- #1
roam
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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/m2. 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/s2. Could anyone please help me? :(