2

Does Newton's second law of rotation only hold for torques about the com?

looking at this: Rotation and center of mass -- A collection of common doubts, in the second part, GRocks asks

"You have a rod (again, free), and now you continuously apply a constant force of equal magnitude and direction, on both ends. Intuitively, you wouldn't expect the rod to rotate. You say 'at all times, the torque about the COM is zero, so the body will not rotate'. But at any instant, about ANY other point, there is a non-zero torque. Why are we only considering torque about the COM? Why not any other point on the rod?"

I read David Hammen's answer in which he states

"Something very nice happens when one chooses to use the center of mass as the point of interest: the translational and rotation equations of motion decouple. The translational acceleration of the center of mass is a function of net force; torque does not come into the picture. The angular acceleration is a function of net torque about the center of mass; force does not come into the picture. "

Isn't the natural conclusion of that that Newton's second law of rotation only holds for rotations about the center of mass because it is the only point where the equations of motion decouple? Does this happen at other points depending on the case, like for example opening a door?

Qmechanic
  • 201,751
Omar Aboutaleb
  • 39

1 Answers1

1

In order for a radius vector to exist, torque must always be defined relative to a specific pivot point. If that pivot point has zero acceleration, or preferably zero motion, then the problem can reduce to a torque-only problem. The force at the zero acceleration pivot does not matter for motion. It's function is to hold the pivot at zero acceleration. This is why the force at a door hinge can be quite complex when a non-constant force is applied to the door handle. If the center of mass is the pivot, then the net torque as a whole contributes zero torque. Net force determines the acceleration of the center of mass. Forces relative to the net force decide torque around the center of mass.

When a pivot point other than the center of mass is accelerating, then you need torque around the pivot point just to keep the object's center of mass aligned with the pivot.
The angular velocity and angular acceleration of the object are the same around any parallel axis in an inertial frame of reference. Net torque and rotational inertia change when that pivot axis changes, but the rotational effect is the same. When the pivot axis accelerates, the frame of reference around the pivot axis is no longer inertial. You need an artificial effect to compensate for this, just as centrifugal acceleration must be added to a rotating reference frame to make Newton's Laws valid in that non-inertial frame.

We could bring the frame of reference into a still pivot by adding opposite of this acceleration to the entire object. This would happen with a single force at the center of mass, equal to total mass times the pivot point acceleration (in the opposite direction). This actually brings the net torque around either end of your example case to zero. Force at the other end is just enough to maintain zero angular acceleration in the accelerating frame of reference.

This is why using the center of mass as pivot is so easy, even when the center of mass accelerates. This artificial force adds zero torque around the center of mass. The center of mass is the only accelerating pivot that doesn't need an extra torque due to the artificial acceleration in order to provide the correct angular acceleration around the pivot.

Ken Mellendorf
  • 1,236