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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?

picture

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1 Answer

up vote 2 down vote accepted

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: $$I=mr^2$$

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In this special case you don't need to look up the moment of inertia. You can see from insection that the speed of the mass in the shell due to rotation is the same as that due to translation. Therefore half the energy is in rotation, which doesn't add to the cylinder's speed. Half the energy means sqrt(2) less speed. –  Olin Lathrop Jun 15 '13 at 22:00
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