Force on pivot when a plank rotates from 60° to horizontal

[shivam jain]

Homework Statement

a thin plank of mass m and length l is pivoted at 1 end.The plank is released at 60 degrees from the vertical. What is the magnitiude and direction of the force on the pivot when the plank is horizontal.

Homework Equations

conservation of energy

The Attempt at a Solution

when horizontal 2 forces mg and force of pivot. mg we know .force of pivot to find.PE=mgl/2cos theta where theta is angle with vertical.conserving energy i get mgl/2cos theta=1/2*I omega^2.then i found net force m^2 g^2+m*omega^2*l/2 whole underroot.but not getting correct answer

 

[Simon Bridge]

What makes you think it is not the correct answer?

reason it out - change in gravitational PE is mgh ... what did you use for h and why?
(note: cos(30)=1/2)

Where would the force on the pivot come from?