A thin uniform rod (X) of mass M and length L is pivoted at a height ((L)/(3)) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this

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Correct answer: D. (3 g)/(L)

The rod is pivoted at a distance (L)/(3) from one end. The centre of mass is at a distance (L)/(6) from the pivot. Using energy conservation: mg( )/(6) = (1)/(2)Iω^2 Moment of inertia about the pivot (using parallel axis theorem): I = (m ^2)/(12) + m(( )/(6))^2 = (m ^2)/(12) + (m ^2)/(36) = (m ^2)/(9) mg( )/(6) = (1)/(2) · (m ^2)/(9) · ω^2 ω^2 = (9g)/(6 · ( )/(2)) = (3g)/( ) ω = (3g)/( )

A thin uniform rod (X) of mass M and length L is pivoted at a...

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