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Digunakan untuk mengetahui posisi dalam representasi 3 dimensi› Posisi Lintang› Posisi Bujur› Ketinggian
Dapat pula mengetahui kecepatan› Arah dan besar kecepatanTerdapat fasilitas tracking› Perjalanan tidak selamanya membentuk garis lurus› Kadang berbelok, menanjak dan menurun
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Pergerakkan umumnya tidak seperti garis lurus melainkan dalam 2 atau 3 dimensi.
Posisi awal
Bergerak
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Tanda panah adalah vektor kecepatan pelari di suatu titik di sepanjang jalanPada waktu sesaat, vektor kecepatan adalah tangen dari linatasan
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Dua kali panah sebelumnya
Vektor digunakan untuk menganalisis gerak pada dua atau tiga dimensi
30 km/jam
60 km/jam
Panah menunjukkan arah sedangkan panjangnya menunjukkan besar atau ukuran
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Pada 1 Dimensi, petunjuk arah digunakan tanda + atau ‐. Contoh, pada kasus jatuh bebas ay = ‐g.Pada 2 atau 3 dimensi, butuh selain tanda untuk menunjukkan arahContohContoh: Dimanakan posisi rr untuk Universitas Indonesia?› Pilih titik asal: MonasMonas› Pilih koordinat
jarak (km), danarah (N,S,E,W)
›› rr adalah suatu vektor yangmenunjukkan 20 km ke selatan.
Monas
UI
r r
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Ada dua cara untuk menunjukkan suatu besaran adalah vektor:
› Notasi tebal: AA
› Notasi “panah” :
A A = Ar
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Jika suatu gerak terjadi pada beberapa dimensi, kita dapat memisahkan gerak pada suatu komponen yang tegak lurus dan penjumlahan dapat membuat
suatu gerak yang berbeda
The net vector shows you the object’s actual direction of motion…
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Komponen rr adalah koordinat (x,y,z)› rr = (rx ,ry ,rz )= (x,y,z)
Pada kasus 2‐D (paling mudah dibuat):› rx = x = r cos θ › ry = y = r sin θ
y
x
(x,y)
θ
rrθ = arctan( y / x )
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Besar (panjang) rr didapatkan dengan teorema Pithagoras :
Panjang suatu vector tidaktidak tergantung dengan arahnya
22 yxr +==rrr
y
x
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A Unit Vector Unit Vector is a vector having length 1 and no unitsIt is used to specify a directionUnit vector uu points in the direction of UU
› Often denoted with a “hat”: uu = û
Useful examples are the Cartesian unit vectors [ i, j, ki, j, k ] › point in the direction of the x, y and z axes
UU
x
y
zii
jj
kk
û û
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Misalkan vektor AA dan BB. Carilah AA + BB
We can arrange the vectors as we want, as long as we maintain their length and direction!!
AA
BB
AA BB AA BB
CC = AA + BB
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Misalkan CC = AA + BB.(a) CC = (Ax ii + Ay jj) + (Bx i i + By jj) = (Ax + Bx)ii + (Ay + By)jj(b) CC = (Cx ii + Cy jj)
Menjumlahkan komponen (a) dan (b):› Cx = Ax + Bx
› Cy = Ay + By CC
BxAA
By
BB
Ax
Ay
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Vector A = {0,2,1}Vector B = {3,0,2}Vector C = {1,‐4,2}
What is the resultant vector, D, from adding A + B + C ?
(a)(a) {3,5,{3,5,‐‐1}1} (b)(b) {4,{4,‐‐2,5}2,5} (c)(c) {5,{5,‐‐2,4}2,4}
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D = (AXi + AYj + AZk) + (BXi + BYj + BZk) + (CXi + CYj + CZk)
= (AX + BX + CX)i + (AY + BY+ CY)j + (AZ + BZ + CZ)k
= (0 + 3 + 1)i + (2 + 0 ‐ 4)j + (1 + 2 + 2)k
= {4, ‐2, 5}
Adhi Harmoko S
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In free fall objects accelerate constantly toward Earth at the rate of g . Objects moving upward slow down until their direction is reversed, and then they accelerate downward.
At the top of their path the upward speed is zero. How long? Only instantaneously. A constant acceleration means the speed is changing all the time, so the speed only passes through the value of zero at the top of the path.
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A dropped ball:› Begins a rest, but soon acquires downward speed› Covers more and more distance each second
A tossed ball:› Rises to a certain height› Comes briefly to a stop› Begins to descend, much like a dropped ball
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Falling Upward, then Downward, with a constant horizontal velocity component
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Here two heavy balls begin “free fall” at the same time.
The red one is dropped, so it moves straight downward.
The yellow ball is given some speed in the horizontal direction as it is released.
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The horizontal lines show that they keep pace with each other in the vertical direction.
Why?
They have the same acceleration, g, downward, and they both started with zero speed in the downward direction.
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The yellow ball’s horizontal speed is not affected by gravity, which acts only in the vertical direction.
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Cannonballs shot horizontally with different speeds from the ship travel different distances.
But each cannonball drops the same distance in the same amount of time, since the vertical acceleration is the same for each.
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A simulated strobe illustration of a plane flying horizontally with constant speed dropping package of food and medical supplies, ignoring air resistance.
The package of food and medical supplies initially has the same horizontal speed of the airplane. Neglecting air resistance, it keeps that horizontal speed as it falls, so it stays beneath the airplane.
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The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as:
rr= x ii + y jj + z kkvv = vx ii + vy jj + vz kk (ii , jj , kk unit vectors )
aa = ax ii + ay jj + az kk
We have already seen the 1‐D kinematics equations:
2
2
dtxd
dtdv
a ==dtdx
v =)(txx =
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For 3‐D, we simply apply the 1‐D equations to each of the component equations
Which can be combined into the vector equationsrr= rr(t) vv = drr / dt aa = d2rr / dt2
2
2
dtxd
ax =
dtdx
vx =
)(txx =
2
2
dtyd
ay =
dtdy
vy =
)(tyy =
2
2
dtzd
az =
dtdz
vz =
)(tzz =
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So for constant acceleration we can integrate to get:›› aa = const›› vv = vv0 + a a t
›› rr= rr0 + vv0 t + 1/2 aa t2
(where aa, vv, vv0, rr, rr0, are all vectors)
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Most 3‐D problems can be reduced to 2‐D problems when acceleration is constant:› Choose y axis to be along direction of acceleration› Choose x axis to be along the “other” direction of motion
ExampleExample: Throwing a baseball (neglecting air resistance)› Acceleration is constant (gravity)› Choose y axis up: ay = ‐g› Choose x axis along the ground in the direction of the throw
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A man on a train tosses a ball straight up in the air.› View this from two reference frames:
Reference frame
on the moving train.
Reference frame
on the ground.
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Mark McGwire clobbers a fastball toward center‐field. The ball is hit 1 m (yo ) above the plate, and its initial velocity is 36.5 m/s(v ) at an angle of 30o (θ) above horizontal. The center‐field wall is 113 m (D) from the plate and is 3 m (h) high.› What time does the ball reach the fence?› Does Mark get a home run?
θvv
h
D
y0
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Choose y axis up.Choose x axis along the ground in the direction of the hit.Choose the origin (0,0) to be at the plate.
Say that the ball is hit at t = 0, x = x0 = 0
› Equations of motion are:
vx = v0x vy = v0y ‐ gtx = vxt y = y0 + v0y t ‐ 1/ 2 gt2
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Use geometry to figure out v0x and v0y :
y
x
g
θvv
v0xv0y
Find v0x = |v| cos θ.and v0y = |v| sin θ.
y0
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The time to reach the wall is: t = D / vx (easy!)We have an equation that tell us
y(t) = y0 + v0y t + a t2/ 2
So, we’re done....now we just plug in the numbers:Find:› vox = 36.5 cos(30) m/s = 31.6 m/s› voy = 36.5 sin(30) m/s = 18.25 m/s› t = (113 m) / (31.6 m/s) = 3.58 s› y(t) = (1.0 m) + (18.25 m/s)(3.58 s) ‐ (0.5)(9.8 m/s2)(3.58 s)2
= (1.0 + 65.3 ‐ 62.8) m = 3.5 mm› Since the wall is 3 m high, Mark gets the homer!!
A long jumper leaves the ground at an angle of 20.0 to the horizontal and at a speed of 11.0 m/s. (a) How long does it take for him to reach maximum height? (b) What is the maximum height? (c) How far does he jump? (Assume that his motion is equivalent to that of a particle, disregarding the motion of his arms and legs.) (d) Find the maximum height he reaches
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(a) Find the time tmax taken to reach maximum height.› Set vy = 0 in and solve for tmax:
Adhi Harmoko S
Motion in a circle with:
› Constant Radius R
› Constant Speed v = |vv|
R
vv
x
y
(x,y)
In general, one coordinate system is as good as any other:› Cartesian:
(x,y) [position]
(vx ,vy) [velocity]› Polar:
(R,θ) [position](vR ,ω) [velocity]
In UCM:
› R is constant (hence vR = 0).› ω (angular velocity) is constant.›› Polar coordinates are a natural way to describe UCM!Polar coordinates are a natural way to describe UCM!
R
vv
x
y
(x,y)
The arc length s (distance along the circumference) is related to the angle in a simple way:› s = Rθ, where θ is the angular displacement.› units of θ are called radians.
For one complete revolution:› 2πR = Rθc
› θc = 2π› θ has period 2π.
R
vv
x
y
(x,y)
θ
1 revolution 1 revolution = 2= 2ππ ��radiansradians
S
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x = R cos θy = R sin θ
π/2 π 3π /2 2π
‐1
1
0
sincos
R
vv
x
y
(x,y)
θ S
In Cartesian coordinates, we say velocity dx/dt = v.› x = vt
In polar coordinates, angular velocity dθ/dt = ω.› θ = ωt› ω has units of radians/second
Displacement s = vt.but s = Rθ = Rωt, so:
v = ωR
R
vv
x
y
(x,y)
θ S
Recall that 1 revolution = 2π radians› frequency (f) = revolutions / second (a)
› angular velocity (ω) = radians / second (b)By combining (a) and (b)› ω = 2π fRealize that:› period (T) = seconds / revolution› So T = 1 / f = 2π/ω
ω = 2π / T = 2πf
ωR
vv
x
y
(x,y)S
This is called This is called Centripetal Acceleration.Now let’s calculate the magnitude
v2
v1v2
Δv
R v1ΔR
R
R
v
v Δ=
ΔSimilar triangles:
But ΔR = vΔt for small Δt
So:R
v
t
v 2
=ΔΔ
R
tv
v
v Δ=
Δ
R
va
2
=
UCM results in acceleration:› Magnitude: a = v2 / R› Direction: circle)of center (toward r̂ ‐
aaR
ω
We know that v = ωR
Substituting for v we find that:
a = ω2R
andRv
a2
=
( )R
Ra
2ω=
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A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ?
gRv
a 92
==2
2
2
2
81.99
90000
9smsm
gv
R×
==
mmR 100081.9
10000≈= mRD 20002 ≈=
2 km
See you on monday