Posted by Cindy on Thursday, May 31, 2012 at 12:00pm.
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.
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