Determine the Angular Acceleration of a Rotating Door

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To determine the angular acceleration of a rotating door made of four glass panes, each weighing 92 kg, a force of 79 N is applied perpendicularly at the edge of one pane. The torque is calculated as T = 79 N * 1.2 m, resulting in 94.8 N·m. The moment of inertia is derived from I = (1/12)ML², where M is the total mass (92 kg x 4). The initial calculation of angular acceleration was incorrect due to not accounting for the total mass of the door. Correcting this leads to the proper determination of angular acceleration.
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Homework Statement



A rotating door is made from four rectangular glass panes, as shown in the drawing. The mass of each pane is 92 kg. A person pushes on the outer edge of one pane with a force of F = 79 N that is directed perpendicular to the pane. Determine the magnitude of the door's angular acceleration.

http://h1.ripway.com/wchvball13/09_34.gif

Homework Equations



\sum T = I\alpha

I = (1/12)ML^{2}

The Attempt at a Solution


T = 79(1.2) = 94.8

So...
94.8 = (1/12)(92)(2.4)^{2} \alpha
94.8 = 44.16\alpha
2.15 = \alpha

the alphas aren't supposed to be superscript...anyway...this is my answer and it's wrong...don't know what I'm doing ...or not doing...
 
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The torque imposed by the person pushing with a force of 79N is correct.
 
Where am I going wrong?
 
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nevermind...I got it..I didn't realize I would take 92 x 4 to use as the weight
 
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