Question

A rod of length `l` whose lower end is fixed along the horizontal plane starts from the vertical position. The velocity of the upper end of the rod when it hits the ground is

Answer 1

When upper end of the rod hits the gound, the distance through which centre of gravity of the rod falls `= L//2`.
`:.` Loss of potantial energy `= Mg (L//2)`, where `M` is mass of the rod.
Gain in rotational `K.E. = (1)/(2)I omega^(2)`
Where `I = ML^(2)//3 and omega = v//L`
`:.` Gain in rotational `K.E.`
`= (1)/(2) ((ML^(2))/(3)) ((v)/(L))^(2) = (M v^(2))/(6)`
According to the law of conservation of energy
`(M v^(2))/(6) = Mg (L)/(2)`
`v = sqrt(3gL)`