Problem: Gman 8/10/03

I'm certain this problem has a mistake in it. The solution should be (20g/11L)^(1/2). The solution given forgot the 1/2 on the calculation for the kinetic energy. K= 1/2 Iw^2

Problem: forscher14 [8/17/03]

I don't understand how this problem was solved by the author. He/she says, " taking as our origin the point of contact between the road and the barrel... ...the angular momentum of the barrel is M * v * R" This doesn't make sense to me, the CM in the center of the barrel should have no angular momentum about it's point of contact with the road as they do not move in relation to each other.

I solved this problem using an energy method and assumed that none of the barrels energy was lost as heat from friction with the road.

1/2M * v^2 = 1/2 M vf^2 + 1/2 I*w^2 where I = M * R^2 w = vf / R

I ended up with vf = (1/2)^(1/2) v Did I mess up somehow?