What is the speed of a cylinder rolling down a hill?

[zeralda21]

Homework 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.

http://i.imgur.com/Ge3x1nu.png

The Attempt at a Solution

The answer is supposed to be v=√(gh) but my calculations give;

At the top the potential energy is E=mgh and at the end(h=0) all energy has become kinetic energy since no friction/air drag is acting. Thus mgh=(1/2)mv^2 <--->v=√(2gh). Why is this wrong??

 

[mfb]

The cylinder is not a point-mass - some of the kinetic energy is required to have it rolling.

 

[zeralda21]

The cylinder is not a point-mass - some of the kinetic energy is required to have it rolling.

I am trying but I am not following. The only acting force is the downward gravitational force mg and normal force from the ground. So if energy isn't the same at the top and bottom, where has it gone.

I also don't understand that kinetic energy is required to have it rolling. The kinetic energy at the top is zero and keeps increasing as h decreases(same rate inversely right?).

 

[darkxponent]

I am trying but I am not following. The only acting force is the downward gravitational force mg and normal force from the ground. So if energy isn't the same at the top and bottom, where has it gone.

I also don't understand that kinetic energy is required to have it rolling. The kinetic energy at the top is zero and keeps increasing as h decreases(same rate inversely right?).

Garvity and Normal the only acting forces, then what makes the cyllinder roll?
HINT: There is one more force that you are missing, draw the diagram and you will get it!

 

[barryj]

Consider the moment of inertia of the cylinder