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