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## Homework Statement

A uniform metre-rule is pivoted to rotate about a horizontal axisthrough the 40-cm mark. The stick has a mass/unit length of [tex]\mu[/tex] kg/m and its rotational inertia about this pivot is [tex]0.093 \mu[/tex] kg/m

^{2}. It is released from rest in a horizontal position. What is the magnitude of the angular acceleration of the rod?

## Homework Equations

An expression for the magnitude of the torque due to gravitational force about an axis through the pivot: [tex] \tau =Mg \left( \frac{L}{2} \right)[/tex]

Angular accleration and torque: [tex]\sum \tau = I \alpha[/tex]

## The Attempt at a Solution

Since [tex]\mu = \frac{M}{L}[/tex], and L=0.4m I get

[tex]\tau = 0.4 \mu g \frac{0.4}{2} = 0.78 \mu[/tex]

[tex]\alpha = \frac{\tau}{I} = \frac{0.78 \mu}{0.093 \mu} = 8.43[/tex]

But my answer is wrong. The correct answer must be 10.5 rad/s

^{2}. Could anyone please help me? :(