Determine the linear acceleration of the tip of the rod

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SUMMARY

The discussion focuses on calculating the linear acceleration of the tip of a uniform rod with a mass of 5.02 kg and a length of 1.08 m, which pivots freely about a hinge. The angular acceleration of the rod at the moment of release is determined to be 13.6 rad/s². The relationship between linear acceleration and angular acceleration is established using the equation a = r(α), where 'r' is the distance from the pivot to the tip of the rod and 'α' is the angular acceleration.

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Homework Statement


Determine the linear acceleration of the tip of the rod. Assume that the force of gravity acts at the center of mass of the rod, as shown


Homework Equations


So there is a first part to this problem that i already got
A uniform rod of mass M = 5.02kg and length L = 1.08m can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as seen in the figure below.The rod is held horizontally and then released. At the moment of release, determine the angular acceleration of the rod. Use units of rad/s^2. So i got the right answer for this and it was 13.6 rad/s^2. i just don't know what to do for the second part

The Attempt at a Solution

 
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Linear acceleration is related to angular acceleration by the eqn. r(\alpha) = a
 

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