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27 March, 08:26

The drive chain in a bicycle is applying a torque of 0.850 N ∙ m to the wheel of the bicycle. You can treat the wheel as a thin uniform hoop (or ring) with a mass of 0.750 kg and a radius of 33.0 cm. What is the angular acceleration of the wheel? A. 10.4 rad/s2B. 3.43 rad/s2C. 5.20 rad/s2D. 1.06 rad/s2E. 20.8 rad/s2

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  1. 27 March, 12:15
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    A) α = 10.4 rad/s²

    Explanation:

    Newton's second law:

    F = ma has the equivalent 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)

    Data

    τ = 0.850 N xm : moment or torque applied to the bicycle wheel

    m : 0.750 kg : mass of the wheel

    R = 33.0 cm = 0.33 m : radius of the wheel

    Moment applied to the wheel (I)

    The moment of inertia of a thin uniform hoop (or ring) is defined as follows:

    I = m * R²

    I = 0.750 kg * (0.33 m) ²

    I = 0.081675 kg*m²

    Angular acceleration of the wheel

    We replace data in the formula (1):

    τ = I * α

    0.850 = 0.081675 * α

    α = 0.850 / 0.081675

    α = 10.4 rad/s²
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