Scenario # 1

I put gunpowder and then a ball bearing in an old musket and fire the bullet.

Scenario # 2

Lets imagine I had a motor with a disk on it and there was the same ball bearing stuck on the edge of the disk somewhere. If I spin the motor up fast and then release the ball bearing somehow, it will fire off perpendicular to the tangent line of the release point. Lets imagine for the purposes of comparison, I spin up the motor with enough rpm such that the ball bearing’s velocity equaled the velocity of the musket shot in scenario 1.

Newton's 3rd law of motions states that "for every action there is an opposite and equal reaction". With that in mind, it shouldn’t really matter how I expelled the hunk of lead, would both systems react in the opposite direction the same at the moment of release? (I’m aware that the spinning system would have a torque while it was spinning, but lets ignore that)

The reason I’m asking that is; intuitively the ball bearing is desperate to escape the faster the disk is being spun. It would seem weird that it would suddenly kick the system backwards when it was already straining to leave the system the whole time. Also, there seems to be an intuitive difference because the expelling force in the spinning system is 90 degrees to the applied force.


If you were actually to do Scenario 2, the motor would indeed be flung backwards. This might seem a bit odd at first glance; if the ball bearing is moving already at the time of release, why would the motor be ejected backwards when it releases the bearing?

The way to resolve this confusion is to note that the motor is actually shaking around when it's spinning with the bearing attached. The easiest way to model it is via a simple picture of a dumbbell, with one weight being the motor and the other weight being the bearing:

enter image description here

Both the motor and the bearing rotate about a pivot point. When the bearing is released (ie, the connecting line in the picture above is severed), it flies in one direction, and the motor flies in the opposite direction, in exact analogy to the musket scenario.

I avoided math in this answer because I figured an intuitive visual explanation was sufficient; other posters will probably give more details, if you want them.

  • $\begingroup$ But based on your answer, wouldn't the motor and bearing system be isolating back and force before the release of the ball anyway? Therefore why would there be an extra 'kick' at the point the ball is released when it would probably already be isolating in the opposite direction? $\endgroup$ – Mike S Feb 26 '14 at 23:26
  • $\begingroup$ @MikeS: You've got it exactly correct. The motor and bearing are wobbling around back and forth like an off-balance washing machine. You're right that there isn't a kick when you release the bearing, so much as they separate from each other and keep moving in opposite directions. In the end, both are moving in opposite directions, just like in the musket case. $\endgroup$ – DumpsterDoofus Feb 27 '14 at 0:59
  • $\begingroup$ They are both moving in opposite directions after release, but not for the same reason as in the musket case. The musket case is an action reaction pair. The reaction force of the ball on the musket kicks the musket backward. In the motor case it is simply because of inertia that the motor "flies" backward. The force of the ball leaving does not kick the system backward. Please, correct me if I'm wrong, but I think the two cases are completely different. $\endgroup$ – wgrenard Feb 27 '14 at 2:33
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    $\begingroup$ @wgrenard: You're right that the mechanism which causes the two to fly in opposite directions is different. I was just highlighting the fact that despite this difference, Newton's 3rd law still holds in both cases. $\endgroup$ – DumpsterDoofus Feb 27 '14 at 4:23
  • $\begingroup$ @DumpsterDoofus fair enough. It seems we have simply taken different approaches and answered two different aspects of this question. $\endgroup$ – wgrenard Feb 27 '14 at 4:49

Newton's third law does not necessarily mean that when one object is kicked forward the other is kicked backward. What it means is that forces always come in pairs. If I press down on the floor with a certain force, then the floor pushes back on me with an equal and opposite force (provided I am not breaking the floor with the force I am exerting, obviously). This is the essence of Newton's third law.

When you fire a musket the gun powder explodes, and this explosion pushes on the ball bearing with a considerable amount of force. The ball bearing, in turn pushes back. You feel this sudden force pressing backward as a kick.

In the second scenario there is no kick because there is no sudden force being applied to the motor that pushes the ball bearing away. Think about the forces involved at the moment just before the ball is released. At this point there is only one thing keeping the ball moving in this circular path, and that is whatever is holding it to the disk. In order to move in a circular direction there must be a force on the ball bearing accelerating it toward the center of the disk. If there is, say, a latch holding the ball in place, then it is the normal force of the latch that achieves this. Thus, we have a reaction pair of forces that behaves according to Newton's third law (the latch pushing on the ball bearing and the ball bearing pushing back on the latch). Now, say that we can release this latch without otherwise disturbing the system. Suddenly, we have removed all forces acting on the ball bearing. Because there is no force present to accelerate the ball toward the center of the disk, it flies off in a tangential direction. There is no kick, because there is no reaction pair to cause a kick. The reason the ball flies off in a tangential direction is because there is a lack of forces keeping it moving in a circular direction.

  • $\begingroup$ The motor is being kicked back due to the application of an extra force in the direction of the movement/kick of the motor, but due to the removal of a force in the opposite direction, namely the normal force of the bullet. Another way to look at this is by looking at the position of the center of mass of the system of the bullet and the motor and the conservation of momentum. $\endgroup$ – fibonatic Feb 26 '14 at 8:38
  • $\begingroup$ But the force that is originally being applied is in the radial direction. Meaning that this removal of force in the opposite direction would have no effect on the angular momentum of the disk. If you were removing a force being applied in the tangential direction then it would be a different story. But this is not the case $\endgroup$ – wgrenard Feb 26 '14 at 17:03
  • $\begingroup$ In DumsterDoofus's example the motor represented by the heavier mass would be carried in the opposite direction, yes, but that has nothing to do with Newton's third law. In this case where we consider the motor "shaking around" as DumpsterDoofus says, the mass is not "kicking" the motor backward. The motor's own inertia carries it backward. $\endgroup$ – wgrenard Feb 26 '14 at 17:11
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    $\begingroup$ @MikeS Correct. I've been thinking a lot about this and even consulted an old physics professor of mine. We concluded that when the ball is released there will be no effect on the angular velocity of the system. Meaning that releasing the ball will not kick in the opposite tangential direction and slow the spinning. I believe what DumpsterDoofus is saying is that according to his model of the motor, the inertia of the system will carry the entire motor backward as the released ball is carried forward. $\endgroup$ – wgrenard Feb 27 '14 at 2:53
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    $\begingroup$ So, essentially I think the both of us answered a different aspect of your question. $\endgroup$ – wgrenard Feb 27 '14 at 2:54

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