Question

a rod of length l is hinged from one end . it is brought to a horizontal position and released. the angular velocity of the rod when it is vertical position is

Answer 1

Answer:

$\sqrt{\dfrac{3g}{L}}$

Explanation:

According to the question initially the rod is in horizontal position and then released from that position . We have to take out angular velocity (ω) of rod when it becomes vertical .

Mass of rod = m

Length of rod = L

Moment of inertia = I

Angular velocity = ω

Applying Energy Conservation -

→ 1/2 I ( 0)^2 + 0 = 1/2(mL^2/3)ω^2 - mg L/2

→ mg * L/2 = 1/2 * ml^2/3 * ω^2

→ Angular Velocity (ω) = $\sqrt{\dfrac{3g}{L}}$

Further Information : -» The angular velocity $\sqrt{\dfrac{3g}{L}}$ is maximum angular velocity of rod .