(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A uniform rod of mass 1.6 kg is 6 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance of 6 m from the center of mass of the rod. Initially the rod makes an angle of 58 degrees with the horizontal. The rod is released from rest at an angle of 58 degrees with the horizontal.

What is the magnitude of the vertical acceleration of the center of mass of the rod at the instant the rod is in a horizontal position? The acceleration of gravity is 9.8 m/s^2 and the moment of inertia of the rod about its center of mass is 1/12 mL^2. Answer in units of m/s^2.

2. Relevant equations

torque = I (moment of inertia) * angular acceleration

tang. acceleration = R*angular acceleration

3. The attempt at a solution

torque = I * ang. acceleration

r mg = 1/12 m l^2 * ang. acc.

6mg = 1/12 m (6^2) * ang. acc.

6mg = 3m* ang. acc.

ang. acc = 2g

tang. acc. = r(ang. acc) = 2(6g)

tang. acc(vertical) = 12g

And it is wrong.

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# Homework Help: Vertical acceleration of a rod rotating around a pivot

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