4. BESARAN – BESARAN FISIKA PADA
GERAK MELINGKAR
Compiled by Rozie SMAN 3 Semarang 4
5. x
y
θ
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
)
θcosrx =
θsinry =
22
yxr +=
Berdasarkan gambar didapatkan:
5Compiled by Rozie SMAN 3 Semarang
Bandul bergerak dari
titik A ke B
A
B
●
6. Contoh:
Partikel bergerak melingkar dengan jari-jari, r = 0,5 m. Saat posisi
sudutnya 30o
tentukan posisi partikel dengan koordinat kartesian
mapun polar!
m325030cos50cosrx o
,, === θ
m25030ins50cosry o
,, === θ
Posisi partikel dalam koordinat kartisius: ( 0,25√3 m , 0,23 m)
Posisi partikel dalam koordinat polar: ( 2 m, 30o
)
Jawab:
7. r
s
)rad( =θ
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 lintasan
s = panjang lintasan/ jarak
tempuh (m)
rs rad .)(θ=
Utk 1 putaran:S = 2πr
Compiled by Rozie SMAN 3 Semarang
Ingat: Hubungan rumus diatas berlaku
jika θ bersatuan radian (rad)
r
θ
S
8. 2. Kecepatan Sudut /angular (ω)
“sudut tempuh dibagi waktu yang dibutuhkan”
tt
θθ
ω =
∆
∆
=
8Compiled by Rozie SMAN 3 Semarang
Untuk satu kali putaran, θ=360o
= 2π rad
dan waktu yang dibutuhkan disebut periode (T), sehingga:
T
π
ω
2
=
ω=kelajuan sudut (angular) /frekuensi sudut (rad/s)
T= periode (s)
f = frekuensi (Hz)
fπω 2=
arah ω
arah gerak
arah gerak
arah ω
9. Hubungan kelajuan Sudut (ω) dg kelajuan
linier/tangensial/translasi (v)
rv
t
rs
t
.
.
=
=
=
ω
ω
θ
ω
rv .ω=
9Compiled by Rozie SMAN 3 Semarang
v = kelajuan linier/tangensial (m/s)
ω = 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/s
1 rps = 1 rotasi/sekon = 2π rad/s
10. 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,
s0,2T
5
1
T
f
1
T.
=
=
=b
5Hzf
8
40
f
t
N
f.
=
=
=a
putaran2,5N
8
20
N
T
t
N.
=
=
=c
rad/s31,4rad/s01
52
2.
==
=
=
πω
πω
πω fd
m/s15,7m/s5
5,0.10
.
==
=
=
π
π
ω
v
v
rve
11. 4. Percepatan Sudut
t
ω
α
∆
=
“Perubahan kecepatan sudut dibagi interval waktu yang dibutuhkan”
α = percepatan sudut (rad.s-2
)
11Compiled by Rozie SMAN 3 Semarang
ω∆ = perubahan kecepatan sudut (rad.s-1
)
t
ot ωω
α
−
=
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/s
ωt = 180 rpm = 180 . π/30 rad/s = 6 π rad/s
t = 40 s
12. Compiled by Rozie SMAN 3 Semarang 12
Penyelesaian:
ωo = 120 rpm = 120 . π/30 rad/s = 4 π rad/s
ωt = 180 rpm = 180 . π/30 rad/s = 6 π rad/s
t = 40 s
Percepatan sudutnya:
2-
rad.s,05π0
40
46
=
−
=
−
=
α
ππ
α
ωω
α
t
ot
13. Compiled by Rozie SMAN 3 Semarang 13
Hubungan Percepatan tangensial (a) dg Percepatan
sudut (a)
α
ω
ω
.
).(
ra
t
ra
t
r
a
t
v
a
=
∆
=
∆
=
∆
=
r.a α=
a = percepatan tangensial (m.s-2
)
α = percepatan sudur (rad.s-2
)
r = jari-jari lintasan (m)
14. • 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 )
14Compiled by Rozie SMAN 3 Semarang
15. Compiled by Rozie SMAN 3 Semarang 15
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 tertentu
Percepatan tangensial (at) : perubahan besarnya kecepatan
linier dalam waktu tertentu
16. Menentukan persamaan percepatan sentripetal
• Berdasarkan gambar di samping:
s
r
θ
v
v
v
-v
Δv θ
o
16Compiled by Rozie SMAN 3 Semarang
v
t.r
s
a
t
v
r
s
t
v
v
r
s
v
r
s
v
v
s
∆
∆∆
∆
∆
∆
=⇒
=⇒
=
=
v
-v
Δvθ
o
17. ra 2
s ω=
r
v
a
2
s = krn v = ω.r
as = percepatan sentripetal (m.s-2
)
17Compiled by Rozie SMAN 3 Semarang
r
v
a
v.v
r
1
a
v
t
s
r
1
a
r
s
s
s
=⇒
=⇒
=⇒
∆
18. 5. Percepatan total (atotal )
Benda yang bergerak melingkar dengan kecepatan
sudut berubah akan memiliki 3 percepatan:
1.Percepatan sudut
2.Percepatan sentripetal
3.Percepatan tangemsial
Resultan dari percepatan sentripetal dengan percepatan
tangensial disebut percepatan total (atot)
at
atot
as
2
s
2
ttot aaa +=
r.at α=
r
r
v
a 2
2
s ω==
Dengan:
atot = percepatan total (m s-
2
)
18
Compiled by Rozie SMAN 3 Semarang
19. Compiled by Rozie SMAN 3 Semarang 19
5. Gaya Sentripetal (Fs)
r..mF
r
v
.mF
a.mF
2
s
2
s
ss
ω=
=
=
v
v
v
v
a
as
atot
Fs
20. 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 Hz
Ditanyakan :
a. v = ?
b. aS = ?
c. FS = ?
Jawab :
a. v = 2πf .R = 2π x 2 x 0,5 = 2π m/s
b. aS = = = 8 π2
m/s2
c. FS = m . aS = 0,25 x 8π2
= 2 π2
N
v2
R
(2π)2
0,5
as
Fs
v
20Compiled by Rozie SMAN 3 Semarang
22. 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 pertama
b. Kapan benda akan berhenti berputar
c. Jumlah putaran benda dari awal hingga berhenti
22Compiled by Rozie SMAN 3 Semarang
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.
23. Compiled by Rozie SMAN 3 Semarang 23
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
24. Compiled by Rozie SMAN 3 Semarang 24
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 s
c.Jumlah putaran benda pada t= 5 s
25. Compiled by Rozie SMAN 3 Semarang 25
HUBUNGAN RODA-RODA
ωB = ωC vA >vB
A
B
B
A
ωB > ωC vA =vB
ωB > ωC vA= vB
1. Sepusat
2. Dihubungkan tali
3. Bersinggungan
A
B
26. 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 cm
RB = 4 cm
RC = 8 cm
fA = 2 Hz
Ditanyakan : vC = ?
Jawab :
Roda A: vA = 2π . RA. fA
= 2π x 6 x 2 = 24 π cm/s
Roda B : vB = vA
ωΒ . RB = vA
ωΒ = vA/ RB = 24 π / 4 = 6 π rad/s
Roda C : ωC= ωB = 6 π rad/s
vC = ωC x RC = 6 π x 8 = 48 π cm/s
C
B
A
ωB = ωC
vA = vB
26Compiled by Rozie SMAN 3 Semarang
27. Compiled by Rozie SMAN 3 Semarang 27
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. 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
33. Hubungan matematik dalam kedua kelompok besaran tersebut juga memiliki
persamaan, yang dapat dilihat dalam hubungan berikut:
d
dt
=
S
v
d
dt
=
θ
ω
2
2
d d
dt dt
= =
v S
a
2
2
d d
dt dt
= =
ω θ
α
Hubungan antara lintasan, kecepatan dan percepatan linier (tangensial)
dengan lintasan, kecepatan dan percepatan sudut diberikan oleh:
rθ=S rω=v rα=a
;
Untuk percepatan sudut konstan berlaku:
t2
tt)t(
t)t(
2
0
2
2
2
1
0
0
αωω
αωθθ
αωω
+=
++=
+=
33Compiled by Rozie SMAN 3 Semarang
34. Dinamika Gerak Melingkar
Menurut Hukum Newton kedua (ΣF = 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
r
v
aF
2
mm sR ==Σ
ΣFR merupakan gaya total dalam arah radial (gaya yang menuju pusat
lingkaran:+ dan dan yamh menjauhi (-)
34Compiled by Rozie SMAN 3 Semarang
37. DINAMIKA GERAK MELINGKAR
Berlaku:
1. Menentukan gaya tegangan tali pada bandul yang diputar melingkar
A. Dengan bidang putar horisontal
W
T
∑ = sradial FF
sFT = R
v
mT
2
= RmT 2
ω=
38. B. Dengan bidang putar Vertikal
W
T
∑ = sradial FF
sFwT =− mg
R
v
mT
2
+=
T
T
T
W
T
W W
W
W cos θ
W
sin
θ
α
θ
W cos α
W sin α
B
A
C
D
E
Ketika bandul ditik A
39. ∑ = sradial FF
sFcoswT =− θ )cosg
R
v
(mT
2
θ+=
Ketika bandul ditik B
)cosgR(mT 2
θ+ω=
∑ = sradial FF
sFT = R
v
mT
2
=
Ketika bandul ditik C
RmT 2
ω=
40. ∑ = sradial FF
sFwT =+ )g
R
v
(mT
2
−=
Ketika bandul ditik D
)gR(mT 2
−ω=
41. 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.
Compiled by Rozie SMAN 3 Semarang
42. 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.
Compiled by Rozie SMAN 3 Semarang
43. 43
Relationship Between Variables of Uniform
Circular Motion
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.
Compiled by Rozie SMAN 3 Semarang
44. 44
Relationship Between Variables of Uniform
Circular Motion
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.
Compiled by Rozie SMAN 3 Semarang
45. 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 = ma
mv
r
=
2
Doubling the speed, quadruples the force (i.e. tension) required to keep the plane in
uniform circular motion.
Compiled by Rozie SMAN 3 Semarang
47. 47
Friction provides the centripetal acceleration
Car Traveling Around a Circular Track
Compiled by Rozie SMAN 3 Semarang
48. 48
Friction Can Yield a Centripetal Acceleration
W
FN
fs
Force X Y
W 0 -mg
FN 0 FN
fs -µsFN
0
Sum ma 0
What is the maximum speed that a car
can use around a curve of radius “r”?
Compiled by Rozie SMAN 3 Semarang
49. 49
Force X Y
W 0 -mg
FN 0 FN
FC -µsFN
0
Sum ma 0
F mg F
F mg
F ma mg
mv
r
mg
v g r
v g r
y N
N
x s
s
s
s
= = − +
=
= = −
− = −
= × ×
= × ×
∑
∑
0
2
2
µ
µ
µ
µ
max
max
max
max
Maximum Velocity
Compiled by Rozie SMAN 3 Semarang
50. 50
F = ma
mv
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?
Compiled by Rozie SMAN 3 Semarang
51. 51
Gravity Can Yield a Centripetal Acceleration:
Hubble Space Telescope
orbits at an altitude of 598 km
(height above Earth’s surface).
What is its orbital speed?
Compiled by Rozie SMAN 3 Semarang
52. 52
Gravity and Centripetal Acceleration:
Centripetal acceleration provided by gravitational force
G m M
R
m v
R
E× ×
=
×
2
2
Compiled by Rozie SMAN 3 Semarang
53. 53
Gravity and Centripetal Acceleration:
G m M
R
m v
R
E× ×
=
×
2
2
Solve for the velocity….
v
G m M R
m R
v
G M
R
v
G M
R
E
E
E
2
2
2
=
× × ×
×
=
×
=
×
Compiled by Rozie SMAN 3 Semarang
54. 54
Hubble Space Telescope:
v
GM
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
Compiled by Rozie SMAN 3 Semarang
56. 56
Banked Curves
Q: Why exit ramps in highways are banked?
Compiled by Rozie SMAN 3 Semarang
57. 57
Banked Curves
Q: Why exit ramps in highways are banked?
A: To increase the centripetal force for the higher exit speed.
Compiled by Rozie SMAN 3 Semarang
58. 58
The Normal Force Can Yield a Centripetal Acceleration:
How many forces are
acting on the car (assuming
no 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.
Compiled by Rozie SMAN 3 Semarang
59. 59
Banked Curves
Why exit ramps in highways are banked?
FN
cosθ = mg
Compiled by Rozie SMAN 3 Semarang
60. 60
Banked Curves
Why exit ramps in highways are banked?
FN
cosθ = mg
Compiled by Rozie SMAN 3 Semarang
61. 61
The Normal Force as a Centripetal Force:
Two: Gravity and Normal
Force X Y
W 0 -mg
FN FNsinθ FNcosθ
Sum ma 0
Compiled by Rozie SMAN 3 Semarang
62. 62
The Normal Force as a Centripetal Force:
F mg F
mg
F F ma
mv
r
mg mv
r
v
gr
y N
x N
= − + =
=
= = =
× =
=
∑
∑
cos
cos
sin
cos
sin
tan
θ
θ
θ
θ
θ
θ
0
2
2
2
FN
Compiled by Rozie SMAN 3 Semarang
63. 63
The Normal Force and Centripetal Acceleration:
tanθ =
v
gr
2
How to bank a curve…
…so that you don’t rely on friction at all!!
Compiled by Rozie SMAN 3 Semarang
67. Speed/Velocity in a Circle
Consider 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.
T
r
T
d
scircle
π2
==
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.
67Compiled by Rozie SMAN 3 Semarang
69. Drawing the Directions correctly
So 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:
r
v
a
T
r
v cc
2
2
==
π
69Compiled by Rozie SMAN 3 Semarang
70. 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
r
mv
FF
r
v
amaF
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.
70Compiled by Rozie SMAN 3 Semarang
71. Examples
T
r
vc
π2
= smvc /26.4
)4*28(.
)76(.2
==
π
The blade of a windshield wiper moves
through an angle of 90 degrees in 0.28
seconds. 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?
2
22
/92.23
76.0
)26.4(
sm
r
v
ac ===
71Compiled by Rozie SMAN 3 Semarang
72. Examples
rg
v
r
mv
mg
r
mv
F
FF
N
cf
2
2
2
=
=
=
=
µ
µ
µ
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), when
the 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(
/524.0
80.1
)15.0(22
sec80.1
555.0
sec1
sec
555.0
sec60
min1
*
min
3.33
22
===
===
==
=
rg
v
sm
T
r
v
T
revrev
revrev
c
µ
ππ
72Compiled by Rozie SMAN 3 Semarang
73. Examples
The 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
smv
m
mgTr
v
mvmgTr
r
mv
mgT
r
mv
maFF ccNET
/74.5
25.0
))8.9)(25.0(14(5.0)(
)( 2
2
2
=
+
=
+
=
=+→=+
===
73Compiled by Rozie SMAN 3 Semarang
75. 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 the
center of the circular path
Uniform Circular Motion
r
v
a
2
=
For uniform circular motion, the velocity is tangential
to the circle and perpendicular to the acceleration 75Compiled by Rozie SMAN 3 Semarang
76. 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
2πr
f =
1
T
T =
2πr
v
The distance traveled in one revolution is
The frequency, f, counts the number of revolutions
per unit time.
r
76Compiled by Rozie SMAN 3 Semarang
77. 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 =
2πr
v
⇒ v =
2πr
T
a =
v2
r
=
4π2
r2
T2
r
=
4π2
r
T2
a = 2.72 ×10−3
m/ s2 g
9.8m/ s2
= 2.78×10−4
g
77Compiled by Rozie SMAN 3 Semarang
78. 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
78Compiled by Rozie SMAN 3 Semarang
79. 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:
r
v
a
2
=The period ofThe period of
the motion:the motion:
v
r
T
π2
=
79Compiled by Rozie SMAN 3 Semarang
80. Uniform Circular Motion
Newton’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=m
v2
r
80Compiled by Rozie SMAN 3 Semarang
81. Demo 1D-5
Does the contact force between the wine
glass and red-water remain constant in
uniform circular motion?
81Compiled by Rozie SMAN 3 Semarang
82. 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.
82Compiled by Rozie SMAN 3 Semarang
86. 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
86Compiled by Rozie SMAN 3 Semarang
87. (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
87Compiled by Rozie SMAN 3 Semarang
88. Demo 1D-2 Conical Pendulum
T
mg
θ
Fy=0⇒∑ Tcosθ=mg
Fr=ma⇒∑ Tsinθ=mv2
/R
tanθ=v2
gR
v= Rgtanθ=Rg/H
Period=T=2πR/v=2πH/g
H
R
The period, T, is independent of mass and
depends only on H.
*as θ 90, v increases.
*v is independent of mass.
88Compiled by Rozie SMAN 3 Semarang
89. Fr∑ =ma⇒fs=mv2
/R
Fy∑ =N=mg
fs=µsN
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
µs=v2
gR
89Compiled by Rozie SMAN 3 Semarang
90. T−Mg=0⇒T=Mg
T=m
v2
r
Mg = m
v2
r
v =
M
m
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.
90Compiled by Rozie SMAN 3 Semarang
91. frictionless
mg
N
Fy=0⇒∑ Ncosθ=mg
Fr=ma⇒∑ Nsinθ=mv2
/R
tanθ = v2
gR
R = v2
gtanθ
θ
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, θ, makes reliance on friction unnecessary?
91Compiled by Rozie SMAN 3 Semarang
92. v=480 km/hr
L
L
W
Fr=0⇒∑ 2Lsin40=M
v2
r
Fy=0⇒∑ 2Lcos40=Mg
2
Mg
2cos40
sin40 =M
v2
r
r=
v2
gtan40
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.)
92Compiled by Rozie SMAN 3 Semarang
97. 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
km18181
4
)3600x12)(25,0(
4
aT
R
T
R4
R
T
R2
R
V
a
s/m25,0ahRRjam12T
B
2
2
2
2
2
2
2
2
2
B
=−=−=
=
π
=
π
=
π
=
π
==
=+==
97Compiled by Rozie SMAN 3 Semarang
98. Circular Motion
Angular velocity
The angular displacement is
if θθθ∆ −≡
if
if
ttt −
−
=≡≡
θθ
∆
θ∆
ωω
Average angular velocity
dt
d
tt
θθ
ω ≡
∆
∆
→∆
≡
0
limit
Instantaneous angular velocity We will worry about the direction
later.
Like one dimensional motion +-
will do. Positive angular velocity
is counter-clock=wise.
98Compiled by Rozie SMAN 3 Semarang
