# Rotation and center of mass — A collection of common doubts

There are plenty of similar questions here, but I will nevertheless add another one because I believe this is the ultimate source of confusion (and will probably provide a one-stop solution for future questions of similar nature) -- what is so special about the center of mass when it comes to the rotation?

## Case 1:

You have a rod in space. You provide an impulse to one of it's ends. It's COM starts translating at a uniform speed, and the rod simultaneously starts rotating (to conserve angular momentum) about its COM (okay, about an axis through COM); why just the COM? Why not any other point? (Same goes for 2 masses connected by a massless rod, an impulse on one of them causes rotation about COM)

Possible answers -- The COM is stationary (if not, choose a frame moving with the same velocity as the COM), so only it can provide an axis of rotation in an inertial frame. Another possible answer (not necessarily correct) -- The rod's motion can be thought of its translation plus rotation about ANY point on the rod; choosing the COM makes the analysis easy. Are any of these explanations correct? Is there a better one?

## Case 2:

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?

Possible answer (not necessarily correct) -- For every point on the rod where the torque is nonzero, there is a symmetrically located point on the other side of the COM with an equal and opposite torque (should you calculate it), so by symmetry we should expect no net rotation.

The idea is, it would be great if someone could clear these common doubts by either elaborating/pointing out the mistakes in the 'possible answers', and providing the right explanations. I believe I have tried to condense all common doubts of similar nature in the question, at the expense of poor readability.

• – David Hammen Jun 8 '18 at 12:32

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.

There are some different ways to look at it.

• Net torque doesn't necessarily mean rotation, it means a change in angular momentum.

Let's suppose the bottom of the rod is at the origin and the top of the rod is on the $$y$$ axis. Both forces push the rod in the $$+x$$ direction. If we analyze the forces about the origin, we can see that the lower force (which goes through the origin) will contribute no torque, while the top one will. But the rod doesn't spin?

A net force yields a change in linear momentum, a net torque yields a change in angular momentum. In many cases that change in angular momentum is accompanied by a change in rotation, but not here.

We can calculate the angular momentum of the rod about the origin as equal to $$mvd$$ where $$m$$ is the mass of the rod, $$v$$ the velocity of the rod, and $$d$$ the vertical distance of the center of mass from the origin. The forces are increasing its velocity, so the (non-rotating) angular momentum is also increasing. When the center of mass isn't moving, we can assume all the angular momentum is in rotation. Here it's not. The unbalanced torque is creating a change in angular momentum, just not rotation.

• Additional forces appear in accelerating reference frames

Rather than analyze rotation about the origin, let's analyze it about the bottom of the rod. Since the rod is accelerating in the $$+x$$ direction, our reference frame is as well. This means that fictitious forces appear. In this case the force will be equal to $$-ma$$. (The force is proportional to the mass of the rod, and in the opposite direction to the acceleration).

If you now compute the torque about the bottom of the rod, you'll discover that the torque from the force at the top of the rod is exactly cancelled by the torque from the frame acceleration applied at the rod's center of mass.

In Case I, the question is asked what is special about the center of mass. The speculation in the question is correct. We can alway pick an inertial frame where (given no net forces) the center of mass is at rest. Therefore any rotation must be about the COM in that frame.

• Great answer @BowlOfRed! For some reason I did not get a notification for this. You could perhaps include something about case 1( in addition to the excellent answer by @DavidHammen) to make this more self contained. – GRrocks Jun 10 '18 at 7:14
• @BowlOfRed Beautiful explanation! It'll be great if you could provide a similar sort of explanation for Case (I) of the original question... – think__tech Oct 8 '18 at 19:54

The answers to your two questions are related. The answer to the first question is, as you have guessed, that the rod's motion can be thought of it's translation plus rotation about ANY point on the rod; choosing the COM makes the analysis easy.

The angular velocity and acceleration of a rigid body are free vectors, meaning that they are the same at every point on the body. The translational velocity and acceleration on the other hand are localized -- they vary as the location of the point of interest changes.

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.

This is not the case for any other point on a rigid body. At such points, the translational acceleration depends on both the net force and on the net torque about that point. The same applies to rotational acceleration. The generalized equations of motion are the Newton-Euler equations. The wikipedia article on this subject is a good starting point if you want to learn more.

Even though the equations of motion simplify so nicely at the center of mass, there are many applications where it makes more sense to use the more generalized form. Introductory physics students oftentimes are taught one such application, a rod with one end fixed at a pivot point. That problem can be solved without using the full-blown Newton-Euler equations. Solving such problems from the perspective of the pivot point is easier than using the center of mass as the point of interest because the pivot point has no translational acceleration (it's fixed).

In more advanced settings, the Newton-Euler equations (or their Lagrangian equivalent) come into play in robotics and in spacecraft. The joints in a robotic arm are subject to constraints. These constraints are significantly easier to express when the formulation is expressed in terms of the joint location.

The equations typically used in introductory to intermediate settings implicitly assume a point mass rocket. A real spacecraft expels roughly 90% of its mass during launch, and its center of mass moves inside the vehicle. Getting the equations of motion right for a real rocket is non-trivial, and it's arguably better to use a fixed point on the rocket structure rather than the center of mass that moves with respect to structure.

• Great answer! Quick clarification- So the other "explanation" for case 1(the one involving the fact that COM remains stationary) " is a happy coincidence? Because in, for example case 2, it doesn't, and I can still assume rotation to be about it, accompanied by translation, just as any other point on the rod.(if I remember correctly, even if the COM happens to be accelerating, the torque about it still neatly comes out as derivative of angular momentum) – GRrocks Jun 9 '18 at 2:57