# Simple question about rotation in a uniform rod pinned on axis in center

Below is the image depicting my confusion. In the top diagram, there is a horizontal rod with X marking the axis of rotation. Since the rod is horizontal, intuitively the rod will not rotate. This is indeed true since when I take the torque at the center of mass I get 0, as no other force is acting on the rod except gravity, which is 0m away from the axis.

However, the 2nd image depicts the same uniform rod, but tilted. Same as before, X is the axis located at the center of mass. Intuitively, I'd expect it to rotate towards to the horizontal state, and then probably swing back and forth. But taking the torque at CM still results in net torque of 0, as I can only identify the force of weight acting along the axis of the rod. This concludes that the rod should not rotate.

So my question is: Am I forgetting other forces present when I'm taking the torque and that's the reason why I missed out on why the 2nd rod would rotate, or is the intuition I have because of some real-life imperfections to the system that exists? • If X is at the centre of mass, there is no potential for gravity to move it from rest. – Tony Stewart Sunnyskyguy EE75 Sep 7 '20 at 4:49
• ok, then I think this is solved. – LHC2012 Sep 7 '20 at 5:13

## 2 Answers

The fact that the torque is zero tells you only that the rod is in equilibrium and that its angular velocity will not change. However, this does not tell you what type of equilibrium the rod is in or how it will respond to small perturbations.

As you have noticed, the torque is zero regardless of the orientation of the rod. This is because, when the centre of mass is fixed, the potential energy does not depend on the orientation of the rod, which tells us that it is in a neutral equilibrium; i.e. there is no favoured orientation.

Your intuition appears to be incorrect. If hung by the centre of mass, it will remain in equilibrium in any position, not just horizontal state. However, this is assuming uniform gravity, which is quite okay for the scale of this question.

If you want to consider the changing gravity of earth due to height, then, contrary to what you thought, the horizontal position is in unstable equilibrium, and when disturbed to an angle, it will rotate towards the vertical state (parallel to the string) which is in stable equilibrium.