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Suppose, a cylinder is placed on its lateral side at the edge of a table, carefully, so that the slightest push makes it fall from the table. We give it a gentle push. And it begins to fall. How is its motion?

My efforts: A slight thought reveals that the rod will initially turn about the edge (the edge is the axis), until it looses contact with the table. Then the only force on it is the force of gravity, so its CM falls down in a straight line. But we cannot ignore the fact that it was rotating about the edge of the table. What happens to that and why? It cannot continue rotating about that axis, 'cause that would make the CM also rotate, which should actually come down in a straight line. But its angular velocity also cannot disappear instantaneously.

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closed as off-topic by John Rennie, Brandon Enright, ACuriousMind, Kyle Kanos, rob Sep 28 '14 at 21:34

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It makes the problem easier to visualize if you collapse to 2 dimensions and imagine a circle in the same situation with homogeneous distribution of mass effectively centered at the center of the circle. Constrained initially by contact with the table's edge, and after your nearly forceless nudge the center will begin to accelerate through an angle under its own weight. Then when the rotation reaches 90 degrees the circle becomes a free falling body with fixed angular velocity that was imparted when it lost contact with the table - at 90 degrees.

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  • $\begingroup$ +1 for a simple solution. I should have thought of the idea of using a circle. Nice. $\endgroup$ – HDE 226868 Sep 28 '14 at 15:03
  • $\begingroup$ At, say 45degrees in your scenario the CofM of the cylinder has a horizontal velocity away from the table. At 90 degrees, the CofM is falling straight down with zero horizontal elocity. How does this happen? The corner can't pull back the cylinder $\endgroup$ – DJohnM Sep 28 '14 at 16:52
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Short answer: The cylinder will fall in a parabola while it continues to rotate.

If the cylinder is given a push, then (regardless of whether or not it is rotating), it will move in the x-axis. This is true for any object, assuming that the applied force overcomes friction, air resistance, etc. For this, I'll assume that there is no friction, because when the cylinder leaves the table, it will lose contact with it. No normal force means no friction.

The cylinder will also move in the y-axis (assuming there is gravity here - and I think you want a solution with that!). It will enter free-fall, accelerating under gravity. Combining its motion in the x and y axis, we find that it will take the path of a parabola.

Finally, there is rotational motion. If the force is applied to somewhere other than the axis containing the center of mass, the cylinder will rotate.

Knowing all of this, we can firmly say that a) The cylinder will continue moving in the x-direction at a constant speed $v$ after the force is applied, b) The cylinder will continue moving in the y-direction as it falls due to gravity, and c) The cylinder will continue rotating. Why? We know this from the principle of conservation of energy, assuming no non-conservative forces: $$E_f=E_o$$ and so $$KE_{tf}+KE_{to}$$ and $$KE_{rf}=KE_{ro}$$ where $KE_t$ is translational kinetic energy and $KE_r$ is rotational kinetic energy.

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