kinetic energy of rotation

kinetic energy of rotation
A cylinder of mass M and radius R rolls (without slipping) down an inclined plane( of height and length L) whose incline angle with the horizontal is Θ.Determine the linear speed of teh cylinder's center of mass when it reaches the bottom of the incline(assuming that it started from rest at the top).
See the inserted graph.
I don't understand the solution to this answer on the book.
Here's the solution:
We will attack this proble using Conservation of Mechanical Energy.(here's the confusing part) As the cylinder wolls down the ramp,its initial gravitational potential energy is converted into kinetic energy,which is a combination of traslational kinetic energy(since the cylinder's center of mass is traslating down the ramp) and totational energy:
K(initial)+U(initial)=K(final)+U(final)
0+Mgh=(1/2Mv^2+1/2Iw^2)+0
...followings omitted...
here's my question.For this problem,the frition between the ball and the ramp should be involved,but isn't that Conservation of Mechanical Energy can only be used in a non-friction situation?Why can they use that?The force of friction should do work.The energy should not be conserved.
Please solve our problem.

There is no dissipation of energy in this process.So,the potential energy changes to the kinetic energy by dropping its position and by friction force (to rotational energy).