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1. Flywheel
Student Name: Rawa Abdullah Taha
Class: Third – Group A2
Course Title: Theory of machines Lab
Department: Mechanic and Mechatronics
College of Engineering
Salddin University - Erbil
Academic Year 2020 – 2021
Date /4/2021

2. Introduction:
The function of the flywheel is to limit the fluctuations of speed during
each cycle. During certain periods of cycle. The engine is developing
more horse power than is being taken from it, while of other period is
develops less power. The flywheel absorbs energy during those periods
when the turning moment is greater than the resistance moment and gives
out the energy during those periods when the resisting moment is greater
than the turning moment. Hence it acts as a reservoir of energy which can
give up and absorb energy as require when the flywheel absorbs energy
its speed increases and when energy is given up its speed decreases. This
accurate at during each stroke.
Capability:
1- Determine the mass moment of inertia of the flywheel using different
periods. 2- Also study the relationship between angular velocity of the
flywheel and the energy stored in it. 3-Comparing the theoretical and
experimental values of the moment of inertia of the flywheel and
studding the transformation of energy through the experiment.
Net force = 𝑚𝑔 – 𝐹
Acceleration = 𝑎
Hence 𝑚𝑎 = 𝑚𝑔 – 𝐹
Provided that 𝑎 is much smaller than 𝐹 = 𝑚𝑔
For the wheel angular displacement 𝑞 = 2𝜋𝑁 [rad]
When N = number of revelations
Average angular velocity = 1/2 (𝜃 + ω𝑁) 𝑟𝑎𝑑/𝑠 Time for N revolution = t
Angular displacement 𝜃 = 1 2 ω𝑁 ∗ 𝑡 and ω𝑁 = 𝛼 𝑡 2 from which according to second
law of motion.
Torque producing acceleration = Fr
𝐹𝑟 = [𝑐] = 𝑘 * 𝛼 = 𝑘 ∗ 1/4𝜋𝑁

3. From which 𝑘 = 1/4𝜋𝑁 ∗ 𝐶𝑡 2
The constant to proportionality k is called the moment of inertia may be calculated
from the dimensions and mass of the flywheel. Note:- C Couple = 𝑚𝑔𝑛 = Torque

4. 𝑘 = 𝜌𝜋𝑅 2
ω ∗ 𝑅 2
/ 2
𝐹𝑟 = 𝑘/ 4𝜋𝑁 𝑡 2
Where:
R = Radius of flywheel
𝜔 = width of flywheel = 2 cm = 0.02 m
𝜌 = density of steel = 7856 kg/𝑚3
R = 12 cm = 0.012 m
r = 1.5 cm = 0.015 m
Hanger width = 70 g
Results:
Tabulate the times for the different values of mass plus hanger and calculate:
𝑘 = 𝜌𝜋𝑅 2
ω ∗ 𝑅 2
/ 2
𝑘 = (7856)(0.12) 2
(0.02) ∗ (0.12) 2
/ 2
𝑘 = 0.05112 Theoretical
Effective
couple (F.r)
N.m
1/t2
Time(s)
Weight
Kg(N)
No.Of turns
N
0.05078
0.00988
10.06
107
8
0.1025
0.0199
7.08
157
8
0.15068
0.0293
5.84
217
8
0.225867
0.04395
4.77
387
8
0.43177
0.0840
3.45
432
8
0.501869
0.09765
3.2
532
8

5. Plot the experimental result on a graph of total load against 1/ 𝑡 2
and
draw the best fit straight line through the points.
Practical Calculations: 𝐹. 𝑟 = 𝑘 ∗ 4𝜋𝑁/ 𝑡 2
∆𝑊. 𝑟 = 𝑘 ∗ 4𝜋𝑁/ 𝑡2
Practical Calculations:

6. Discussion:
1- What is the relation between angular velocity and kinetic energy?
2-If two flywheels are going to be use in an engine, do you think the
additional flywheel is going to be useful or not?
3-Comparing the theoretical and experimental of flywheel
Answers:
1.kinetic energy increases quadratically with angular velocity.
Ker = 1/2 I * w2
When the angular velocity of a spinning wheel doubles, its
kinetic energy increases by a factor of four.
2- If two flywheels are going to be use in an engine, do you
think the additional flywheel is going to be useful or not? It's
useful, because Flywheels is a rotating mechanical device that is
used to provide continuous energy (rotational energy) in systems
where the energy source is not continuous, the rate that is
compatible with the energy source, and then releasing that
energy at a much higher rate over a relatively short time
3.generally the differences are between theoretical and practical
data with some causes such as that giving data with our mind
and not directly and not rightness and maybe caused by the
engine because the age of engine that is old and maybe giving
wrong data .