Angular Acceleration- Moment of Inertia Homework help?

Question

A rotating door is made from four rectangular glass panes, as shown in the drawing. The mass of each pane is 74 kg. A person pushes on the outer edge of one pane with a force of F = 57 N that is directed perpendicular to the pane. Determine the magnitude of the door's angular acceleration

I am not sure what I am doing wrong because I following the notes given in class. I came up with .32 as my answer and that is apparently incorrect. Could someone help point me in the right direction?

Answer 1

Do you have the dimensions of the rectangular glass pane? To the best of my knowledge you have to know where the force was applied with respect to the center of rotation, which means that you have to, at least, know the width of the glass pane. Recall: The sum of all moments/torque about a fixed point equals to the moment of inertia of the mass * the angular acceleration of the rigid body, in other words,

ΣM = I * α

=> F * r = I * α

where, F is the applied force (57 N), and r is the distance from the center of rotation at which the force is applied.

=> α = F*r / I

Knowing the moment of inertia of the glass pane, (which can be determined with radius of gyration, I = mk², where k is the width of the glass), and the distance at which the force was applied, you can easily determine the angular acceleration of the glass panes about the center of rotation.