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If the ratio of maximum and minimum speeds of a flywheel is 1.11 and the kinetic energy at mean speed is 250 kJ. Maximum fluctuation of energy is ___________ kJ
(Important - Enter only the numerical value in the answer)
    Correct answer is '50,53'. Can you explain this answer?
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    If the ratio of maximum and minimum speeds of a flywheel is 1.11 and t...
    e = 52.13kJ
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    If the ratio of maximum and minimum speeds of a flywheel is 1.11 and t...
    Given information:
    - Ratio of maximum and minimum speeds of a flywheel = 1.11
    - Kinetic energy at mean speed = 250 kJ

    Calculating maximum and minimum speeds:
    Let the maximum speed of the flywheel be Vmax and the minimum speed be Vmin.
    We know that the ratio of maximum and minimum speeds is 1.11.

    Therefore, Vmax/Vmin = 1.11

    Calculating kinetic energy at maximum and minimum speeds:
    The kinetic energy of a rotating object can be calculated using the formula:

    K.E. = (1/2) * I * ω^2

    Where K.E. is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.

    The moment of inertia of a flywheel is given by:

    I = (1/2) * m * r^2

    Where m is the mass of the flywheel and r is the radius.

    Since the mass and radius of the flywheel are constant, the moment of inertia remains the same at all speeds.

    The angular velocity can be calculated using the formula:

    ω = V / r

    Where V is the linear velocity and r is the radius.

    Substituting the values of moment of inertia and angular velocity into the kinetic energy formula, we get:

    K.E. = (1/2) * (1/2) * m * r^2 * (V / r)^2
    = (1/8) * m * V^2

    Calculating maximum and minimum kinetic energies:
    Let Kmax be the kinetic energy at maximum speed and Kmin be the kinetic energy at minimum speed.

    We know that the kinetic energy at mean speed is 250 kJ.

    Therefore, Kmean = (1/8) * m * Vmean^2 = 250 kJ

    Since the mass and radius of the flywheel are constant, the mean kinetic energy is directly proportional to the square of the mean speed.

    Similarly, Kmax = (1/8) * m * Vmax^2
    And Kmin = (1/8) * m * Vmin^2

    Calculating the maximum fluctuation of energy:
    The maximum fluctuation of energy can be calculated using the formula:

    Max fluctuation = (Kmax - Kmin) / 2

    Substituting the values of Kmax and Kmin into the formula, we get:

    Max fluctuation = [(1/8) * m * Vmax^2 - (1/8) * m * Vmin^2] / 2
    = (1/16) * m * (Vmax^2 - Vmin^2)

    Since Vmax/Vmin = 1.11, we can substitute Vmax = 1.11 * Vmin into the equation:

    Max fluctuation = (1/16) * m * [(1.11 * Vmin)^2 - Vmin^2]
    = (1/16) * m * (1.11^2 * Vmin^2 - Vmin^2)
    = (1/16) * m * (1.11^2 - 1) * Vmin^2

    Simplifying further, we get:

    Max fluctuation = (1/16) * m * (0.2321
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    If the ratio of maximum and minimum speeds of a flywheel is 1.11 and t...
    52.13
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    If the ratio of maximum and minimum speeds of a flywheel is 1.11 and the kinetic energy at mean speed is 250 kJ. Maximum fluctuation of energy is ___________ kJ(Important - Enter only the numerical value in the answer)Correct answer is '50,53'. Can you explain this answer?
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