7 Jan 2008 ... Pada 1 Dimensi, petunjuk arah digunakan tanda + atau -. Contoh, pada ... Pada
2 atau 3 dimensi, butuh selain tanda untuk menunjukkan arah.
7/1/2008
Digunakan untuk mengetahui posisi dalam representasi 3 dimensi › Posisi Lintang › Posisi Bujur › Ketinggian Dapat pula mengetahui kecepatan › Arah dan besar kecepatan Terdapat fasilitas tracking › Perjalanan tidak selamanya membentuk garis lurus › Kadang berbelok, menanjak dan menurun
Posisi awal
Bergerak
Pergerakkan umumnya tidak seperti garis lurus melainkan dalam 2 atau 3 dimensi.
Tanda panah adalah vektor kecepatan pelari di suatu titik di sepanjang jalan Pada waktu sesaat, vektor kecepatan adalah tangen dari linatasan
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Vektor digunakan untuk menganalisis gerak pada dua atau tiga dimensi
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Panah menunjukkan arah sedangkan panjangnya menunjukkan besar atau ukuran
30 km/jam 60 km/jam
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 arah Contoh: Dimanakan posisi r untuk Universitas Indonesia? Contoh › Pilih titik asal: Monas Monas › Pilih koordinat Monas x jarak (km), dan r x arah (N,S,E,W) › r adalah suatu vektor yang
Dua kali panah sebelumnya
menunjukkan 20 km ke selatan.
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UI
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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
Ada dua cara untuk menunjukkan suatu besaran adalah vektor: › Notasi tebal: A A
r A = A A › Notasi “panah” :
The net vector shows you the object’s actual direction of motion…
Komponen rr adalah koordinat (x,y,z) › r = (rx ,ry ,rz ) = (x,y,z)
Pada kasus 2‐D (paling mudah dibuat): › rx = x = r cos θ › ry = y = r sin θ
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(x,y)
y r
θ = arctan( y / x )
θ
x
Besar (panjang) rr didapatkan dengan teorema Pithagoras :
r
y
r = r = x2 +y2
A Unit Vector Unit Vector is a vector having length 1 and no units It is used to specify a direction Unit vector u u points in the direction of U U › Often denoted with a “hat”: u u = û
U
û
x Panjang suatu vector tidak tidak tergantung dengan arahnya
Useful examples are the Cartesian unit vectors [ i, j, k ] › point in the direction of the x, y and z axes
y j
z 11
k
i
x
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Misalkan vektor A A dan B B. Carilah A A + B B
A
A
B
Misalkan C C = A + B. (a) C = (Ax i + Ay j ) + (Bx i + By j ) = (Ax + Bx)i + (Ay + By)j (b) C = (Cx i + Cy j )
Menjumlahkan komponen (a) dan (b): › Cx = Ax + Bx › Cy = Ay + By
B
A
C= A + B
B
By B
C
We can arrange the vectors as we want, as long as we maintain their length and direction!!
A
Ay
Bx
Ax 13
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 ?
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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}
{3,5,‐‐1} (a) {3,5,
(b) {4, {4,‐‐2,5}
(c) {5, {5,‐‐2,4}
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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. 18
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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.
The yellow ball’s horizontal speed is not affected by gravity, which acts only 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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Cannonballs shot horizontally with different speeds from the ship travel different distances.
A simulated strobe illustration of a plane flying horizontally with constant speed dropping package of food and medical supplies, ignoring air resistance.
But each cannonball drops the same distance in the same amount of time, since the vertical acceleration is the same for each.
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. 25
The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as:
(ii , j , k unit vectors )
We have already seen the 1‐D kinematics equations: x = x (t )
For 3‐D, we simply apply the 1‐D equations to each of the component equations x = x (t )
r = x i + y j + z k v = vx i + vy j + vz k a = ax i + ay j + az k
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v=
dx dt
a=
dv d2 x = dt dt 2
y = y(t )
z = z(t )
vx =
dx dt
vy =
dy dt
vz =
dz dt
ax =
d 2x dt 2
ay =
d2 y dt 2
az =
d 2z dt 2
Which can be combined into the vector equations v = dr / dt a = d2 r / dt2 r = r(t)
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So for constant acceleration we can integrate to get: › a = const › v = v 0 + a t › r = r0 + v 0 t + 1 /2 a t2 (where a a, vv , vv 0 , 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
Example Example: 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.
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?
v y0
h
θ D
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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
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Use geometry to figure out v0x and v0y :
y
g v
› Equations of motion are:
vx = v0x x = vxt
y0
θ v0x
vy = v0y ‐ gt y = y0 + v0y t ‐ 1 / 2 gt2
v0y
x Find and
v0x = |v | cos θ. v0y = |v | sin θ.
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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 m › Since the wall is 3 m high, Mark gets the homer!!
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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 = |v v|
y v (x,y)
R
x
In general, one coordinate system is as good as any other: › Cartesian: y [position] x (x,y) v x (vx ,vy) [velocity] (x,y) R › Polar: [position] x (R,θ) [velocity] x (vR ,ω) In UCM: › R is constant (hence vR = 0). › ω (angular velocity) is constant. › Polar coordinates are a natural way to describe UCM!
x
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. y
For one complete revolution: › 2πR = R θc › θc = 2π › θ has period 2π.
v R
(x,y)
θ
S x
1 revolution = 2 = 2π π radians
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y
v
x = R cos θ y = R sin θ
R
(x,y)
θ
S x
In Cartesian coordinates, we say velocity dx/dt = v. › x = vt In polar coordinates, angular velocity dθ/dt = ω. › θ = ωt y › ω has units of radians/second
1 cos
0
sin π/2
π
3 π /2
2π
v
Displacement s = vt. but s = Rθ = Rωt, so:
(x,y)
R
θ
S
v = ωR
x
‐1
Recall that 1 revolution = 2π radians › frequency (f) = revolutions / second › angular velocity (ω) = radians / second By combining (a) and (b) › ω = 2π f y Realize that: › period (T) = seconds / revolution › So T = 1 / f = 2π/ω ω
UCM results in acceleration: › Magnitude: a = v2 / R › Direction: ‐ rˆ (toward center of circle)
This is called Centripetal Acceleration. This is called Now let’s calculate the magnitude
Δv v2
R
But ΔR = vΔt for small Δt
R
ΔR
Δv v 2 = Δt R
So: Δ v = v Δ t
v2
(x,y) S
Δv ΔR = v R
Similar triangles: v1
v
x
ω = 2π / T = 2πf
(a) (b)
v
R
v1 a=
We know that a =
v2 R
v2 R
and v = ωR
Substituting for v we find that:
a
ω
a=
R
(ωR )2 R
a = ω 2R
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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 9 times 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 ?
m2 90000 2 v2 s R= = 9g 9 × 9 .81 m s2
v2 a = = 9g R R=
10000 m ≈ 1000m 9.81
D = 2R ≈ 2000m
2 km
See you on monday
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