Torque and paralle-axis theorem - question

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The discussion revolves around calculating the angular acceleration of a seesaw-like rod with point masses attached at either end. Participants confirm that torque can be calculated around the pivot or one of the ends, but emphasize that using the pivot point is generally simpler. One user expresses a desire to validate their understanding by calculating torque around the left end and applying the parallel-axis theorem. The consensus is that while both methods are theoretically valid, careful consideration of the pivot's torque is necessary. The conversation ultimately reinforces the importance of clarity in applying these physics concepts.
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Homework Statement


The figure shows a simple model of a seesaw and/or valves used in car engines. These consist of a plank/rod of mass m_r and length 2L allowed to pivot freely about its center (or central axis), as shown in the diagram. A small sphere of mass m_2 is attached to the left end of the rod, and a small sphere of mass is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward. The magnitude of the acceleration due to gravity is equal to g.

Suppose that the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, your answer may include the symbol , the moment of inertia of the assembly, whether or not you have answered the first part correctly.)

What is the angular acceleration of the rod immediately after it is released?
________

Ok, I can find the torque around the pivot and then the only forces relevant here will be the weights of the two masses in the ends. From this I can find alpha.

But on the way home in the bus I was wondering if I can find alpha when calculating the torque around on of the ends and then use the parallel-axis theorem? Is this operation allowed and will it give the same result?

Thank you in advance.
 

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Yes, in theory this could work. Just be careful handling the torque created by the pivot now. In my opinion, it's easier the way you did it.
 
dwintz02 said:
Yes, in theory this could work. Just be careful handling the torque created by the pivot now. In my opinion, it's easier the way you did it.

Thanks. I think you are right. If we calculate the torque around the left end, is this correct (I am only doing this to prove to myself that I have understood the material):

(Positive counterclockwise):

T_z = -m_l*g*L-m_2*g*2L = I_{total}*alpha. From this I can find alpha?
 
Can you guys confirm what I wrote in #2, please?
 
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