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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.

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    $\begingroup$ First of all if you let something fall in the atmosphere, the air affects it in a rather complicated way. So letting 2 connected boxes of different weights fall, it is almost impossible to say exactly what they will do. But fundamentally I don't understand your question. What do you mean when you say why things spin? Why do things move linearly? Can you explain this? The answer to this question and yours are essentially the same. $\endgroup$ – tst Aug 13 '18 at 12:48
  • $\begingroup$ Please see the following old question: physics.stackexchange.com/questions/12140/… $\endgroup$ – David Bar Moshe Aug 13 '18 at 13:53
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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:

impulse separating into two waves

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.

reflected waves

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.

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  • $\begingroup$ Thank you Mark, that is a beautiful answer (in fact that is exactly the way I had first understood what the center of mass was)! This makes sense to me. However, for some reason when I think of putting the force on the actual boxes instead of on the bar, my intuition fails me. In your explanation, the reason it doesn't spin is because 2M gets twice the force. With the rockets (or gravity), we are putting twice the force on 2M regardless of where we put the pivot point, thus the boxes have the same acceleration regardless of where we fix the bar. So why is it that it spins? Thank you! $\endgroup$ – Joshua Ronis Aug 13 '18 at 14:13
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    $\begingroup$ @JoshuaRonis: I think what you might be missing is that the pivot itself exerts a force on the bar. If you apply a force on both boxes, the bar will try to move, but the pivot will exert a force to resist this. This force will follow the description in the answer, and its effect will depend on where the pivot is along the bar. $\endgroup$ – Meni Rosenfeld Aug 13 '18 at 17:59
  • $\begingroup$ OH! And the resistive force of the pivot is exactly the force you described in your answer: it will be equal to the sum of the forces for each of the boxes! $\endgroup$ – Joshua Ronis Aug 13 '18 at 18:12
  • $\begingroup$ So if we have it fixed on a pivot above the surface of the earth, the total resistive force of the pivot would be 3Mg. And, if we put it on the center of mass, that means the momentum per second provided upwards by the pivot would be twice as much on the 2M than on the M, because the 2M is closer so the momentum gets there faster, meaning the entire thing doesn't spin! $\endgroup$ – Joshua Ronis Aug 13 '18 at 18:13
  • $\begingroup$ However, if we place the pivot right in the middle, half of the total reaction force of 3Mg would go to the M and half to the 2M, since the momentum is getting to each at the same time. That means we would have a net force of .5mg downwards on the 2M and a net force upwards of .5mg on the M (thats 2Mg - 1.5Mg and 1Mg -1.5Mg, downards being positive and upwards being negative), giving us a total force of Mg to spin the bar. Except, we don't call it force, we just remember that the rate of transfer of momentum is inversely proportional to the length and call it torque! $\endgroup$ – Joshua Ronis Aug 13 '18 at 18:15
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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.

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  • $\begingroup$ OK OK! Now it's really starting to make sense! So what you are saying is that since the pole transfers the reaction force both ways (from one box to the other) if we put the pivot right in the middle the reaction force on the smaller box will be greater than the force it exerts downwards on the bar? $\endgroup$ – Joshua Ronis Aug 13 '18 at 16:35
  • $\begingroup$ If this answer and the one before it could somehow be connected, like how the force spreads along the bar implemented into this answer, I feel like that would be ultimate understanding. Thanks so much V.F. that was an awesome answer! $\endgroup$ – Joshua Ronis Aug 13 '18 at 16:38
  • $\begingroup$ @JoshuaRonis "... if we put the pivot right in the middle the reaction force on the smaller box will be greater than the force it exerts downwards on the bar?". Although, it could be the case, I did not say that. My answer focuses on two points: a) unopposed force causes acceleration b) if body is pinned, it cannot move straight (accelerate linearly), but it can rotate (accelerate rotationally). (a)+(b) causes rotation. I'd add that the direction of rotation will depend on which side of pivot point the net (equivalent) force is acting on. I'd be useful to check out how to find COM. $\endgroup$ – V.F. Aug 13 '18 at 17:38
  • $\begingroup$ Ok, not reaction force, how about: force upwards from the bar which was the reaction force for the heavier one that got transferred across the bar to the lighter one...? $\endgroup$ – Joshua Ronis Aug 13 '18 at 17:57
  • $\begingroup$ @JoshuaRonis Not sure which force you are talking about and what your question regarding this force is. Could you please state the whole question? $\endgroup$ – V.F. Aug 13 '18 at 18:09
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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.

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  • $\begingroup$ Hey Alex, I just got back to this after a long while. I'm a bit confused on your first paragraph, could you perhaps clarify why we would have to exert twice the force on the smaller mass? $\endgroup$ – Joshua Ronis Oct 1 '18 at 0:52
  • $\begingroup$ The idea is that the way you distribute the mass of an object doesn't affect its rotational dynamics. For that, it suffices to consider the center of mass and something called moment of inertia. The center of mass tells you the point about which the object rotates, and the moment of inertia tells you how hard it is to get the object to rotate. Applying force near the pivot point is more efficient to produce rotation (think about swinging a door open from the hinge versus the edge), and the center of mass is closer to the heavier object, so you need less force on that side of the rod. $\endgroup$ – Alex Buser Oct 5 '18 at 16:37

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