A rotating 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