Posted by Jim on .
A rotating door is made from four rectangular glass panes, as shown in the drawing. The mass of each pane is 88 kg. A person pushes on the outer edge of one pane with a force of F = 53 N that is directed perpendicular to the pane. Determine the magnitude of the door's angular acceleration.
In the drawing the radius is 1.2m and F is counterclockwise.
Net external torque=moment of Inertia X Angular acceleration
Moment of Inertia = sum of mass X r^2
= (88kg X 4)(1.2m^2)
I transpose the above formula making angular acceleration the subject of the equation.
Angular acceleration= sum of external torque / Moment of inertia
angular acceleration= 63.6 / 506.88
This answer is incorrect please tell me where I went wrong.
The moment of inertia depends upon where the glass panes are placed, which your question does not explain. it is unfortunate that this web site cannot handle figures. In any case, the moment of inertia is not just the sum of masses of the panes multipied by some R^2. You have to use the parallel axis theorem. See