Conservation of energy principle
[itex]T1 + \sum U1-2 = T2[/itex]
The Attempt at a Solution
I have the correct answer for part a).
[itex]T1 = 0[/itex] (since the system starts from rest).
[itex]T2 = (1/2)m(vG)^2 + (1/2)I\omega^2[/itex]
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 [itex]vG = \omega * GC[/itex]
[itex]GC = r - (4r)/(3\pi)[/itex]
From here I get the correct [itex]\omega[/itex], 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:
[itex]\sum Fy = m(ay): N - mg = m(rG)\omega^2[/itex]
But the problem is that I use [itex]rG = GC = r-(4r)/(3\pi)[/itex], but in the solutions they use [itex]rG = (4r)/(3\pi)[/itex], why is that??? Is to half-cylinder rotating about O? If so why are we using [itex]rG = GC[/itex] in part a) of the problem?
Thank you in advance for the help.
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