4. Two and Three Dimensional Motion
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1 4. Two and Thee Dimensional Motion 1 Descibe motion using position, displacement, elocity, and acceleation ectos Position ecto: ecto fom oigin to location of the object. = x i ˆ + y ˆ j + z k ˆ Displacement: 1D: x = x 2 x 1 3D: = 2 1 = ( x 2 x 1 )ˆ i + ( y 2 y 1 )ˆ j + ( z 2 z 1 )ˆ k xˆ i + y j ˆ + z k ˆ Velocity: 1D: = dx
2 3D: = d ( ) = d x i ˆ + y ˆ j + z k ˆ = dx i ˆ + dy ˆ j + dz k ˆ xˆ i + y ˆ j + ˆ z k x = dx y = dy z = dz Acceleation: 1D: a = d = d2 x 2 2
3 3D: a = a xˆ i + a y ˆ j + a ˆ z k = d = d2 2 a x = d x a y = d y = d2 x 2 = d2 y 2 a z = d z = d2 z 2 The definition of indiidual components of a 3D ecto is the same as 1D. Witing the equation in ecto fomat (i.e. = d ) is a compact way of witing thee equations, one fo each component. On most cases, wok with each component sepaately; no need to sole eeything at once. Poblem 12 (p. 73) 3
4 The position of a paticle is gien by = ( 2.00t t) i ˆ + ( t 4 ) ˆ j, whee is in metes and t in seconds. Calculate (a), (b), (c) a at t = 2.00 s, (d) What is the oientation of a line tangential to the paticle's path at t = 2.00 s? (a) ( t = 2.00 s) = 2.00( 2.00) ( 2.00) i (b) ( t) = dx [ ] ˆ + [ ( 2.00) 4 ] j ˆ = 6.00 i ˆ 106 ˆ j i ˆ + dy j ˆ = d [ 2.00t3 5.00t] i ˆ + d [ t ] ˆ j = ( 6.00t ) i ˆ + ( 28.0t 3 ) ˆ j 4
5 [ ] i ˆ ( t = 2.00 s) = 6.00( 2.00) (c) a ( t) = d x 28.0( 2.00) 3 ˆ j = 19.0ˆ i 224 j ˆ ˆ i + d y ˆ j 5 = d [ 6.00t2 5.00] i ˆ + d [ 28.0t3 ] j ˆ = 12.0t i ˆ 84.0t 2 ˆ j a ( t = 2.00 s) = 24.0 i ˆ 336 j ˆ (d) The oientation is along the elocity, which is tangent to the path of the paticle. Constant Acceleation: If acceleation is constant then each component of acceleation is also constant, i.e. a = constant a x = constant a y = constant a z = constant
6 6 Equation of Motion: Vecto Fom: = t = + a t 2 = a Component Fom: x - diection x = x 0 + ox t a x t2 x = ox + a x t a t 2 ( ) 0 y - diection y = y 0 + 0y t a yt 2 y = 0y + a y t z - diection z = z 0 + 0z t a zt 2 z = 0z + a z t 2 x = 2 0x + 2a x ( x x 0 ) 2 y = 2 0y + 2a y ( y y 0 ) 2 z = 2 0z + 2a z ( z z 0 ) Time is the only common aiable between motion in diffeent diections Equation of motion fo each diection is independent of any othe quantities fom othe diections Motion in each diection is independent of motion in othe diections
7 Example: Spacecaft P131 is flying initially in a staight line with a speed of 15 km/s. Suddenly, it fies thustes which poduce an acceleation of 0.1 km/s2 in the diection pependicula to the diection of motion fo 100 s. Afte the thustes ae tuned off, (a) what is the elocity of the spacecaft with espect to the oiginal diection of motion? (b) What is the speed? (c) How fa is it fom the point it stated fiing the thustes? y 7 t = 100 s =? =? 2 a = 0.1 km/s t = 0 0 = 15 km/s x
8 (a) y = 0y + a y t 8 = 0 + ( 0.1 km / s 2 )100 ( s) =10 km / s x = 0x =15 km / s (b) = x 2 + y 2 = ( 15 km / s) 2 + ( 10 km / s) 2 = 18 km / s (c) x = x x t a x t2 = 0x t = ( 15 km / s) ( 100 s) = 1500 km
9 y = y y t a y t 2 9 = 1 2 a y t2 = 1 2 ( 0.1 km / s2)100 ( s) 2 = 500 km = x 2 + y 2 = ( 1500 km) 2 + ( 500 km) 2 = 1600 km Soled each component sepaately and combined then at the end when necessay. Questions: (a) Which path best epesents the path of the spacecaft? y A B C x (b) If spacecaft P132 is flying next to P131 with same initial elocity, but does not fie thustes,
10 which plot best epesents the position of the two cafts at 0, 25, 50, 75, and 100 s? A B C 10 P131 P132 Pojectile Motion 2D motion: etical and hoizontal Hoizontal motion is independent of etical motion. Time is the only aiable that connects the two motions. Acceleation due to gaity acts only in the etical diection. Hoizontal elocity emains constant (neglecting ai esistance) At any gien height, the magnitude of the etical elocity is the same whethe ising o falling. At the maximum height, etical elocity
11 is zeo. The tajectoy is paabolic. V y 0 y θ 0 V 0 x V 1 y V 0 x V = 0 2 y g V 0 x -V 1 y V 0 x V -V 0 x 0 y x 11 Hoizontal Motion: x = x x t = x ( cosθ 0 )t Vetical Motion:
12 12 y = y y t 1 2 gt2 = y ( sinθ 0 )t 1 2 gt2 y = 0 y gt = 0 sinθ 0 gt 2 y = 2 0 y 2g( y y 0 ) = ( 0 sinθ 0 ) 2 2g( y y 0 ) Question: If a bullet is fied diectly towad a ball that has just been eleased, will the bullet hit the ball? y d H 0 θ 0 R x
13 Plan: Calculate the bullet flight time t and then compae the etical locations of bullet and ball x = x ( cosθ 0 )t R = ( cosθ 0 )t t = R 0 cosθ 0 13 y bullet = y ( sinθ 0 )t 1 2 gt2 R = sinθ cosθ g R cos 2 θ 0 gr 2 = Rtanθ cos 2 θ 0 = R H R gr cos 2 θ 0 gr 2 = H cos 2 θ 0 y ball = y t 1 2 gt2 = H g R 2 = y bullet!!! 0 2 cos 2 θ 0
14 14 The bullet will hit the ball fo any H, R, 0, and θ 0. Example: A pojectile is fied with an initial speed 0 = 30.0 m / s at a taget R = 20.0 m away. Find (a) the two possible fiing angles and (b) the maximum height fo the highe tajectoy. One unknown (θ), but two equations of motion (etical and hoizontal). Use time that connects the two equations as the second unknown. y V 0 V 0 R (a) x = x ( cosθ)t R = t = ( cosθ)t R 0 cosθ x
15 y = y ( sinθ)t 1 2 gt 2 0 = ( sinθ)t 1 2 gt 2 0 ( sinθ)t = 1 2 gt 2 0 sinθ = 1 2 gt gr 0 sinθ = 2 0 cosθ 2sinθ cosθ = gr 2 0 sin2θ = gr
16 (b) 2θ = sin 1 gr 2 0 θ = 1 2sin 1 gr 0 2 ( ) 20.0 m 9.8 m / s 2 ( ) = 1 2 sin 1 ( 30 m / s) 2 = 1 2sin 1 ( 0.218) = 1 ( o o 167 o ) = 6.4 o o 83.7 o ( ) 2 = 2 0y 2g y y 0 0 = ( 0 sinθ) 2 2gh max h max = ( sinθ 0 )2 2g = 30 m / s ( )sin83.7o [ ] m / s 2 = 45.4 m 16
17 Conceptual Question: 1. A P131 student thows a ball staight up and the othe staight down with same initial speed. Neglecting ai esistance, the ball to hit the gound with geate speed is the one thown: (a) upwad (b) downwad (c) neithe - they both hit at the same speed 2. The student thows one ball hoizontally and at the same time eleases anothe ball fom est. Which hits the gound fist? (a) the ball thown hoizontally (b) the ball eleased at the same time (c) both hit at the same time 3. The figue below shows thee paths fo a kicked football. Ignoing ai esistance, ank the paths accoding to (a) time of flight, (b) initial etical elocity, (c) initial hoizontal elocity, and (d) initial speed. 17
18 y 18 a b c x Unifom Cicula Motion: An object moes in a cicle with a constant speed (not elocity!): Since elocity is tangent to the path, the elocity is always tangent to the cicle. The speed is constant, i.e. the length of the elocity ecto does not change but the diection does change a motion with a constant speed but changing elocity! Since the elocity ecto is changing
19 thee is an acceleation, i.e. the object is acceleating but the speed is constant! Centipetal Acceleation: y #1 θ #2 θ θ θ x 19
20 t = distance = (2θ) = cosθ 1x 1y 2 x 2y = sinθ = cosθ = sinθ 20 a x a y = x t cosθ cosθ = t = 0 = y t sin θ sinθ = 2θ = 2 sin θ θ
21 a y = lim t 0 lim θ 0 = 2 θ θ a y a y 21 = 2 The acceleation always points diectly towad the cente centipetal acceleation, i.e. "cente seeking" a a a a a Like the elocity ecto, the diection of acceleation is always changing. Acceleation is not constant but the magnitude is a constant:
22 22 a c = 2 Peiod of Reolution: T = cicumfeence speed = 2π
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