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Lab No 9

This experiment aims to determine the moment of inertia of a flywheel using the falling weight method. A flywheel stores rotational kinetic energy based on its moment of inertia and rotational speed. The experiment involves attaching weights to a cord wrapped around a flywheel axle and measuring the flywheel's slowing rotational speed. Using equations that relate the potential and kinetic energy of the falling weights to the work done in slowing the flywheel's rotation, the moment of inertia can be calculated for different experimental trials using varying masses, initial rotational speeds, and final rotational speeds. The results show some deviation from theoretical values likely due to energy losses and human error in measurement.

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Saad AliKhan
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185 views3 pages

Lab No 9

This experiment aims to determine the moment of inertia of a flywheel using the falling weight method. A flywheel stores rotational kinetic energy based on its moment of inertia and rotational speed. The experiment involves attaching weights to a cord wrapped around a flywheel axle and measuring the flywheel's slowing rotational speed. Using equations that relate the potential and kinetic energy of the falling weights to the work done in slowing the flywheel's rotation, the moment of inertia can be calculated for different experimental trials using varying masses, initial rotational speeds, and final rotational speeds. The results show some deviation from theoretical values likely due to energy losses and human error in measurement.

Uploaded by

Saad AliKhan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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LAB NO 9

Objective

To determine moment of inertia of a Fly wheel by falling weight method.

Introduction
Fly wheel
It is a mechanical device which specifically designed to efficiently store
rotational energy which is also known as kinetic energy. To resist the motion of flywheel by their
moment of inertia. The amount of the energy stored in a fly wheel is proportional to the square of
its rotational speed and its mass.
E = IꞶ2 / 2.
In machine it is used as a reservoir which stores energy during the period when the supply of
energy is more than its requirement, and releases it during the period when the requirement of
energy is more than the supply. While in the case of steam engines, internal engines,
reciprocating compressors and pumps the energy is developed during one stroke and the engine
is to run for the whole cycle on the energy produced during this one stroke. 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.

The way to change a flywheel stored energy without changing its mass by increasing or
decreasing its rotational speed. A little consideration will show that when the flywheel absorbs
energy, its speed increases and when it releases energy, the speed decreases. Hence a fly wheel
does not maintain a constant speed, it simply reduces the fluctuation of speed in other words a
flywheel controls the speed variations cause by the fluctuation of the engine turning moment
during each cycle of operation.

Inertia
             
Moment of inertia is the property of the body. In rotational motion when a body rotates
about its axis of rotation. Each particle in the body moves in a circle with a linear velocity, each
particle moves with an angular acceleration, it’s a property of body due to which it resists
angular acceleration which is the sum of the products of the mass of each particles in a body with
a square of its distance from the axis of the rotation.
Derivation
mgh = ½ mv2 + ½ IꞶ2 +n1E……………… eq (1)

When flywheel stops

½ IꞶ2 = n2 E

E = ½ IꞶ2 / n2

Put the value of E in equation (1)

mgh = ½ mv2 + ½ IꞶ2 n1/n2 [ ½ IꞶ2]

mgh = ½ mv2 + ½ IꞶ2 + [ 1+ n1/n2]

mgh = ½ mv2 + ½ IꞶ2 +[ (n1+n2)/n2]

I = 2[mgh – ½ mv2]/ Ꞷ2(n1/n1+n2)


It’s a formula to calculate moment of inertia

Observations and calculations

Sr. No m n1 n2 h t1 v ω I
(Kg) (m) (sec) (m/s) (rad/sec) (kgm2 ¿

1 0.4 1.5 70 1.14 1.7 1.3411 9.579 1.88x10-3

2 0.8 1.5 144 1.14 1.18 1.932 13.8 9.58x10-4

3 1.2 1.5 225 1.14 0.98 2.326 16.6 6.42x10-4

Conclusion
 From the analysis the snapshot of inactivity of fly wheel had been concentrated in which
the outcomes are the reliance of mass and span of wheel.

 The test the snapshot of inactivity of flywheel is found to have immense deviation from
the hypothetical one.

 The gigantic deviation is because of wellsprings of blunder and abatement in the


effectiveness proportion of the machine in reasonable procedure.
 We reason that the blunder was finished by human errors and furthermore may be a direct
result of vitality misfortune because of grinding along these lines it is exceptional with
the hypothetical.

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