Why do things spin?

Question

Let's say I have 2 boxes, one of mass M and one of mass 2M. They are connected by a rigid rod. 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. Keep in mind that they're connected by a rigid rod. I have 3 questions:

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Question 1

If I let them go from an airplane, with the rod connecting one to the other, will they spin as they fall through the air?

I think not, because, without the rod, they would have the same acceleration.

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Question 2

If I attach the rod to a wall with the pivot point (point of attachment from the rod to the wall) halfway in between the two masses, can someone please explain (without torque please, since torque states they spin but doesn't really explain why) why it is that they spin around that pivot point?

What changes, physically, that causes them to spin? I know it must have something to do with the atoms of the rod...but I don't know how to explain it. See my other question here:

Law of the lever - Explained Physically

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Question 3

We're in space.

The two masses are connected by the rigid rod, but the rigid rod isn't attached to anything.

Attached to each of the masses are little rockets that fire in the same direction.

My intuition tells me in order to keep the entire system from spinning, the rocket on the smaller mass M must be exerting half the force than the rocket on the bigger mass 2M.

This is intuitively obvious because if they were not attached to the rod, this ratio of forces would give them both the same acceleration.

However, if I then fix the bar on a pivot point in space, why is it that if I put the pivot point closer to 2M the bar doesn't spin, but if I put it smack in the middle or closer to M the bar spins, when without the bar, both masses would have the same acceleration?

Again, an explanation without torque would be greatly appreciated, although if we can get to torque from a more physical explanation, that would also be awesome.

It must have something to do with the way momentum spreads through the bar...or the bending of the bar...but I don't know how to describe it

Answer 1

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.

Answer 2

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.

Answer 3

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.