BlogIIT-JEEMoment of Inertia of Flywheel – JEE

Moment of Inertia of Flywheel – JEE

Moment of Inertia of Flywheel: Flywheels seem to be circular disc-shaped objects that are used to store energy in machines. The flywheel has been made up of a large circular disc/massive wheel with a strong axle protruding on either side. The axle would be supported by two fixed supports and is mounted on ball bearings. Just on the axle, there is a small peg. The weight-hanger is carried by one end of a cord that is loosely looped around the peg.

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    The formula that is used to calculate the moment of inertia of a flywheel is:

    Formula for moment of inertia of flywheel

    In which, I = moment of inertia of the flywheel. Here, the symbols denote;

    m = rings’ mass.

    N = flywheel rotation.

    n = number of windings of the string.

    h = height of the weight assembly.

    g = acceleration due to gravity.

    r = radius of the axle.

    Similarly, we can also use the following expression;

    mgh = ½ mv2 + ½ Iω2 + n1E

    Determining the Moment of Inertia of Flywheel

    A few important factors must be considered when determining the moment of inertia of a flywheel. Initially, we must assemble a flywheel, as well as equipment such as a weight hanger, slotted weights, a metre scale, and a stopwatch.

    moment of inertia of flywheel

    Afterwards, we make some assumptions. We’ll use mass (m) for both the weight hanger and the hanging ring. The maximum height will be (h). Just consider a situation in which the mass will descend to a new height. There’ll be some potential energy loss, which we can express as follows in the equation:

    Ploss = mgh

    In the meantime, when the flywheel and axle rotate, there is a gain in kinetic energy. We say it this way:

    Kflywheel = (½) Iω2

    I = moment of inertia

    ω = angular velocity

    Correspondingly, the kinetic energy of descending weight assembly is written as;

    Kweight = (½) Iv2

    Here, v = veocity

    We must also consider the effort expended in overcoming the friction. It can be determined by;

    Wfriction = nWf

    In this particular instance,

    n = number of windings of the string

    Wf = work done in overcoming frictional torque

    When we state the law of conservation of energy then we obtain;

    Ploss = Kflywheel + Kweight + Wfriction

    Now, We’ll substitute the values and the equation will now become;

    mgh = (½)Iω2 + (½) mv2 + nWf

    Shifting on to the next phase, we examine the kinetic energy of the flywheel assembly as it rotates (N) times against the frictional torque. We obtain;

    NWf = (½ ) Iω2 and Wf = (1 / 2N) Iω2

    Moreover, we establish a relationship between the weight assembly’s velocity (v) and the axle’s radius (r). The equation is as follows:

    v = ωr

    Now, we need to substitute the values for Wf and v.

    mgh = (½) Iω2 + (½ )mr2ω2 + (n / N) x ½ Iω2

    When we solve the equation for finding the moment of inertia of flywheel, we obtain;

    FAQs

    What is meant by MOI of the flywheel?

    The moment of inertia of a body has been defined as the object's resistance to rotational changes. The flywheel flexibly rotates about a horizontal axis. A calliper could be used to measure the radius of the flywheel's axile.

    What is the main focus of the flywheel experiment?

    The moment of inertia of a flywheel has been investigated by varying the flywheel's point of mass. The experiment has been carried out by recording the time taken by a fixed load until the point at which the load escapes from the flywheel, as well as the number of rotations performed afterwards, which is independent of the load.

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