What Is the Angular Acceleration of a Released Uniform Plate?

AI Thread Summary
The discussion revolves around calculating the angular acceleration of a uniform plate after a connection is released. The initial attempt used torque and moment of inertia but yielded an incorrect angular acceleration of 1.2g/c instead of the expected 2.4g/c. A revised approach using the center of mass provided the correct answer, indicating a mistake in the first method's lever arm calculation. The correct lever arm was identified as c/4 when choosing point A. The importance of accurately determining the lever arm in torque calculations is emphasized.
Himanshu_123
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Homework Statement



A uniform plate of mass m is suspended as shown.What will be the angular acceleration immediately after connection at B has been released?

http://i.stack.imgur.com/1hMuk.png

2. My attempt-

I used torque=M.I*angular acceleration

I chose the axis to be perpendicular to the plane at A where M.I=5mc^2/12 and at a distance of c/2 from line of action of mg by putting it in equation we get angular acc.=1.2g/c but the given ans. is 2.4g/c

But when I solve choosing the axis as centre of mass -
The tension/force by which string will pull= mg/2 and distance of line of action from axis =c/2 and M.I about c.m=5mc^2/48 and using torque =M.I*angular acceleration i get the right ans. I don't know what's the mistake in my 1st approach.
 
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Where is the diagram?
 
Jonathan Yang said:
Where is the diagram?
Thanks for pointing. I corrected it.
 
I think you calculated the lever arm incorrectly. If you choose point A, the lever arm that is perpendicular to the line of action is (c/2)/2 or c/4
 
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