Newtons three laws of motion appears to apply only for linear motion:

  1. An object remains at rest or moves in a straight line at uniform velocity unless a force is applied.

  2. Force is mass times acceleration.

  3. Every action causes an equal and opposite reaction.

Is there a rotational equivalence? For example:

1'. Every body rotates around a fixed axis at uniform angular velocity unless a torque is applied

2'. Torque is Moment of Inertia times angular acceleration

3'. When one body exerts a torque on another; there is an equal and opposite torque applied on the first body by the second.

First are these actually correct; if not, what are the correct equivalence; and who formulated them?

  • 1
    $\begingroup$ Everything that's true in classical mechanics derives from the basic formulations of classical mechanics, of course. In your rotation laws, you speak of objects which are no just points. Maybe have a look at Newton–Euler equations and Euler's equations (Wikipedia). $\endgroup$
    – Nikolaj-K
    Commented May 6, 2015 at 13:12
  • $\begingroup$ Mechanics has a long history; I recall reading somewhere that ideas of Torque go back to Archimedes work on levers. $\endgroup$ Commented May 6, 2015 at 13:17
  • $\begingroup$ That's a trivial "No". The first sentence is simply not true, as even a cursory look at the physics of rigid rotating bodies will show. Most irregular bodies will actually present a chaotic tumbling motion. $\endgroup$
    – CuriousOne
    Commented May 6, 2015 at 13:19
  • $\begingroup$ @curiousOne: what about conservation of angular momentum? $\endgroup$ Commented May 6, 2015 at 13:22
  • $\begingroup$ @CuriousOne: You say "not true", but I feel you can bend the meaning of the words in the formulation to make it fit. It might be worth it to point out that there are funky things like the tennis racket theorem. That's why I said we must keep in mind that however we (re-)formulate the statements here, it must correspond to some truths in classical mechanics. The interesting question would be that if we take the stance to take the resulting statements as axioms, how much must we add to get Newton back. PS, Ullah: I'd not trouble myself with Newton too much. $\endgroup$
    – Nikolaj-K
    Commented May 6, 2015 at 13:26

1 Answer 1


There is a rotational equivalence, but it is not what you stated. The problem, as pointed out by @curiousOne, is that conservation of angular momentum does NOT imply rotation about the same (fixed) axis. But I think a simple restatement like this could work:

  • if no torque acts on a body, its angular momentum will remain unchanged
  • rate of change of angular momentum is proportional to applied net torque
  • when two bodies interact, the torque that A applies to B is equal and opposite to the torque that B applies to A, so that the angular momentum of the combined system (A+B) is preserved.

I believe that addresses the objections raised to your earlier version. Note that "axis of rotation is unchanged" is fundamentally different from "angular momentum is unchanged".

  • $\begingroup$ Possibly, but all of that, and much more, is trivially covered by Noether. The bigger problem is that I am not sure that there is an easy way to derive the actual dynamics of rotating bodies from the above axioms. The third statement is also problematic, because of the coupling of orbital angular momentum and spin, which makes for an extremely ugly analysis of the dynamics of two coupled rigid rotators. It's probably the latter that prevents us from axiomatizing rotating motions this way. $\endgroup$
    – CuriousOne
    Commented May 6, 2015 at 13:58
  • $\begingroup$ @CuriousOne - this question is tagged "classical-mechanics". What coupling of angular momentum and spin are you talking about in the classical context? $\endgroup$
    – Floris
    Commented May 6, 2015 at 14:03
  • $\begingroup$ The one that increases the length of the day on Earth while moving the moon further away. That's the classical equivalent of orbital angular momentum and spin coupling and it's totally non-trivial as the astronomers can tell. $\endgroup$
    – CuriousOne
    Commented May 6, 2015 at 14:06
  • $\begingroup$ Does "axis of rotation" in the first and the last paragraph mean the axis in the reference frame of the rotating object (e.g. a line between two painted dots on the object) or in the reference frame of an external non-rotating observer (e.g. a line between two painted dots on the lab where the object is rotating)? $\endgroup$
    – JiK
    Commented May 6, 2015 at 14:29
  • $\begingroup$ It would have to be "in an external frame" - but I can't read the mind of the person who wrote it. Which is why my answer doesn't explicitly speak of an axis - just of "angular momentum" (which, being a vector, is only constant if both magnitude and direction are constant). $\endgroup$
    – Floris
    Commented May 6, 2015 at 14:45

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