gerak melingkar 2016ok

Compiled by Rozie @ SMAN 3 Semarang

GERAK MELINGKAR (circular motion)

1Compiled by Rozie SMAN 3 Semarang

Gerak Melingkar (Rotasi) merupakan gerak benda yang berputar terhadap sumbu putar atau sumbu rotasi


BESARAN – BESARAN FISIKA PADA GERAK MELINGKAR

Compiled by Rozie SMAN 3 Semarang 4

x

y

r

v

1. Sudut tempuh(θ)Posisi partikel yang bergerak melingkar dapat dinyatakan dalam : 1. Koordinat kartesius: (x,y) atau (x,z) atau (y,z)2. Koordinat polar: (r, θ) dengan r= jari-jari (m) dan θ= sudut tempuh(o)

cosrx sinry

22 yxr

Berdasarkan gambar didapatkan:


Bandul bergerak dari titik A ke B

A

B

●

Contoh:Partikel bergerak melingkar dengan jari-jari, r = 0,5 m. Saat posisi sudutnya 30o tentukan posisi partikel dengan koordinat kartesian mapun polar!

m 325030cos50cos rx o ,,

m 25030ins50cos ry o ,,

Posisi partikel dalam koordinat kartisius: ( 0,25√3 m , 0,23 m)

Posisi partikel dalam koordinat polar: ( 2 m, 30o)

Jawab:

rs

)rad(

Hubungan Sudut tempuh (θ) dg Panjang lintasan (s)

1 putaran = 360o = 2π rad

1 π rad = 180o 1 rad = 180o/π = 180o/3,14 = 57,3o

1o = 1/57,3= 0,01745 rad

θ = posisi sudut (rad)r = jari-jari lintasans = panjang lintasan/ jarak tempuh (m)

rs rad .)(

Utk 1 putaran:S = 2πr

Compiled by Rozie SMAN 3 Semarang

Ingat: Hubungan rumus diatas berlaku jika θ bersatuan radian (rad)

r

θS

2. Kecepatan Sudut /angular (ω) “sudut tempuh dibagi waktu yang dibutuhkan”

tt


Untuk satu kali putaran, θ=360o = 2π raddan waktu yang dibutuhkan disebut periode (T), sehingga:

T 2

=kelajuan sudut (angular) /frekuensi sudut (rad/s)T= periode (s)f = frekuensi (Hz)

f 2

arah ω

arah gerak

arah gerak

arah ω

Hubungan kelajuan Sudut (ω) dg kelajuan linier/tangensial/translasi (v)

rvtrs

t

.

.

rv .


v = kelajuan linier/tangensial (m/s) = kelajuan sudut (rad.s-1)r = jari-jari lintasan(m)

Satuan ω selain rad/s juga sering dinyatakan dengan:rpm (rotasi per menit)rps (rotasi per sekon)

1 rpm = 1 putaran/menit = 2π rad/ 60 s = π/30 rad/s1 rps = 1 rotasi/sekon = 2π rad/s

Contoh:Sebuah bola kecil diikat tali sepanjang 50 cm kemudian diputar horisontal. Jika dalam 8 s bola dapat berputar 40 kali. Tentukan: a. frekuensi gerak bola!b. periode gerak bola !c. banyaknya putaran gerak bola selama 20 s !d. Kecepatan sudut bola!e. Kecepatan linier bola!

Compiled by Rozie SMAN 3 Semarang 10,

s 0,2T 51T

f1T.

b

5Hzf 840f

tNf .

a

putaran 2,5N 820N

TtN.

c

rad/s 31,4 rad/s 01 52

2.

fd

m/s 15,7 m/s 5 5,0.10

.

vv

rve

4. Percepatan Sudut

t

“Perubahan kecepatan sudut dibagi interval waktu yang dibutuhkan”

α = percepatan sudut (rad.s-2)


= perubahan kecepatan sudut (rad.s-1)

tot

t = waktu yang dibutuhkan (s)

Contoh:Partikel yang berputar pada lintasan melingkar berubah kecepatan sudutnya dari 120 rpm menjadi 180 rpm dalam 40 sekon. Berapakah percepatan sudut gerak partikel itu?

Penyelesaian:ωo = 120 rpm = 120 . π/30 rad/s = 4 π rad/sωt = 180 rpm = 180 . π/30 rad/s = 6 π rad/st = 40 s


Penyelesaian:ωo = 120 rpm = 120 . π/30 rad/s = 4 π rad/sωt = 180 rpm = 180 . π/30 rad/s = 6 π rad/st = 40 sPercepatan sudutnya:

2-rad.s ,05π040

46

t

ot


Hubungan Percepatan tangensial (a) dg Percepatan sudut (a)

.

).(

rat

ra

tra

tva

r.a

a = percepatan tangensial (m.s-2 )α = percepatan sudur (rad.s-2 )r = jari-jari lintasan (m)

• Sebuah benda yang bergerak melingkar, meskipun bergerak dengan laju (besar v)konstan, akan memiliki percepatan karena arah kecepatannya (arah v) selalu berubah

• Percepatan yang disebabkan karena perubahan arah kecepatan linier v ini disebut percepatan sentripetal, yang arahnya ke pusat lingkaran

5. Percepatan sentripetal (as )



Jadi benda yang bergerak melingkar memungkinkan memiliki tiga percepatan berikut:

Percepatan sudut (α) : perubahan kecepatan sudut dalam waktu tertentu

Percepatan sentripetal (as ) : perubahan arah kecepatan linier dalam waktu tertentuPercepatan tangensial (at) : perubahan besarnya kecepatan linier dalam waktu tertentu

Menentukan persamaan percepatan sentripetal

• Berdasarkan gambar di samping:s

rθ

v

v

v

-v

Δv θ

o


vt.r

sa

t

vrs

tv

vrs

v

rs

vv

s

v-v

Δv

θ

o

ra 2s

rv

a2

s krn v = ω.r

as = percepatan sentripetal (m.s-2 )


rv

a

v.vr1

a

vt

sr1

a

r

s

s

s

5. Percepatan total (atotal )Benda yang bergerak melingkar dengan kecepatan sudut berubah akan memiliki 3 percepatan:1.Percepatan sudut2.Percepatan sentripetal 3.Percepatan tangemsial

Resultan dari percepatan sentripetal dengan percepatan tangensial disebut percepatan total (atot)

at

atot

as

2s

2ttot aaa

r.a t

rr

va 2

2

s Dengan:

atot = percepatan total (m s-

2 )

18



5. Gaya Sentripetal (Fs)

r..mFr

v.mF

a.mF

2s

2

s

ss

vv

v

v

a

as

atot

Fs

Sebuah benda massanya 0,25 kg, diikat pada ujung tali yang panjangnya 0,5 m dan diputar mendatar dengan 2 putaran tiap sekon. Hitunglah :

a. Laju linier benda

b. Percepatan sentripetal benda

c. Gaya sentripetal pada benda

Penyelesaian :Diketahui :m = 0,25 kg ; R = 0,5 m ; f = 2 HzDitanyakan :a. v = ? b. aS = ?c. FS = ?

Jawab :

a. v = 2f .R = 2 x 2 x 0,5 = 2 m/s

b. aS = = = 8 m/s2

c. FS = m . aS = 0,25 x 8 = 2 N

v2

R(2

0,5

as

Fs

v


JENIS GERAK MELINGKAR

kecepatan)arah mengubah yangn(percepata0)(a lsentripetapercepatan memiliki Hanya

0)(a l tangensiaPercepatan 0 )(sudut Percepatan

konstan)(sudut tanecepak

s

t

sialdan tangen lsentripetapercepatan memiliki 0dan konstan )(sudut percepatan

konstantidak)(sudut tanecepak

t.2

αt ωωtetapα

t0

0t

Gerak Melingkar Beraturan (GMB):

Gerak Melingkar Berubah Beraturan (GMBB):

2αωω

αt21

tωθ

20

2t

20

t

t

atau


Contoh Soal

1. Sebuah benda bergerak rotasi dengan percepatan sudut tetap - 2 rad/s2 . Jika mula-mula benda memiliki kecepatan 10 rad/s dan posisi sudut awalnya sama dengan nol. Tentukan:

a. Sudut yang ditempuh selama 2 detik pertamab. Kapan benda akan berhenti berputarc. Jumlah putaran benda dari awal hingga berhenti


2. Dari keadaan diam sebuah benda berotasi sehingga dalam waktu 2 s kecepatannya menjadi 4 rad/s. Tentukan percepatan total titik pada benda tersebut yang terletak 50 cm dari sumbu rotasi benda setelah benda berotasi selama 5 s.


Titik A berada di pinggir sebuah roda. Jika roda ditarik dengan tali yang berkecepatan 2 m/s seperti gambar di samping maka berapakah kecepatan sudut roda tersebut? Berapa pula kecepatan titik A?

Sebuah benda melakukan gerak melingkar beraturan sebanyak 300 rpm. Jika diameter lingkaran 80 cm, Tentukan percepatan sentripetal benda tersebut!

Sebuah satelit komunikasi mengorbit di atas permukaan bumi pada ketingian 600 km. Jika waktu yang diperlukan satelit tersebut untuk menempuh satu kali putaran adalah 1,5 jam, Tentukan kecepatan satelit tersebut!

Terdapat tiga buah roda A, B, dan C yang memiliki jari-jari berturut-turut 25 cm, 15 cm, dan 40 cm. Roda A dab B dihubungkan oleh rantai, sedangkan roda B dan C seporos. Jika roda C memerlukan waktu 2 menit untuk menempuh 120 putaran, maka kecepatan sudut roda A dan B adalah


Seseorang bersepeda dengan kecepatan 18 km/jam. Saat melewati sebuah tikungan yang berjari-jari 100 cm, ia mengerem dan mengurangi kecepatannya 2 m/s tiap detiknya. Tentukan percepatan total pada detik ke-2 setelah pengereman. Sebuah roda berputar dengan kecepatan sudut tetap 120 rpm. Jari-jari roda 50 cm. Tentukan:a.sudut yang ditempuh roda dalam waktu 5 sekon,b.panjang lintasan yang ditempuh titik di tepi roda dalam waktu 5 detik,c.kecepatan linier titik yang berada di tepi roda!d.percepatan sentripetal titik di tepi roda!

Sebuah benda mula-mula diam kemudian bergerak melingkar hingga kecepatan angulernya 60 rad/s dalam waktu 2,5 s. Tentukan:a.Percepatan anguler yang dialami benda b.Kecepatan benda pada t= 2 sc.Jumlah putaran benda pada t= 5 s


HUBUNGAN RODA-RODA

B = C vA >vB

A

B

B

A

B > C vA =vB

B > C vA= vB

1. Sepusat

2. Dihubungkan tali

3. Bersinggungan

AB

Tiga roda A, B, dan C dirangkai seperti pada gambar. Masing-masing berjari-jari 6 cm, 4 cm dan 8 cm. Roda A dan B dihubungkan dengan rantai dan roda C seporos dengan roda B. Jika roda A berputar 2 putaran tiap detik, tentukan kecepatan linier roda C.

Penyelesaian :Diketahui :RA = 6 cmRB = 4 cmRC = 8 cmfA = 2 HzDitanyakan : vC = ?Jawab :Roda A: vA = 2 . RA. fA

= 2 x 6 x 2 = 24 cm/sRoda B : vB = vA

. RB = vA

= vA/ RB = 24 / 4 = 6 rad/s

Roda C : C= B = 6 rad/s

vC = C x RC = 6 x 8 = 48 cm/s

C

B

A

B = C

vA = vB



Dua buah roda dihubungkan dengan rantai. Roda yang lebih kecil dengan jari-jari 8 cm diputar pada 100 rad/s. Berapakah kelajuan linier kedua roda tersebut? Jika jari-jari roda yang lebih besar 15 cm, berapa rpm roda tersebut akan berputar?

Sebuah roda berdiameter 1 m melakukan 120 putaran per menit. Tentukan Kecepatan linier titik pada tepi roda!

Empat buah roda A, B, C, dan D masing-masing berjari-jari RB = 3 cm, RC = 50 cm, dan R= 5 cm dihubungkan satu sama lain seperti pada gambar. Jika periode A sama dengan 2 sekon, tentukan:a. kecepatan sudut roda C, b. kecepatan linier roda D!

28

Direction of Centripetal Force, Acceleration and Velocity

Without a centripetal force, an object in

motion continues along a straight-line path.

Without a centripetal force, an object in

motion continues along a straight-line path.Compiled by Rozie SMAN 3 Semarang

29

Direction of Centripetal Force, Acceleration and Velocity


30

What if velocity decreases?


31

What if mass decreases?


32

What if radius decreases?


Hubungan matematik dalam kedua kelompok besaran tersebut juga memiliki persamaan, yang dapat dilihat dalam hubungan berikut: ddt

Sv d

dt

2

2d ddt dt

v Sa

2

2d ddt dt

Hubungan antara lintasan, kecepatan dan percepatan linier (tangensial) dengan lintasan, kecepatan dan percepatan sudut diberikan oleh:

rS rv ra

;

Untuk percepatan sudut konstan berlaku:

t2

tt)t(

t)t(

20

2

221

0

0


Dinamika Gerak Melingkar

Menurut Hukum Newton kedua (F = ma), benda yang mengalami percepatan pasti resultan gaya yang bekerja pada benda tidak sama dengan nol.

Benda yang bergerak melingkar pasti memiliki percepatan sentripetal yang disebabkan adanya resultan gaya yang radial

rvaF

2

mm sR

FR merupakan gaya total dalam arah radial (gaya yang menuju pusat lingkaran:+ dan dan yamh menjauhi (-)


APLIKASI GERAK MELINGKAR BERATURAN


lanjutan


DINAMIKA GERAK MELINGKAR

Berlaku:

1. Menentukan gaya tegangan tali pada bandul yang diputar melingkarA. Dengan bidang putar horisontal

W

T

sradial FF

sFT RvmT

2 RmT 2

B. Dengan bidang putar Vertikal

W

T

sradial FF

sFwT mgRvmT

2

T

T

T

W

T

W W

W

W cos θW

sin θ

α

θ

W cos α

W sin α

B

A

C

D

E

Ketika bandul ditik A

sradial FF

sFcoswT )cosgRv(mT

2

Ketika bandul ditik B

)cosgR(mT 2

sradial FF

sFT RvmT

2

Ketika bandul ditik C

RmT 2

sradial FF

sFwT )gRv(mT

2

Ketika bandul ditik D

)gR(mT 2

41

What provides the centripetal force?

• Tension• Gravity• Friction• Normal Force

Centripetal force is NOT a new “force”. It is simply a way of quantifying the magnitude of the force required to maintain a certain speed around a circular path of a certain radius.


42

Relationship Between Variables of Uniform Circular Motion

Suppose two identical objects go around in horizontal circles of identical diameter but one object goes around the circle twice as fast as the other. The force required to keep the faster object on the circular path is

A. the same as B. one fourth of C. half of D. twice E. four times the force required to keep the slower object

on the path.

The answer is E. As the velocity increases the centripetal force required to maintain the circle increases as the square of the speed.


43


Suppose two identical objects go around in horizontal circles with the same speed. The diameter of one circle is half of the diameter of the other. The force required to keep the object on the smaller circular path is

A. the same as B. one fourth of C. half of D. twice E. four times the force required to keep the object on the larger path.

The answer is D. The centripetal force needed to maintain the circular motion of an object is inversely proportional to the radius of the circle. Everybody knows that it is harder to navigate a sharp turn than a wide turn.


44


Suppose two identical objects go around in horizontal circles of identical diameter and speed but one object has twice the mass of the other. The force required to keep the more massive object on the circular path is

A. the same as B. one fourth of C. half of D. twice E. four times

Answer: D.The mass is directly proportional to centripetal force.


45

Tension Can Yield a Centripetal Acceleration:

If the person doubles the speed of the airplane, what happens to the tension in the cable?

F = mamv

r

2

Doubling the speed, quadruples the force (i.e. tension) required to keep the plane in uniform circular motion.


46

Friction Can Yield a Centripetal Acceleration:


47

Friction provides the centripetal acceleration

Car Traveling Around a Circular Track


48

Friction Can Yield a Centripetal Acceleration

W

FN

fs

Force X Y

W 0 -mg

FN 0 FN

fs -sFN 0

Sum ma 0

What is the maximum speed that a car can use around a curve of radius “r”?


49

Force X Y

W 0 -mg

FN 0 FN

FC -sFN 0

Sum ma 0

F mg F

F mg

F ma mg

mvr

mg

v g r

v g r

y N

N

x s

s

s

s

0

2

2

max

max

max

max

Maximum Velocity


50

F = mamv

r

2

Centripetal Force: Question

Smaller radius: larger force required to keep it in uniform circular motion.

A car travels at a constant speed around two curves. Where is the car most likely to skid? Why?


51

Gravity Can Yield a Centripetal Acceleration:

Hubble Space Telescopeorbits at an altitude of 598 km(height above Earth’s surface).What is its orbital speed?


52

Gravity and Centripetal Acceleration:

Centripetal acceleration provided by gravitational force

G m MR

m vR

E

2

2


53

Gravity and Centripetal Acceleration:

G m MR

m vR

E

2

2

Solve for the velocity….

vG m M R

m R

vG M

R

vG M

R

E

E

E

22

2


54

Hubble Space Telescope:

vGM

R km

v

v

E

E

598

6 67 10 974 10

7 600

11 24( . ) (5.

,

m kg s kg)6,976,000 m

m / s

3 -1 -2


55

Banked Curves

Why exit ramps in highways are banked?


56

Banked CurvesQ: Why exit ramps in highways are banked?


57

Banked CurvesQ: Why exit ramps in highways are banked?

A: To increase the centripetal force for the higher exit speed.


58

The Normal Force Can Yield a Centripetal Acceleration:

How many forces areacting on the car (assumingno friction)?

Engineers have learned to “bank” curves so that cars can safely travel around the curve without relying on friction at all to supply the centripetal acceleration.


59

Banked CurvesWhy exit ramps in highways are banked?

FN cos = mg


60

Banked CurvesWhy exit ramps in highways are banked?

FN cos = mg


61

The Normal Force as a Centripetal Force:

Two: Gravity and Normal

Force X Y

W 0 -mg

FN FNsin FNcos

Sum ma 0


62

The Normal Force as a Centripetal Force:

F mg F

mg

F F mamv

r

mg mvr

vgr

y N

x N

cos

cos

sin

cossin

tan

0

2

2

2

FN


63

The Normal Force and Centripetal Acceleration:

tan vgr

2

How to bank a curve…

…so that you don’t rely on friction at all!!


64

Artifical Gravity


65

Vertical Circular Motion


66

Vertical Circular Motion


Speed/Velocity in a CircleConsider an object moving in a circle around a specific origin. The DISTANCE the object covers in ONE REVOLUTION is called the CIRCUMFERENCE. The TIME that it takes to cover this distance is called the PERIOD.

Tr

Tdscircle

2

Speed is the MAGNITUDE of the velocity. And while the speed may be constant, the VELOCITY is NOT. Since velocity is a vector with BOTH magnitude AND direction, we see that the direction o the velocity is ALWAYS changing.

We call this velocity, TANGENTIAL velocity as its direction is draw TANGENT to the circle.


Centripetal Acceleration

metersin length arc

srs

vv

rvt

vtsvv

rs

Suppose we had a circle with angle, between 2 radaii. You may recall:

vo

v

v

vov

onacceleratilcentripetaa

atv

rv

c

c

2

Centripetal means “center seeking” so that means that the acceleration points towards the CENTER of the circle


Drawing the Directions correctlySo for an object traveling in a counter-clockwise path. The velocity would be drawn TANGENT to the circle and the acceleration would be drawn TOWARDS the CENTER.

To find the MAGNITUDES of each we have:

rva

Trv cc

22


Circular Motion and N.S.L

Recall that according to Newton’s Second Law, the acceleration is directly proportional to the Force. If this is true: ForcelCentripetaF

rmvFF

rvamaF

c

cNET

cNET

2

2

Since the acceleration and the force are directly related, the force must ALSO point towards the center. This is called CENTRIPETAL FORCE.

NOTE: The centripetal force is a NET FORCE. It could be represented by one or more forces. So NEVER draw it in an F.B.D.


Examples

Trvc2

smvc /26.4)4*28(.)76(.2

The blade of a windshield wiper moves through an angle of 90 degrees in 0.28seconds. The tip of the blade moves on the arc of a circle that has a radius of 0.76m. What is the magnitude of the centripetal acceleration of the tip of the blade?

222

/92.2376.0

)26.4( smrvac


Examples

rgv

rmvmg

rmvF

FF

N

cf

2

2

2

What is the minimum coefficient of static friction necessary to allow a penny to rotate along a 33 1/3 rpm record (diameter= 0.300 m), whenthe penny is placed at the outer edge of the record?

mg

FN

Ff

Top view

Side view 187.0)8.9)(15.0(

)524.0(

/524.080.1

)15.0(22

sec80.1555.0

sec1sec555.0

sec60min1*

min3.33

22

rgv

smT

rv

Trevrev

revrev

c


ExamplesThe maximum tension that a 0.50 m

string can tolerate is 14 N. A 0.25-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have (a) the top of the circle, (b)at the bottom of the circle?

mgT

smvm

mgTrv

mvmgTrr

mvmgT

rmvmaFF ccNET

/74.525.0

))8.9)(25.0(14(5.0)(

)( 22

2


ExamplesAt the bottom?

smvm

mgTrv

mvmgTrr

mvmgT

rmvmaFF ccNET

/81.425.0

))8.9)(25.0(14(5.0)(

)( 22

2

mg

T


An object moving in a circle with constant speed, v, experiences a centripetal acceleration with: *a magnitude that is constant in time and is equal to

*a direction that changes continuously in time and always points toward thecenter of the circular path

Uniform Circular Motion

rva

2

For uniform circular motion, the velocity is tangential to the circle and perpendicular to the acceleration 75Compiled by Rozie SMAN 3 Semarang

A circular motion is described in terms of the period T, which is the time for an object to complete one revolution.

Period and Frequency

2r

f 1T

T 2rv

The distance traveled in one revolution is

The frequency, f, counts the number of revolutions per unit time.

r


The moon’s nearly circular orbit about the earth has a radius of about 384,000 km and a period T of 27.3 days. Determine the acceleration of the Moon towards the Earth.

Example of Uniform Circular Motion

T 2rv

v 2rT

a v2

r

4 2r2

T 2r

4 2rT 2

a 2.7210 3 m / s2 g9.8m / s2

2.7810 4 g


Moon…...

*So we find that amoon / g = 0.000278*Newton noticed that RE

2 / R2 = 0.000273

*This inspired him to propose that Fgravity 1 / R2

(more on gravity in future lectures)

amoon

R RE

g


Uniform Circular Motion**motion in a circle or circular arc at constant

speed

**the acceleration changes the direction of the velocity, not the magnitude

**the “center-seeking” or centripetal acceleration is always orthogonal to the velocity and has magnitude:

rva

2

The period of The period of the motion:the motion:

vrT 2


Uniform Circular MotionNewton’s 2nd Law: : The net force on a body is equal to the product of the mass of the body and the acceleration of the body.

*The centripetal acceleration is caused by a centripetal force that is directed towards the center of the circle.

F ma mv2

r80Compiled by Rozie SMAN 3 Semarang

Demo 1D-5Does the contact force between the wine glass and red-water remain constant in uniform circular motion?


Consider the glass directly overhead. Choose the correct statement:

a. The water doesn’t fall because the centripetal force on the water cancels the force of gravity.

b. The water doesn’t fall because there isn’t enough time for it to fall.

c. The water doesn’t fall because of the horizontal force applied to it by the glass, plus friction with the glass.

d. The water is falling, but the glass is falling faster than it would under free fall.


v

mg

N

Fy N mg ma

N m(a g)

N m(v2

r g)

Top

y

x

mac = mv2/r = mg + Ny

or

ac = g N/m

When N=0, the centripetal acceleration is just g.


v

v

mg

N

Fy N mg ma

N m(a g)

N m(v2

r g)

Top

mg N

Bottom

Fy N mg ma

N m(a g)

N m(v2

r g)

y

x


mg

N

v

Fy N mg ma

N m(a g)

N m(v2

r g)

Top

What speed is needed to lose contact between wine glass and red-water?

v rg


2) A person riding a Ferris Wheel moves through positions at (1) the top, (2) the bottom and (3) midheight. If the wheel rotates at a constant rate, rank (greatest first) these three positions according to...

(a) the magnitude of persons centripetal acceleration

(a) 2,1,3

(b) 1,2,3

(c) 3,2,1

(d) all tie

(b) The magnitude of the Normal force?

(1) Top

(2) Bottom

(3) Middle


(b) the magnitude of the net centripetal force on the person

1. 1,2,3

2. 3,1,2

3. 3,2,1

4. all tie

(c) the magnitude of the normal force on the person

1. all tie

2. 2,3,1

3. 3,2,1

4. 1,2,3


Demo 1D-2 Conical Pendulum

T

mg

Fy 0 T cos mg

Fr ma T sin mv2 / R

tan v2gR

v Rg tan R g / H

PeriodT 2R / v2 H / g

H

R

The period, T, is independent of mass and depends only on H.

*as 90, v increases.*v is independent of mass.


Fr ma fs mv2 / R

Fy N mg

fs sN

A car of mass, m, is traveling at a constant speed, v, along a flat, circular road of radius, R. Find the minimum µs required that will prevent the car from slipping

s v2gR


T Mg0 T Mg

T mv2

r

Mgm v2

r

v Mm

gr

A mass, m, on a frictionless table is attached to a hanging mass, M, by a cord through a hole in the table. Find the speed with which m must move in order for M to stay at rest.


frictionless

mg

N

Fy 0 N cos mg

Fr ma N sin mv2 / R

tan v2gR

Rv2g tan

A car of mass, m, is traveling at a constant speed, v, along a curve that is now banked and has a radius, R. What bank angle, , makes reliance on friction unnecessary?


v=480 km/hr

L

L

W

Fr 0 2L sin 40 Mv2

r

Fy 0 2L cos40Mg

2 Mg2 cos40

sin 40 M v2

r

r v2

g tan 40

L Mg

2cos40

An airplane is flying in a horizontal circle with a speed of 480 km/hr. If the wings of the plane are tilted 40o to the horizontal, what is the radius of the circle in which the plane is flying? (Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface.)


12

x1x2x tt

vva

12

y1y2y tt

vva

vstt 12

v

v cos

v sin

v

v cos

v sin

R

12

s

GERAK MELINGKAR


0tt

cosvcosva12

x

12y tt

sinvsinva

sinRv

R2sinv2a

vR2

vstt

2

2

y

12

V

V cos

V sin

V

V cos

V sin

R

12

s


Rvsin

Rvlimalima

22

00y

V

ay

R

V

ax

Percepatan centripetal (menuju pusat)

Rva

2


V

a

R

V

a

Rva

2

f60rpmT1f

VR2T

T = Perioda [s]

f = Frekuensi [c/s, Hz]

rpm = Siklus per menit


Contoh Soal 1.10

Sebuah satelit direncanakan akan ditempatkan di ruang angkasa sedemikan rupa sehingga ia melintasi (berada di atas) sebuah kota A di bumi 2 kali sehari. Bila percepatan sentripetal yang dialami olehnya adalah 0,25 m/s2 dan jari-jari bumi rata-rata adalah 6378 km, pada ketinggian berapa ia harus ditempatkan ?

Jawab :v

a

RB

h

km5440637811818RRh

km181814

)3600x12)(25,0(4aTR

TR4

RT

R2

RVa

s/m25,0ahRRjam12T

B

2

2

2

2

2

2

2

2

2B


Circular MotionAngular velocity

The angular displacement is

if

if

if

ttt

Average angular velocity

dtd

tt

0limit

Instantaneous angular velocity We will worry about the direction later.

Like one dimensional motion +- will do. Positive angular velocity is counter-clock=wise.


Circular MotionCoordinate System


Circular MotionSo, is there an acceleration?


Student Workbook


Student Workbook

bankF

w

T a


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