$A$ uniform rod of length $2L$ is placed with one end in contact with a horizontal surface. The other end is released from an angle $\alpha$ with the horizontal,such that the end in contact does not slip. What will be its angular velocity when it becomes horizontal?

  • 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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