Problem statement:
A cylinder rolls without slipping down a hill. It is released from height h. What is its speed when it come down? The cylinder mass may be completely concentrated on the radius R, which is the radius of the cylinder.
My thoughts:
At the top(hight h) the potential energy of the cylinder is E=mgh and at the bottom(h=0) all energy has become kinetic energy since friction and air drag is neglected in this context.(I assumed this). Thus:
$mgh=\frac{1}{2}mv^2 \Leftrightarrow v=\sqrt{2gh}$
Correct answer is however $v=\sqrt{gh}$ which means that energy must have been lost,correct?
What have i missed?