# Will a free-falling rod (without drag) rotate?

When we consider a bicycle is turning on a flat plane, we know that there is friction, which provide centripetal force on the bicycle. And we know that the bicycle is no longer perpendicular to the flat plane so as to reach equilibrium. What I want to ask why we always choose the center of gravity of the bicycle be the rotational center?

Just like when a rod is free falling and parallel to the ground, if I choose the center of gravity as the rotational center the rod will not rotate, because there is no net moment acting on the rod. But if I choose the points other than the center of gravity as the rotational center, there would be net moment due to gravity and the rod will rotate. Will the rod rotate or not? Why we always choose center of gravity as the rotational center?

• A rod horizontally parallel to the ground while falling isn't in a stable configuration. Random turbulence is always going to cause the rod to rotate away from parallel to the ground. Nov 24 '13 at 8:46
• @BrandonEnright Normally you assume vacuum in such experiments Nov 24 '13 at 13:11
• @BrandonEnright The question clearly specifies "without drag". Mar 19 '18 at 11:45
• @PianoLand "without drag" was added to the question in an edit made after my comment. Mar 19 '18 at 21:17

why we always choose the center of gravity of the bicycle be the rotational center.

We do not do that always, sometimes it is better to use the point in contact with the ground or some other point. We use center of mass when it leads to simpler equations than the other points. In problems dealing with torques or rotations we use the theorem:-

"The sum of torques around any stationary point $$O$$ equals the rate of change of the angular momentum of the body with respect to the point $$O$$."

In uniform gravitational field, e.g. near the ground of the Earth, the gravitational forces have zero net torque around the center of mass, so if we choose for $$O$$ the center of mass, the contribution to net torque due to the gravitational field vanishes and we can focus on other parts of the problem.

It is important to realize that the angular momentum with respect to point $$O$$ depends on where this point is. The angular momentum can be expressed as

$$\mathbf L = \mathbf r\times M\mathbf v_{CM} + I\boldsymbol \omega$$

where $$\mathbf r$$ is the radius vector of the center of mass with respect to the point $$O$$, $$M$$ is the mass of the rigid body, $$\mathbf v_{CM}$$ velocity of the center of mass, $$I$$ moment (tensor) of inertia and $$\boldsymbol \omega$$ angular velocity of the rotation of the body.

But if I choose the points other than the center of gravity as the rotational center, there would be net moment due to gravity

Yes.

and the rod will rotate. Will the rod rotate or not?

No. The angular momentum of the rod with respect to the point $$O$$ will increase, but not due to change in rotation of the rod, but just due to the fact that the velocity of the rod's center of mass $$\mathbf v_{CM}$$ is increasing. The rod will not change its angular velocity $$\boldsymbol \omega$$, since for this there would have to be non-zero net torque with respect to its center of mass (if we choose CM for $$O$$, the contribution $$\mathbf r\times M\mathbf v_{CM}$$ to angular momentum vanishes and non-zero torque then implies change in angular velocity.)