A uniform rod of length `L` and mass `M` is pivoted freely at one end and placed in vertical position.
a. What is angular acceleration of the rod when it is at an angle `theta` with the vertical?
b. What is the tangential linear acceleration of the free end when the rod is horizontal?

Explanation

Figure shows the rod at ann angle `theta` to the vertical. If we take torques about the pivot we need not be concerned with the force due to the pivot.
The torque due to the weight about `O`.
`tau_(mg)=(mgL)/2sintheta` ltbr. Using tourqe equation about `O`. Here we apply torque equation about fixed axis passing through `O`. We can apply torque equation about fixed axis or about centre of mass only.

a. `tau=Ialphaimplies(mgL)/2sintheta=(ML^(2))/3alpha`
Thus,`alpha=(3gsintheta)/(2L)`
when the rod is horizontal `theta=(pi)/2` and `alpha=(3g)/(2L)`
b. The tangenital linear acceleration of the free end is `a_(t)=alpha=(3g)/2`