v y gt = 0. v y = v sin θ = gt. So, the time to reach maximum height is t = v sin θ. g If the take-off and landing are in the same horizontal plane, the total time of flight is: T = 2 v sin θ g. i.e., R = v 2 g sin 2θ. Maximum range when θ = 45. ut sin 2θ = 2sin θcosθ and sin θ = cos(90 θ) sin 2θ = 2cos(90 θ)cosθ i.e., R(90 θ) = R(θ). So... R(30 ) = R(60 ), etc. 75 60 15 30 45
lternative derivation of range: When x = R 2, i.e., at mid-range, dy dx = 0. y R 2 R x Earlier we derived the equation of the trajectory: g y = x tanθ 2(v cosθ ) 2 x2. dy dx = tanθ g ( ) 2 x, v cosθ g R so tanθ ( v cosθ ) 2 = 0. 2 Question 3.8: The drawing shows the trajectories of two golf balls, and. Which ball had the greater initial velocity? Ignore air resistance. : all. : all. : They had the same initial velocity. R = 2v 2 tanθ cos 2 θ g = v 2 g sin 2θ.
The balls reach the same height, so the vertical components of the initial velocities are equal, i.e., v sin θ = v sin θ, where v and v are the initial velocities of and, ( ) is respectively. learly, the take-off angle for θ greater than that for ( θ ), then sin θ > sin θ, so v > v, i.e., the initial velocity of is greater than that of. Question 3.9: pitcher throws a fastball at 140 km/h towards home plate, which is 18.4 m away. Neglecting air resistance, how far does the ball drop because of the effect of gravity by the time it reaches home plate?
(y y ) y y 18.4m x x Take up as the positive direction, then: ( ) = 18.4 m : v x = 140 km/h : a x = 0 x x ( y y ) =? : v y = 0 : a y = 9.81 m/s 2. onvert 140 km/h m/s... 140 1000 = = 38.9 m/s (~87 mi/h) 60 60 The time of flight: t = x x v x = 18.4 m = 0.473 s. 38.9 m/s Question 3.10: cannonball is fired from the top of a tower 40 m tall with an initial speed of 42.2 m/s at angle of 30 above the horizontal. (a) How long is the cannonball in the air? (b) How far is it from the base of the tower when it strikes the ground? = 1 2 ( y y ) = v y t + 1 2 a yt 2 ( 9.81 m/s2 ) ( 0.473 s) 2 = 1.10 m.
( x, y ) 40 m 30 v = 42.2 m/s : θ = 30 : (x, y ) = (0,40 m) : a x = 0 a y = 9.81 m/s 2. R v x = v cos30 = 36.5 m/s v y = v sin 30 = 21.1 m/s DISUSSION PROLEM [3.2]: warship fires two shells simultaneously at enemy ships X and Y. The shells follow the parabolic paths shown... which ship is hit first? Ignore air resistance. (a) ( y y ) = v y t + 1 2 a yt 2, where t is time of flight. i.e., ( 0 40 m) = ( 21.1 m/s) t + 1 ( 9.81 m/s2) t 2. 2 Warship X Y 4.91t 2 21.1t 40 = 0 (quadratic equation) t = 21.1± ( 21.1)2 4 4.91 ( 40) 2 4.91 : Ship X, : Ship Y, : oth at the same time, t = +5.72s or t = 1.42s s?? D: Need more information. (b) Range R = v x t = ( 36.5 m/s) ( 5.72 s) = 209 m.
Question 3.11: During the Roman invasion of ritain, a group of Roman soldiers (at ) locate a ritish camp on the top of a hill (at ). Using a catapult, the Romans want to lob a rock from onto the ritish camp a distance of R = 500 m away. If the maximum angle of elevation of the catapult is θ = 60, and the height of the hill is 16 m, (a) what initial speed of the rock is required? (b) What is the flight time? (c) With what speed does the rock strike the camp? (d) What is the maximum height (h) achieved by the rock? v 60 500 m (a) R = v x t, i.e., t = lso, ( y y ) = v y t 1 2 gt2, ( ) 1000 i.e., 16 m = 0.866v 16 m v x = v cos60 = 0.50v, v y = v sin 60 = 0.866v, ( x, y ) = ( 0,0), y = 16 m. R = 500 m = 1000 m. v x 0.50v v v 4.91 m/s 2 = 866 4.91 106 v 2. ( ) 1000 v 2 v 60 h 500 m 16 m i.e., v 2 = 4.91 106 850 v = 76.0 m/s. = 5776.5 ( m/s) 2 (b) From above t = 1000 m v = 1000 m 76.0 m/s = 13.16 s
v 60 500m h 16m (c) v y v x " v = ( v x, v y ) The rock strikes the camp at, with speed v = v x 2 + v y 2. Now v 2 y = v 2 y 2g( y y ) ( ) 16 m 0 = ( 0.866 76.0 m/s) 2 2 9.81 m/s 2 ( ) = 4017.8 ( m/s) 2 i.e., v y = 63.4 m/s. v = v x 2 + v y 2 = v x 2 + v y 2 = ( 0.5 76.0 m/s) 2 + ( 63.4 m/s) 2 = 73.9 m/s. (d) Let the maximum height be h, where v y = 0. v 2 y 2g( h y ) = 0, i.e., v 2 y = 2gh, h = v y 2 2g 0.866 76.0 m/s = ( )2 2 9.81 m/s 2 = 220.8 m.