CHAPTER 3 MOTION IN TWO AND THREE DIMENSIONS General properties of vectors displacement vector position and velocity vectors acceleration vector equations of motion in 2- and 3-dimensions Projectile motion path of a projectile time of light range Examples of motion in 2- and 3-dimensions: any non-linear motion (circular motion, orbits of planets, etc) flight of projectiles (artillery shells, golf balls, footballs, etc) Motion in 2- and 3-dimensions requires the use of vectors Basically, there are two main types of physical quantities we deal with: Scalars have size (magnitude) only and they can be added, subtracted, multiplied and divided according to the rules of simple math, ie, +,,, Examples include time, ordinary numbers, mass, distance (not displacement) and speed (not velocity)
Examples of vector quantities: Displacement (The result of going E W is not the same as going W E) Force (The result of pushing an object up is not the same as pushing it down) Velocity Acceleration You cannot use the rules of simple math with vectors As an example take the displacement vector BBQ at Spanish River Park Follow the map Military Trail US 1 A1A Spanish River D Blvd F C A Glades Road Palmetto Park Road E B You have several choices: Military Trail US 1 A1A Spanish River D Blvd F C A A Glades Road Palmetto Park Road Displacement (resultant) 73 km 35 60 km E B C 42 km Note that: AB + BC = 102 km AC " " " but AB + BC = AC vector equation The displacement vector is 73 km, 35 N of E B
Military Trail US 1 A1A Spanish River D Blvd F C Glades Road All parallel vectors with the same lengths are equivalent A Palmetto Park Road E B These two vectors are not equivalent D C A 55 F " " " AD + DC = AC ( 55 E of N) C ADDITION A + B = C " AE " + EF " + FC " = AC SUBTRACTION A E In all three cases the displacement vectors are the same and so are the distances A + ( B) = D ie, A B = D
Multiplication by a scalar When working with vectors we will either draw them to scale, or use trigonometry and components: y A y A θ x A vector A multiplied by a scalar n is another vector C parallel to A but with magnitude (size) n times as large: ie, C = n A and C = n A In component form: The components are: where tanθ = A y A x A x A = (A x,a y ) A = A x 2 + Ay 2 and A x = A cosθ and A y = A sinθ,
Adding two vectors in component form: A + B = C y Question 31: A yacht sails from a port (A) for 707 km in B y C A y θ A B x a north-easterly direction to point B Then, it sails in a south-easterly direction for a further 424 km to point C A x B x In component form: A = (A x,a y ) and B = (B x,b y ) The resultant is C = (C x,c y ), where C = (A x + B x ) 2 + (A y + B y ) 2 (a) What is the displacement of the ship relative to the port when it reaches point C? (b) If the yacht travels at 100 km/h, what is its average velocity for the journey from A to C? Also C x = (A x + B x ) and C y = (A y + B y ) and tanθ = C y C x
A y A θ C B A x B x (a) We have to determine A and B are: C = A + B B y The components of A x = A cos45 " = (707 km)(0707) = 50 km A y = A sin 45 " = (707 km)(0707) = 50 km B x = B cos( 45 " ) = (424 km)(0707) = 30 km B y = B sin( 45 " ) = (424 km)( 0707) = 30 km But C = (C x,c y ) = (A x + B x,a y + B y ) = (80 km,20 km) (b) Total distance traveled by yacht was 707 km + 424 km = 1131 km, and the speed was 100 km/h So, the time taken was 1131 km = 131 h 100 km/h The average velocity is defined as displacement 824 km = = 629 km/h time 131 h Velocity is a vector (in the same direction as the displacement) so to be precise the average velocity of the yacht is 629 km/h in a direction 140 N of E C = (80 km) 2 + (20 km) 2 = 824 km and θ = tan 1 C y C x = tan 1 20 km 80 km = 140
Often, when we use vectors in component form, it is more convenient to use unit vectors Unit vectors indicate direction only (and have magnitude = 1) ĵ unit vectors ˆk î î = ĵ = ˆk = 1 Question 32: If N = î 4 ĵ 4ˆk, L = 6î + 3ĵ ˆk, what are: (a) L + 2 M N, and (b) L + 2 M N? M = 4î 5ĵ+ 8ˆk and In vector addition if A = (A x,a y ) and B = (B x,b y ) then A = A x î + A y ĵ and B = B x î + B y ĵ So: A ± B = (A x ± B x )î + (A y ± B y ) ĵ+ (A z ± B z )ˆk
(a) L + 2 M N = (6î + 3ĵ ˆk) + 2(4î 5ĵ+ 8ˆk) (î 4 ĵ 4ˆk) = 6î + 3ĵ ˆk + 8î 10 ĵ+16ˆk î + 4 ĵ+ 4ˆk = 13î 3ĵ+19ˆk Question 33: If A = 38î +15ĵ what is the angle between and A and B? B = 14î + 35ĵ, (b) L + 2 M N = (13) 2 + ( 3) 2 + (19) 2 = 539 = 232
From earlier, tanθ A = A y Ax, where θ A is the angle between A and the ˆ i direction (x-axis) θ A = tan 1 A y Ax = tan 1 15 ( 38 ) = 215 " Similarly, from the ˆ i direction (x-axis), θ B = tan 1 B y Bx = tan 1 35 ( 14 ) = 682 " ˆ j 4 3 2 1 B 682 " So, the angle between A and B is θ B θ A = 682 " 215 " = 467 " I will show you another way to solve this problem later A 215 " 0 1 2 3 4 ˆ i Position and velocity vectors The position vector of a point with coordinates (x, y,z) is r = x î + yĵ+ zˆk Consider an object with position vector r = x î + yĵ+ zˆk that 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 (parallel to Δ r ) Δt The instantaneous velocity vector at any point r is v = Limit Δt 0 ie, Δ r Δt = d r r dt = dx dt î + dy dt v = v x î + v y ĵ+ v z ˆk ĵ+ dz dt ˆk,
Acceleration vector The velocity vector at the point r is the tangent to the position-time graph at r and is parallel to the instantaneous direction of motion The instantaneous speed is the magnitude of the instantaneous velocity Consider an object with the velocity vector that 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 (parallel to Δ v ) Δt The instantaneous acceleration vector is a = Limit Δt 0 Δ v Δt ie, v = v x î + v y ĵ+ v z ˆk = d v dt = dv x dt î + dv y dt a = a x î + a y ĵ+ a z ˆk ĵ+ dv z dt ˆk,
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 (ie, speed) may be constant v a Shown here are the position and velocity vector diagrams of an object in uniform circular motion, ie, 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, ie, as θ 0 Then the radial acceleration is 2v 2 sin ( θ a r = Limit 2) θ 0 rθ = 2v2 sin ( θ Limit 2) r θ 0 θ Put φ = θ, then a 2 r = 2v2 sinφ Limit r φ 0 2φ sin φ But Limit φ 0 2φ = Limit 1 sinφ φ 0 2 φ = 1 2 Limit sin φ φ 0 φ = 1 2 a r = 2v2 r 1 2 = v2 r Note that the direction of the acceleration is parallel to Δ v, ie, it is radial It is called the centripetal (ie, centerdirected) acceleration Question 34: An object experiences a constant acceleration of a = (6î + 4 ĵ) m/s2 At time t = 0, it is at rest at r" = (10,0) m Find (a) the velocity and position vectors at any later time t, and (b) the equation describing the path of the object in the xy-plane We ll return to this topic in chapter 5
When t = 0 : r " = (10,0) m and v " = (0,0) m/s, a = (a x,a y ) = (6î + 4 ĵ) m/s2 (a) From chapter 2: v = v + at v " = v " + at " Also: (b) But v = (6î + 4 ĵ)t = (6tî + 4tĵ) m/s ( r r " ) = v " t + 1 at 2 r = r 2 " + v " t + 1 at 2 2 r = 10î + 1 2 (6î + 4 ĵ)t2 = (10 + 3t 2 )î + 2t2 ĵ m r = (x,y) = x î + yĵ x = (10 + 3t 2 ) and y = 2t 2 Hence, t 2 = y so x = 10 + 3 y 2 2, ie, y = 2 3 x 20 3 Question 35: A ship departs from point A heading to a port 52 km due north at point B The speed of the boat relative to the water is 10 km/h If there is a steady current flowing at 2 km/h in a north-westerly direction, (a) what is the proper heading to make the trip from A to B? (b) How long does the journey take? (c) HOME PROBLEM: If the captain of the ship had failed to take into account of the wind, how far due west of B would the ship had been?
(a) Draw the velocity vector diagram The resultant path " B AB is the result of the ship setting " N out along but being re-directed 45 AC C " D by the current along CB So, v AB = v AC + v CB A θ Since the resultant heading is due north, the resultant velocity component in the E-W direction is zero (10 km/h)sinθ (2 km/h)sin45 = 0 ie, (10 km/h)sinθ = (2 km/h)cos45 θ = sin 1 (2 km/h)cos45 = 81 east of north 10 km/h (b) The resultant speed of the ship is v AB = v AD + v DB = v AC cosθ + v CB sin45 " = (10 km/h)cos81 + (2 km/h)sin45 = 113 km/h Projectile motion thrown objects (baseball, arrows, shells, etc) bouncing balls, etc Investigated first by Galileo Historically important for the military Since AB = 52 km, the journey time is 52 km = 460 h 4 h 36 min 113 km/h
Analysis of projectile motion a x = 0, Since v x (t) constant = v x v y v v x v v y v x v Since a y = g, v y (t) = v y gt The components of the displacements are: y v y θ (x,y ) v x v " x and x(t) = x + v t y(t) = y + v y t 1 2 gt2 These two equations represent a parabola Important concept: we can treat the x and y components separately The initial velocity components are: v x = v " cosθ and v y = v " sinθ If we ignore air resistance and other drag forces, the acceleration components are: a x = 0 and a y = g The four equations listed above are the equations of motion for a projectile (in the absence of air resistance) Although separate, the equations for the horizontal and vertical components are connected by time (t)
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 an assuming the ground is horizontal and level, the bullet and the rock will hit the ground at the same time, since all objects fall the same distance in the same time, ie, a distance of 1 2 gt2 in t seconds
v yˆ j θ v xˆ i " v At every point on the path: Range tanθ = v y v x At 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 If the take-off and landing are in the same horizontal plane the horizontal range is: R = v x T where T is the total flight time R = (v cosθ ) 2 v sin θ g = v 2 2sin θ cosθ g ie, R = v 2 g sin 2θ Maximum range when θ = 45 But sin 2θ = 2sin θcosθ and sin θ = cos(90 θ) sin 2θ = 2cos(90 θ)cosθ ie, R(90 θ) = R(θ) So R(30 ) = R(60 ), etc 75 15 60 30 45
A B Question 36: The sketch above shows the trajectories of two golf balls, A and B If the balls reach the same height, which one had the greater initial velocity? Ignore air resistance The balls reach the same height, so the vertical components of the initial velocities are equal, ie, v A sin θ A = v B sin θ B, where v A and v B are the initial velocities of A and B, ( ) is respectively Clearly, the take-off angle for A θ A greater than that for B ( θ B ), then sin θ A > sin θ B, so v B > v A, ie, the initial velocity of B is greater than that of A
Question 37: A cannonball is fired from the top of a 40 m high tower with an initial speed of 422 m/s at an angle of 30 above the horizontal (a) How long is the cannonball in the air before it hits the ground? (b) How far is it from the base of the tower? (c) What is its speed when it hits the ground? 491t 2 211t 40 = 0, t = 211± ( 211)2 4 491 ( 40) ie, 2 491 t = +572 s and t = 142 s The appropriate solution is t = +572 s (b) Range R = v x t = (365 m/s)(572 s) = 209 m
(c) V x The final speed is v = 2 vx + v 2 V y y V v x = v x = 365 m/s, and v 2 y = v y 2 + 2( g)(y y ) = (211 m/s) 2 + 2( 981 m/s 2 )( 40 m) = (1230 m/s) 2 v y = 351 m/s DISCUSSION QUESTION: Hence, v = (365 m/s) 2 + (351 m/s) 2 = 506 m/s
Question 38: During the Roman invasion of Britain, a group of Roman soldiers (at A) locate a British camp on the top of a hill (at B) Using a trebuchet they want to lob a rock from A onto the British camp, a horizontal distance of 500 m away If the maximum angle of elevation of the trebuchet is 60, and the height of the hill is 16 m, (a) what initial speed of the rock is required in order to hit the camp? (b) What is the time of flight of the rock? (c) With what speed does the rock strike the camp? (d) What is the maximum height (h) achieved by the rock? (a) The horizontal range is v x t, ie, Also, (b) From above, ie, (y y ) = v y t 1 2 gt2 v x = v cos60 = 050v, v y = v sin60 = 0866v, (x, y ) = (0,0); y = 16 m (1000 m) 16 m = 0866v (491 m/s 2 ) v = (866 m) (491 m/s2 )(10 6 m 2 ) 2 v v 2 = (491 m/s 2 )(10 6 m 2 ) 850 m t = 1000 m v v = 760 m/s t = 500 m 050v = 1000 m v = 57765 (m/s) 2, (1000 m) v = 1000 m = 1316 s 760 m/s 2
(c) The rock strikes the camp with speed v = v x 2 + vy 2 But v x = v x = 050v = 380 m/s, and 2 v y = 2 vy 2g(y y ) = (0866 760 m) 2 2(981 m/s 2 )(16 m 0) = 40178 (m/s) 2, ie, v y = 634 m/s v = (380 m/s) 2 + ( 634 m/s) 2 = 739 m/s (d) The maximum height is h, where v y = 0 2 v y 2g(h 2 y ), ie, v y = 2gh h = v y 2 2g (0866 760 m/s)2 = 2(981 m/s 2 ) = 2208 m