Click here to load reader
Upload
sandeep-patil
View
32
Download
3
Embed Size (px)
Citation preview
Oscillation of the lever caused by the swinging of the pendulum
Introduction
Two-armed lever has a weight on one arm, and the pendulum on the other, which is
oscillating forcibly because of the additional force moment created by the swinging of the
pendulum. “Gradual“ oscillations of the lever are also possible, but for the selection of the
masses on lever arms and the selection of the force arm and the weight arm, there are no
spontaneous oscillations.
The first part will contain the theory of the mathematical pendulum, and then all the
results will be used for the solution of the basic problem: oscillation of the lever due to the
oscillation of the pendulum.
1. Swinging of the mathematical pendulum
Swinging of the pendulum under the influence of the active component of the weight:
Picture 1.
Equation for the swinging of the pendulum is:
)sin(θmgsm −=&& , θls = (1)
From that, it follows:
0)sin(
)sin(
=+
−=
θθ
θθ
l
g
mgml
&&
&&
(2)
It is presumed that the oscillations are small, so the approximation is:
θθ ≈)sin( (3)
)sin(θmg
l
θ
θ )cos(θmg
mg
Therefore, differential equation (2) becomes:
0)sin(2 =Ω+ θθ&& , l
g=Ω (4)
Equation (4) will be solved for border cases:
0)2(,0)0( θ
πθθ =
Ω= (5)
In the domain of small oscillation which is being researched, it could be considered
that 523.06
0 ==π
θ , because 5.06
=π
m , so the approximation (3) is correct.
Solution for the equation (4) is:
)sin()cos()( tBtAt Ω+Ω=θ (6)
Therefore:
0)2
sin()2
cos()2(,0)0( θ
πππθθ ==+=
Ω== BBAA (7)
And,
)sin()sin()( 00 tl
gtt θθθ =Ω= (8)
Oscillation period is:
g
lT π
π2
2=
Ω= (9)
So that (8) can be shown like this:
)2
sin()( 0 tT
tπ
θθ = (10)
At the end, we will get the oscillating speed and the middle value of its square. This way, we
can get medium centrifugal force which affects the pendulum during swinging. The
differentiation (10) gives us:
)2
cos(2
)( 0
.
tTT
tπ
θπ
θ = (11)
To get the tangential speed, we multiply the corner speed with L and get:
)2
cos(2
)()( 0
.
tTT
ltltv
πθ
πθ ==
And we also get:
)2(cos
4)( 22
02
222 t
TT
ltv
πθ
π= (12)
From that, we get:
∫=
4/
0
22
02
222 )
2(cos
44T
dttTTT
lv
πθ
π (13)
And we have:
))4
cos(1(2
1)
2(cos2 t
Tt
T
ππ+=
After that, we get:
gll
gl
T
lt
T
T
TT
lv
T
2
022
2
0
222
02
224/
0
2
02
222
2
1
42
2)
4sin(
1
8(
44θ
πθπθ
ππ
πθ
π===+⋅=
Since the centrifugal force isl
vmFz
2
= , the final result is:
mgFz2
02
1θ= (13)
2. Oscillation of the lever under the influence of the swinging of the
pendulum
The point A is the weight with mass M, and the point B is the pendulum with mass m.
Picture 2.
Forced oscillations of the lever can be caused under the influence of the moment of the force
(see picture 2):
)cos(αFmgRM = (14)
The equation of free oscillations would be:
0)cos( =+ αα FmgRJ && (15)
Where J is the moment of the inertia of the lever part OAC. In further text, we will
approximate:
2
FmRJ = (16)
The pendulum creates an additional moment of the force by swinging.
αα
GR O
FR
Mg
)cos(αmg
mg
)sin(αmg
Picture 3.
We get the moment of force due to swinging of the pendulum, as in the picture where the
force is )cos(θmg and zF is:
FFzad RmgmgRFmg ))cos(2
1)(cos()90sin())cos((
2
0
2 θθθθθµ +=−+= o
Total moment of the force created due to the swinging of the pendulum is:
FGadGTOT RmgmgMgRMgR ))cos(2
1)(cos(
2
0
2 θθθµµ +=−=
This equation of the forced lever osculation (it is forced because of the pendulum
oscillations) we get:
)cos(2
1)(cos)cos(
2
0
2 θθθαα FFGF mgRmgRMgRmgRJ −−=+&&
That is:
)cos(2
1)(cos)cos(
2
0
2 θθθααJ
mgR
J
mgR
J
MgR
J
mgR FFGF −−=+&& (17)
In further text we will use a series of approximations. If we take the approximation (16), then
(17) becomes:
)cos(2
1)(cos)cos(
2
0
2
2θθθαα
FFF
G
F R
g
R
g
R
Rg
m
M
R
g−−=+&& (18)
Next series of approximations is based on the fact that angles α and θ are small. We will take:
222222
2
11)cos(;1)
2
11()(cos;1
2
11)cos( θθθθθαα −≈−≈−≈≈−≈ (19)
Include (18) into (19) and we will get:
22
0
2
2
0
2
1
2
1
2θθθ
θα
FFFFF
G
FF R
g
R
g
R
g
R
g
R
R
m
M
R
g
R
g++−−+−=&&
mgFz2
02
1θ=
O
θ GR
)cos()90sin( θθ FF RR =−o
Mg
)cos(θmg
θ
Or:
)2
2()4
11(
2
022
0
F
G
FF R
R
m
M
R
g
R
g−+−+=
θθθα&& (20)
Then, based on (9):
)(sin 22
0
2
l
gtθθ = (21)
Include that into (20) and we will get:
)2
2()(sin)4
11(
2
022
0
2
0
F
G
FF R
R
m
M
R
g
l
gt
R
g−+−+=
θθθα&& (22)
The equation (22) is integrated by time with initial conditions:
00 )0(,)0( αααα == & (23)
And we get:
tR
R
m
M
R
g
l
gt
R
glt
F
G
FF
)8
2()2sin(4)
4
11()(
4
02
0
2
0
θθθα −−−+−=& (24)
After integration by time, we get:
2
4
02
0
2
00 )
82(
2)2cos()
4
11(
8)( t
R
R
m
M
R
g
l
gt
R
lt
F
G
FF
−−−++=θ
θθ
αα (25)
Finally, we will take:
)2
(sin)2
(2
222
FFF R
gt
R
gtt
R
g≈= (26)
For more simple calculation, we will take that it is:
l
glRF == ω;2 (27)
Under the condition of:
)(sin)8
2()2cos()4
11(
4)( 2
4
02
0
4
00 t
R
R
m
Mtt
F
G ωθ
ωθθ
αα −−−+=− (28)
Since:
)(sin21)(sin)(cos)2cos( 222 tttt ωωωω −=−= (29)
It follows that:
)(sin)22
12()
4
11(
4)( 2
2
02
0
2
00
l
gt
l
R
m
Mt G θ
θθ
αα +−−+=− (30)
Function 0)( αα =t has zeros which are:
l
R
m
Ml
gt
G
2
1
22
)4
11(
4
1)(sin
2
0
2
0
2
02
−+
+
=θ
θθ
l
R
m
Ml
gt
G
2
1
22
)4
11(
2)sin(
2
0
2
00
−+
+
=θ
θθ
Finally, zeros are:
...2,1,0,22
1
22
)4
1(
2arcsin
2
0
2
0
0
=+
−+
+
= kk
l
g
l
g
l
R
m
M
t
G
K
π
θ
θθ
In the moments Kt is 0)( 0 =−αα t and the weight hits the surface.
Numbering:
192.0)arcsin(
1908.0
222
41
22618.0
2
0121.222
20865.14
1
0685.04
2618.02
125.08.0
1.0
22.0
9523.44.0
2
0
2
0
00
2
0
2
0
2
0
2
0
=
=
−+
+
==
=−+=+
==
====
===
X
R
R
m
MX
R
R
m
M
R
R
m
MmMmR
l
gml
F
G
F
G
F
GG
θ
θθθ
θθ
θθ
ω
...5,3,16345.00425.0 =+= kktK
7530.5
1185.5
4850.4
8495.3
2150.3
5805.2
9460.1
3115.1
6770.0
0425.0
9
8
7
6
5
4
3
2
1
0
=
=
=
=
=
=
=
=
=
=
t
t
t
t
t
t
t
t
t
t
0980.12
4635.11
8290.10
1945.10
5600.9
9255.8
2910.8
6565.7
0220.7
3875.6
19
18
17
16
15
14
13
12
11
10
=
=
=
=
=
=
=
=
=
=
t
t
t
t
t
t
t
t
t
t
4430.18
8085.17
1740.17
5395.16
9050.15
2705.15
6360.14
0015.14
3670.13
7325.12
29
28
27
26
25
24
23
22
21
20
=
=
=
=
=
=
=
=
=
=
t
t
t
t
t
t
t
t
t
t
7120.19
0775.19
31
30
=
=
t
t
In Novi Sad, 2000
Academician professor Bratislav Tošić