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29 November, 23:04

8. A baseball batter angularly accelerates a bat from rest to 20 rad/s in 40 ms (milliseconds). If the bat's moment of inertia is 0.6 kg m2, then findA) the torque applied to the bat andB) the angle through which the bat moved.

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  1. 29 November, 23:21
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    A) τ = 300 N*m

    B) θ = 0.4 rad = 22.9°

    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

    I = 0.6 kg*m² : moment of inertia of the bat

    Angular acceleration of the bat

    We apply the equations 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 (s)

    Data

    ω₀ = 0

    ωf = 20 rad/s

    t = 40 ms = 0.04 s

    We replace data in the formula (2):

    ωf = ω₀ + α*t

    20 = 0 + α * (0.04)

    α = 20 / (0.04)

    α = 500 rad/s²

    Newton's second law to the bat

    τ = (0.6 kg*m²) * (500 rad/s²) = 300 (kg*m/s²) * m

    τ = 300 N*m

    B) Angle through which the bat moved.

    We apply the equations of circular motion uniformly accelerated:

    ωf² = ω₀ ² + 2α*θ Formula (3)

    Where:

    θ : Angle that the body has rotated in a given time interval (rad)

    We replace data in the formula (3):

    ωf² = ω₀² + 2α*θ

    (20) ² = (0) ² + 2 (500) * θ

    400 = 1000*θ

    θ = 400/1000

    θ = 0.4 rad

    π rad = 180°

    θ = 0.4 rad * (180°/π rad)

    θ = 22.9°
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