# Energy of a cylinder rolling down a path

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?

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You're missing the rotational kinetic energy at the bottom, $\frac{1}{2}I \omega ^2$. The key word in the problem is, 'rolling without slipping'. Also remember the equation, $$v=r\omega$$ The cylinder's moment of inertia should be looked up in a table or given.
On wikipedia, the moment of inertia of a thin cylindrical shell is given as: $$I=mr^2$$
"insection": n., the intersection between inspection, inception, perception, and introspection. ;-) –  rubenvb Jun 27 '14 at 9:25