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Angular momentum and newtons' law relation

  1. Jul 6, 2008 #1
    Hi i'm obsessed with this angular momentum conservation thing.From second law we can understand the linear momentum conservation easily, but in angular momentum i can't understand the logic.There comes a cross R which is the position vector.I searched a lot about this and couldn't find a reason.Is it observed and experimentally accepted ?Like opening a door we can experimentally observe that we can open it easily as we push it from the longest positon to the hinge.so r cross is coming from here?But this time i can not relate it to the second law, it's like an independent law, and 2nd law only relates with linear momentum.I hope i could tell my trouble with my limited english.
  2. jcsd
  3. Jul 6, 2008 #2
    Ohh i foget.I can guess that linear monetum deals with the change of speeds' magnitude and angular momentum is coming from the change of direction of speed(although magnitude remains constant) but scientificly just using the second law how can we get to the angular monetum?
  4. Jul 6, 2008 #3

    Andy Resnick

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    Newton's law is not a conservation law, so I can see why you are having trouble. Personally, I best understand conservation laws as a consequence of symmetry laws (Noether's theorem).

    If I may copy from the website:


    "For illustration, if a physical system behaves the same regardless of how it is oriented in space, its Lagrangian is rotationally symmetric; from this symmetry, Noether's theorem shows the angular momentum of the system must be conserved. The physical system itself need not be symmetric; a jagged asteroid tumbling in space conserves angular momentum despite its asymmetry - it is the laws of motion which are symmetric. As another example, if a physical experiment has the same outcome regardless of place or time [...], then its Lagrangian is symmetric under continuous translations in space and time; by Noether's theorem, these symmetries account for the conservation laws of linear momentum and energy within this system, respectively."
  5. Jul 7, 2008 #4
    the point i really had difficulty to get is the moment arm thing.Last night i found i pseudo-logic.For example for linear momentum an observer from an initial reference frame sees the objects speed change and says that according to newtons 2nd law if we simlify the equation the force acts on the object is ma.The observer sees the change in the magnitude of the speed.And in angular momentum he should see the difference in the direction of speed not the magnitude of it.The whole angular rotation stuff is about the change of direction of speed.So he can get the idea of changing direction from the change of the vertical distance to the object(if there's no rotation to that point vertical distance remains same) and the FR which is torque is like a potential of rotation( its unit is already an energy unit) i'm not fully satisfied and still searching but that make sense to me.
  6. Jul 7, 2008 #5
    Uhm... I am a bit puzzled by your questions.

    Newton's law is formalized as


    however I always seen this as being the same law as


    The angular momentum which is conserved is the one describing the spinning of the object around its center of mass (more precisely, around any axis passing through its center of mass), i.e.


    much like the linear/translational momentum is


    Newton's law and its angular equivalent describe how the momentums change due to an external force or torque.

    But what you have in mind is not the spinning of an object, but the deviation of a (point-like, for convenience) object. That is still described by F=ma, except that F is not directed as v.

    In that case, you can still calculate an angular momentum related to a point in space other than the center of mass (more precisely, related to an axis not passing through its center of mass). However this "momentum" is not the one which is conserved!
  7. Jul 7, 2008 #6
    ok i got the f=ma but its rotational equivelant comes from where?From the 2nd law?Special case of linear momentum?2nd law doesn't say anything about rotation.Where's this vectoral cross come from?I nearly understand it as i tell in my previous message, but scientifically still i'm not satisfied.
  8. Jul 7, 2008 #7
    Unsatisfied by the fact that the angular version is not included in Newton's treatise? :D

    IIRC, Newton laws describe only objects approximated as point-like particles, hence no I is defined.

    But I am sure you can derive [tex]\tau=I\alpha[/tex] from F=ma by simply considering the extended object as the sum of tiny point-like particles each of which still obeys Newton's 2nd law.
  9. Jul 7, 2008 #8
    yeah but i couldn' find this derivation anywhere.
  10. Jul 7, 2008 #9
    Something along the lines of this...

    (a) definition of torque:

    [tex]\tau=r [/tex]X[tex] F[/tex]

    (b) application of Newton:

    [tex]\tau=r [/tex]X[tex] F = r[/tex]X[tex](ma)[/tex]

    (c) now consider the above applied to each particle dm the object is made of:

    [tex]d\tau=dm r[/tex]X[tex]a[/tex]

    (d) replace linear acceleration with linear velocity a=dv/dt:

    [tex]d\tau=dm r[/tex]X[tex]dv/dt[/tex]

    (e) consider that v here will be only the tangential velocity (v_t) since we're considering only rotations around one object's axis, therefore

    [tex]d\tau=dm |r|dv_t/dt[/tex]

    (e) replace tangential velocity with angular velocity [tex]v_t=\omega |r|[/tex], which is now identical for all particles:

    [tex]d\tau=dm |r|^2 d\omega/dt[/tex]

    (f) replace angular velocity with angular acceleration [tex]\alpha=d\omega/dt[/tex], which is still identical for all particles:

    [tex]d\tau=dm |r|^2 \alpha[/tex]

    (g) find the total torque by summing all the particles:

    [tex]\tau = \int d\tau= \int |r|^2 dm \alpha[/tex]

    where \alpha is outside the integral because it's the same for all particles making up the object.

    Now, the quantity [tex] \int |r|^2 dm [/tex] is something new and unseen before. Let's give it a name: ANGULAR MOMENTUM I.

    And here it is:

    [tex] \tau = I\alpha[/tex]
  11. Feb 23, 2009 #10
    Hi very long time after i finally understand this a little i think.Angular momentum and torque are things that we define and doesn't come from anywhere else.First we define rotation and then we see the "thing" what we need for rotation, we see the conserved quantity in rotational motion and relate them to each other.That makes our work easy.Now i can see according to a refernce point o we need to apply such a force(point particle i am talking about) that perpendicular to the position vector, i can get this intuitively.All the other components of force will contribute to the translational motion.But how do we come up with the idea of multiplying each other, according to equal time equal areas concept i can get this mathematically, but i can not get this multplication thing intuitvely.I mean you get f=ma straightly, in words İf you apply a force to a particle you will get an accelaration inversly proprtional to the particles mass and same direction as the force.On the other hand if we want to translate the rotational analog of this equation, i can't express in words.In my expression i would say intuitively the tangential part of the force relative to the position vector provides a change of angle of position vector but why do we multiplicate these things?That's not clear for me.
  12. Feb 23, 2009 #11
    and one more thing in the origin of this angular momentum concept.How does Newton come up with this idea?I mean after discovering f=ma, did he say i should vector product both sides with r?:) Of course this is not the way but in the way most physics books this rotational analog comes up like this.I think he first analyse the rotational motion and as i mentioned in my previous message he saw the conserved quantity and after some manipulatons he evaluates this rotational analog.But books don't even mention about this, ok may be people don't need to learn where ,it comes form they just use it, but i can not accept the dogmatic formulas.May be just because i don't get it:).Excuse my sillyness again.But anyone shares this point of view and wants to fully understand the real meaning of this rather than some cold mathematical formulas, please contact with me or advise some books(i already got feynman's).
  13. Feb 24, 2009 #12
    yes it is understood that F=ma is linear equivalent of t=Ia
    yes.. each particle is integrated to be I in some circular orbit.
    so what is the problem in accepting t=Ia
  14. Feb 25, 2009 #13


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    The tricky part about angular momentum is that tangental velocity isn't conserved if the angular inertia is changed by moving mass inwards or outwards.

    Angular momentum is conserved in a closed system, but the angular energy isn't if internal energy is used to move masses inwards and outwards. If a mass is moved inwards or outwards, then the mass follows a spiral path during the transition, with some work done, because the radial component of force isn't perpendicular to the spiral path.

    One apparent exception is the case of a mass on a string that winds or unwinds around a pole. In this case, the path is involute of circle, and no work is done. Angular energy of the mass is conserved, because the force from the string is perpendicular to the spiral path. Angular momentum doesn't appear to be conserved, if the torque the string exerts on the pole is ignored. If the torque on the pole is taken into account, and the angular momentum of the pole and whatever the pole is attached to is taken into account, then angular momentum is conserved, and the angular energy of pole and what it is attached to changes.
  15. Feb 25, 2009 #14

    D H

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    Conservation of linear momentum for a single particle is Newton's first law. Conservation of angular momentum for a single point mass particle is a direct consequence of Newton's first law. Point masses have a null inertia tensor, so the angular momentum of a point mass is simply [itex]{\boldsymbol r}\times {\boldsymbol p}[/itex]. Since the cross product of a pair of parallel or anti-parallel vectors is zero, angular momentum of a constant mass point mass is trivially conserved whenever linear momentum is conserved.

    To derive conservation of linear momentum for a system of particles you need to look at Newton's third law as well as Newton's second law. The weak form of Newton's third law -- the force exerted by particle B on particle A is equal in magnitude but opposite in direction to the force exerted by particle A on particle B -- suffices for deriving conservation of linear momentum, but does not suffice for deriving conservation of angular momentum for a system of particles. To derive the latter you need the strong form of Newton's third law -- forces are equal-but-opposite and are directed along the line connecting the particles.

    You can't derive that because in general [itex]{\boldsymbol \tau}={\mathbf I}{\boldsymbol \alpha}[/itex] is not true. The rotational analog of Newton's second law is [itex]{\boldsymbol \tau} = {\mathbf I}{\boldsymbol \alpha}+ {\boldsymbol \omega}\times({\mathbf I}{\boldsymbol \omega})[/itex].

    Strictly speaking, Newton's laws are only valid in an inertial reference frame. In particular, [itex]{\boldsymbol F}= d{\boldsymbol p}/dt[/itex] is only valid in an inertial frame. Fictitious forces (e.g., centrifugal force, coriolis force) must be introduced to make Newton's laws appear to be valid in a rotating frame. Fictitious torques must similarly be introduced to make the rotational analog of Newton's second law, [itex]{\boldsymbol \tau}= d{\boldsymbol L}/dt[/itex] appear to be valid in a rotating frame.

    The angular momentum due to some composite body rotating around some axis expressed as [itex]{\boldsymbol L}={\mathbf I}{\boldsymbol \omega}[/itex]. Taking the time derivative, [itex]d{\boldsymbol L}/dt=(d{\mathbf I}/dt){\boldsymbol \omega}+{\mathbf I}(d{\boldsymbol \omega}/dt)[/itex][itex]d{\boldsymbol L}/dt[/itex]. The problem: While the inertia tensor I of some constant mass, solid object is constant in a body frame defined by the object, it is anything but constant from the perspective of an inertial frame.

    To have the time derivative of the inertia tensor be zero, which is exactly what is going on when one claims [itex]d{\boldsymbol L}/dt={\mathbf I}d{\boldsymbol \omega}/dt[/itex], one must express the angular momentum, inertia tensor, and angular velocity in a rotating frame. A fictitious torque term, [itex]{\boldsymbol \omega}\times({\mathbf I}{\boldsymbol \omega})[/itex], must be introduced to make the rotational analog of Newton's second law appear to be valid in this rotating frame.
  16. Feb 26, 2009 #15
    Had a quick look at some posts.

    Angular momentum , like linear momentum is a consequence of Newton's third law - forces occurs in pairs which are etc etc.

    Cons. of ang. momentum can be derived from N's laws.
  17. Feb 26, 2009 #16

    D H

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    That is true only if you assume the strong form (forces are equal-but-opposite and are central forces) of Newton's third law. The weak form, forces are equal-but-opposite, does not suffice.
  18. Feb 26, 2009 #17
    Sorry new to this forum so I couldn't change my previous post before correcting it.

    By extending the derivation of linear momentum from N's 3rd law - a deribvation of cons. of ang. momentum can be obtained.
  19. Feb 26, 2009 #18
    Cant see my post for some reason.

    so apologies if this is repeated

    By extending the derivation of linear momentum (from N's laws), a derivation of angular momentum can be obtained
  20. Feb 26, 2009 #19
    The general approach looks right but maybe not!:smile:

    Both laws 3rd and 2nd is used in deriving linear momentum, not just F=ma

    and using torque(moments) is the right approach for ang. mom.
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