A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. a) How high would the marble go if both sides were as rough as the left side? b) How far up the smooth side will the marble go, measured vertically from the bottom? My attempt - I am able to attain the correct answers but concept wise could do with some help ... a) via consevation of energy I conclude h. However I do not really understand why energy is conserved in such cases 'roll without slipping'. ( I have seen a basic derivation which considers the two components of the velocity of the marble - the rotational and the translational, and setting this sum to zero to deduce that v=wr ) However I do not understand why, in the conservation of energy equation , we do not include the work done via friction due to the rough surface. b) I first of all calculate the speed at the bottom via consevation of energy - equating the loss of GPE = KE ROTATIONAL + KE TRANSLATIONAL, and using the above deduction  which follows from the nature of the surface and marble's mutual. behaviour. Then using this speed, equated the KE at the bottom = GPE GAIN, knowing KE at max height = 0. However again, I do not really understand the principles behind this- why is rotational kinetic energy instantly eliminated as soon as the body slips - How does slipping - no frition - cause motion to be purely translational .? (That's the cnclusion I reach as this led me to the right answer). Anyone who could shed some light on this, really aprreciated, thanks alot !