As shown in the figure, two blocks, each of mass `m`, suspended from the ends of a rigid light rod of length `L`. The rod is held horizontally on the fulcrum and then released.
(`a`) Find the initial angular acceleration of the rod.
(`b`) If the mass of rod is also `m` uniformly distributed. find the tension in the strings attached to blocks.

Explanation

(`a`)
Consider (block s `+` rod) as system.
Taking torque about `O`
`tau_(0)=mgxx(2L)/(3)-mgxx(L)/(3)=I_(0)alpha`
`=[m((2L)/(3))^(2)+m((L)/(3))^(2)]alpha`
`(mgL)/(3)=(5)/(9)mL^(2)alpha implies alpha=(3g)/(5L)`
(`b`)
Now the rod has mass `m`. Its moment of inertia about `O`
`I'=(mL^(2))/(12)+m((L)/(2)-(L)/(3))^(2)=(mL^(2))/(12)+(mL^(2))/(36)=(mL^(2))/(9)`
Considering (block `+` rod) as a system and taking torque about `O`
`tau_(0)=mgxx(2L)/(3)-mgxx(L)/(3)=I_(0)alpha`
`=[m((2L)/(3))^(2)+m((L)/(3))^(2)+(mL^(2))/(9)]alpha`
`(mgL)/(3)=(2)/(3)mL^(2)alpha implies alpha=(8)/(2L)`
`a_(1)=(2L)/(3) alpha=(2L)/(3)xx(8)/(2L)=(8)/(3)`
`mg-T_(1)=ma_(1) implies T_(1)=m(g-a_(1))=(2mg)/(3)`
`a_(2)=(L)/(3)alpha=(L)/(3)xx(g)/(L)=(8)/(6)`
`T_(2)-mg=ma_(2) implies T_(2) = m(g+a_(2))=(7mg)/(6)`