The moment of inertia about the axis will be given by the formula;
I0=(1/3)mh², where m is the mass of a section and h is the distance to outer edge.
Therefore, with four sections combined the inertia will be;
4×I0 = (4/3)mh²
But Tor(M)que applied to the door is
Torque = F × h
Therefore,
M = I0 × x
x = M/I0 (But M =Fh
= (Fh)/(4/3 × mh² ( simplifying the equation)
= 3F/4mh
but F = 80 N, m= 86 kg and i assume the height is 1.2 m
Therefore;
= (3×80)/(4×86×1.2)
= 0.581 rad/s²
