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Consider a thin circular ring of mass $m$, radius $r$ rolling without slipping with velocity $v$ towards a step of height $h$ $(<r)$. Assume no rebound and no slipping at the time of contact. What is the condition on $v$ to ascend the step?

First equation I have is by conserving energy. The kinetic energy of the ring at the bottom has to be greater than the gravitational potential energy at the top of the step. For the second equation, can I conserve angular momentum about the point of contact between ring and step since the point of contact is at rest? How do I involve the impulse provided by the step to the ring in the form of equations, if at all required? Advice will be appreciated.

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Firstly what you did wrong was conserve energy. You see, if we ignore rebound then that means the collision is inelastic. Thus there will be net loss of energy whose amount we do not know.

Anyways, to work around the problem, you have to conserve angular momentum about the point of collision. The reason being that about this point, resulting torque due to mass, normal reaction and friction would not matter since they will not be impulsive IN THIS CASE and the torque due to impulse at stair would also cancel since we have conserved angular momentum about that point.

torque due to mass, normal reaction and friction would not matter because the normal force due to stair would have component along positive y axis

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