For one section the moment of inertia about the axis of rotation is
Iₒ =ma²/3 = 86•(1.2)²/3 =41.28 kg•m²
where: m = mass of section, a= distance to outer edge
There are 4 sections so the combined inertia is
I= 4•Iₒ = 4•41.28 = 165.12 kg•m²
The Torque (M) applied to the door is
M = F•a =61 •1.2 =73.2 N•m.
The Newton’s 2 Law for rotation
M=I•ε,
ε = M/I = 73.2/165.12 = 0.44 rad/s.
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 = 61 N that is directed perpendicular to the section. Determine the magnitude of the door's angular acceleration.
The picture shows the revolving door having four separate parts with a radius of 1.20
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