classical physics - rod anchored to ground, rotational inertia

[TomCowley]

I'm sure there's an easier way for the first one, but using the initial PE = final KE method, initial PE is 1/2 Mg. Since velocity at the end is proportional to distance from the anchored part of the rod, we can call velocity at distance r kr
Final KE = Int [r on 0,1] 1/2 M (kr)^2 dr = 1/6 M k^2 r^3 evaluated at 1 and 0 = 1/6 M k^2. So 1/6M k^2 = 1/2 M g.

So 1/3 k^2 = g, k=(3g)^0.5, so velocity at the end = k*1 = (3g)^0.5 which is the same answer lifes3ps got.