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

[Macroer]

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

t_net=Ia_angular

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

t_net= mg(L/2) = Ia_angular
mg(L/2)=(1/3)mL²(a_linear/0.5L) as the centre of mass is 0.5m away from rotation axis.
mg/2=(2/3)ma_linear
g/2=(2/3)a_linear
(3/4)g =a_linear

Thats not the answer, where did i go wrong?

 

[tiny-tim]

Hi Macroer!

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 … wrong L !

(1/12 of 3L squared)

 

[Macroer]

Thanks! I would have never found that.