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SIMPLE PENDULUM. SIMPLE PENDULUM. SIMPLE PENDULUM. SIMPLE PENDULUM. SIMPLE PENDULUM. ENERGY. U = mgy = mg ( L – L cos q ) E = K + U = ½ mv ² + mg ( L – L cos q ) E = mg ( L – L cos q m ) mg ( L – L cos q m ) = ½ mv ² - PowerPoint PPT Presentation
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Physical Modeling, Fall 2006 1
SIMPLE PENDULUMSIMPLE PENDULUM
Physical Modeling, Fall 2006 2
SIMPLE PENDULUMSIMPLE PENDULUM
F mg
md s
dtmg
s
Ls L
mLd
dtmg
sin
sin
,
sin
2
2
2
2
or
F mg
md s
dtmg
s
Ls L
mLd
dtmg
sin
sin
,
sin
2
2
2
2
or
Physical Modeling, Fall 2006 3
SIMPLE PENDULUMSIMPLE PENDULUM
mLd
dtmg
d
dt
g
L
d
dt
g
L
2
2
2
2
2
2
sin
sin
For small ,
mLd
dtmg
d
dt
g
L
d
dt
g
L
2
2
2
2
2
2
sin
sin
For small ,
Physical Modeling, Fall 2006 4
SIMPLE PENDULUMSIMPLE PENDULUM
d
dt
g
L
d x
dt
k
mx
t x A t
g
L
k
m
m
2
2
2
2
cos( ) cos( )
d
dt
g
L
d x
dt
k
mx
t x A t
g
L
k
m
m
2
2
2
2
cos( ) cos( )
Physical Modeling, Fall 2006 5
SIMPLE PENDULUMSIMPLE PENDULUM
T fT
Tm
k
TL
g
2 1
2
2
2
Spring Mass
Simple Pendulum
T fT
Tm
k
TL
g
2 1
2
2
2
Spring Mass
Simple Pendulum
Physical Modeling, Fall 2006 6
ENERGYENERGY
U U == mgy mgy == mg mg((L – L cos L – L cos ))
E E == K + U K + U == ½½ mv mv² + ² + mgmg((L – L cos L – L cos ))
EE = = mgmg((L – L cos L – L cos mm))
mgmg((L – L cos L – L cos mm) =) = ½½ mv mv² ²
+ + mgmg((L – L cos L – L cos ))
– – gL cos gL cos mm = = ½½ v v²² – gL cos – gL cos
Physical Modeling, Fall 2006 7
ENERGYENERGY
½½ v v²² = gL cos = gL cos – gL cos – gL cos mm
vv²² = 2gL = 2gL ((cos cos – cos – cos mm) and ) and ss = = L L
ds
dtv L
d
dtv L
d
dt
Ld
dtgL m
so 2 22
22
2 (cos cos )
ds
dtv L
d
dtv L
d
dt
Ld
dtgL m
so 2 22
22
2 (cos cos )
Physical Modeling, Fall 2006 8
ENERGYENERGY
Lddt
gL
ddt
gL
ddt
m
m
m
22
2
22
2
2
2
(cos cos )
(cos cos )
(cos cos )
Lddt
gL
ddt
gL
ddt
m
m
m
22
2
22
2
2
2
(cos cos )
(cos cos )
(cos cos )
Physical Modeling, Fall 2006 9
AmplitudeAmplitude
2
020
2
2
22
21
coscos
coscos2
1
)cos(cos2
dtd
dtd
dtd
m
m
m
2
020
2
2
22
21
coscos
coscos2
1
)cos(cos2
dtd
dtd
dtd
m
m
m
Physical Modeling, Fall 2006 10
TimeTime
)cos(cos2
)cos(cos2
)cos(cos2 22
m
m
m
ddt
dtd
dtd
)cos(cos2
)cos(cos2
)cos(cos2 22
m
m
m
ddt
dtd
dtd
Physical Modeling, Fall 2006 11
TimeTime
m m
m
dt
ddt
)cos'(cos2'
)cos(cos2
m m
m
dt
ddt
)cos'(cos2'
)cos(cos2
Physical Modeling, Fall 2006 12
NEWTON'S LAW OFNEWTON'S LAW OFUNIVERSAL GRAVITATIONUNIVERSAL GRAVITATION
F Gm m
r
G
1 22
11 26 67 10. N m /kg2
F Gm m
r
G
1 22
11 26 67 10. N m /kg2
Physical Modeling, Fall 2006 13
At the surface ofAt the surface ofthe Earththe Earth
F GMmR
F mg
gGMR
2
2
F GMmR
F mg
gGMR
2
2
Physical Modeling, Fall 2006 14
POTENTIAL ENERGYPOTENTIAL ENERGY
U U U W
W d F dr
b a ab
ab a
b
a
b
F s
U U U W
W d F dr
b a ab
ab a
b
a
b
F s
Physical Modeling, Fall 2006 15
POTENTIAL ENERGYPOTENTIAL ENERGY
Choose Choose UU = 0 at = 0 at rr = =
U F dr
UGMm
r
r
U F dr
UGMm
r
r
Physical Modeling, Fall 2006 16
Conservation ofConservation ofMechanical EnergyMechanical Energy
When dropped fromWhen dropped from
rr = = RR
with with vv00 = 0. = 0.
GMm
Rmv
GMm
r12
2 GMm
Rmv
GMm
r12
2
Physical Modeling, Fall 2006 17
Conservation ofConservation ofMechanical EnergyMechanical Energy
v GMr R
v GMr R
2 21 1
21 1
v GMr R
v GMr R
2 21 1
21 1
