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.
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