# Rod Falling on Frictionless Surface

A rigid rod with mass $$m$$ is initially held at an angle with respect to a frictionless plane by a string. Then the string is cut. What happens to the rod? My intuition suggests that the rod should slide to the left but I cannot figure out how this is supposed to happen when the only forced involved are vertical (reaction force and weight). Moreover, while initially the reaction force should balance the weight, the reaction force causes the rod to rotate about its centre of mass. This would cause the end of the rod in contact with the surface to leave it. But then the reaction force disappears and the weight of the rod places it back on the surface again. I am not sure how to deal with this "jerky motion" either.

• An image / diagram would help. I do think this is conceptual enough to avoid being closed as homework. – Brandon Enright Feb 12 '14 at 16:39

I'm guessing from your description that the left end of the rod is the end on contact with the ground, and the right end is the in the air at the point that the string is cut.

I think that what will happen is this: When the string is cut, gravity will cause the rod to fall, such that its centre of gravity falls vertically. As the rod falls, it will rotate to become more horizontal (it must be horizontal when it lands on the ground). That will of course cause the left end of the rod, which is already on the ground, to move to the left (your intuition is correct in that regard), but the right end will move to the right, ensuring no horizontal movement of the centre of gravity.

What are the forces that cause this movement of the ends? They will be internal forces within the rod (caused because the external forces are acting to compress the rod, and the rod reacts to that). Those internal forces will act along the rod and so have a horizontal component. If you sum/integrate these horizontal components along the whole rod, they'll cancel out and so add up to zero, but at any particular point along the rod, the horizontal force acting on that bit of rod will be nonzero.

The left end of the rod will not leave the ground while the rod is falling: The rod is simultaneously falling and rotating, and the vertical component of the two motions will exactly cancel at the left end, so that the left end will only move horizontally along the ground. (The entire rod may of course bounce off the ground immediately after falling, as a result of the collision with the ground, but I don't think that's what you were asking about).

• "The left end of the rod will not leave the ground while the rod is falling: The rod is simultaneously falling and rotating, and the vertical component of the two motions will exactly cancel at the left end, so that the left end will only move horizontally along the ground." The rod may leave the ground, if the impulse of the normal force is great enough. This requires calculation, but I think there are initial angles (perhaps all of them) for which the end will leave the ground. – Ján Lalinský Feb 12 '14 at 21:49
• Hmmm. I don't want to say definitely no without doing a detailed calculation, so maybe you're correct. But to my mind, it doesn't seem very plausible: What force could lift the left end up? The rod is likely to be in compression, not tension, and therefore internal forces in the rod would be pushing that end downwards. The only other force with any upwards component is the normal reaction force on the ground, but a normal reaction force (in the absence of a collision) would normally be exactly enough to just stop the rod from falling into the ground, no more. Or am I missing something? – Simon Robinson Feb 13 '14 at 13:52
• Think of a ball rolling from the top of a spherical surface. At certain point, the ball will separate from the surface, as the velocity gets so great that gravity is not able to curve the trajectory according to the surface. Similar thing can happen with the rod - when its angular velocity due to normal force from the ground is great enough, the lower point of the rod can lift off the ground. – Ján Lalinský Feb 13 '14 at 19:03

The only force applied is vertical. So centre of mass's horizontal position can't change. Say left edge doesn't slide to left(in this case center of mass will shift to right). It says it all.

If (the surface on which the rod is falling/or the rod) is elastic, only then there will be a jerky motion and a rebound and hence possible displacement of horizontal position of centre of mass.