Question

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.

Answer 1

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