One end of a straight uniform 1m long bar is pivoted on horizontal table. It is released from the rest when it makes an angle `30^(@)` from the horizontal (see figure). Its angular speed when it hits the table is given as `sqrt(n)s^(-1)`, where n is an integer. The value of n is ..............

Explanation

Loss of gravitation potential energy = Gain in kinetic energy
`:. Mg""(l)/(2) sin 30^(@) = ( I)/(2) Iomega^(2) ` ` :' mg ""(l)/( 2) ((1)/(2))= ( 1)/( 2)((ml^(2))/( 3)) omega^(2)` `rArr ( mgl)/( 4) =( ml^(2))/( 6) omega^(2) rarr = omega = sqrt(( 6g)/( 4l))`
Velocity of end point `=omega l = sqrt(( 6gl )/(4)) = sqrt((60)/( 4)) = sqrt(15 ) m//s ` `:. n = 15 `