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Mechanics of a Pendulum: Derivation of Lagrangian and Energy Equations, Study notes of Classical Mechanics

A detailed derivation of the lagrangian and energy equations for a pendulum, using the principles of physics and calculus. It covers the transformation of coordinates, the application of lagrange's equations, and the determination of the kinetic energy and potential energy of the pendulum. The document also includes solutions for the equations of motion and the identification of trigonometric functions.

Typology: Study notes

2021/2022

Uploaded on 03/24/2024

puji-lestari-3
puji-lestari-3 🇮🇩

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bg1
Bandul
Pegas
IIIIIIIIIIIIIIIIIII
I
V
=
O
#
of
panjang
awa)
pegas
:
lo
-
y
3
i
M
Persamaan
gerak
bandul
?
----------
*
Menentukan
derajat
kebebasan
-
ketika
bandul
bergerak
tidak
hanya
-
y
berubah
,
ruya
juga
berubah
.
Shy
derajat
Kebebasannya
ada
2
-
&
*
Persamaan
Lagrange
dg
2
DK
(i
&8)
*
Ten t u k an
Koordinat
/polar
-
>
r
&
6)
*
Transfor m asi
koordinat
#(d)
=
=
r
*
=
sin
A
y
=
Co5
-
x
=
(rsino)
=
Sino
+
rAsino
(rIKonstan)
Ed
=
Elym (r
+
r's
e
dt
=
i sin
o
+
r
asin t
de
-Emp
=
Mr
O
d
dt
·
=
mr
*
=
sin
o
+
8
Cost
i
=
(rcost)
=
r
CO50
+
rdco
e
·
26
=
limlrtrf
-
[K(r
=
2r
+
ei)
+
mgr
(050)
ar
dt
=
cost
+
r d cost
do
=
EMro2-k(r-bi)
=
MrE-b(r-10)
d
dt
i
=
ICO50-rEsin
t
maka
*
masukkan
energi
kinetik
I(84)
=
&
·
T
=
jM(
*
+
+
y
T
=
( m)(aino
+
2 coro
+
[custo
+
rain2])
mr
=
mrt-klr-lot
--
M
-1
~
Z
=
-
M(r
(sin
=
cos"
o)
+
r(05t
+
<in 2
-)
i
=
r
8
-
*
(U-lo
w
:
M
L
*
=
W
=
m
·
2)
i
=
82 -W
Ir-lo2)
T
=
jm/r
=
+
r
#
Up
:
12K
OX
=
=
/K
(r-10)
Solusi
untuk
bagian
r
·
<
Plegas
faci
V
=
- mgr
cost
V
=
<
K
(r-lo)
-
mgr
cost
-(86)
=
=
Solun
y
*
Masukkan
he
fungsi
Lagrange
1
=
T-V
·
&
=
Mr
E.
=
-
Mgrsi
1
=
[m(r
+
r
2)
-
Ek(r-lo)"
+
mgrcost
·
-
2)
=
Mr
.
shgMr
=
-gran
o
solusi
u
·
&
-
=
-sino
*
bagian
o
-kr
-er
+
l
pf3
pf4
pf5

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Bandul Pegas

IIIIIIIIIIIIIIIIIIII V= O

-^ #^ of^ panjang^ awa)^ pegas^ :^ lo y 3 i M Persamaan (^) gerak bandul^?


  • Menentukan^ derajat kebebasan
  • ketika bandul (^) bergerak tidak (^) hanya - (^) y berubah (^) , (^) ruya juga berubah. Shy derajat^ Kebebasannya^ ada 2 - &

* Persamaan Lagrange dg 2 DK^ (i^ &8)

  • (^) Tentukan Koordinat (^) /polar - > (^) r& (^) 6)
  • (^) Transformasi koordinat #(d) = = (^) r
  • = sin^ A y =^ Co5^ - x =(rsino)^ = (^) Sino + rAsino^ (rIKonstan)^ Ed = Elym (r
  • r'se dt = i sin o + r asin t^ de^ -Emp = Mr O d^ dt^ · = (^) mr
  • (^) = sin o (^) + 8 Cost i = (rcost) = r (^) CO50 + rdco e · (^26) = limlrtrf

[K(r

2r+ ei) + mgr (050) dt ar = (^) cost + r d cost do^ = d EMro2-k(r-bi)^ =^ MrE-b(r-10) dt i =^ ICO50-rEsin^ t maka

  • masukkan (^) energi kinetik I(84) = · &

T = jM( *

  • (^) y T = ( m)(aino
  • (^) 2 coro + (^) [custo + rain2]) mr (^) = (^) mrt-klr-lot

-- M

Z^ -1~

M(r (sin

= cos"

o) +^ r(05t +^ <in 2^ -) i (^) = (^) r 8 -

  • (U-lo w :^ M

L

  • =^ W^ = · m

T = i^ =^ 82 -W^ Ir-lo2) jm/r = (^) + (^) r

Up :^ 12K^ OX^

·^ /K^ (r-10)^ Solusi^ untuk^ bagian^ r

< (^) Plegasfaci (^) V =

  • mgr cost V= < K (^) (r-lo) - mgr cost -(86) =^ =^ Solun^ y
  • Masukkan he^ fungsi Lagrange 1 =^ T-V^

& = (^) Mr E. = - Mgrsi 1 = [m(r +^ r^ 2)^

  • Ek(r-lo)" + mgrcost (^) · - 2) = Mr. shgMr = -gran o · & solusi (^) u ↓ -^ = -sino

bagiano -kr-er+^ l

* Energi Kinetik

Devajat Kebebasannya - &^ Ac^ T = <M , 8 ,^ +^ -M, V

e (^2) , &^ &2^ =^ Konstan

jm/xI +^ y^ )^ +^ 5 M(x= +^ y,2)

  • 1 > (^) x ,^ x,x^ I = Em lit (^) + (^) M2[lit + (^) lit+ (^24) , 28 , (05(0, - A2)]
  • =^ I, sin O^ Xc :^ l, sin^8 ,^ +^ ha sin^2 T = MitMa)l" + Emelit +^ Malilit,^ E,^ (05(81^ -Oc) I,^ =^ - l, Cos (^) O (^) , Yc = -^ I, (^185 8) ,^ - le (^) CO5 Oz v = x (^) V = (^) M , (^) gY (^) , +^ M29Y

= m , g [ - 2 , 20581) + Mc

9 (

  • l, <850 (^1) - lc(05E2) · (^) X = (2, cint1)^ = A sin (^) ti + e (^) ,^ drino^ e dt (^) = - (^) M, gl,^2058 ,^ -^ McGl,^ COS^ O^ ,^

M2g 2228582 = i = in (^) , + (^1) , Osint ,o^ e

Konstan

dt V = - (m , + M2) gl, COSO ,^ - Meg IcCO5 Oz

= 0 + (^1) , (^8) , 10581 = (^) & (^) , (^8) , C058 (^) , masukan he (^) fungsi Lagrange · i , =

1 - 2,^ 20581)^ =.

d 2058 -1, dist

dt^ At t^ L^ =^ T-^ V

E(MitM2) "^ +^ Emelit^ +^ Malilit,^ E,^ (05(0,^

  • O2)
  • (^0) + 1 , , sin (^) O (^) , = (^1) , (^) , sin (^) O,

+ (m , + M2) 90 , COSO , + Meg IcCO5 Oz

· xc =* )^ (^1) , <in t, +Ic sin to (^) e jika (^) detanya berapa (^) fungsi kecepatannya = (^1) , 8 , (^2058) , +^ 12G2 20582 · (^) i = -1, 1058 , - &c C85 (^) & (2)^ =^2 ,

  • (^) E

  • (^2) , 8 , sin^ A, + l,^ sin^ z

= Ed =e

↑ +?)^ =^ It,^ "^ cos'o,^ +^ 2,^ ",sino^ ,^ )

= (^) , (x2 +^ j?)^ =^ Cit,^ Cost, +^24 , 12 8^ , 22058 ,^ CO5O2+^ &c^ <COs" Ozt esin, +^21 , l it in^ - , sin 2 +^ lesinO = 1 ,^8 , (Co5"t,^ +^ sin201) +^ lit (cost, +^ sin^

-2) +^2 lilze, Es 2 2058 , 20582 +^ <in^8 , <inO2) identifas (^) trigonometri cos[8 (^) , -82) =^ CO58, 28582 + (^) SinO, sin (^) Oz = (^) lit, (^) +l + 2 lilz - (05 (^) (t 1 - E2)

/VI

M V^ =^ O

Et

=) migy^ ,^ -Megy^ ,^

  • My one L - Y (^) , = (^) M (^) = 9 - Mzg
  • M

= g(M , - Mc - M3) ·I^

M

C-Y V^ =/E) = V

m -

i( m (^) , +^ M2 + (^) M3) + i (^) (M3 -M2) = G (M , - M2 - M3)

Y ,^ =^ Y ,^ L^ &^ C^ =^ Konstant^ is = g) M , - M2 -M3) - y z(M3 - M2)

Y2 =^ ( L^ - Y (^) , ) +^ Y Y3 = (- y (^) , ) +^ (l - Yc) 1M^ ,^

  • (^) Mc +^ M3) untuk (^) Y i =^ i, · 26 =^ (TM,^ ee & / ↓ 2i, ;+ (^5) M3 (^) y-2) · i=^ - y Ye^ I^ I ,^ +^ y^ =^ jz- y,^ /^ 24 , - / M i (^) = - j, -y, = McG -M2y (^) , + (^) M3 Y (^) , + M i, (i-Y^ ,^ )^ = (32-^ Y^ , ) (^) (3- yi) (^) = Y (^) , (Ms -M2) + (^) y (M3 + (^) M2) =^ - Y2 Y (^) ,^ - By (^) , + (^) it,^2 = (^) Y - (^2424) , + (^) y,

/j,^ (Ms^ -^ M2)^ +^ Y^2 (M^ Me e) (-Y (^).^ - (2) = (- Y - (2) (-Y,^ - Yz) = (^) jj(Ms- M2) + (M3 + (^) M2) = i + y (^) , y +^ y (^) , y+ (^) y = Y +^ 2y (^) , y +^ y=

Ed

EG) M (^) - (^942) -Mye T = <M , y^ , + (^) EM2(32 - 3) (^) + (^5) M3( - Y - Y2) = =^ M29^ -^ M jm,^ y,^ + +M2 (i-2y, 32 +^ y, ) + (^) -Ms (y ,+^24 , y+^ Y2)

= g(M2- M3)

  • dibawah V^ =^ O V = -(m , (^) gy , + (^) M29(y2+ (^) (L- Y (^) , )) + m ,g((L-^ Y^ ,^ ) +^ (l- ya) = - M,9y ,^ - M2g(k+^ L^

M3g(l -^ y^ , +^ l-^ Yz)^ =JB d :^ T-V (^) jj(Ms- M2) + (M3 + (^) M2) = g(Mc - M3) = Em ,^ y,^ +TM2(i -2y,^32 +^ y,^ )^ +^ 5 Mz(y ,+^24 , y+^ Y2) i (^) =^9 /Mc^ -^ Mz)^ -^ y^ ,^ (M3^ -^ M2)

m (^) , (^) gy ,^ - M2g(y+^ L - y (^) , ) - M3g(2 - y (^) , +^ l- Yz)) (M3^ +^ M2) L=^ m^ ,^ y,^ + jm2 (i-2y, 32 +^ y, ) + (^) 5 Mz(y ,+^24 , y+^ Y2)

+ M,gy , + M2g(Y2+^ L - y , ) + M39(1-y, + 1 - Yz)

untuk (^) y (^) ,

(^8)! =^ Itm

, y^ +^ McFey, y +^ imay, + im, y

+ty Ms zyi/,the = M (^) , G (^) , -Mayc +^ M2Y (^) , + (^) M3Y (^) , + (^) M3Y · ) = /m , y - May + (^) Mey (^) , + (^) M3y + (^) M (^) iis = M^ , jj, -M2iz +^ Mij,^ +^ M3 (^) i, + (^) M3 (^) i = (^) Y (^) (M , + (^) M2 + (^) M3) + i (^) (M3 -M2)

y n

Y,^ =^0 D2 =^ l^ CUS # X (^) ,^ =^ X, (^) X2 = l sin o (^) + X (^) , = O is (^) ,^ =^0 is (^) al (^) cost (^) + y I I dest^ de

  • =^ X (^) ,^ dt^ do dt i =^ -^1 sint 2 ·^ =^ 1 cost^ +^ &^ ,
  • (^) = +^ (205) (, +^ 16 (05t) = + 2 1 x, #^ CO50 +^1 Cost is =^12 :^2 sin (i+^ y) =^ <+ 21 x, (058 + 12052 - + 2 : (^) sin + = + (^21) ,C85t + l(2052 - Fsin^

= * + (^) 21 x (^) , C050 + 1

T = jMX ↑ +

jMc(+^ ya) =E MX,^ + [M/ ,+^ 21 x^ ,(050^ +^ 2)^ / = & =D untu o =

"MX,^ +,M2X,^ +^ Mc^ Ix,(05t^ +^ Melt

V =^ - M291 Cost =^ Mlx,^ Co5 O^ +^ Mr: L =^ T-V indiamitmimimi -4) = 28 = Untuk^ e^ =-M2lx,^ sino-Meglan

2

E,^ IEMX,^ +^ -M2X,+^ M2lx,^ Ecos e^ ·^ &C ) = m1 <, Cost +^ m &^ / = MX (^) , +^ M2X (^) , + Mr ECO5t^ (tx,^1058 +^2 )^ = -Mano(,+ (^) ge)

l & ee

· (^) X

· Co58 + = -^ X f^ sint g Of

  • e

sin t

f (^) l Komponen X. Larangiannya Independen (x,^ (M^

  • (^) M) + M2(5) =^0 Karena -^ x^ =^ Konstan^ ,^ X^ =^0
  • =^ constant^ :. Cost + sin + = (^0) E + r E =^ - sin