1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Newtons 2nd law for rotating bodies

  1. Aug 12, 2012 #1
    Is shown like this in my book:
    Consider a rotating body with an angular acceleration α. There must be a tangential force component if it is rotating:
    For a general point on the body we can write:
    Ftan = mi * ai = mi * ri * α (1)
    Multiply by ri and sum up you get:

    Ʃτ = Iα (2)

    I just don't understand why you multiply by ri and thus force yourself to work with "torques" rather than just forces:

    Why don't you just sum the equation for a single point (1). Then you get:

    ƩFtan = Ʃmiri * α

    Then you could call Ʃmiri whatever you like, not the center of mass though since the ri are not vectors.
    But then doing that must be wrong. Because then it seems like you lose that part in (2), which says that the acceleration is bigger the further away from the centre of rotation you apply your force. So what is wrong with my version, i.e. equation (3)?

    Also I have now struggled alot with rotating bodies, but one question always pops into my mind: Can you prove that rotation exists mathematically, or does all these mathematical derivations for rotating bodies just assume that you physically can observe rotations in nature. hmm.. maybe this last question doesn't make sense for you, and so if not, don't bother answering it.

    Cheers :)
     
  2. jcsd
  3. Aug 12, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Makes it easier to keep track of what is happening in more complicated systems.

    Force is a vector. The motion is circular - each of the tangential forces may be pointing in different directions but they all contribute to rotation, not translation. You need the torques to add them up properly.

    eg. o a see-saw the tangential forces point in the same direction - but they oppose each other by being on opposite sides of the fulcrum. How much they oppose depends also on how far away from the fulcrum they act. Thus - formulating the problem in torques. Similarly, for an arbitrary body with arbitrarily placed forces, the contribution that a force makes to the rotation depends also on the distance from the center as well as the strength of the force.

    I think the description hides a lot by just dealing with the magnitudes. The actual torque relation is a cross product of vectors.
     
  4. Aug 12, 2012 #3
    I see. But what does wrong in writing the equation as I do? that is:

    ƩFtan = Ʃmiri * α

    Certainly it seems right but then it doesn't seem to incorporate that the angular acceleration should get bigger the farther from the rotational centre the force is applied?
     
  5. Aug 12, 2012 #4

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    You seem to be asking about the motion of a rigid body. A flexible body will in general deform (change shape) as it rotates, and there isn't really any concept of a single "rotation".

    A rigid body doesn't change shape by definition, and that is why you can describe the motion of the whole body by a single "rotation speed". The reason why the motion of a rigid body can only be a translation of the center of mass plus a rotation about the CM is basically a question about geometry, and it is true for Euclidean geometry, but false if the theory of relativity becomes important. The concept of a "rotating rigid body" is inconsitent with relativity.

    You don't "have to" work with torques for a rigid body. You will get the same answer if you work with individual forces, but this is usually harder because the forces and accelerations of different points are in different directions as well as having different magnitudes.
     
  6. Aug 12, 2012 #5
    So would the equation:

    ƩFtan = Ʃmiri * α

    work just as well? But then it doesn't see that α is dependent on the radius of which Ftan is applied (saying there is just ONE external tangential force).
     
  7. Aug 13, 2012 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Aside: It helps to use LaTeX to represent equations... we do it a lot here and as you advance in your studies you will need it more and more. The sooner you learn it the bigger your advantage over the others at your level. Compare what you wrote with:
    [tex]\sum_i F_{i\perp} = \sum_i m_i r_i \alpha \left ( = \sum_i m_i a_{i\perp} \right )[/tex]... isn't this a lot easier to read? It's not as hard as it appears - you learn it as you need it: just hit the "quote" button at the bottom of this post to see how I did it.

    Back on topic:
    All you've done is summed the magnitudes of the various forces, discounting their distances from the center of rotation : a tangential force far from the center of rotation contributes more to the angular acceleration than the same magnitude tangential force close to the center of rotation.

    Looking at it from the math:
    With [itex]\sum r_i F_{i\perp} = \sum r_i^2 m_i \alpha[/itex] it is tempting to try to simplify by dividing through by [itex]r_i[/itex] and write [itex]\sum F_{i\perp} = \sum m_i r_i \alpha[/itex] isn't it?

    It does not work because each [itex]r_i[/itex] may have a different values. If they were all the same then that would be different.

    Making it explicit - you have said that:
    [itex](r_1F_1 + r_2F_2)=(m_1 r_1^2 + m_2 r_2^2)\alpha[/itex] means that [itex]F_1 + F_2 = (m_1r_1 + m_2r_2)\alpha[/itex]
    ... do you see the problem? This is only true if [itex]r_1=r_2[/itex].

    You've basically disobeyed the rules of algebra.
     
    Last edited: Aug 13, 2012
  8. Aug 14, 2012 #7
    Well that is not what I meant. I simply suggested that you never multiplied the equation of the force on a single point of mass by ri.
    We had for a start:

    Ftan-i = miai
    I would just sum up all these:
    ƩFtan-i = Ʃmiai = (Ʃmiri

    This is a completely different equation, not a mathematically equivalent version to the correct one. You can see that, as there is nothing that makes a force applied at a greater distance from the rotational centre cause a larger angular acceleration.
    My question is: Why is this one wrong and the other one correct? I have thought about it myself and came to the conclusion that they are both correct, but that only the torque version works for only external torques. I know you can make an argument that the sum of internal torques cancel, but the same can't be said about the sum of tangential forces? Why is that?
     
  9. Aug 15, 2012 #8

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    OK - I'll put it another way:
    <Jedi wave>That is not the angular acceleration you are looking for <observes subject to see if convinced> ... <sigh>

    That equation is not "wrong" - it just does not describe the same thing. It is the sum of the magnitudes of the tangential forces... that's only right or wrong depending on what you expect to be able to do with it.

    You cannot, for instance, use it to determine the angular acceleration of the object due to all the tangential forces. That is what the other one does.

    You have an alpha in there and it is an angular acceleration - but it is not the angular acceleration of the object. It is what the angular acceleration of the object would have been if all the forces acted at unit distance from the pivot.

    But don't take my word for it - do an experiment. Work it out both ways and see which one agrees with the experiment.
    Bottom line is that physics is an empirical science - our equations are mathematical models that have to conform with observed reality.
    An equation is right or wrong to the extent that it agrees with experiments.
     
    Last edited: Aug 15, 2012
  10. Aug 15, 2012 #9
    Exactly. But I don't understand why it only works for forces acting at a unit distance. How do you see that?
    Let me stipulate once again that the version above is NOT to be thought of as a mathematically equivalent version of Ʃτ = Iα. It is a version which relates all the internal tangential forces to the total angular acceleration of the body. MY QUESTION: Why does this not work like the correct version? i.e. why can you not sum up all the external tangential forces and use the above equation to get a correct number for the angular acceleration?
    My answer: Something goes wrong in the derivation (not some basic math rule neglection), rather I think it is because you can not say always that the sum of tangential forces add in such away that all the internal ones cancel leaving out only external ones. Because THAT is the smart thing about Ʃτ = Iα. So my question is now, if my predictions are correct: How can you see that the tangential forces inside the body in general do NOT sum up in such away that all external ones cancel out?
     
  11. Aug 15, 2012 #10

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Because that is just not how the Universe works.
    actually it's only for when all the (tangential) forces act at the same distance from the pivot.

    Probably what you need to do is work the whole thing out for a simplified rigid body - 2 different masses at different distances from a pivot point, joined to each other and the pivot by massless incompressable rods. Apply a different tangential force to each. Do the free body diagram for each one, apply the rigidity rule, define alpha as rate of change of angle rather than in terms of tangential acceleration, and work out what happens. I'm too tired :)
     
    Last edited: Aug 15, 2012
  12. Aug 15, 2012 #11

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    It doesn't work because it's wrong. Suppose you apply two forces of equal magnitude and direction at diametrically opposed points on a uniform density sphere. If the two forces are purely tangential, your formula would indicate that a good amount of angular acceleration would result. The correct formula says there is no angular acceleration from this pair of forces. Now suppose the forces are of equal magnitude but opposite in direction: a couple. Your formula once again yields the incorrect result.
     
  13. Aug 15, 2012 #12

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I thought of that approach but figured that a + or - directional system could be assigned.

    You are half-way to defining a torque right there of course.
     
  14. Aug 15, 2012 #13
    I have tried to do this but somehow always end up assuming that the internal forces cancel out - because why shouldn't they? Consider my drawing of a rigid body attached.

    The pivot is distanced r from m1 and r' from m2. Let m1be subject to an external tangential force, Fext, aswell as an internal force, which is a counterforce from the axis. We then have for m1:

    m1αr = Fext - Fint

    And for m2:

    m2αr' = Fint

    But this gets me nowhere since I beforehand assume that the force that pulls m1 down is equal to the force that pulls m2 just as I normally would in translational mechanics.
    How did you propose to do it?

    It all boils down to the fact that I can't see why the internal tangential forces wouldn't cancel. After all Newtons third law is still valid. You, however, seem to view it as quite obvious that they don't cancel out. So I ask you to point out how you see that they cant cancel out.
     

    Attached Files:

  15. Aug 15, 2012 #14
    okay now I did a thought experiment, where I was able to convince myself that there is no reason that internal tangential forces should in general cancel out - I don't know why that was so hard for me to see.
    Now I just have one question:

    In the above derivation of newtons second law for rotating rigid bodies, you kind of have to come op with the convention of the direction of torque yourself. It is very intuitive how you should make that, but my question is, if you can actually derive this convention of how torque should be defined mathematically or if it is an assumption that τ = r x F. Either way it would be madness to think that a tangential force did not rotate the body in the direction of itself, but I still want to know if the definition of the torque can be "derived" from a vectorial version of the above derivation.
     
  16. Aug 15, 2012 #15

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    That would work for rotation in a plane, but not for rotations in three dimensional space.

    For rotations in three dimensional space, even the freshman physics [itex]\sum_i \vec r_i \times \vec F_i = \mathbf I \vec \alpha[/itex] is not correct, or at least it's not universal. This version of the rotational analog of Newton's second law only works in the cases of an object with a spherical mass distribution or rotational motion that is along one of the eigenvectors of the inertia tensor.
     
  17. Aug 15, 2012 #16

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Well since aaaa202 appears to think that working the directions is easy or obvious or something re:
    ... ah yes but when you have several all pointing in different directions in 3D, how would they add up? Not as a flat sum.

    <has flashback to college>
    <gets headache>
    Oh yeah.... I remember...

    aaaa202's analyses all have co-planar forces so what counts as a the turning direction is clearer.
    (That's not precise is it? lets see, all the examples have rxF pointing along the same vector.)
    Don't know why you put "derived" in scare-quotes? Do you think such a derivation would be suspect?

    Anyway - I don't actually follow your reasoning in your derivation prev. Fint? You seem to be saying that internal forces oppose the tangential forces?

    Don't know what you are going on about. You need an arbitrary angle between the masses - the simple geometry you'd picked is a special case - you are looking for a general relationship: don't pick a simple geometry.

    It's tough going ... I'll leave you to it.
    I can't help it if you won't believe me or the Universe. Your original question has been answered. <checks> Oh hang on - seems we missed one:

    Rotations exist both as a mathematical construction and in nature. That's how come we can use math to describe them, and why we would want to.
     
  18. Aug 15, 2012 #17
    "Don't know what you are going on about. You need an arbitrary angle between the masses - the simple geometry you'd picked is a special case - you are looking for a general relationship: don't pick a simple geometry."

    Dont know quite what I was doing either but it was the best I could do. I dont know how to setup the situation and how to do the freebody diagrams without assuming that there must be a tangential force that connects the two masses to have same angular acceleration by weighing one down at the expense of the other. Maybe, if you have time at some point in the day, can sketch in more detail how I should go about convincing myself? i.e. make the first part of the setup and then tell my step by how to arrive at the right conclusion.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Newtons 2nd law for rotating bodies
  1. Newtons 2nd law? (Replies: 9)

Loading...