A uniform rod of length l is free to rotate in a vertical plane about a fixed horizontal axis through B . The rod begins rotating from rest from its unstable equilibrium position. When, it has turned through an angle , its angular velocity is given by

Step-by-step solution

When the rod rotates through an angle , the centre of gravity falls through a distance h. From B C G^ aligned & = (l / 2)-h ₂ & or, h= 1 2 (1- ) aligned Decrease in PE =m g 1 2 (1- )...(i) The decrease in P.E. is equal to the kinetic energy of rotation ( 1 2 I ^2 )( KE )_ rotational = 1 2 ( m l^2 3 ) ^2...(ii) From Eqs. (i) and (ii), we get aligned 1 2 ( m l^2 3 ) ^2 & =m g 1 2 (1- ) & = 6 g l 2 aligned