Simple angular acceleration problem

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The problem involves a 4-meter rod hinged at one end, initially horizontal, and released to fall. The calculated angular acceleration using the formula provided was 2.45 rad/s², but the correct answer is 3.6 rad/s². The discrepancy arises from not considering the torque about the pivot point, which is essential to determine the angular acceleration accurately. Gravity is the only force acting on the rod, but its effect on torque must be factored in. Understanding the relationship between torque and angular acceleration is crucial for solving this problem correctly.
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Homework Statement


A 4 meter long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end is allowed to fall. What is the angular acceleration as it is released?


Homework Equations


Ang. Accel. = (Tan. Accel.)/(radius)


The Attempt at a Solution



(9.8m/s^2)/(4m)= 2.45rad/s^2

The answer is 3.6 rad/s^2.

I don't know what I'm doing wrong. Gravity is the only force acting on the rod, but there is something that I'm not taking into account. Can someone point me in the right direction please?
 
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Consider the torque about the pivot.

What is the force acting to create the angular acceleration.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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