COMMENT: In this exercise the whole system was considered, 3 situations were analyzed
when m2 was at a height h, the instant when m2 contacts m1 and when m2 and m1 move as a
single body. It took the initial conditions to find the constants of the equation of motion at
time 0. TIME: 5 HOURS
COMMENT: In this demonstration we used energy method which we consider the fluid as a
constant system throughout its section which allowed us to express the mass as a function of
density and area since the tubular area throughout the section is constant while the height is
the variable that changes over time. TIME: 1 HOUR
COMMENT: The effective mass depends on several variables such as initial conditions and load
conditions which for our problem we use a bar in a simple situation. It was necessary to
integrate the K so that we can find the effective mass which depends directly on the mass of
the object and the length, that while either of these two increases its effective mass will
increase, in the same way the effective k depend on the charge and the maximum deformation
in the condition that is established. TIME:6 HOURS
COMMENT: We have a system with two springs that will depend on two masses for their
angular frequency this means that if we add much more mass to our system we will have a
lower frequency and the more spring the system has, the frequency will increase in the same
way it is necessary the analysis of the free body of the disk and the mass to know what forces
act on them. TIME: 4 HOURS
Five Cycle and Small Angle
period 2
period 3
period 4
period 5
Average
1.7
1.33
1.55
1.48
1.67
1.546
1.28
1.53
1.62
1.46
1.75
1.528
1.55
1.47
1.38
1.73
1.55
1.536
1.74
1.42
1.37
1.54
1.65
1.544
1.54
1.64
1.44
1.52
1.55
1.538
period 1
test 1
test 2
test 3
test 4
test 5
An analysis was carried out taking 5 tests with 5 periods each with an angle of 30 °
M. AXIS
17.25 R AXIS
0.025
M. BAR
0.185 L
0.44
M.P.
1
M. PROPELLER
10.48 R PROPELLER
0.1175
We have the masses of the system to later draw their respective inertia.
TEST 1
JP (ML^2)
JB (ML^2/3)
JA (MR^2/2)
F1
Jo-Exper.
Kk
KK/R
Jo-Theory
% ERROR
TEST 2
0.2007632 JP (ML^2)
0.01193867 JB (ML^2/3)
0.00539063 JA (MR^2/2)
0.29486568 F1
0.07407788 Jo-Exper.
0.08407437 Kk
0.71552656 KK/R
0.07234475 Jo-Theory
2.39565100 % ERROR
TEST 3
0.2007632 JP (ML^2)
0.2007632
0.011938667 JB (ML^2/3) 0.01193867
0.005390625 JA (MR^2/2) 0.00539063
0.288039441 F1
0.29106346
0.067251637 Jo-Exper.
0.07027565
0.080107059 Kk
0.08188829
0.681762204 KK/R
0.69692161
0.07234475 Jo-Theory
0.07234475
7.04005931 % ERROR
2.86005261
TEST 4
TEST 5
JP (ML^2)
0.2007632 JP (ML^2)
0.2007632
JB (ML^2/3) 0.01193867 JB (ML^2/3) 0.01193867
JA (MR^2/2) 0.00539063 JA (MR^2/2) 0.00539063
F1
0.29410326 F1
0.29182193
Jo-Exper.
0.07331546 Jo-Exper.
0.07103412
Kk
0.0836406 Kk
0.08232901
KK/R
0.71183489 KK/R
0.70067239
Jo-Theory
0.07234475 Jo-Theory
0.07234475
% ERROR
1.34178181 % ERROR
1.81164035
We use for experimental and theoretical inertia the following expressions
𝐽𝑂−𝑒𝑥𝑝 = (
𝜏2
𝑀𝐵
∗(
+ 𝑀𝑝 ) ∗ 𝑙𝑔) − 𝐽𝑃 − 𝐽𝐵 − 𝐽𝐴 − 𝑀𝐴 𝑅𝐴 2
2
4𝜋
2
𝐽𝑂−𝑡ℎ𝑒𝑜 =
𝑀𝑃𝑅𝑂 𝑅𝑃𝑅𝑂 2
2
COMMENT: we perform 5 tests taking 5 periods from the second until the system stabilizes
our results show that the theoretical and experimental inertia do not differ much being the
maximum error of 7%. Although you cannot have a good approximation since considering the
helix as a disk will differ greatly the results of real inertia by not being a regular figure and
despising many things. This process was carried out by the energy method, and it was also
necessary to consider the inertia of each object and express the inertia in terms of period in
our case will be the one that will change in each of our tests. We have a turning radius of
between 69 and 71 percent of the propeller radius which is a good approximation.
TIME: 6 HOURS