I am trying to deepen my intuition: If in outer space is a rod of length 2 meters standing still. At the both ends it has some heavy wheels of equal mass and the associated motors. If a motor starts spinning one of the wheels ... will the center of mass of the rod stand still or move? I think that the following are possible:

  1. the center of mass will slowly rotate about the end of the rod with the wheel spinning. That end of the rod will stand still. So the c.o.m of the rod moves in a circle

  2. the center of mass will stand still and the ends will slowly rotate about the center of mass of the rod. So the c.o.m will not move, but the ends of the rod will rotate about it.

  3. neither ...

The motors are attached (welded) to the rod. Of the motor's rotors are attached the heavy wheels. (One motor at one end of rod)

I used to think that option 1 is the correct one ...

  • $\begingroup$ Which of the 3 options do you think is correct? You seem to be hedging your bets. $\endgroup$ – sammy gerbil Feb 24 '17 at 18:16
  • $\begingroup$ I appreciate a positive feedback, logic and math. Please if possible give that $\endgroup$ – C Marius Feb 24 '17 at 19:24
  • $\begingroup$ @ sammy gerbil please if you have a thought on this I would really like to see you explanation on the subject. It helps me a lot! $\endgroup$ – C Marius Feb 24 '17 at 19:35
  • $\begingroup$ It can help to explain your thought process in addition to the question. One of the frustrating things about Physics is that it's all easy if you understand it all perfectly. The hard part is when there' something you don't understand and don't even realize you've misinterpreted. If we see some of your thought process, we can better tailor the answer to provide clarity on the particular issues you might be having. Otherwise we have to approach it as a general question, and then the answer may or may not help you, even if its right! $\endgroup$ – Cort Ammon - Reinstate Monica Feb 24 '17 at 20:41

To accelerate the centre of mass you need to apply an external force applied to your system of rod, wheels and motors.
There is no external force so the centre of mass does not accelerate.

There are no external torques acting on the system about the centre of mass and so the angular momentum of the system about the centre of mass must be conserved.
If the initial angular momentum was zero then the net angular momentum of the system after the motor was started must stay at zero.
If the rotor in the motor is rotating then it has angular momentum, so the rest of the system must rotate in the opposite sense with an angular momentum to compensate.

To explain how the system might conserve angular momentum consider the following simplified diagram.

enter image description here

The rotor has a clockwise torque on it which causes it to rotate clockwise.
The Housing has Newton third law forces acting on in (red) due to the forces on the rotor (grey).
Those forces on the housing produce a net anticlockwise torque of $FR - Fr$ about the centre of mass of the system which causes the whole system to rotate in an anticlockwise direction about the centre of mass of the system
The clockwise angular momentum of the rotor is balanced by the anticlockwise angular momentum of the system.

  • $\begingroup$ So you are saying: there are not external forces, hence the com will not move, hence eventually the whole system will rotate about the center of mass. $\endgroup$ – C Marius Feb 24 '17 at 16:29
  • $\begingroup$ There is distinction here as the OP asks about the CM of the rod only and your answer describes the entire system. I think your answer is correct overall, but note that the CM of the rod will move, but the CM of the system will not. $\endgroup$ – ja72 Feb 24 '17 at 16:29
  • $\begingroup$ @CMarius - please be specific when talking about an individual part CM or the system combined CM. $\endgroup$ – ja72 Feb 24 '17 at 16:30
  • $\begingroup$ @ja72 Ok thank you!. The wheel are of rotating about their center of mass. $\endgroup$ – C Marius Feb 24 '17 at 16:34
  • $\begingroup$ @Farcher you are saying that the rod will rotate about it's center of mass. Right? $\endgroup$ – C Marius Feb 24 '17 at 16:37

Your intuition is correct.

If no external torque is applied, then the angular momentum of the thing doesn't change. That means, if one or both wheels start to rotate, the entire assembly will start to rotate in some way so that the total angular momentum does not change.

You can sometimes see this if you watch dirtbikes taking jumps. They can be in the air, not touching anything, and suddenly rotate forward so that the front wheel dips down. This happens if they hit one of the brakes in the air. One component (the wheel) stops rotating and loses angular momentum, then the entire assembly (bike and rider) starts rotating so that the net angular momentum of the entire assembly doesn't change.

  • 1
    $\begingroup$ Thank you for shearing your thoughts. So if you think is one of the three outcomes that I have enumerated, please specify it $\endgroup$ – C Marius Feb 24 '17 at 16:23
  • $\begingroup$ How can you say "Your intuition is correct" when 3 mutually exclusive options have been given? $\endgroup$ – sammy gerbil Feb 24 '17 at 18:19
  • $\begingroup$ Sorry, I was not clear. The center of mass will not move since no external force was applied. The entire assembly will rotate about the center of mass. $\endgroup$ – Daddyo Feb 24 '17 at 22:22

Farcher has already given a good explanation : the centre of mass of the system does not move; the total angular momentum of the system remains zero unless ther is an external torque.

If the centre of the rod coincides with the centre of mass of the sytem, the centre of the rod will not move. If the centre of the rod does not coincide with the CM of the system, it could move around the CM of the system, depending on how other parts rotate.

If the wheels rotate in the same sense (eg both anti-clockwise, as in Farcher's diagram) then there is a net torque on the rod which turns it clockwise. If the wheels rotate in opposite directions there is zero net torque on the rod so it does not rotate.

Total angular momentum must be conserved, for the same reason that the CM of the whole system does not move : because of Newton's 1st and 3rd Laws. So if the wheels both rotate anticlockwise, then the whole system rotates clockwise with an equal amount of angular momentum in the opposite direction. The total amount of clockwise angular momentum must equal the total amount of anticlockwise angular momentum at all times.

  • $\begingroup$ Only one wheel rotates. If both wheel rotate then if both rotate cw or ccw then the rod will not rotate whereas if the wheel rotate in different directions, then the rod will rotate faster, right? $\endgroup$ – C Marius Feb 24 '17 at 20:29
  • 1
    $\begingroup$ If only one wheel rotates, the rod will rotate in the opposite direction. If both wheels rotate, it is the opposite of what you suggest - read my 3rd paragraph again. Angular momentum must be conserved. $\endgroup$ – sammy gerbil Feb 24 '17 at 20:55

The system is going to rotate about the combined center of mass, which just happens to be at center of the rod for this case.

There are forces acting on the rod, from the connections to the motors, but they happen to be equal and opposite if the masses are symmetric. Anyway the rule is for any system of connected bodies (or gravitationally bound bodies) in the absence of external forces the system is going to rotate about the Barycenter.

This can be advantageous, as in a freely floating complicated mechanism, if a force couple is applied on one of the bodies (a pure torque) and the instant center of rotation is observed, then this point is the effective center of mass of the body. Any force applied through this point will result in a pure translation.

  • $\begingroup$ You mean any force applied to that point will result in a pure translation, right? $\endgroup$ – C Marius Feb 26 '17 at 16:10
  • $\begingroup$ Yes. That is what I mean. Each body in a serial chain of connected bodies will have such a point. $\endgroup$ – ja72 Feb 26 '17 at 17:59
  • $\begingroup$ I think therefore you may consider modifying the last word in you answer "rotation" -> "translation" ... However, thank you so much for your support sir! $\endgroup$ – C Marius Mar 1 '17 at 9:34
  • $\begingroup$ @CMarius that you for pointing it out. I did not see it originally. $\endgroup$ – ja72 Mar 1 '17 at 13:05
  • $\begingroup$ Sir, I am having another question about mathematical modelling: physics.stackexchange.com/questions/316169/… . Would you please be kind to have a look at it? $\endgroup$ – C Marius Mar 3 '17 at 21:37

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