1. of the force of elasticity and elongation

1. AIM

·     
To
prove the theory of energy in a spring and flywheel

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2. INTRODUCTION

Part 1: energy in a spring

Hooke’s Law is the basic law of elasticity theory. It
was discovered by the English scientist Robert Hooke in 1660. Hooke’s Law
states that the elastic force arising from the elastic deformation of the
stretching or compressing of the spring is proportional to the absolute value of
the change in spring length.

If the elongation of the spring is denoted by x, and
the elastic force as F, then Hooke’s law can be written in the form of the
following mathematical formula:

F =
-kx

k – coefficient of proportionality, called as a stiffness
of the spring. The minus sign in front of the right side of the equation
indicates opposite direction of the force of elasticity and elongation unit of
rigidity in SI is Newton per meter (1 N / m)

F – force applied to spring (N)

x – the change in the length of the spring under the
force action (m)

Energy that is stored in a stretched or compressed
spring can be calculated using the equation

SE=1/2kx^2

(SE- spring energy) The k value represents the “spring
constant’ for any particular spring and tells about the stiffness of the
spring. Each body has its own stiffness. The greater the stiffness of the body
(springs, wires, rods, etc.), the less it changes its length under the action
of a given force.

It should be noted that Hooke’s law is valid only for
elastic deformation. Hooke’s law is well satisfied only for small deformations.
For large deformations, the change in length ceases to be directly proportional
to the applied force, and for very large deformations the body collapses.

Hooke’s conclusions formed the basis of the modern
theory of elasticity which plays an important role in modern world. As for
example, almost all the details of modern technology are fastened together with
threaded joints, the principle of which is that when the pins, bolts and screws
are tightened, they are elastically deformed and reliably fix the parts,
pressing them against each other. A good example of this is the car wheel
bolts. They do not have fixing nuts on the back side, but securely fasten the wheel
due to the elastic deformation of the bolt pin and the frictional force.

This
experiment aims to determine the k value of a coil spring.
By applying forces to a spring, measuring the extensions (x) and plotting the
results in a graph, koefficient k can be calculated by the gradient of the
trend line. The energy stored in a spring can also be calculated by the work
done to compress or stretch the spring. In this experiment SE is found by
calculating the work done by Potential Energy PE = mgh (mgh with h being
calculated from when the hanging spring is in equilibrium position).

 

Part 2: energy in a flywheel

In this part of the experiment, the task is to
determine the moment of inertia of a system consisting of a flywheel of an axis
and a shaft.

A small flywheel that can rotate with very small
friction near the horizontal axis is located at a height of h0 from the floor.
The axis of rotation passes through the center of gravity of the flywheel. On
the axis of the wheel is a small diameter shaft, which winds the cord.

If hang a weight of m0 on the cord and screw the shaft
so that the weight rises to a height h, the weight and the entire flywheel
system will acquire potential energy:

PE=mgh

where m is the weight of the weight, g = 9.81 m/s^2 is
the acceleration due to gravity, h is the height of the weight above the floor.

In the dynamics of the rotational motion of a rigid
body, the concepts of “force” and “mass” are replaced by
the concepts “moment of force” and “moment of inertia”.

When a solid body rotates about an axis, the effect of
the force depends not only on the magnitude of the force, but also on the
distance of the point of application of the force to the axis of rotation.
Therefore, instead of force, a moment of force is introduced, equal to the
product of the value of the acting force on the arm (the shortest distance from
the axis of rotation to the direction of the acting force)

M = F x r

 

Rotational motion of a solid body is characterized by
angular velocity ? and
angular acceleration ?.

 

The angular velocity of rotation is a vector
numerically equal to the first derivative of the angle of rotation of the
radius vector with respect to time and directed along the axis of rotation so
that from its end the rotation can be seen going counter-clockwise:

 

 

The
angular acceleration is a vector ? (rad/s^2) equal to the first derivative of
the angular velocity ? with respect to time:

The
moment of inertia of a material point relative to the axis of rotation is a
scalar physical quantity numerically equal to the product of the mass of this
point by the square of its distance to the axis of rotation:

 

I = mr^2

Dependence
of angular acceleration ? on the moment M acting on the body of force and
moment of inertia I the body is expressed by the second law of Newton for
rotary motion (basic law of rotational motion):

                                

? = M/I

Flywheels
can be used to store energy and to produce very high electric power pulses for
experiments, where drawing the power from the public electric network would
produce unacceptable spikes. A small motor can accelerate the flywheel between
the pulses.

 

 

 

 

 

 

 

 

 

 

 

3.
METHOD

3.1
Materials

No

Equipment

Quantity

1

Work
panel

1

2

Plummet arm

1

3

Weight
hanger

1

4

Guide

1

7

Coil spring

1

8

String

1

9

Set of weights (10g each)

30

10

Pulley Block

1

11

Ruler

1

12

Flywheel

1

13

Stopwatch

1

 

3.2
Progress

Part
1.1

1.
Set up the work panel for beginning of the experiment

2.
Use an ‘S’ hook to hang the bottom of the weight hanger

3.
Use ruler to measure the height from the floor to the bottom of spring whilst
it is hanging stretched in equilibrium position

4.
Add a 10 gram weight disc until the total mass of the disks is 60 grams, then
add 20 grams from 60g to the total weight of 280 grams. During each phase of
the weight change, measure the extension of the spring with a ruler. The
results are recorded in the table

5.
Make a graphic representation of the load (N) against extension (m) in excel.

6.
Determine the stiffness k value of a coil spring

 

Part
1.2

1.
Slightly support and lift the weight hanger until its hook just touches the
spring loop and release the weight in the free fall, observing how the spring
elongate.

2.
 The task is to choose the appropriate
weight value at which weight hook will touch the table once after falling

3.
After several attempts, the weight of 340 grams showed the desired result at which
the weights hanger touched the table once

4.
While holding the weight hanger, measure the height from the bottom of the
weight hanger to the desk

5.
Using the results obtained, calculate the potential and spring energy

 

Part
2

1.
Put the metal disk into the pivot with the twisting element. The disc turns
easily so assume its movement frictionless

2.
Fit the plastic guide to the bottom of the weight hanger. Add weight discs to
get 350 grams in total

3.
Measure the height of the lifted weight hanger and take the result obtained for
the point of the beginning of the fall

4.
Use the stopwatch to measure how many seconds it take for weight hanger to fall
from 238 mm position to the 0 mm position. Repeat several times to get a good
average value fro the time

5.
Calculate potential energy, kinetic energy (angular and linear)

6.
Get the total moment of inertia I for the flywheel

 

4.
Results

Part 1.1

table 1

 

Mass (g)

Load (N)

Scale Reading (mm)

Extension
x(m)

k
(N/m)

0

0

0

0

0

10

0.0981

150

0.005

19.62

20

0.1962

148

0.007

28.03

30

0.2943

145

0.01

29.43

40

0.3924

144

0.011

35.68

50

0.4905

142

0.013

37,74

60

0.5886

139

0.016

36.79

80

0.7848

136

0.019

41.31

100

0.981

131

0.024

40.88

120

1.1772

128

0.027

43.6

140

1.3734

122

0.033

41.62

160

1.5696

117

0.038

41.31

180

1.7658

114

0.041

43.07

200

1.962

109

0.046

42.7

220

2.1582

104

0.051

42.32

240

2.3544

99

0.056

42.04

260

2.5506

94

0.061

41.82

280

2.7468

90

0.065

42.26

k (average) = 41.62

 

 

 

 

 

Part 1.2

 

table 2

Drop height (mm)

Drop height (m)

Mass (g)

Mass (kg)

Potential energy

151
mm

0.151
m

340

0.34

0.504J

 

 

Part 2

 

table 3

Mass
m
(kg)

PE
mgh
(J)

Falling time
(seconds)

Acceleration
(ms^-2)

Velocity
? (ms^-1)

Velocity
? (rad.s^-1)

KE
linear

KE
angular

h
(m)

0.35

0.82

0.99

0.51

0.48

73.85

0.0804

0.77

0.238

0.35

0.82

0.94

0.57

0.51

78.46

0.0809

0.78

0.238

0.35
 
 

0.82
 

1.04
 
 
 

0.53
 
 
 

0.46
 

70.77
 
 
 

 
0.0805
 
 

0.79
 
 
 

 
0.238
 
 

 

 

5. CALCULATION

 

Part1.1

 

Calculating load (N) : using formula F = mg (m – mass
in kg; g – acceleration of gravity = 9.81 m/s^2)

 

F1 = 0.01*9.81 = 0.0981 N

F2 = 0.02*9.81 = 0.1962 N   same calculation process with other weights.

 

Determine the stiffness k. Rigidity is the
coefficient of proportionality between the force of elasticity and the change
in the length of the spring under the action of the force applied to it.
According to Newton’s third law, the force applied to the spring is modulo
equal to the elastic force arising in it. Thus, the spring stiffness can be
expressed as:   k = F/x

 

 

k1 = 0

k2 = 0.0981 N/ 0.005 m = 19.62 N/m

k3 = 0.1962 N/ 0.007 m = 28.03 N/m

k4 = 0.2943 N/ 0.01 m = 29.43 N/m

k5 = 0.3924 N/ 0.011 m = 35.68 N/m

k6 = 0.4905 N/ 0.013 m = 37.74 N/m

k7 = 0.5886 N/ 0.016 m = 36.79 N/m

k8 = 0.7848 N/ 0.019 m = 41.31 N/m

k9 = 0.981N / 0.024 m = 40.88 N/m

k10 = 1.1772 N/ 0.027 m = 43.6 N/m

k11 = 1.3734 N/ 0.033 m = 41.62 N/m

k12 = 1.5696 N/ 0.038 m = 41.31 N/m

k13 = 1.7658 N/ 0.041 m = 43.07 N/m

k14 = 1.962 N/ 0.046 m = 42.7 N/m

k15 = 2.1582 N/ 0.051 m = 42.32 N/m

k16 = 2.3544 N/ 0.056 m = 42.04 N/m

k17 = 2.5506 N/ 0.061 m = 41.82 N/m

k18 = 2.7468 N/ 0.065 m = 42.26 N/m                             k(average)
= 41, 62

 

 

Part 1.2

Determine potential and spring
energy.

PE = mgh  SE =kx^2/2 PE=SE

PE = 0.34kg * 9.81m/s^2 * 0.151m =
0.504J

SE => 0.504J

 

 

Part 2

 

PE = mgh

PE = 0.35kg * 9.81m/s^2 * 0.238m =
0.82J

 

If let go the weight, it will start
to fall, spinning the cord and rotating the entire system.

The potential energy of the weight
will be converted into the kinetic energy of the falling weight and the
rotating system, the flywheel-axis-shaft. Consequently, neglecting
friction, according to the law of conservation of energy this equation can be
written:

 

PE =   KEang. +KElin.

mgh = mv^2/2 + I?^2/2

 

where:

v – is the speed of the weight drop
at the moment of landing,

? – the angular
velocity of rotation of the system at the same moment,

 I – moment of inertia of the
flywheel system – axis – shaft.

 

 

Assuming the motion of the dumbbell
is uniformly accelerated get:

h = at^2/2 => a = 2h/t^2  => a1 = 2*0.238/0.99^2 = 0.51 ms^-2

                                         a2
= 2*0.238/0.94^2 = 0.57 ms^-2        

                                         a3
= 2*0.238/1.04^2 = 0.53 ms^-2

As dumbell start moving from state of rest => v = at      

Substitute the acceleration value in the velocity formula: V = 2h/t

V1 = 2*0.238/0.99 = 0.48 ms^-1

V2 = 2*0.238/0.94 = 0.51 ms^-1

V3 = 2*0.238/1.04 = 0.46 ms^-1

 

The linear speed of rotation of the point on the outer
surface of the shaft is equal to the rate of fall of the weight. Relationship
between linear and angular velocity of rotation: v = ?*r

where r is the radius of the shaft and ? = v/r or ? = 2h/rt.

?1 = 0.48/(6.5*10^-3) = 73.85 rad.s^-1

?2 = 0.51/(6.5*10^-3) = 78.46 rad.s^-1

?3 = 0.46/(6.5*10^-3) = 70.77 rad.s^-1

 

Substitute the values of v and ? into the PE =   KEang. +KElin and get: 

mgh = 2mh^2/t^2 + 2Ih^2/r^2t^2

from which I = mr^2(gt^2-2h)/2h

 

I1 = 0.35*(6.5*10^-3)^2 (9.81(0.99)^2-2*0.238)/2*0.238
= 2.84*10^-4

I2 = 0.35*(6.5*10^-3)^2 (9.81(0.94)^2-2*0.238)/2*0.238
= 2.54*10^-4

I3 = 0.35*(6.5*10^-3)^2 (9.81(1.04^2-2*0.238)/2*0.238
= 3.15*10^-4

 

KE angular = 1/2I?^2

KE1 = (2.84*10^-4*73.85^2)/2 = 0.77

KE2 = (2.54*10^-4*78.46^2)/2 = 0.78

KE3 = (3.15*10^-4*70.77^2)/2 = 0.79

 

KE linear =1/2 mV^2

KE1 = 1/2 * 0.35*0.48 = 0.084

KE2 = ½*0.35*0.51 = 0.089

KE3 = ½ 0.35*0.46 = 0.0805

 

6. Discussion/conclusion

 

In part 1.1 in each of the experiment, stiffness is determined for
different values of the elastic and elongation forces, so that the experimental
conditions vary. Therefore, to find the average stiffness value, it is not
possible to calculate the arithmetic mean of the measurement results. The
graphical method was used in finding the mean value. ?fter plotting
the graph, take the value of the elastic force and the tension of the spring of
the point located in the middle part of the graph. So that k(average) = 41, 62
N/m

 

The potential energy was derived using the area under Force vs Extension
graph. This is based on the assumption that a spring has ideal elastic
properties and the area under the graph is a triangle. The results prove that
this does apply to the spring used as when we added the first 2 masses of
0.01kg then 0.02 kg there was an extension table 1, meaning that the
relationship between Force and Extension was directly proportional.

 

The accuracy of these results cannot be proven further by taking into consideration,
the limitations in precision of the ruler and weights, and the percentage
uncertainty of the measurements of the maximum extension of the spring because
all was done in not ideal conditions

 

In the
second part of the laboratory, the moment of inertia of the flywheel was found
by means of multiple expressions of formulas based on the law of conservation
of energy. In three experiments, the total time value for the same weight was
different, but the resulting values of the moment of inertia are not
significantly different. Same situation about linear and angular kinetic
energies.

 

Flywheels are
very common especially when used in punching and riveting machines. For
combustion engine applications, the flywheel is a heavy wheel mounted on the
crankshaft. The main function of it is to maintain a near constant angular
velocity of the crankshaft.

 

To varying
degrees in both laboratories, the results were satisfying the main provisions. 

x

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