## Angular Acceleration (I don't see HOW I can be wrong?)

My friend in introductory physics came to me for help today. He has a test on angular momentum, acceleration etc. I figured, with me in upper-division physics courses, I'd be able to help him out. Well, I guess I was dead wrong. In half an hour I couldn't figure out a SINGLE THING. I don't see how my answers could be wrong. Everything I tried did not line up with that his professors answers were.

1. The problem statement, all variables and given/known data

We have a horizontal board of length 2.4 meters and mass 1.8 kg connected on the left side to a pivot point, and was suspended by some string at the other side. The string is cut, and we are to find the initial angular acceleration of the board. Sounds easy enough... as the initial acceleration is just due to gravity, the only force now acting on the board.

2. Relevant equations

A lot of formulas were provided, a tangential = r*a angular, t=I(angular a) t=rF. Standard equations for angular rotation.

3. The attempt at a solution

At first I tried simple a tangential = r a angular. I used g for the tangential acceleration and the r I used was both the full length of the board, and the cm length (l/2). Either way, I didn't get the 6.13 rad/s that the answer supposedly was.

Next I tried using I(a)=rF, where F is due to gravity, the r was at the cm length. The I was 1/3ML^2, where L is the length of the board, M is the mass of the board (1.8 kg). I STILL did not get the correct answer.

I'm out of ideas now, and even though this isn't my class, I still want to know why can't I get basic physics right? It's times like these that I feel like I've learned absolutely nothing as a physics major over the two-three years I've been in the department.
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 Isn't it just the torque divided by the moment of inertia? I think the torque is (gML)/2 and the moment is (ML^2)/3 which would give 3g/2L as the answer.
 Yes, 3g/2L it is.

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## Angular Acceleration (I don't see HOW I can be wrong?)

 Quote by Lucretius Sounds easy enough... as the initial acceleration is just due to gravity, the only force now acting on the board.
Gravity is not the only force on the board--you still have a force from the pivot. Accordingly, the acceleration of the center of mass does not equal g.
 At first I tried simple a tangential = r a angular. I used g for the tangential acceleration and the r I used was both the full length of the board, and the cm length (l/2). Either way, I didn't get the 6.13 rad/s that the answer supposedly was.
This won't work, for reasons stated above.
 Next I tried using I(a)=rF, where F is due to gravity, the r was at the cm length. The I was 1/3ML^2, where L is the length of the board, M is the mass of the board (1.8 kg). I STILL did not get the correct answer.
This is perfectly correct, so you must have made an error somewhere.

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