$A$ uniform rod of length $l$ and mass $m$ is pivoted at point $A$. The rod is released from a horizontal position. If the moment of inertia of the rod about point $A$ is $ml^2/3$,then its initial angular acceleration is:

  • A
    $\frac{mg}{2l}$
  • B
    $\frac{3}{2}gl$
  • C
    $\frac{3g}{2l}$
  • D
    $\frac{2g}{3l}$

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