## 1. of the force of elasticity and elongation

1. AIM

·

To

prove the theory of energy in a spring and flywheel

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.