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*A heavy uniform hoop of mass M and radius R rolls in a straight line on level ground without slipping. It encounters a step of height h, where h<R perpendicular to its path. On striking the corner of the step, the hoop does not rebound and has sufficient energy to just surmount it. If the angular velocity of the hoop about the corner just after impact is omega, show that:*

omega = ((g*h)^0.5)/R

omega = ((g*h)^0.5)/R

I've solved this problem using conservation of energy, saying that the sum of the linear and angular kinetic energy before the impact is equal to the potential energy the hoop gains by surmounting the corner. The thing is, while the maths works out fine, I'm wondering while the hoop is still moving at all? If it's just got enough energy to surmount the corner, surely it would be stationary and not have any angular velocity?

Additionally, I'm wondering whether the omega that I've calculated is just the omega before the impact; the angular velocity that's necessary to surmount the corner.