Inclined plane problem
Consider two cylinders that start down an incline from rest. One rolls without slipping, while the other slides frictionlessly without rolling. They both travel a short distance at the bottom and then start up another incline.
Show that they both reach the same height on the other incline, and that this height is equal to their original height.
Note that (1/2) MR^2.
For "without slipping," I found that the starting height, h_1, and the "ending" height, (when ball changes direction), h_2, are the same by virtue of mgh_1 = 1/2 mv^2 + 1/2 Iomega^2 = mgh_2. That is, I used the transitive property of equality (no clue if this is what the book wants).
Then, using h=h_1=h_2, I solved for h. I got h = (3v^2)/4g
For "frictionlessly without rolling," I used the same principles, but instead mgh = 1/2 mv^2, since there is no rotational kinetic energy here. Solving for h (same way as above), I got (v^2)/2g.
This is as far as I got. I would genuinely appreciate help. Thank you!
