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notes_3 - Motion in 2- and 3-dimensions Examples: •...

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Unformatted text preview: Motion in 2- and 3-dimensions Examples: • non-linear motion CHAPTER 3 (circles, etc.) MOTION IN TWO & THREE DIMENSIONS • General properties of vectors o the displacement vector o position and velocity vectors o acceleration vector o equations of motion in 2 and 3-dimensions • flight of projectiles (shells, golf balls, etc.) Motion in 2- and 3-dimensions means we have to deal with VECTORS ... • Projectile motion o path of projectile o time of flight o range HOORAY !! “What are you so happy about? What the heck is a vector?” Basically there are two main types of physical Examples of vectors: quantities we have to deal with: SCALARS VECTORS • displacement (The result of going E → W is not the same as going W → E .) Size Size Direction • force (The result of pushing something UP is not the same as Scalars have size (magnitude) only, e.g., time, and they pushing it DOWN.) can be added, substracted, multiplied and divided • velocity according to the rules of simple math, i.e., ( + , − , × , ÷ ). • acceleration Other examples: You cannot use the simple rules of math with vectors! • ordinary numbers • mass Let’s look more closely • distance (NOT displacement) at some examples ... • speed (NOT velocity) Vectors have the additional complication of direction to take into account. The displacement vector Military Trail Party at Spanish River Park Y’all invited! Follow the map ... Military Trail D US1 Spanish River Blvd. F C E Spanish River Blvd. A1A B Glades Rd. A1A F D US1 C Palmetto Park Rd. A Glades Rd. Alternatively: Palmetto Park Rd. 35o N of E D Displacement (Resultant) 7.3km 55o E of N C B A 4.2km 6.0km A C B E A B Note that: AB + BC = 10.2 km ≠ AC →→→ but AB + BC = AC Vector equation 35o N of E →→→ Note that: AD + DC = AC Vector equation Military Trail D US1 A1A F C E Spanish River Blvd. All parallel vectors of the same length are equivalent. B Glades Rd. Palmetto Park Rd. A but these two are not equivalent ... WHY? Note, yet another alternate route will give the same final displacement! F C • ADDITION r C rrr A+B=C r B r A A E Vector equation • SUBTRACTION →→→→ AE + EF + FC = AC Is the distance travelled the same as before? r B r A r D r −B r r r A + ( − B) = D r i.e., r r A−B=D Multiplying a vector by a scalar: r A Working with vectors: r C r A r r rr C = 2A ⇒ A + A r A r A r C r r rrrr C = 4A ⇒ A + A + A + A r A Draw them to scale, or Use components and trigonometry y r A Ay r A r A vector ( A ) multiplied by a scalar (n) is another vector r r (C), parallel to A but with magnitude (size) n times as large, • • r r i.e., C = nA r r and C = n A . means magnitude x θ Ax In component form: r A = (A x , A y ), r where A = A x2 + A y2 . r r The components are: A x = A cos θ and A y = A sin θ , Ay . where tan θ = Ax Adding vectors in component form: rrr A+B=C y By r C Ay θ r B r A Ax x Bx In component form: r r A = (A x , A y ) and B = (B x , B y ) r The resultant is C = (C x , C y ), r where C = ( A x + B x )2 + ( A y + B y )2 . Also C x = ( A x + B x ) and C y = ( A y + B y ) Cy and tan θ = . Cx (Similarly in 3-dimensions) Question 1: A bear walks north-east for 12m then walks east for another 12m. (a) Draw the displacement graphically and find the resultant displacement. (b) If the bear walks at 0.50m/s, what is his average velocity? rrr A+B=C N r B r A r displacement C =. (b) Average velocity ⇒ total time t r C θ r B E r (a) In component form: A = ( A x , A y ). Ax r = cos 45o ∴ A x = 12 × 0.707 = 8.49 m. A Ay r = sin 45o ∴ A y = 12 × 0.707 = 8.49 m. A r Similarly: B = (B x , B y ) = (12 m, 0). r ∴ C = (C x , C y ) where and Then: and: C x = A x + B x = 20.49 m C y = A y + B y = 8.49 m. 8.49 = 22.5o (N of E) θ = tan −1 20.49 r C = C x2 + C y2 = 491.9 = 22.2 m. 12m = 24.0s , 0.5m/s 12m = 24.0s . and time for B → C is 0.5m/s Time for A → B is ∴ t = 48.0s . 22.2m ∴ vav = = 0.46m/s . 48.0s As we will see below, velocity is a vector and so to be precise the average velocity is 0.46m/s in a direction 22.5o N of E. Often when we use vectors in component form, we will Position and velocity vectors use “unit vectors”: z ˆ j They indicate direction only unit vectors ˆ i P2 ( t 2 ) r r2 ijˆ So ˆ = ˆ = k = 1. y Ay x Ax r A r A x = A xˆ i r Ay = Ayˆ j x Use in vector addition ... r r if A = (A x , A y ) and B = (B x , B y ), r r i j i j then A = A x ˆ + A y ˆ and B = B x ˆ + B y ˆ . So: rr ˆ A ± B = ( A x ± B x )ˆ + ( A y ± B y )ˆ + ( A z ± B z ) k i j in 3-dimensions ˆ k ˆ i r ∆r r r1 y P1 ( t1 ) ˆ j A particle with position vector at any instant r ˆˆ ˆ r = xi + yj + zk moves from P1 ( t1 ) to P2 ( t 2 ). Then, if rrr ∆ r = r2 − r1 and ∆t = t 2 − t1 the average velocity vector is: r ∆r r v av = . ∆t r The instantaneous velocity at any point r is: r r r ∆ r = dr = dx ˆ + dy ˆ + dz k ˆ v = Limit i j dt dt dt ∆t→0 ∆t dt r ˆ i.e., v = v x ˆ + v y ˆ + v z k i j Acceleration vector z r v2 ˆ k P2 ( v 2 , t 2 ) P2 ( t 2 ) r r2 r ˆˆ ˆ r = xi + yj + zk x r ∆r y r r1 ˆ i rr r ∆v = v 2 − v 1 P1 ( t1 ) y z r 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 r instantaneous velocity, v . r v1 P1 ( v 1, t1 ) r v2 ˆ j r r ∆ r , the velocity Since the velocity vector is v = Limit ∆t→0 ∆t r at any point r : r ˆ v=v ˆ+v ˆ+v k i j x r v1 A particle with velocity vector at any instant r ˆ v=v ˆ+v ˆ+v k i j x y z moves from P1 ( v1, t1 ) to P2 ( v 2 , t 2 ). Then, if rr r ∆v = v 2 − v1 and ∆t = t 2 − t1 the average acceleration vector is: r ∆v r a av = . ∆t Thus, the instantaneous acceleration is: r r r ∆v = dv = dv x ˆ + dv y ˆ + dv z k ˆ a = Limit i j dt dt dt dt ∆t→0 ∆t r ˆ i j i.e., a = a x ˆ + a y ˆ + a z k r r a is parallel to ∆v r v DISCUSSION PROBLEM [3.1]: r a A ski-jumper “flies” off the edge of a cliff. What is the direction of his acceleration at A, B, X, Y ? r v r a r r Since a is parallel to ∆v , the direction of the acceleration A is towards the “inside” of a curve. X B Note: on a curve or turn, there is always an acceleration even though the magnitude of the velocity (i.e., speed) may be constant. Y When t = 0 : r ro = ( x o , y o ) = 10ˆ m = (10 m, 0) i r v o = ( v o x , v o y ) = (0, 0) m/s r a = (a x , a y ) = (6, 4) m/s2 = 6ˆ + 4ˆ m/s2 i j rrr (a) From Ch 2: v = v o + at ⇒ v = v o + at rrr ˆ ˆ ∴ v = v + at = (6ˆ + 4ˆ ) t = 6 ti + 4 tj m/s. i j ( ( o Question 2: An object experiences a constant r i j acceleration of a = 6ˆ + 4ˆ m/s2. At time t = 0, it is at r rest and its position vector is r = 10ˆ m. (a) Find the i ( ) 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. ) ) 1r 1r rr r rrr r − ro = v o t + at 2 ∴ r = ro + v o t + at 2 2 2 1 r i i j i j i.e., r = 10ˆ + (6ˆ + 4ˆ ) t 2 = (10 + 3t 2 )ˆ + 2 t 2 ˆ m. 2 r ˆˆ (b) But r = ( x, y ) = xi + yj. ∴ x = 10 + 3t 2 and y = 2 t 2 . Hence t 2 = y 2 , so x = 10 + 3 2 y , i.e., y = 2 3 x − 6.67 () () y( m) 10 Slope = 2 3 x( m) 0 −6.67 m −5 20 10 −6.67 This is a “relative velocity” problem. C 45o N B θ ˆ j 520 km ˆ i A → Question 3: A small plane departs from point A heading for an airport 520km due north at point B. The airspeed of the plane, i.e., relative to the air, is 240km/h. There is a steady wind blowing at 50km/h from northwest to southeast. (a) What is the proper heading to make the trip from A to B? (b) How long does the journey take? (a) The resultant flight path (AB) is the result of the → plane setting out along AC but being “redirected” by a → wind along BC. Then r r r rAB = rAC + rBC. r r But v = dr dt . r r r ∴ v AB = v AC + v BC = (0, v y ) . r Since v AC = −240 sin θ ˆ + 240 sin θ ˆ i j r v BC = 50 cos 45o ˆ − 50 sin 45o ˆ , i j and r v AB = ( −240 sin θ + 50 cos 45o , 240 cos θ − 50 sin 45o ) . ∴ −240 sin θ + 50 cos 45o = 0, 50 cos 45o ⇒ θ = 8.47o (W of N). i.e., sin θ = 240 Projectile motion • thrown objects (baseballs, arrows, shells) • bouncing balls C 45o N B θ ˆ j 520 km ˆ i A vy = 0 r (b) ∴ v AB = (0, 240 cos θ − 50 sin 45o ) = (0, 202) km/h ⇒ Effective velocity of plane. vo So, time to travel 520 km 520 km ⇒ = 2.57 h (2h 34min). 202 km/h • First investigated by Galileo. • Historically important for the military. Analysis of projectile motion: r v vy Assume no component of acceleration in the x-direction r v (i.e., no air resistance/drag), then: vx r v vy vx * v x ( t ) ⇒ constant = v o x. But in the y-direction: r v vy * v y ( t ) = v o y − gt . vx The components of the displacement are: y vy v ox r vo * x( t ) = x o + v ox t 1 * y ( t ) = y o + v o y t − gt 2 . 2 v oy θo ( xo , y o ) vx x We can consider the x-y components separately: Important concept!! (As we show below, these two equations represent a “ parabola”). These four equations (*) are the equations of motion for a projectile (with no air resistance). The initial velocity components are: v o x = v o cos θo and v o y = v o sin θo The acceleration components are: a x = 0 and a y = − g ** We ignore air resistance/drag forces ** NOTE: the equations of motion for the horizontal (x) and vertical (y) components are separate! The only link between them is ... time (t). We have * x( t ) = x o + v ox t , and 1 * y ( t ) = y o + v o y t − gt 2 . 2 ∆t ∆t x − xo From the first equation we have t = . vox ∆t ∆t Substituting in the second equation we find x − xo 1 x − xo y − yo = voy − g vox 2 vox = voy vox (x − xo ) − 2 g 2 2 (x − xo ) . 2( v o x ) If we shift the origin so that X = x − x o and Y = y − y o , then we get Y = aX + bX2, where a and b are constants. This is the equation of a parabola. Note, if x o = y o = 0 , g 2 y = x tan θo − 2x . 2( v o cos θo ) Equal vertical distances ∆y traveled in the same time interval ∆t . ∆t ∆t Flash photographs taken at equal intervals of time show these two objects fall at the same rate even though one has a horizontal velocity component. The x- and y-components of the motion can be treated quite separately ... they are “connected” only through time. BANG !! Here’s some really weird stuff !! 12 gt 2 BANG !! 12 gt 2 12 gt 2 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 If a rock is dropped from the same height as a rifle at the same vertical rate under gravity ... 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 ... 1 i.e., ... a distance gt 2 in t seconds, 2 so if the ground is flat, they will hit the ground at the same time! BANG !! 12 gt 2 ... even if you fire upward or downward! If the take-off and landing are in the same horizontal r v plane the horizontal range is: vyˆ j R = voxT θ v xˆ i where T is the total flight time. Range v sin θo ∴ R = ( v o cos θo ) × 2 o g v 2 2 sin θo cos θo =o . g At every point on the path, vy tan θ = . vx At maximum height v y = v o y − gt = 0 ∴ v o sin θo = gt So, the time to reach maximum height is v sin θo t= o . g If the take-off and landing are in the same horizontal plane, the total time of flight is: v sin θo T=2 o . g v2 i.e., R = o sin 2θo . g Maximum range ⇒ when θo = 45o . But sin 2θ = 2 sin θ cos θ and sin θ = cos(90o − θ) ∴ sin 2θ = 2 cos(90o − θ)cos θ i.e., R (90o − θ) = R (θ). So ... R (30o ) = R (60o ), etc. 75o 60o 15o 30o 45o Alternative derivation of range: When x = R 2, i.e., at mid-range, dy dx = 0. y x R R 2 Earlier we derived the equation of the trajectory: g 2 y = x tan θo − 2x . 2( v o cos θo ) ∴ dy g x, = tan θo − dx ( v o cos θo )2 so tan θo − ∴R = g R 2 2 = 0. ( v o cos θo ) 2 v o2 tan θo cos2 θ o g v2 = o sin 2θo. g Question 4: A pitcher throws a fastball at 140km/h towards home plate, which is 18.4m away. Neglecting air resistance, how far does the ball drop because of gravity by the time it reaches home plate? ( y − yo ) yo 18.4m ( x − x o ) = 18.4 m : v o x = 140km/h : a x = 0 ( y − y o ) = ? : v o y = 0 : a y = −9.81m/s2. Convert 140 km/h ⇒ m/s ... 140 × 1000 = = 38.9 m/s (~87 mi/h) 60 × 60 tower 40m tall with an initial speed of 42.2m/s at angle of 30o above the horizontal. How far is the cannonball from the tower when it strikes the ground? The time of flight: t= Question 5: A cannonball is fired from the top of a x − x o 18.4 m = = 0.473s. 38.9 m/s vox 1 1 ∴ ( y − y o ) = v o y t + a y t 2 = × ( −9.81) × (0.47)2 2 2 = −1.10 m. Earlier we saw that the maximum range is obtained when 30o the initial angle is θ = 45o , independent of v o. Since we 40 m have an algebraic expression for the trajectory of a R o v o = 42.2 m/s : θo = 30 : ( x o , y o ) = (0, 40m) : a x = 0 a y = −9.81 m/s2. v o x = v o cos 30o = 36.5m/s v o y = v o sin 30o = 21.1m/s 1 From notes y − y o = v o y t + a y t 2 2 1 i.e., 0 − 40 = 21.1 × t + × ( −9.81) × t 2 2 ∴ 4.91t 2 − 21.1t − 40 = 0 (quadratic equation) t= 21.1 ± ( −21.1)2 − 4 × 4.91 × ( −40) 2 × 4.91 projectile, i.e., y = f ( x ), we can use calculus to determine other properties. For example, what initial angle results in the longest path length (L max ) of the projectile? Sarafian (The Physics Teacher, 37, 86 (1999)) derived an expression for the path length as a function of θ , 1 + sin θ v2 . L(θ) = o sin θ + cos2 θ ⋅ ln cos θ g The maximum path length occurs when dL(θ) = 0 . Now, dθ 1 + sin θ dL(θ) v2 , = 2 o cos θ 1 − sin θ ⋅ ln cos θ dθ g ∴ t = +5.72s or t = −1.42s s ?? so the maximum path length occurs when 1 + sin θ sin θ ⋅ ln = 1. cos θ Range R = v o x t = 36.5 × 5.72 = 209 m. Note that this expression is independent of v o and g. 30 The most convenient way to solve this expression is dL(θ) graphically; so, we plot L(θ) and vs θ ... dθ dL(θ) dθ 40 v o = 15m/s L(θ) 30 L(θ) m dL(θ) dθ 20 θ 0 L(θ) 100 10 dL(θ) dθ 80 60o 30o 60 θ 56.46o 90o o The value of θ that maximizes the path length is 56.46 . Although the path length (L) depends on v o and g, the angle that maximizes the path length does not, as shown in the following plots. θ 60 Below are the plots with v o = 10m/s : −10 L(θ) 0 1 + sin θ dL(θ) v2 = 2 o cos θ 1 − sin θ ⋅ ln . cos θ dθ g 0 v o = 10m/s 10 v2 1 + sin θ L(θ) = o sin θ + cos 2 θ ⋅ ln , cos θ g 20 L(θ) m dL(θ) dθ 20 v o = 20m/s L(θ) 40 20 θ 0 0 56.46o 90o Path length L(θ) for various initial velocities v o . Note that the maximum path length is independent of v o . ...
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