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6 June, 06:50

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80 N is applied to the rim of the wheel. The wheel has radius 0.12 m. Starting from rest, the wheel has an angular velocity of 1.2 rad/s after 2 s. What is the rotational inertia of the wheel?

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  1. 6 June, 09:16
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    I = 16 kg*m²

    Explanation:

    Newton's second law for rotation

    τ = I * α Formula (1)

    where:

    τ : It is the moment applied to the body. (Nxm)

    I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

    α : It is angular acceleration. (rad/s²)

    Kinematics of the wheel

    Equation of circular motion uniformly accelerated:

    ωf = ω₀ + α*t Formula (2)

    Where:

    α : Angular acceleration (rad/s²)

    ω₀ : Initial angular speed (rad/s)

    ωf : Final angular speed (rad

    t : time interval (rad)

    Data

    ω₀ = 0

    ωf = 1.2 rad/s

    t = 2 s

    Angular acceleration of the wheel

    We replace data in the formula (2):

    ωf = ω₀ + α*t

    1.2 = 0 + α * (2)

    α * (2) = 1.2

    α = 1.2 / 2

    α = 0.6 rad/s²

    Magnitude of the net torque (τ)

    τ = F * R

    Where:

    F = tangential force (N)

    R = radio (m)

    τ = 80 N * 0.12 m

    τ = 9.6 N * m

    Rotational inertia of the wheel

    We replace data in the formula (1):

    τ = I * α

    9.6 = I * (0.6)

    I = 9.6 / (0.6)

    I = 16 kg*m²
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