Finding angular acceleration of 2 masses on a rod

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SUMMARY

The discussion focuses on calculating the angular acceleration of a uniform rigid rod with a mass of 5.9 kg and a length of 2.9 m, which rotates about a frictionless pivot. Two point masses, 5.7 kg and 1.8 kg, are attached to the ends of the rod. The calculation requires the use of the moment of inertia formula, I = 1/12 ML², and the application of torque principles at an angle of 42.7° with the horizontal. The participants emphasize the importance of drawing a free-body diagram to identify forces acting on the rod.

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Homework Statement



A uniform rigid rod with mass Mr = 5.9 kg, length L = 2.9 m rotates in the vertical xy plane about a frictionless pivot through its center. Two point-like particles m1 and m2, with masses m1 = 5.7 kg and m2 = 1.8 kg, are attached at the ends of the rod. What is the magnitude of the angular acceleration of the system when the rod makes an angle of 42.7° with the horizontal?

Homework Equations



all i know is that I= 1/12ML^2

The Attempt at a Solution



I honestly don't know where to star, could anyone guide me towards the right way
 
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Start by drawing a free-body diagram and identifying the forces on the rod.
 

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