A solid cylinder of mass $M$ and radius $R$ is rolling without slipping on a horizontal surface. It encounters a step of height $h$ where $h<R$. What is the minimum horizontal velocity $v$ required for the cylinder to climb the step?

Solution

1. Angular momentum about the edge of the step is conserved during the impact.
2. Initial angular momentum $$\begin{array}{rl}{L}_i& =Mv(R-h)+I\omega \\ & =Mv(R-h)+(1/2M{R}^2)(v/R)\\ & =Mv(R-h+R/2)\\ & =Mv(3R/2-h)\end{array}$$.
3. Final angular momentum $$\begin{array}{rl}{L}_f& =I_{edge}{\omega }_f\\ & =(3/2M{R}^2){\omega }_f\end{array}$$.
4. Energy conservation: $1/2M{v}^2+1/2I{\omega }^2=Mgh+1/2I_{edge}{\omega }_f^2$.
5. Solving for $v$ by substituting ${\omega }_f=L_i/I_{edge}$ gives the required velocity.
6. Hence the answer is (B).