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Let's say I have 2 boxes, one of mass M and one of mass 2M. They are separated by a distance of 1 meter. I drop them from the same height and see that due to the earth's gravity, they accelerate at the same rate towards the ground. Now, I connect them with a meter long bar. I have 3 questions:

1. If I let them go from an airplane, with the bar connecting one to the other, will they spin as they fall through the air? I think not, because they have the same acceleration.




2. If I attach them to a wall so that the bar isn't allowed to fall, and I put the pivot point right in the middle of them, can someone please explain (without torque please, since torque states they spin but doesn't really explain why) why is it that they spin? (Read the question below, I think its really similar to this one but states my confusion better).




3. And just one more thing (this is exactly like the earth question but I think it shows my confusion more clearly). If I'm in space, with the boxes connected by the bar but the bar not attached to anything, and I attach little rockets to each of the boxes so that the rockets fire in the same direction, my intuition tells me in order to keep it from spinning the rocket on the smaller mass M must be exerting half the force than the rocket on the bigger mass 2M in order to give them the same acceleration and keep them from spinning. However, if I then fix the bar on a pivot point in space, what changes depending on where on the bar I put the pivot point if regardless of the pivot point the boxes have the same acceleration? Why is it that if I put it closer to 2M the bar doesn't spin, but if I put it smack in the middle or closer to M the bar spins? Again, an explanation without torque would be greatly appreciated.

I'm going to start with the barred masses in space example (#3) since that is the simplest. Also, instead of a continuous force from a rocket, let's consider discrete impulses, as if the bar was being hit by a hammer. In order to ignore torque, we have to look at what the bar material is actually doing to transmit the force to the masses.

So, we have two masses, $M$ and $2M$ at opposite ends of a bar. Imagine we tap the middle of the bar with the hammer. This creates a momentary deformation at the site of the impact. This deformation causes a wave pulse (more accurately, a stress wave) to travel in both directions along the bar, away from the point of impact. See below for an illustration:

Picture from: https://www.acs.psu.edu/drussell/Demos/Membrane-vs-String/Membrane-vs-String.html

Now, what happens when these pulses reach the masses? They will put equal forces on the masses since the waves are identical. So, the larger masses will accelerate more slowly and lag behind the lighter mass. The bar is stiff, so it doesn't change length, meaning that it must rotate for the larger mass to lag behind.

Now, if the bar is hit closer to the larger mass, the wave pulse reaches the larger mass first due to the shorter distance. So, the larger mass gets a head start to move ahead before the smaller mass catches up due to its larger acceleration. There's a whole analysis to be done here regarding reflected waves (see picture below) and the timing of such to get Newton's Third Law to work out and to show that the proper point to hit for no rotation is the same as that derived from torque equations.

From here: https://en.wikipedia.org/wiki/Wave_equation#Stress_pulse_in_a_bar

A continuous force can be thought of as an infinite sum of infinitesimally small discrete impulses. In other words, an extremely rapid series of very small taps that adds up to a given force. So, in case #2 with the wall mount and case #3 with the single rocket, imagine a rapid series of tiny impacts and we get the same result as with the hammer.

In summary, torque is an abstraction that allows us to ignore all these internal forces in the bar since, by Newton's Third Law, they always come in pairs and don't affect linear or angular momentum.

In case #1, where the barred mass assembly is dropped from an airplane (assumminmg no air resistance), the bar assembly does not rotate since gravity pulls on both masses in proportion to their masses, so they have the same acceleration.

The moment you introduce a pivot, you have to take into account the reaction force from the pivot - not just the weight of the boxes or the thrust of the rockets.

The applied forces acting on the boxes in both cases are equivalent to one force applied to the COM of the whole assembly.

So, if the pivot is aligned with the COM, the reaction force can completely cancel the applied force and prevent the acceleration.

If not, there would be a net (unopposed) force applied to the COM and, therefore, the assembly would have to accelerate, i.e., would have to start moving. But, since it cannot just move forward (translationally), ignoring the pivot, it will start spinning with the pivot point as the center of rotation.

In general, we can say that a body starts spinning, if one of its points (or a number of points located on the same line) is pinned and there is a force, applied to the body, which does not pass through that point or that line.

In case #3, the situation is actually the opposite of what your intuition tells you! You must exert twice the force on the smaller mass as on the larger mass in order to stop the bar from spinning. This can be seen easily from a torque argument, since the center of mass is closer to the larger box.

Intuitively, notice that in space the pivot point about which the bar rotates is its center of mass. Since the smaller mass is further from the center of mass it sweeps out an arc with twice the radius as that of the larger mass as it rotates, so the lighter end must accelerate twice as fast to prevent rotation. The actual mass of either end is irrelevant, since the composite system acts as if it were a single object constrained by the rigidity of the bar.

You can generalize this conceptual picture to understand what happens when you apply forces to the masses with an arbitrary pivot point - think about how much acceleration you need to counteract the natural tendency of one side to spin when you apply an unbalanced force to the other.

For what it's worth I strongly encourage you to come to terms with the validity of the torque/angular momentum approach. When you fully understand the derivations of such concepts they can be just as intuitive as any other explanation and often greatly simplify problems, allowing you to understand much more complicated phenomena! These ideas are in fact just as fundamental as F = ma, and shouldn't be discounted as explaining only the how and not the why any more than Newton's laws.