Dynamics: half-cylinder rolls without sliding.

[etotheix]

Homework Statement

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

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.

 

[ideasrule]

I think you made a mistake in the solution for a, despite getting the right result. For pure rotation, you can either choose the point of contact or the center of mass as your reference point. If you choose the center of mass, you must use the body's rotational speed with respect to its center of mass, as seen in an inertial reference frame. The contact point is not inertial, so you can't use its reference frame. Rotational kinetic energy would be given be the usual 1/2*I*w^2, and translational kinetic energy would be 1/2*m*(GC*w)^2 because w is how fast GC is rotating with respect to the horizontal.

So in the end, we get the same equation you initially got for part a. Note that w was with respect to an inertial frame, and represents the rate at which GO is rotating.