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
