Finding the Acceleration of a Pivoted Rod Released from a Horizontal Position

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The discussion focuses on calculating the acceleration of the center of mass of a uniform rod of length 3L, pivoted at one end and released from a horizontal position. The moment of inertia is initially calculated using the parallel axis theorem, leading to an incorrect expression due to a miscalculation of the length. After identifying the error, the correct moment of inertia is established as I=(1/12)(3L)² + m(3L/2)², which is crucial for determining the net torque and subsequently the linear acceleration. The final correct expression for linear acceleration is derived as (3/4)g after correcting the length in the calculations. The discussion emphasizes the importance of accurate parameters in physics problems.
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Homework Statement



A uniform rod of length 3L is pivoted as shown and is resting at a horizontal position.
What is the acceleration of the center of mass of the rod the instant it is released?
The diagram is added to the attachment.

Homework Equations



tnet=Iaangular

The Attempt at a Solution


From parallel axis theorem the moment of inertia I, around the rotation axis is I=(1/12)mL2 + m(L/2)2=(1/3)mL2

tnet= mg(L/2) = Iaangular
mg(L/2)=(1/3)mL2(alinear/0.5L) as the centre of mass is 0.5m away from rotation axis.
mg/2=(2/3)malinear
g/2=(2/3)alinear
(3/4)g =alinear

Thats not the answer, where did i go wrong?
 

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Hi Macroer! :smile:
Macroer said:
From parallel axis theorem the moment of inertia I, around the rotation axis is I=(1/12)mL2 + m(L/2)2=(1/3)mL2

erm :redface: … wrong L !

(1/12 of 3L squared)
 
Thanks! I would have never found that. :smile:
 
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