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7/1/2008
1
Berbelok
String & Spring
Accelerator
1
Ketika naik kendaraan yang sedang berbelok ke kiri› Tubuh kita terdorong ke kanan
2
Gerak Melingkar Beraturan mempunyai› Magnitude:
› Direction: circle)of center (toward r̂ ‐
aaR
ω
Rv
a2
=
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Sebuah bola massa m diikat dengan sebuah benang panjang r dan diputar dengan kecepatan konstan pada lintasan mendatar
Mengapa bola dapat berputar melingkar?
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Karena inersia bola› Bola cenderung bergerak lurus› Benang menahan bola
› Timbul gaya pada benang› Gaya inilah yang menyebabkan bola berputar
Hukum Newton 2
Gaya tersebut mengubah arah kecepatan bola
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R
vmmaF rr
2
==∑
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A ball of mass 0.500 kg is attached to the end of a cord 1.50 m long. The ball is whirled in a horizontal circle as was shown in Figure. If the cord can withstand a maximum tension of 50.0 N, what is the maximum speed the ball can attain before the cord breaks? Assume that the string remains horizontal during the motion.
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Because the force causing the centripetal acceleration in this case is the force T exerted by the cord on the ball,
Solving for v, we have
The maximum speed the ball can have corresponds to the maximum tension.
Rv
mmaF rr
2
==∑Rv
mT2
=
mRT
v.
=
( )( )( ) m/s 12.2
kg 0.500
m 1.50N 50.0
m
T.Rv ===
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A small object of mass m is suspended from a string of length L. The object revolves with constant speed v in a horizontal circle of radius r, as shown in Figure. (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.) Find an expression for v.
9
In the free‐body diagram shown in Figure, the force T exerted by the string is resolved into a vertical component T cos θ and a horizontal component T sin θ acting toward the center of revolution. Because the object does not accelerate in the vertical direction,
0==∑ yy maF mgT =θcos
Rv
mmaTF rr
2
sin ===∑ θ
rgv2
tan =θ θtanrgv =
θθ tansinLgv =10
Bagaimana bila pengamat berada dalam ruang tidak inersiaMisalkan sebuah benda bergerak lintasan berputarKetika mobil berbelok ke kiri
Penumpang terdorong ke kanan› Gaya sentrifugalAda gaya misterius (fictitious force)› Mendorong ke kanan› Ruangan dipercepat
Bagaimana penjelasannya?
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Ketika mobil mau berbelok, penumpang bergerak lurusKetika mobil berbelok, penumpang cenderung bergerak lurus› Sesuai hukum Newton 1
› Benda bergerak cenderung untuk tetap bergerakGaya gesek penumpang dengan bangku menahan penumpang tidak bergeser› Bila gaya gesek kecil pasti penumpang akan bergeser
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Jika partikel bergerak dengan percepatan relatif terhadap pengamat dalam ruang inersia› Pengamat boleh menggunakan Hukum Newton 2Jika pengamat dalam ruang dipercepat, boleh menggunakan Hukum Newton 2 tetapi harus menyebutkan gaya misterius agar benarGaya tersebut ditemukan oleh pengamat dalam ruang dipercepat dan bukan merupakan gaya yang benar‐benar ada pada partikel
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Sebuah bola bermassa m digantung dengan menggunakan benang pada langit‐langit kereta yang dipercepat ke kanan seperti pada gambar. Carilah percepatan kereta bila sudut yang terbentuk sebesar 30°!
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Bila diamati oleh pengamat inersia
maTFx ==∑ θsin
0cos =−=∑ mgTFy θ
22 m/s 5,6630 tan m/s 9,81tanθ ga =°==
15
Bila diamati oleh pengamat tidak inersia
0sin' =−=∑ misteriusx FTF θ
0cos' =−=∑ mgTFy θ
mamaF inersiamisterius ==
22 m/s 5,6630 tan m/s 9,81tanθ ga =°==
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Can be used to pull from a distance.TensionTension (T) at a certain position in a rope is the magnitude of the force acting across a cross‐section of the rope at that position.› The force you would feel if you cut the rope and grabbed the ends.
› An action‐reaction pair
cut
TT
T
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Consider a horizontal segment of rope having mass m:› Draw a free‐body diagram (ignore gravity).
Using Newton’s 2nd law (in xx direction): FNET = T2 ‐ T1 = ma
So if m = 0 (i.e. the rope is light) then T1 = T2
T1 T2
m
a x x
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An ideal (massless) rope has constant tension along the rope.
If a rope has mass, the tension can vary along the rope› For example, a heavy rope
hanging from the ceiling...
We will deal mostly with ideal massless ropes.
T = Tg
T = 0
T T
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The direction of the force provided by a rope is along the direction of the rope:
mg
T
m
Since ay = 0 (box not moving),
T = mg
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A fish is being yanked upward out of the water using a fishing line that breaks when the tension reaches 180 N. The string snaps when the acceleration of the fish is observed to be is 12.2 m/s2. What is the mass of the fish?
m = ?a = 12.2 m/s2
snap ! (a) 14.8 kg
(b) 18.4 kg
(c) 8.2 kg
22
Draw a Free Body Diagram!!Use Newton’s 2nd lawin the upward direction:
T
mg
m = ?a = 12.2 m/s2 FTOT = ma
T ‐mg = ma
T= ma + mg
= m(g+a)
ag
Tm
+=
( )kg
sm
Nm 2.8
2.128.9180
2 =+
=
23 24
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Used to change the direction of forces› An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:
FF1ideal peg
or pulleyFF2
| FF1 | = | FF2 |
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Used to change the direction of forces› An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:
mg
T
m T = mg
FW,S = mg
27 28
Hooke’s Law: Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
› FX = ‐k x› Where x is the displacement from the elaxed position and k is the constant of proportionality
relaxed position
FX = 0
x
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Hooke’s Law: Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
› FX = ‐k x› Where x is the displacement from the elaxed position and k is the constant of proportionality
relaxed position
FX = ‐kx > 0
xx < 0
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Hooke’s Law: Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
› FX = ‐k x› Where x is the displacement from the elaxed position and k is the constant of proportionality
FX = ‐ kx < 0
xx > 0
relaxed position
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Springs can be calibrated to tell us the applied force.› We can calibrate scales to read Newtons, or...› Fishing scales usually read weight in kg or lbs.
02468
1 lb = 4.45 N
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A block weighing 4 lbs is hung from a rope attached to a scale. The scale is then attached to a wall and reads 4 lbs. What will the scale read when it is instead attached to another block weighing 4 lbs?
m m m
(a) (a) 0 lbs. (b) (b) 4 lbs. (c) (c) 8 lbs.
(1) (2)
?
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Draw a Free Body Diagram of one of the blocks!!Use Newton’s 2nd Lawin the y direction:› a = 0 since the blocks are stationary
mg
T
m T = mg
FTOT = 0
T ‐mg = 0
T = mg = 4 lbs.
34
A spring with spring constant 40 N/m has a relaxed length of 1 m. When the spring is stretched so that it is 1.5 m long, what force is exerted on a block attached to the end of the spring?
x = 0
Mk k
M
x = 0 x = 1.5
(a) ‐20 N (b) 60 N (c) ‐60 N
x = 1
35
Recall Hooke’s law: › FX = ‐kx › Where x is the displacement from equilibrium.
FX = ‐ (40) ( .5)
FX = ‐ 20 N
(a) ‐20 N (b) 60 N (c) ‐60 N
36
A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizon‐tal road, and has an acceleration a in the x direction. The string makes an angle θwith respect to the vertical (y) axis. Solve for θ in terms of a and g.
aθ
i i
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Draw a free body diagram for the mass:› What are all of the forces acting?
TT (string tension)
mgg (gravitational force)
θ
i i
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Using components (recommended):i: FX = TX = T sin θ = ma
j: FY = TY − mg = T cos θ − mg = 0
TT
mgg
θ
maa
jj
ii
θ
39
Using components :
i: T sin θ = ma
j: T cos θ ‐mg = 0
Eliminate T : TT
mgg
θ
maa
jj
iiT cos �= mg
T sin �= ma g
a=θtan
40
Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as systematic):systematic):Find the total vector force FFNET:
TTmgg
FFTOT
θ
TT (string tension)
mgg (gravitational force)
θ
41
Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as systematic):systematic):Find the total vector force FFNET:Recall that FFNET = ma:
g
a
mg
ma==θtan
g
a=θtan
TTmgg
FFTOT
θTT (string tension)
mgg (gravitational force)
θ
42
Let’s put in some numbers:
Say the car goes from 0 to 60 mph in 10 seconds:
› 60 mph = 60 x 0.45 m/s = 27 m/s.› Acceleration a = Δv/Δt = 2.7 m/s2.› So a/g = 2.7 / 9.8 = 0.28 .
› θ = arctan (a/g) = 15.6 deg
aθ
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