Arotating door is made from four rectangular sections, as indicated in the drawing. the mass of each section is 86 kg. a person pushes on the outer edge of one section with a force of f = 80 n that is directed perpendicular to the section. determine the magnitude of the door's angular acceleration.

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

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