# Determining the equilibrium position of a uniform rod pivoting about its center

here is the question:

A uniform rod pivots about a frictionless, horizontal axle through its center. It is placed on a stand, held motionless in the position shown below, and then gently released. On the right side of the figure, draw the final, equilibrium position of the rod. There are some online solutions showing that the rod will not moved at all as you release it, because it is already in equilibrium. However, I don't think that solution is correct. My intuition tells me that the rod will be horizontal in equilibrium, and I even conducted an experiment by hand-holding a uniform rod to verify my prediction. (As I released the tilted rod, the rod rotates and becomes horizontal) I can't come up with a good explanation on this problem. Can anyone help? Thank you!

In the diagram the axle is said to be (exactly) through the centre of mass of the rod and there was no friction present.

When you did your experiment are you sure this condition was satisfied?
In the real world it is not possible to satisfy that condition and so you have to produce an answer for a hypothetical ideal situation.

That being so you can say that for every position that the rod takes there is no net force or torque acting on the rod and its gravitational potential energy does not change.
So the rod is a position of "neutral" equilibrium for all orientations.
If left alone the rod would stay in the same position.

In a standard experiment a metre rule is balanced on a knife edge. You will see from the diagram unless the rule is horizontal there will be a net torque (couple) on the rule.

When the rule is horizontal it is in a position of unstable equilibrium which explains why it is so difficult to find the balance condition.

• That makes sense. Thank you! Apr 22 '20 at 16:07