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Moment of Inertia of Flywheel
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Aim:

  

To determine the moment of inertia of a flywheel.

 

Apparatus:

 

Fly wheel, weight hanger, slotted weights, stop watch, metre scale.

 

Theory:

 

The flywheel consists of a heavy circular disc/massive wheel fitted with a strong axle projecting on either side.The axle is mounted on ball bearings on two fixed supports. There is a small peg on the axle. One end of a cord is loosely looped around the peg and its other end carries the weight-hanger.

                                                    

Let "m" be the mass of the weight hanger and hanging rings (weight assembly).When the mass "m" descends through a height "h", the loss in potential energy is

 

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»P«/mi»«mrow»«mi»l«/mi»«mi»o«/mi»«mi»s«/mi»«mi»s«/mi»«/mrow»«/msub»«mo»=«/mo»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«/math»

The resulting gain of kinetic energy in the rotating flywheel assembly (flywheel and axle) is

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»K«/mi»«mrow»«mi»f«/mi»«mi»l«/mi»«mi»y«/mi»«mi»w«/mi»«mi»h«/mi»«mi»e«/mi»«mi»e«/mi»«mi»l«/mi»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»I«/mi»«msup»«mi»§#969;«/mi»«mn»2«/mn»«/msup»«/math»

Where

I -moment of inertia of the flywheel assembly

ω-angular velocity at the instant the weight assembly touches the ground.

 

The gain of kinetic energy in the descending weight assembly is,

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»K«/mi»«mrow»«mi»w«/mi»«mi»e«/mi»«mi»i«/mi»«mi»g«/mi»«mi»h«/mi»«mi»t«/mi»«moȤnbsp;«/mo»«/mrow»«/msub»«mo»=«/mo»«moȤnbsp;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math» 

Where v is the velocity at the instant the weight assembly  touches the ground.

 

The work done in overcoming the friction of the bearings supporting the flywheel assembly is

 

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»W«/mi»«mrow»«mi»f«/mi»«mi»r«/mi»«mi»i«/mi»«mi»c«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«/mrow»«/msub»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»n«/mi»«msub»«mi»W«/mi»«mi»f«/mi»«/msub»«/math»

Where

n - number of times the cord is wrapped around the axle

Wf - work done to overcome the frictional torque in rotating the flywheel assembly completely once

Therefore from the law of conservation of energy we get

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»P«/mi»«mrow»«mi»l«/mi»«mi»o«/mi»«mi»s«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«/mrow»«/msub»«mo»=«/mo»«msub»«mi»K«/mi»«mrow»«mi»f«/mi»«mi»l«/mi»«mi»y«/mi»«mi»w«/mi»«mi»h«/mi»«mi»e«/mi»«mi»e«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«/mrow»«/msub»«mo»+«/mo»«msub»«mi»K«/mi»«mrow»«mi»w«/mi»«mi»e«/mi»«mi»i«/mi»«mi»g«/mi»«mi»h«/mi»«mi»t«/mi»«/mrow»«/msub»«mo»§nbsp;«/mo»«mo»+«/mo»«msub»«mi»W«/mi»«mrow»«mi»f«/mi»«mi»r«/mi»«mi»i«/mi»«mi»c«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«/mrow»«/msub»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mn»1«/mn»«mo»)«/mo»«/math»


On substituting the values we get

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»I«/mi»«msup»«mi»§#969;«/mi»«mn»2«/mn»«/msup»«mo»§nbsp;«/mo»«mo»+«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«mo»§nbsp;«/mo»«mo»+«/mo»«mi»n«/mi»«msub»«mi»W«/mi»«mi»f«/mi»«/msub»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mn»2«/mn»«mo»)«/mo»«/math»

Now the kinetic energy of the flywheel assembly is expended in rotating N times against the same frictional torque. Therefore

 Ãƒâ€šÃ‚«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»N«/mi»«msub»«mi»W«/mi»«mi»f«/mi»«/msub»«moȤnbsp;«/mo»«mo»=«/mo»«moȤnbsp;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»I«/mi»«msup»«miȤ#969;«/mi»«mn»2«/mn»«/msup»«moȤnbsp;«/mo»«/math»and Ãƒâ€šÃ‚«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»W«/mi»«mi»f«/mi»«/msub»«moȤnbsp;«/mo»«mo»=«/mo»«moȤnbsp;«/mo»«mfrac»«mn»1«/mn»«mrow»«mn»2«/mn»«mi»N«/mi»«/mrow»«/mfrac»«mi»I«/mi»«msup»«miȤ#969;«/mi»«mn»2«/mn»«/msup»«/math»

If r is the radius of the axle, then velocity v of the weight assembly is related to r by the equation

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»v«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»§#969;«/mi»«mi»r«/mi»«/math»

Substituting the values of v and Wf we get:

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»I«/mi»«msup»«mi»§#969;«/mi»«mn»2«/mn»«/msup»«mo»§nbsp;«/mo»«mo»+«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«msup»«mi»§#969;«/mi»«mn»2«/mn»«/msup»«mo»§nbsp;«/mo»«mo»+«/mo»«mfrac»«mi»n«/mi»«mi»N«/mi»«/mfrac»«mo»§#215;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»I«/mi»«msup»«mi»§#969;«/mi»«mn»2«/mn»«/msup»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mn»3«/mn»«mo»)«/mo»«/math»
 

 Now solving the above equation for I

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»I«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mrow»«mi»N«/mi»«mi»m«/mi»«/mrow»«mrow»«mi»N«/mi»«mo»+«/mo»«mi»n«/mi»«/mrow»«/mfrac»«mfenced»«mrow»«mfrac»«mrow»«mn»2«/mn»«mi»g«/mi»«mi»h«/mi»«/mrow»«msup»«mi»§#969;«/mi»«mn»2«/mn»«/msup»«/mfrac»«mo»-«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfenced»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mn»4«/mn»«mo»)«/mo»«/math»

 Where, I = Moment of inertia of the flywheel assembly

             N = Number of rotation of the flywheel before it stopped

             m = mass of the rings

             n = Number of windings of the string on the axle

             g = Acceleration due to gravity of the environment.

             h = Height of the weight assembly from the ground.

             r = Radius of the axle.

 

Now we begin to count the number of rotations, N until the flywheel stops and also note the duration of time t for N rotation. Therefore we can calculate the average angular velocity  Ãƒâ€šÃ‚«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«miȤ#969;«/mi»«mrow»«mi»a«/mi»«mi»v«/mi»«mi»e«/mi»«mi»r«/mi»«mi»a«/mi»«mi»g«/mi»«mi»e«/mi»«moȤnbsp;«/mo»«/mrow»«/msub»«/math»in radians per second.

 Ãƒâ€šÃ‚«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«miȤ#969;«/mi»«mrow»«mi»a«/mi»«mi»v«/mi»«mi»e«/mi»«mi»r«/mi»«mi»a«/mi»«mi»g«/mi»«mi»e«/mi»«/mrow»«/msub»«moȤnbsp;«/mo»«mo»=«/mo»«mfrac»«mrow»«mn»2«/mn»«miȤ#960;«/mi»«mi»N«/mi»«/mrow»«mi»t«/mi»«/mfrac»«/math»

Since we are assuming  that the torsional friction Wf is constant over time and angular velocity is simply twice the average angular velocity

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mi»§#969;«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«mn»4«/mn»«mi»§#960;«/mi»«mi»N«/mi»«/mrow»«mi»t«/mi»«/mfrac»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»(«/mo»«mn»5«/mn»«mo»)«/mo»«/math»

 

Applications:

 

Flywheels can be used to store energy and used 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.

The phenomenon of precession has to be considered when using flywheels in moving vehicles. However in one modern application, a momentum wheel is a type of flywheel useful in satellite pointing operations, in which the flywheels are used to point the satellite's instruments in the correct directions without the use of thrusters rockets.

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

 

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