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7 December, 20:53

Part c if exena applies a force of 220 n at the edge of the door and perpendicular to it, how much time t does it take her to close the door?

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  1. 8 December, 00:15
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    The initial formula is J·α = F·w but we need to find the angular acceleration, so the other formula is α = F·w/J

    where

    J moment of inertia, α angular acceleration, w width of the door ( = length of lever arm of the force F)

    Moment of inertia of plane rectangular object, rotating around axis through on edge (see link)

    J = (1/3) ·m·w

    since mass equals weight W divided by acceleration due to earth gravity g

    J = (1/3) · (W/g) ·w²

    So the angular acceleration of the door is:

    α = F·3·g / (W·w)

    = 220N · 3 · 9.81m/s² / (750N · 1.25m)

    = 6.90624 s^-2

    Integrating twice with initial conditions

    ω (t=0) = 0; and

    α (t=0) = 0

    look for the angular displacement of the door:

    α = dω / dt will be ω = ∫ α dt = α·t

    ω = dφ / dt will be φ = ∫ ω dt = ∫ α·t dt = (1/2) ·α·t²

    So time elapsed until door closed at an angle of

    φ' = 90° = (1/2) ·π

    is

    t' = √ (2·φ'/α)

    = √ (π / 6.90624 s^-2)

    = 0.67s
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