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Introductory Physics Homework Help
A small cylinder in a pipe
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[QUOTE="asdfghhjkl, post: 4964766, member: 486811"] Ok so I went thought through this again, and I came to the following conclusions; I) To get the relation between $$\omega$$ and derivative of $$\psi$$, you assume that there is no slipping. Hence the center of mass of the small cylinder should move with the same velocity as the edge of the cylinder. Now the CM of the cylinder moves with $$\psi * (R-r)$$ and the edge of the cylinder moves with $$\omega *r$$ equating this and differentiating gives the required expression. II) For the second part use the energy method. There are Pe, Ke, Kr. Potential energy is $$mgh=mg(R-r)(1-cos(\psi))$$, the velocity of CM, $$v=( \dot{\psi})(R-r)$$, therefore the Ke due to the motion of CM, $$Ke = 0.5*v^2*m$$. The last component of the energy is rotational kinetic energy Kr, $$Kr=0.5*I*\omega^2$$. This yields the equation of motion; $$\dots{\psi}=\dfrac{2*g*sin(\psi)}{3(R-r)}$$ and then for small angles this simplifies to; $$\omega = \sqrt{\dfrac{2*g}{3*(R-r)}}$$ I don't have the solutions to the problem, so I would appreciate if someone could tell me if this is correct. [/QUOTE]
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A small cylinder in a pipe
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