Consider a uniform rod of mass `M` and length `L` is hinged at one end as shown. Initially the rod is vertical, it is slightly pushed and released. Find the angular velocity of rod and liner velocity of its one end when the rod becomes horizontal. Also calculate angular velocity of rod when it has rotated by an angle `theta`(`thetalt90^(@)`).

Explanation

This is the case of pure rotation

Applying energy consevation between (`1`) and (`2`), taking `O` as reference level
`Mg(L/2)=(1)/(2)I_(0)omega^(2)=(1)/(2)((ML^(2))/(3))omega^(2)`
`omega=sqrt((3g)/(L))`
Linear velocity of end `A`
`v_(A)=omega_(L)=sqrt(3gL)`

`Mg(L/2)=Mg(L/2)costheta+(1)/(2)I_(0)omega^(2)`
`Mg(L/2)(1-costheta)=(1)/(2)xx(ML^(2))/(3)omega^(2)`
`omega=sqrt((3g(1-costheta))/(L))`
Another method : When a rod makes an angle `theta` with vertical, torque acting on the rod about `O`
`tau_(0)=Mg(L/2)sintheta=I_(0)alpha=(ML^(2))/(3)alpha`
`alpha=(3g)/(2L)sintheta`
`omega(d omega)/(d theta)=(3g)/(2L)sintheta`
`int_(0)^(omega)omegadomega=(3g)/(2L)int_(0)^(omega)sintheta d theta`
`(omega^(2))/(2)=(3g)/(2L)|-costheta|_(0)^(theta)`
`(3g)/(2L)(1-costheta)`
`omega=sqrt((3g(1-costheta))/(L))`