# Dynamics: half-cylinder rolls without sliding.

## Homework Statement

[PLAIN]http://img62.imageshack.us/img62/6531/capturefma.jpg [Broken]

## Homework Equations

Conservation of energy principle
$T1 + \sum U1-2 = T2$

## The Attempt at a Solution

I have the correct answer for part a).
$T1 = 0$ (since the system starts from rest).
$T2 = (1/2)m(vG)^2 + (1/2)I\omega^2$

I believe that the instantaneous center is located at the point of contact of the half-cylinder and the ground after the 90 degree rotation, let's call this point of contact C. So $vG = \omega * GC$
$GC = r - (4r)/(3\pi)$

From here I get the correct $\omega$, the real problem comes when I try to solve part b), which asks for the normal force at point C.

My strategy is to use the force and acceleration principle in this way:
$\sum Fy = m(ay): N - mg = m(rG)\omega^2$

But the problem is that I use $rG = GC = r-(4r)/(3\pi)$, but in the solutions they use $rG = (4r)/(3\pi)$, why is that??? Is to half-cylinder rotating about O? If so why are we using $rG = GC$ in part a) of the problem?

Thank you in advance for the help.

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