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Lab Report

KB5008

Title: Flywheel Lab Report.

Introduction This experiment is meant for find the mass moment of inertia of flywheel and comparing the calculated and the experimental value of the flywheel’s moment of inertia. A flywheel is made to rotate by attaching a mass to a string that is wrapped around the flywheel axle. The falling mass exerts a force that is related to the Torque, T, and the rate of change of the wheel's angular velocity, which is also known as the angular acceleration,

α ,

of the flywheel. The constant of proportionality between the angular acceleration and angular velocity is called the moment of inertia, I. The the moment of inertiais the inertia that is dependant on the totating object’ s effective radius This law is verified using a mass and taking measurements of the resulting acceleration. This is one of the objectives of this experiment. A flywheel is a device that has a sufficient moment of inertia and is whose purpose is storage of rotational energy. Flywheel resists rotational speeds changes; this helps to steady the shaft’s rotation, in cases where a changing torque is exerted on it. For a flywheel with a solid disk, its mass moment of inertia is found using the formula below.

I=

m r2 2

Where, M= mass r= radius of the flywheel For rotational motion, Newton’s second law of motion is used to describe the relationship between the applied torque and the angular acceleration. The equation is shown below.

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Lab Report

KB5008

T =I × α When the angular acceleration is constant, the angular displacement of the rotating flywheel can be obtained using the formula, 1 θ=ω0 + α t 2 2 The analysis of flywheel is important since flywheels are used in many engineering applications such as in punching machines, riveting machines and in moving vehicles.

System Diagram The diagram below shows the System Diagram.

Figure 1: System Diagram

Free Body Diagram The diagram below shows a hand-drawn free body diagram of the flywheel.

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Lab Report

KB5008

Figure 2: Free Body Diagram

Where, ω

= angular acceleration.

r= radius of the flywheel T= torque and W = weight.

Application of Conservation of Momentum Given that the m, represents the mass of the hanger, when the mass falls through the height “h” there is a loss of potential energy. This loss of potential energy can be equated as shown below. Ploss=mgh

The flywheel’s Kinetic energy as the mass falls is given as:

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Lab Report

KB5008

1 K flywheel = Iω2 2

Where, I = is the flywheel’s assembly moment of inertia. ω = is the weight’s angular velocity at the point when it reaches the ground. The falling weight gains kinetic energy on its way down, the kinetic energy is calculated as; 1 K weight = m V 2 2 Where, v is the velocity of the weight at the time, it reaches the ground. From the law of conservation of energy, the work needed to handle the friction torque in the rotating flywheel is; Ploss=K flywheel +K weight +W friction (1) After substituting the values, we get; 1 2 1 mgh= I ω + I v 2 +n W f (2) 2 2 When the flywheel rotates N times, against the same frictional torque, the kinetic energy becomes, 1 1 N W f = I ω 2 And W f = I ω2 2 2N Given that the axle’s radius is r, the weight’s velocity is calculated by the following equation; V =ωr

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The replacing the values of

W r ∧V

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gives the equation,

1 2 1 2 n mgh= I ω + mr2 ω2 + × I ω (3) 2 N 2 Then solve the equation above for I,

I=

(

)

Nm 2 gh 2 −r (4) N + n ω2

Where; I is the flywheel’s moment of inertia. N is the flywheel’s umber of rotations before it stops. m is the rings’ mass. n is the number of the string windings on the shaft.. g is the gravitational acceleration of the Environment. h is the height ,from the ground, of the weight assembly. r is the radius of the axle. Then while counting the number of rotations, N, for as long as the flywheel rotates it is paramount to note the time it takes to make N number of rotations. The average angular velocity, ω average=

ω average , can therefore be calculated in radians per second.

2 πN t

There is an assumption that the torsional friction

Wf

is constant with time and the angular

velocity two times as much as the average angular velocity, then,

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ω=

Lab Report

4 πN (5) t

The flywheel has a mass, m, The density is ρ , The radius of the flywheel is R, By definition, dI=ρ r 2 dV

ρ=

M V

= disk

M π r2 h

2

V =π r h

dV =2 π r h dr R

I=

M 2 r (2 πhrdr ) 2 ∫ πr h 0 R

¿

M ⋅2 πh ∫ r3 πr 2 h 0

¿

2M r4 [ ] r2 4

¿

1 M R2 2

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Lab Report

KB5008

Results Calculated Total Inertia I: = Load

Load

mg

−5

1.8926 ×10 Kg .m

2

α

Angular Acceleration

Linear Mas s (g)

Mass

( rad . s Test Test

m(Kg)

−2

) acceleration,

ma

mg-ma

1273.3333

0.0286

0.9523

3

5

5

0.1113

1.8507

0.2401

2.7028

7

4

4548.6666

0.4093

3.5146

7

8

2

0.6162

4.2887

8

3

a/ r 2

Test Average a (m/s2)

100

0.1

1

2

3

0.98

19.

19.

19.

1

1

1

1

1.96

37.

37.

37.

19.1 0.2865

200

0.2

37.1

2473.3333 0.5565

300

0.3

2

0

0

3

3

2.94

53.

53.

53.

53.37 0.80055

400

0.4

3

3

5

3

3.92

68.

68.

68.

4

2

3

2

4.90

82.

82.

81.

68.23

3558

1.02345 500

0.5

82.17 1.23255

5

1

5

5478

9 Figure 3: Table of Results

Calculations The theoretical mass moment of inertia is found using the formula below.

I=

Where, I is the mass moment of inertia. m is the mass r is the radius

m r2 2

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The mass of inertia for the flywheel and the shaft is given as,

Figure 4: Flywheel and shaft

Part A :r=4 mm , M =35.94 I =2.8752× 10−7 kg . m2 Part B :r =48 mm , m =568.2 g Part C :r =5 mm ,m =78.9 g F=Missing hole :r =5 mm ,m=6.2 g ×14 G=Missing Hole :r=5 mm , m =7.1 g Inertias : B− ( F ×14 )+ C−G

I =5.3111 ×10−4 kg . m2 Total Shaft Inertia= A+ B−( F ×14 ) C−G −4

I =5.314 ×10 kg . m

2

The mass moment of inertia for the pulley is calculated as,

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Figure 5: Pulley

Part A (Brass Collet ):inner r=4 mm , outer r =6 mm , m=7.53 g Part B ( Solid Body ) :r =15 mm ,m =169.65 g F=Missing Hole :r =6 mm , m =13.57 g Totalinertia= A + B − F −5

I =1.8926 ×10 Kg .m

2

A simple approach to calculating the linear acceleration is shown below. linear acceleration=angular acceleration× theradius For the first mass, 100 g, The linear acceleration is calculated as; Angular acceleration = 19.1 rad . s−2 Radius = 15 mm = 0.015 m linear acceleration=19.1 ×0.015

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KB5008

linear acceleration=0.2865 m/m 2 The a/ r 2 is calculated as, 0.2865 =1273.33333 0.0152 ma is calculated as, 0.1 ×0.2865= 0.02865 mg – ma is calculated as, 0.981−0.02865 =0.95235 All the other values for the table are calculated similarmannerly. Graphs

mg- ma

A graph of mg- ma against a/� ^2 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 1000

1500

2000

2500

3000

3500

4000

4500

a/�^2

Figure 6: Graph of mg-ma against a/r^2

Discussion The gradient from the graph can be calculated as shown below

Gradient=

change∈( mg−ma) change∈( a/r 2)

5000

5500

6000

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Gradient =

Lab Report 3.51462−1.8507 4548.66667−2473.33333

=

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1.66392 2075.333

Gradient = 0.80176 ×10−3 The gradient is equivalent to I. The experimental flywheel’s mass moment of Inertia is calculated as shown below, The experimental value of themass moment of inertia=gradient × the radius −3

¿ 0.80176 ×10 × 0.015

¿ 1.20264 ×10−5 Kg . m2 The value of the theoretical and the Experimental mass moment of inertia is not equal. The value calculated from the experiment slightly lower than the value calculated from the theoretical approach. Percentage error

Percentage error =

experimentalmoment of Inertia−theoretical moment of Inertia theoretical moment of Inertia

Percentage error =

1.20264 × 10 −1.8926 × 10 × 100 % −5 1.8926 × 10

Percentage error =

−0. 3646× 100 % =36.46 %

−5

x 100%

−5

Some of the reason for the difference between the experimental and the theoretical values are due to assumptions during the experiment as well as errors in the experiment. During the experiment, there were several assumptions that were not taken into consideration when calculating the mass moment of inertia value. One of the assumptions in the experiment is friction force. There are obviously friction forces between the various moving parts in the setup. These forces were however not take in consideration for possible compensation when

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Lab Report

KB5008

calculating the value of inertia. Another cause of the difference is due to errors both in the measuring equipment and human sources. There also could be errors resulting from a wrong transfer of values from the measuring instruments on to the paper. Other errors are encountered when reading values from the measuring instruments, some of them being as a result of parallax errors. The flywheel is made to rotate by attaching a mass to a string that is wrapped around the flywheel axle. The falling mass exerts a force that is related to the Torque, T, and the rate of change of the wheel's angular velocity. The change of the angular velocity with respect to time is the angular acceleration,

α , of the flywheel. The constant of proportionality

between the angular acceleration and angular velocity is called the moment of inertia, I. The moment of inertia is the inertia that is depends on the rotating object’s effective radius This law is verified using a mass and taking measurements of the resulting acceleration We can find that the moment of inertia increases with an increase in the weight of the fly wheel that will make it store more energy but also make it take more energy to rotate and that obey the law of conservation of energy. And it’s easy now to measure the amount of energy that the flywheel can store.

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Lab Report

KB5008

References Ashby, M. (2011). Material Selection in Mechanical Design. 3rd edn. New Jersey :Burlington. Hibbeler, R. (2011). Engineering Mechanics Mathematics Dynamics. 11th edn. New Jersey : Prentice- Hall International. Meriam, J. (2015). Engineering Mechanics Dynamics. 6th edn. New Jersey : John Wiley & Sons Inc....