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etotheix

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## Homework Statement

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

## Homework Equations

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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