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27 August, 12:21

16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?

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  1. 27 August, 15:36
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    Given Information:

    Torque = τ = 3.0 N. m

    Moment of inertia = I = 5.0 kg. m²

    Time = t = 8 seconds

    Required Information:

    Rotational kinetic energy = Erot = ?

    Answer:

    Rotational kinetic energy = 57.6 Joules

    Explanation:

    We know that the rotational kinetic energy of the wheel is the energy due to its rotation and is part of its total kinetic energy and is given by

    Erot = ½Iω²

    Where ω is the angular velocity and I is the moment of inertia of the wheel.

    We also know the relation between torque and moment of inertia is

    τ = Iα

    Where α is the angular acceleration of the wheel.

    α = τ/I

    From the equations of kinematics, we know that final angular velocity is given by

    ω = ω₀ + αt

    Where ω₀ is the initial angular velocity of the wheel and since wheel starts from rest, ω₀ is zero.

    ω = 0 + αt

    ω = αt

    ω = (τ/I) t

    ω = τ*t/I

    Finally the equation of rotational kinetic energy becomes

    Erot = ½Iω²

    Erot = ½I (τ*t/I) ²

    Erot = ½*5 * ((3*8) / 5) ²

    Erot = ½*5 * (23.04)

    Erot = ½ * (115.2)

    Erot = 57.6 J

    Therefore, the wheel's rotational kinetic energy at the end of 8 s is 57.6 Joules.
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