$A$ uniform rod of length $2L$ has one end on a horizontal floor. It is inclined at an angle $\alpha$ to the horizontal floor. It falls without slipping,rotating about the point of contact. Its angular velocity when it hits the horizontal floor will be:

  • A
    $\omega = \sqrt{\frac{3g\sin\alpha}{2L}}$
  • B
    $\omega = \sqrt{\frac{2L}{3g\sin\alpha}}$
  • C
    $\omega = \sqrt{\frac{6g\sin\alpha}{L}}$
  • D
    $\omega = \sqrt{\frac{L}{g\sin\alpha}}$

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